Functional analysis IIProf. Dr. Thomas Sørensen

winter term 2014/15

Marcel Schaub

February 4, 2015

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Contents

0 Motivation and repetition 3

1 Spectral Theory for compact operators 101.1 Spectral Theorem for compact operators . . . . . . . . . . . . . . . . . . . . . . . . . 141.2 Consequences of the spectral Theorem for compact operators . . . . . . . . . . . . 191.3 The finite-dimensional case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.4 Spectral Theory for normal compact operators on Hilbert spaces . . . . . . . . . . 21

1.4.1 The Spectral Theorem for normal compact operators on a Hilbert space . 26

2 Spectral Theory for bounded, self-adjoint operators on a Hilbert space 312.1 The Continuous Functional Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.1.1 Properties of the Continuous Functional Calculus . . . . . . . . . . . . . . . 342.2 Outlook on the Measurable Functional Calculus . . . . . . . . . . . . . . . . . . . . 352.3 A recall on Measure Theory and Integration Theory . . . . . . . . . . . . . . . . . . 362.4 Statements aboutMb(D) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382.5 The Measurable Functional Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

2.5.1 Projection Valued Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422.5.2 Characterization of the Measurable Functional Calculus . . . . . . . . . . . 45

2.6 Spectral Theorem for self-adjoint bounded operators . . . . . . . . . . . . . . . . . 48

3 Unbounded operators 553.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553.2 Unbounded operators and their adjoint . . . . . . . . . . . . . . . . . . . . . . . . . . 553.3 Closed operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573.4 Essential self-adjointness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 583.5 Existence of self-adjoint extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603.6 The spectrum of an unbounded operator . . . . . . . . . . . . . . . . . . . . . . . . . 62

4 Spectral Theory for unbounded self-adjoint operators 634.1 Spetral Theorem – Multiplication operator version . . . . . . . . . . . . . . . . . . . 634.2 Spectral Theorem – Spectral measure version . . . . . . . . . . . . . . . . . . . . . . 654.3 Consequences of the Spectral Theorems . . . . . . . . . . . . . . . . . . . . . . . . . 704.4 Outlook on other approaches to the Spectral Theorem (for bounded/unbounded

self-adjoint operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 724.5 Another outlook: Spectral theory of bounded operatos; Generalisation to Ba-

nach algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 744.5.1 Banach algebras setup (B(X )) . . . . . . . . . . . . . . . . . . . . . . . . . . 744.5.2 Adding a unit to a unit-less algebra . . . . . . . . . . . . . . . . . . . . . . . 754.5.3 Functional Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

4.6 Outlook on the Functional Calculus and Spectral decompositions: Usefullness?!? 784.7 (Partial) Differential Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

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0 Motivation and repetition

Recall

A Banach space is a vector space (mostly over C) with a norm ‖ · ‖ such that (s.t.) the metricspace (X , d) with d(x , y) := ‖x − y‖ is a complete metric space. A Hilbert space is a Banachspace where the norm is given by a scalar product (inner product) ⟨·, ·⟩, and ‖x‖=

p

⟨x , x⟩.

Example. L2(R) with

f , g

:=

∞∫

−∞

f (x)g(x)dx

is a Hilbert space (the Hilbert space). So is `2(N) = x = (x1, x2, . . .) ∈ CN |∑∞

i=1 |x i|2 <∞.But the norm ‖·‖p (p 6= 2) with

‖x‖p =

∞∑

i=1

|x i|p!1/p

turnes `p = x ∈ CN | ‖x‖p < ∞ into a Banach space, but not a Hilbert space (p 6= 2).The space B(X , Y ) is the set of bounded (i.e. continuous) linear operators from X to Y , i.e.T ∈ B(X , Y ) iff T : X −→ Y is linear and bounded (∃C > 0 : ‖T x‖ ¶ C ‖x‖ for every x ∈ X(this definition needs only normed spaces – not completeness). Set B(X ) := B(X , X ). IfY = C (if X is a C-vector space (v.s.); if X is some R-vector space then Y = R) we call theelements in B(X ,C) linear bounded functionals and write X ′ :=B(X ,C), X ′ is called the dual(or, the dual vector space) of X .

Note: For a Hilbert space, “H = H ′” (Riesz representation theorem) in the sense that ∃φ :H −→ H ′, y 7−→

y, ·

such that φ is an isometry (bijection, conserves norm).

• However: φ is not linear – it is conjugate linear, i.e. φ(αx+ y) = αφ(x)+φ(y), α ∈ C,x , y ∈ H.

One has “X ⊆ X ′′” (X ′′ is the bi-dual) in the sense that there is a canonical embedding ι : X −→X ′′ that embeds X isometrically in X ′′ (ι is injective, but not always surjective). If X ′′ = X (inthe sense that ι is surjective), then X is called reflexive (so, all Hilbert spaces are reflexive).

• So are `p, Lp (1< p <∞).

Definition 0.1. Let T ∈B(X ) (X C-Banach space, X 6= 0). We define the resolvent set of T by

ρ(T ) = x ∈ C | N(T −λI) = 0 and R(T −λI) = X ⊆ C

Here N(S) = x ∈ X | Sx = 0 is the kernel and R(S) = y ∈ X | ∃x ∈ X = Sx = y is therange of the operator S ∈ B(X ). Also I : X −→ X , x 7−→ x is the identity on X (I = IX ). Wedefine the spectrum of T ∈B(X ) by

σ(T ) := C \ρ(T )⊆ C

The spectrum can be split in to 3 types (3 subsets).

• The point spectrum (“Punkt-Spektrum”)σp(T ) = λ ∈ σ(T ) | N(T−λI) 6= 0 ⊆ σ(T )

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• The continuous spectrum (“Stetiges Spektrum”)

σc(T ) = λ ∈ σ(T ) | N(T −λI) = 0, R(T −λI) 6= X , R(T −λI) = X

• The rest spectrum/Residual spectrum:

σr(T ) = λ ∈ σ(T ) | N(T −λI) = 0, R(T −λI) 6= X

Remark 0.2. (1) Note that λ ∈ ρ(T ) iff T − λI : X −→ X is bijective. This is equivalent tothe existence of

Rλ(T ) := (T −λI)−1 !∈B(X )

Rλ(T ) is also continuous and called the resolvent of T at λ.

(2) λ ∈ σp(T ) iff ∃x 6= 0, x ∈ X : T x = λx (or (T−λI)x = 0). Then λ is called an eigenvalue,and x an eigenvector. However, when X is some space of functions, e.g. Lp(R), we oftencall x an eigenfunction. N(T − λI) is the eigenspace of T belonging to the eigenvalueλ ∈ σp(T ). It is a T -invariant (linear) subspace of X . If Y ⊆ X is a linear subspace, thenY is called T -invariant iff T (Y )⊆ Y .

Proposition 0.3. Let T ∈ B(X ), then ρ(T ) ⊆ C is an open subset of C (hence σ(T ) is closed)and the resolvent map (or, function) ρ(T ) 3 λ 7−→ Rλ(T ) ∈B(X ) is complex analytic map with‖Rλ(T )‖−1 ¶ dist(λ,σ(T )). Here, “complex analytic” means: ∀λ0 ∈ ρ(T ) ∃r0(λ0) > 0 andc j(λ0) ∈B(X ):

Rλ(T ) =∞∑

j=0

c j(λ−λ0)j

for λ ∈ Br0(λ0) = µ ∈ C | |λ0 − µ| < r0 with the series convergent in the (operator) norm on

B(X ).

The aim of the course is (mainly) to study the spectrum (and its properties) for variousclasses/types of operators and prove theorems in analogy to one on “diagonalisation of sym-metric/self adjoint/hermitian matrices” in linear algebra.

We would like to see how/why this is “useful”. The theory here (in infinite dimensional case)is, however, much, much more rich.

Definition 0.4. The compact (linear) operators from X to Y are defined by

K(X , Y ) = T ∈B(X , Y ) | T (B1(0))⊆ Y is compact ⊆B(X , Y )

Remarks 0.5. (1) As for B(X ), we write K(X ) if X = Y .

(2) If Y is Banach, then “T (B1(0) is compact” can be replaced with “T (B1(0)) is pre-compact”

(3) That is, T ∈ K(X , Y ) iff T maps bounded sequences (in X ) into sequences in Y whichhave a convergent subsequence. (Note: T ∈ B(X , Y ) =⇒ T maps bounded sequencesinto bounded sequences.)

(4) Example: I = [0, 1]⊆ R, k ∈ C [I × I], f ∈ C (I) =: X . For x ∈ I set

(Tk f )(x) :=

1∫

0

k(x , y) f (y)dy

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and let X be equipped with ‖ f ‖= ‖ f ‖∞ = supx∈I | f (x)|. Then Tk : X −→ X is a bounded,linear operator which is compact (i.e. Tk ∈ K(X ) ⊆ B(X )). We shall see other examplesand further properties of compact operators soon.

Definition 0.6 (Banach space adjoint). Let T ∈B(X , Y ) and define for y ′ ∈ Y ′ (y ′ : Y −→K)and x ∈ X :

(T ′ y ′)(x) := y ′(T x)

Then T ′ : Y ′ −→ X ′, y ′ 7−→ T ′ y ′ is linear and bounded (i.e. T ′ ∈ B(Y ′, X ′). It is called the(Banach space) adjoint of T (or, the adjoint operator of T).Furthermore: ‖T ′‖ = ‖T‖ (the first norm is in B(Y ′, X ′) and the second one in B(X , Y )). Infact,

′ : B(X , Y )−→B(Y ′, X ′)

T 7−→ T ′

is an isometric, linear embedding (i.e. injective).

NB. In the above “definition” there are in fact things to prove – exercise!

• Linearity: Let α ∈K, u′, v′ ∈ Y ′, x ∈ X , then:

T ′(αu′+ v′)(x) = (αu′+ v′)(T x) = αu′(T x) + v′(T x)

= α(T ′u′)(x) + (T ′v′)(x) = (αT ′u′+ T ′v′)(x)

So T ′(αu′+ v′) = αT ′u′+ v′.

• Boundedness comes by |T ′ y ′(x)|= |y ′(T x)|¶ ‖y ′‖‖T‖‖x‖ and then

‖T ′ y ′‖= supx∈X

|T ′ y ′(x)|‖x‖

¶ ‖y ′‖‖T‖ =⇒ ‖T ′‖= supy ′∈Y ′

‖T ′ y ′‖‖y ′‖

¶ ‖T‖

• For “‖T ′‖= ‖T‖” see Theorem 4.18 in [FA1].

• ′ is isometric by the above and linear by 0.8 (1).

Definition 0.7 (Hilbert space adjoint). Let H be a Hilbert space1 and let φ : H −→ H ′ be themap y 7−→

y, ·

, identifying H and its dual H ′, and let T ∈B(H) (T : H → H) (T ′ : H ′→ H ′).Then T ∗ = φ−1T ′φ : H −→ H is called the Hilbert space adjoint of T ∈B(H). It satisfies:

T ∗x , y

=

x , T y

x , y ∈ H

T is called selfadjoint (s.a.) iff2 T ∗ = T .

We shall see other examples and further properties of the adjoint and self adjoint operator(s)soon.

1We could do this for T ∈B(H1, H2), H1, H2 do different Hilbert spaces. We need φH1,φH2

then.2Note that T ∈ B(H). For unbounded operators (typical examples occur in quantum mechanics or partial

differential equations) there is a more complicated definition of “(self)adjoint”. We will see this later.

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Program of the course (mainly...)

• Spectral theory for compact operators (in X ).

• Spectral theory for bounded, self adjoint operators (in H).

• Unbounded operators (in particular, symmetric operators and quatratic forms).

• Spectral theory for unbounded self adjoint operators.

Lemma 0.8 (Algebraic properties of adjoints). Let X , Y be Banach spaces, let H be a Hilbertspace. Let T, T1, T2 ∈B(X , Y ), S, S1, S2 ∈B(H) and α ∈K. Then the following statements hold:

(1) (αT1+ T2)′ = αT ′1+ T ′2 and (αS1+ S2)∗ = αS∗1 + S∗2.

(2) (IX )′ = IX ′ (IX : X −→ X , IX ′ : X ′ −→ X ′).

(3) (T2T1)′ = T ′1T ′2 and (S1S2)∗ = S∗2S∗1

(4) For ιX : X −→ X ′′, ιY : Y −→ Y ′′ the canonical embedding: T ′′ ιX = ιY T : X −→ Y ′′ andS∗∗ = S.

Proof. Simple algebraic manipulations from the definitions. Exercise.

Proposition 0.9. Let X , Y be Banach spaces and T ∈ B(X , Y ). Then T−1 ∈ B(Y, X ) exists iff(T ′)−1 ∈ B(X ′, Y ′) exists and, in this case, (T ′)−1 = (T−1)′. If X = Y = H is a Hilbert space,then (T−1)∗ = (T ∗)−1.

Remark. For T ∈ B(X ) we have: T−1 ∈ B(X ) exists iff (T − 0 · I)−1 ∈ B(X ) exists iff0 ∈ ρ(T ). I.e. invertability of T is a “spectral question” related to “0 ∈ ρ(T ) or 0 ∈ σ(T )?”.The proposition 0.9 says (for T ∈B(X )):

0 ∈ σ(T )⇐⇒ 0 ∈ σ(T ′)

This is therefore a property of the spectrum.

The proof uses [FA1]:

Theorem (Open mapping theorem). Let X , Y be Banach spaces. Let T ∈B(X , Y ) be onto. ThenT is open.

Proof of Proposition 0.9. Note that T : X → Y , T ′ : Y ′ → X ′, T−1 : Y → X , (T−1)′ : X ′ → Y ′,(T ′)−1 : X ′→ Y ′.

(⇒) If T is invertible, then (by 0.8 (2) + (3))

IX ′ = (IX )′ = (T−1T )′ = T ′(T−1)′

andIY ′ = (IY )

′ = (T T−1)′ = (T−1)′T ′

Hence: (T ′ is invertible, and) (T ′)−1 = (T−1)′ ∈B(X ′, Y ′).

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(⇐) Assume now T ′ is invertible. By the above, T ′′ : X ′′ −→ Y ′′ is invertible – in particularit is a bijection and continuous. Therefore, by the open mapping theorem, it mapsclosed subsets in closed subsets. Recall that by 0.8 T ′′ιX = ιY T (ιX , ιY are the canonicalembeddings) and that ιX , ιY are isometries, hence:

R(ιY T ) =R(T ′′ιX ) = T ′′(R(ιX ))⊆ Y ′′

is closed in Y ′′ because: Take a convergent sequence in R(ιX ) ⊆ Y . Then it’s Cauchyand also Cauchy in X (ιX isometry). Therefore it is convergent with limit in X . Then (ιXisometry) the limit in X ′′ of the sequence in R(ιX ) is in R(ιX ).

It follows that R(T ) = ι−1Y (R(ιY T )) is closed in Y . Since T ′ is injective, it follows that

0 = N(T ′). By Lemma 0.10 we get 0 = N(T ′) = Annil(R(T )). It now follows fromthe proposition 0.11 (a consequence of the Hahn-Banach Theorem) that

Y =R(T )

which is equal to R(T ) since R(T ) is closed, hence T is surjective. Since T ′′ is injective(since T ′′ = (T ′)′ is invertible) we have 0 = N(T ′′ιX ) = N(ιY T ) = N(T ). So T isinjective. Hence T−1 exists (and is linear) and is automatically bounded (i.e. continuousby the open mapping theorem), i.e. T−1 ∈B(X ).

Lemma 0.10. Let W be a Banach space. For a subspace Z ⊆W let

Annil(Z) = w′ ∈W ′ | w′(z) = 0 ∀z ∈ Z ⊆W ′

be the set of bounded linear functionals on W which vanish on all of Z ⊆ W. For T ∈ B(X , Y )we have

N(T ′) = Annil(R(T ))

Proof. Let y ′ ∈ N(T ′), i.e. T ′ y ′ = 0 (∈ X ′) iff (T ′ y ′)(x) = 0 ∀x ∈ X (definition of a mapbeing 0) iff y ′(T x) = 0 ∀x ∈ X (definition of T ′) iff y ′ ∈ Annil(R(T )).

Proposition 0.11. Let W be a normed space, Z ⊆W a closed subspace and w0 /∈ Z. Then thereexists w′ ∈W ′ such that w′ ≡ 0 on Z, ‖w′‖= 1 and w′(w0) = dist(w0, Z)> 0

The proof uses [FA1]:

Theorem (Hahn-Banach; complex version). Let X be a C-vector space. Let p : X → R with

p(αx + β y)¶ |α|p(x) + |β |p(y)

for every x , y ∈ X and α,β ∈ C with |α|+ |β | = 1. Let Y ⊆ X be a subspace and λ: Y → Clinear with |λ(x)| ¶ p(x) for every x ∈ Y . Then there exists a linear extention Λ: X → C of λ(i.e. Λ|Y = λ ) with

|Λ(x)|¶ p(x) ∀x ∈ X .

Proof of 0.11. Set U := span(Y ∪ w0) and define f : U −→ K as f (w0) := dist(w0, Z) > 0because Z is closed and f (z) = 0 ∀z ∈ Z . For some u ∈ U , we decompose as u = λx0 + z forsome λ ∈K and z ∈ Z and extend f linearly on U by setting f (u) := λ f (w0)+ f (z) = λ f (w0).We have

| f (u)|= | f (λw0) + f (z)|= dist(w0, Z) = infy∈Z‖w0− y‖¶ ‖λw0− z‖= ‖u‖

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i.e. | f (u)|¶ ‖u‖ for every u ∈ U . By Hahn-Banach with p = ‖·‖ we obtain a linear extension `of f on W with |`(w)|¶ ‖w‖ ∀w ∈W , so ‖`‖¶ 1. By closeness of Z there is some (yn)n with

‖w0− yn‖n→∞−−−→ dist(w0, Z)

and then

dist(x0, Z) = |`(w0)− `(yn)|¶ ‖`‖‖w0− yn‖n→∞−−−→ ‖`‖dist(w0, Z)

hence ‖`‖¾ 1 yielding ‖`‖= 1.

Proposition 0.12 (Neumann-series). Let X be a Banach space, T ∈B(X ), with ‖T‖< 1. Then(I − T )−1 ∈B(X ) and

(I − T )−1 =∞∑

n=0

T n (∗)

in B(X ), i.e. convergence in the operatornorm in B(X ).

Proof. See [FA1], Lemma 5.21.

Remark 0.13. (i) The series in (∗) is called the Neumann-series (for T) (when applied foroperators)

(ii) Of course

(T − 1 · I)−1 = (T − I)−1 =−(I − T )−1 =−∞∑

n=0

T n ∈B(X )

Hence: ‖T‖< 1 implies 1 ∈ ρ(T ) (“spectral property”).

(iii) “Perturbation result”: The identity I : X −→ X is invertible. If one adds something nottoo big (a “small perturbation”, i.e. “add −T”) then the result (I − T ) is also invertible(“perturbing I a little preserves invertability”).

Corollary 0.14. Let X , Y be Banach spaces. Then the subset (of B(X , Y )) of invertible operatorsin B(X , Y ) is an open subset (in B(X , Y )). More precisely: For S, T ∈B(X , Y ): If T is invertibleand ‖S− T‖< ‖T−1‖−1, then S is invertible.

Note: Again, a “perturbative result”: If S is close to T and T is invertible, then so is S.

Remark 0.15. “Functions of an operator”:

(1) For T ∈ B(X ), we have the obvious definition T2 = T T = T T , T n+1 := T T n, T1 = T ,T0 := I (already used), and T−1 is the inverse of T (if exists!) and (exercise) thenT−n := (T−1)n (Note: T kT m = T k+m).

(2) More generally: If p : C−→ C is a polynomial, then, if

p(z) = a0+ a1z+ · · ·+ anzn =n∑

i=0

aizi

we can define p(T ) ∈B(X ) the obvious way, by

p(T ) :=n∑

i=0

ai Ti ∈B(X )

Exercise: For p, q polynomials as above: p(T )q(T ) = (pq)(T ).

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(3) We have already seen other “functions of T”, namely the resolvent of T at λ:

Rλ(T ) = (T −λI)−1 ∈B(X ) λ ∈ ρ(T )

and, if ‖T‖< 1, the Neumann-series:

(I − T )−1 =∞∑

n=0

T n ∈B(X )

i.e. f (T ) for f : C \ 1 −→ C, z 7−→ 11−z= (1− z)−1.

(4) More generally: Let f (z) =∑∞

n=0 anzn be a powerseries in K (being R or C) with conver-gence radius r > 0. Let X be a Banach space and T ∈ B(X ). Then, by the same type ofargument that goes into proving Prop 0.12 (see [FA1]),

∑∞n=0 anT n converges (in B(X ),

i.e. in operatornorm) for all T ∈B(X ) which satisfy ‖T‖< r (r = 1 in Prop 0.12).

NB. In fact, both here, and in 0.12, it is enough (exercise) to have

limsupm→∞

T m

1/m< r

Conclusion: For T ∈B(X ) we can define f (T ) ∈B(X ) for functions f which are powerseries(analytic functions).Hence a part of spectral theory is the goal of enlarging the class of functions f for which wecan make sense of “ f (T )” (possibly at the price of not making sence for all T ∈ B(X ), asabove for example (‖T‖< r needed).

Why do so?!? - We shall see later what this is “good for”. Also, we shall study the relationshipbetween spectra of T and f (T ).

Examples 0.16. (1) For all T ∈B(X ) we define the exponential function by

exp(T ) := eT :=∞∑

n=0

1

n!T n ∈B(X )

(in ODE for matrices...). For S, T ∈B(X ): If ST = TS then eT+S = eT eS (otherwise mayfail).

(2) For T ∈ B(X ), define A(s) := esT ∈ B(X ) for s ∈ R (R 3 s 7−→ A(s) ∈ B(X ); A ∈C∞(R,B(X )) then

d

dsA(s) = TA(s) = A(s)T

A(0) = I

(i.e. A′ = TA, A(0) = I . Note the analogy to f (t) = eat , f ′ = a f , f (0) = 1).

(3) For T ∈B(X ) with ‖I − T‖< 1 let

ln(T ) = log(T ) =−∞∑

n=1

1

n(I − T )n ∈B(X )

(4) For T ∈B(X ) with ‖T‖ < 1 and s ∈ R, s < 1 define A(s) = log(I − sT ) (as in (3)). ThenA∈ C∞((−1,1),B(X )) with

d

dsA(s)

!=−T (I − sT )−1 =−(I − sT )−1T

and (!) exp(A(s)) = exp(log(I − sT )) = I − sT = (exp log)(I − sT ).

NB. This says f (g(T )) = ( f g)(T ). We already saw this for f (x) = x−1, g(x) = xn.

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1 Spectral Theory for compact operators

For a while we will be interested in the point spectrum σp(T ) for operators T ∈B(X ) (or, forsome of them...), i.e. we look at the eigenvalue problem for T .

Given y ∈ X , λ ∈K, x ∈ X we look for all solutions to

(T −λI)x = T x −λx = y (∗)

If λ ∈ ρ(T ) (so that (T − λI)−1 ∈ B(X ) exists), then (∗) has a unique solution x0 (for ally ∈ X , x0 = x0(y)) given by x0 := (T −λI)−1 y .

If, on the other hand, λ ∈ σp(T ) then a solution x0 to (∗) – if such exists – is not uniqueanymore. If x ∈ N(T −λI) (6= 0, when λ ∈ σp(T )) then

(T −λI)(x + x0) = (T −λI)x + (T −λI)x0 = 0+ y = y

So the number of “degrees of freedom” for solutions x equals (!) the dimension of N(T −λI).On the other hand, for a solution x0 to (∗) to exist we must have that y ∈ R(T −λI). This canbe thought of as a “constraint”. If we had/have a scalar product in X : y has to be orthogonalto the orthogonal complement (U := R(T − λI)⊥) of R(T − λI). So y ∈ R(T − λI) iff y ⊥ Uiff ⟨y, ui⟩ = 0 for all ui ∈ u j j ONB for U . I.e. the number of u′js in the basis is equal to thenumber of constraints on y . An important class of operators are those when both the “numberof degrees of freedom for x” (= dim N(T − λI)) as well as the “number of constraints on y”are finite.

Definition 1.1. An operator A∈B(X , Y ) is called a Fredholm operator iff

(i) dimN(A)<∞

(ii) R(A)⊆ Y is closed

(iii) codimR(A)<∞

The index of a Fredholm operator is then defined as:

ind(A) = dimN(A)− co dimR(A)<∞

Remark 1.2. co dimR(A)<∞ means

Y =R(A)⊕ Y0 (#)

with Y0 ⊆ Y a linear subspace with dim Y0 <∞ where the claim is that in this case,

codimR(A) := dim Y0

is independent of the choice of Y0 such that (#) holds. I.e. if also Y1 ⊆ Y is a linear subspacewith dim Y1 <∞ and Y =R(A)⊕ Y1 holds, then dim Y0 = dim Y1 holds aswell1.

A large and important class of Fredholm operators (the most important class?!?) is whenX = Y and A is a “compact perturbation” of the identity I :

1In a Hilbert space we can replace this with Y =R(A)⊕R(A)⊥. We see that “R(A) is closed” is important.

10

Theorem 1.3. Let X be a Banach space and T ∈ K(X ). Then A := I − T is a Fredholm operatorwith index zero. In particular

(1) dim N(A)<∞

(2) R(A) is closed.

(3) N(A) = 0 implies R(A) = X .

NB. “close to finite dimensions”!!!

(4) codimR(A)¶ dimN(A)

(5) dimN(A)¶ co dimR(A)

NB. The information in (1) - (5) is redundant, but as we refer to them later, we list all thepoints.

Proof. (1) Note that Ax = 0 is equivalent to x = T x . So B1(0) ∩N(A) ⊆ T (B1(0)) which ispre-compact in X since T ∈ K(X ). Hence N(A) is finite-dimensional.

(2) Let x ∈ R(A) then there is a sequence (yn)n∈N ⊆ R(A), with ynn→∞−−−→ x . I.e. there is

a sequence (xn)n∈N ⊆ X with Axn = yn ∈ R(A) such that Axnn→∞−−−→ x . Claim: We can

assume, with dn := dist(xn,N(A)), that ‖xn‖ ¶ 2 · dn. True because if not, we change xn:Choose an ∈N(A) with

‖xn− an‖¶ dist(xn,N(A)) + dist(xn,N(A))

= 2 dist(xn,N(A))≡ 2dn

NB. This is just the property of inf being the biggest lower bound. The right numbercould have been any b > 0. We chose it to be dist(xn,N(A)).

Let then xn := xn − an then Axn = Axn and ‖ xn‖ ¶ 2 dist(xn,N(A)) = 2dist( xn,N(A))(since an ∈ N(A)). Then the sequence (dn)n ⊆ R is bounded, because assume for con-tradiction that not, i.e. dn

n→∞−−−→ ∞ for a subsequence (called the same!). Let zn := xndn

,

then Azn =Axndn

n→∞−−−→ 0. Now ‖zn‖ ¶ 2, because ‖xn‖ ¶ 2dn, so (zn)n is bounded and T

is compact, so there is a subsequence (called the same) such that Tznn→∞−−−→ z for some z.

Hence (A= I − T),

zn = Azn+ Tznn→∞−−−→ z (∆)

and, A is continuous, soAz = lim

n→∞Azn = 0

(see (∆)). So z ∈N(A). Hence

‖zn− z‖¾ dist(zn,N(A))

= dist

xn

dn,N(A)

=1

dndist(xn,N(A)) = 1

11

which is a contradiction to (∆). Hence, (dn)n is a bounded sequence. So, since ‖xn‖ ¶2dn ∀n ¾ 1, it follows that (xn)n is bounded. Since T ∈ K(X ), there exists a subsequence(called the same) such that T xn

n→∞−−−→ u for some u. Hence

R(A) 3 xn→∞←−−− yn = Axn = A(A+ T )xn = A(Axn+ T xn)

n→∞−−−→ A(x + u)

So x = A(x + u) (uniqueness of limit). So x ∈R(A) and R(A) is closed.

(3) Assume that N(A) = 0 and that, for contradiction, there exists an x ∈ X \R(A). Claim:Then

An x ∈R(An) \R(An+1) (∆∆)

for every n ¾ 1. True because, if not, then there exists n ∈ N, such that An x ∈ R(An+1),then An x = An+1 y for some y , so

A(An−1(x − Ay)) = An(x − Ay) = 0

i.e. An−1(x − Ay) = 0, since N(A) = 0 and An−1 x = An y . By induction x − Ay = 0. Butthen x = Ay ∈ R(A), contradiction (we chose x ∈ X \R(A)). R(An+1) is closed2. RecallAn x ∈R(An) \R(An+1) for every n¾ 1. So then there exists an+1 ∈R(An+1)

0< ‖An x − an+1‖¶ 2 dist(An x ,R(An+1)) ()

Let now

xn :=An x − an+1

‖An x − an+1‖∈R(An)

For y ∈R(An+1) we have

‖xn− y‖=‖An−

∈R(An+1)︷ ︸︸ ︷

(an+1+ ‖An x − an+1‖y)‖‖An x − an+1‖

¾dist(An x ,R(An+1)‖An x − an+1‖

¾1

2(by ())

So dist(xn,R(An+1))¾ 12. Now, for m> n we have R(Am)⊆R(An):

‖T xn− T xm‖= ‖xn− (Axn+ xm− Axm)︸ ︷︷ ︸

∈R(An+1)

‖¾1

2(∗∗)

But ‖xn‖= 1 by definition, for all n ∈ N, so (xn)n is bounded. Since T is compact, (T xn)nshould at least have one convergent subsequence, but (∗∗) says it does not. Contradiction!Hence X =R(A), so A is surjective.

(4) By (1) we have that n := dimN(A) < ∞ is finite. Let x1, . . . , xn be a basis for N(A)and assume for contradiction that codimR(A) > dimN(A). Then there exist linearlyindependent vectors y1, . . . , yn ⊆ X such that

spany1, . . . , yn ⊕R(A)& X (∗)

2This is by (2), An+1 is still Fredholm according to () in the proof of Theorem 1.4 (3)

12

Then, by Theorem 1.5 there exists x ′1, . . . , x ′n ⊆ X ′ such that x ′`(xk) = δk`, k,` =1, . . . , n. For x ∈ X define

T x := T x +n∑

k=1

x ′k(x)yk

This defines a compact operator T ∈ K(X ). Let A= I − T then N(A) = 0. True becauseif Ax = 0, then

0= Ax = (I − T )x︸ ︷︷ ︸

=Ax∈R(A)

−n∑

k=1

x ′k(x)yk

︸ ︷︷ ︸

∈spany1,...,yn

and by (∗) it follows that Ax = 0 and x ′k(x) = 0 for k = 1, . . . , n (since y1, . . . , yn islinearly independent). But x ∈N(A) and then x =

∑nk=1αk xk for some αk ∈K. Hence

0= x ′`(x) =n∑

k=1

αk x ′`(xk) = α` `= 1, . . . , n

So x = 0. Hence N(A) = 0. Now use (3) on A, then R(A) = X . Since Ax` = −y` for`= 1, . . . , n and

A

x +n∑

`=1

x ′`(x)x`

= Ax ∀x ∈ X

so X = R(A) ⊆ spany1, . . . , yn ⊕ R(A). This contradicts (∗). Hence codimR(A) ¶dimN(A).

(5) Let m := co dimR(A) and n := dimN(A). Then by (4) m ¶ n. We first reduce to the casem= 0. As in (4), choose x1, . . . , xn ∈N(A) and x ′1, . . . , x ′n ∈ X ′ and y1, . . . , ym ∈ X with

X = spany1, . . . , ym ⊕R(A)

As in the proof of (4) it follows that the map

T x := T x +m∑

k=1

x ′k(x)yk ∈ K(X )

is compact and A= I − T is surjective with N(A) = spanx i | m < i ¶ n. It then remainsto prove that N(A) = 0 (i.e. dimN(A) = 0). This means the statement is reducedto be proved for the case m = 0. In the case m = 0 we have R(A) = X . Assume, forcontradiction, that there exists x1 ∈N(A)\0. Since A is surjective, we can – inductively– choose a sequence xk ∈ X , k ¾ 2 with Axk = xk−1. Then xk ∈ N(Ak) \N(Ak−1). By theRiesz-Lemma there exists zk ∈ N(Ak) with ‖zk‖ = 1 and dist(zk,N(Ak−1)) ¾ 1

2. Then for

` < k

‖Tzk − Tz`‖= ‖zk − (Azk + z`− Az`)︸ ︷︷ ︸

∈N(Ak−1)

‖¾1

2

So (as earlier...) the sequence (Tzk)k does not have any convergent subsequence. How-ever, ‖zk‖= 1 is a contradiction to the compactness of T .

This result is important in itself, but also essential in proving the following:

13

1.1 Spectral Theorem for compact operators

Theorem 1.4 (Spectral Theorem for compact operators, Riesz-Schauder). Let X be a Banachspace. For every compact T ∈ K(X ) one has

(1) σ(T )\0 consists of countably (finite or infinitely) many eigenvalues, with 0 as only possibleaccumulation point. If σ(T ) contains infinitely many points, then it follows

σ(T ) = σp(T )∪ 0

(= σ(T ), since σ(T ) closed).

NB. 0 might not be an eigenvalue!

(2) For λ ∈ σ(T ) \ 0 one has

1¶ nλ :=maxn ∈ N | N((T −λI)n−1) 6=N((T −λI)n)<∞

nλ is the order (or index3...) of λ (as an eigenvalue) and dim(N(T −λI)) is the multiplicityof λ.

(3) (Riesz decomposition). For λ ∈ σ(T ) \ 0 one has

X =N((T −λI)nλ)⊕R((T −λI)nλ) (∗)

Both subspaces are closed, T -invariant and N((T−λI)nλ) is finite-dimensional (“generalisedeigenspace”).

(4) For T |R((T−λI)nλ ) : R((T−λI)nλ)−→R((T−λI)nλ) we haveσ(T |R((T−λI)nλ )) = σ(T )\λ.

(5) Let, for λ ∈ σ(T ) \ 0, Eλ : X −→ X be the projection on N((T − λI)nλ) according to (∗),then EµEλ = δµλEλ for λ,µ ∈ σ(T ) \ 0.

Proof. (1) Let λ /∈ σp(T ), λ 6= 0. Then N(I − T/λ) = N(λI − T ) = 0 since λ /∈ σp(T ). So,since T/λ is compact, it follows from Theorem 1.3 (3) that

R(T −λI) =R(λI − T ) =R(I − T/λ) = X

So λ ∈ ρ(T ) (i.e. λ /∈ σ(T )). In other words

σ(T ) \ 0 ⊆ σp(T ) (∆)

If σ(T ) \ 0 is not finite, choose λn ∈ σ(T ) \ 0, n ∈ N with λn 6= λm for n 6= m, andeigenvectors en 6= 0 corresponding to λn (Ten = λnen, see (∆)) and define

Xn := spane1, . . . , en

(dim Xn ¶ n<∞). Claim: Then the en’s are linearly independent (in particular, dim Xn =n). True because: By induction in n. n = 1 is trivial. Assume the claim is true for n− 1

3This is no good choice since we use “index” for Fredholm operators.

14

(i.e. e1, . . . , en−1 is linearly independent). Assume that en =∑n−1

k=1 αkek (for n ∈ N) andαk ∈ C. Then

0= Ten−λnen = T

n−1∑

k=1

αkek

−λn

n−1∑

k=1

αkek

=n−1∑

k=1

αk(Tek −λnek) =n−1∑

k=1

αk(λkek −λnek)

=n−1∑

k=1

αk(λk −λn)ek =:n−1∑

k=1

βkek

where βk := αk(λk − λn) ∈ C. Hence, βk = 0 for all k, so αk = 0 for every k (sinceλn 6= λk ∀k!). So en =

∑n−1k=1 αkek = 0 . Hence, dim Xn = n and Xn−1 & Xn is a proper

subspace (meaning Xn−1 6= Xn). Also, Xn−1 is closed (since finite-dimensional). Then4

there exists an xn ∈ Xn, with ‖xn‖= 1 and dist(xn, Xn−1)¾ 1/2. Note that all the Xk ’s areT -invariant, for let x ∈ Xk, then x =

∑ki=1 βiei for some βi ∈ C. Then

T x =k∑

i=1

βi Tei =k∑

i=1

(βiλi)ei ∈ Xk

Also, since xn ∈ Xn, Xn−1 & Xn, and Xn = spane1, . . . , en we have xn = αnen + xn forsome αn ∈ C, xn ∈ Xn−1. Then

T xn−λn xn = αnTen+ T xn−αnλnen−λn xn

= αn(Ten−λnen︸ ︷︷ ︸

=0

) + T xn−λn xn = T xn−λn xn ∈ Xn−1

So, for m< n

T

xn

λn

− T

xm

λm

=

xn+1

λn(T xn−λn xn)−

1

λmT xm

︸ ︷︷ ︸

∈Xn−1 by calc just done m<n

¾1

2

Hence, the sequence(

T

xk

λk

)

k∈N

has no convergent subsequences (i.e. no accumulation points). Since T is compact, itmaps bounded sequences into sequences with convergent subsequences. So, the sequence

¨

xn

λn

«

n∈N

cannot have any bounded subsequence. So (‖xn‖= 1) we have that

1

|λn|= ‖xn/λn‖

n→∞−−−→∞

that is: λnn→∞−−−→ 0. This proves that 0 is the only possible accumulation point for σ(T ) \

0. In particular σ(T ) \ Br(0) is finite for all r > 0, so σ(T ) \ 0 is countable.4This is Lemma 2.20 in [FA1]: If X is normed, Y ⊆ X is a closed, proper subspace and θ ∈ (0,1) then ∃xθ ∈ X

with ‖xθ‖ = 1 and θ ¶ dist(x0, Y ) ¶ 1. Note: If X is an inner product space (Hilbert for example), one can findxθ such that dist(x0, Y ) = 1, take it in the orthogonal complement!

15

(2) Let A = λI − T = −(T − λI) Then N(An−1) ⊆ N(An) for all n (always!). Assume (forcontradiction) that N(An−1) & N(An) for all n ¾ 1. Note: N(S) = S−1(0) is always aclosed subspace when S is bounded. So N(An−1) is a closed, proper subspace of N(An)so we an choose xn ∈ N(An) such that ‖xn‖ = 1, dist(xn,N(An−1)) ¾ 1

2. Then for m < n

(T = λI − A)

T xn− T xm

=

λxn− (Axn+λxm− Axm)

= |λ| ·

xn−1

λ(Axn+λxm− Axm)

Claim: Axn+λxm−Axm ∈N(An−1). In this case, also 1λ(Axn+λxm−Axm) ∈N(An−1) so

‖T xn − T xm‖ ¾ |λ|dist(xn,N(An−1)) ¾ |λ|2> 0. To see the claim, we have xm ∈ N(Am),

m < n (by construction) and N(Ak−1) & N(Ak). So xm ∈ N(An−1) (n > m). Henceλxm ∈ N(An−1) and An−1(Axm) = A(An−1 xm) = A · 0 = 0, so Axm ∈ N(An−1), similarly:An−1(Axn) = An xn = 0 since xn ∈N(An), i.e. Axn+λxm− Axm ∈N(An−1).

But ‖xn‖ = 1 ∀n ∈ N, so (xn)n∈N is bounded and (T xn)n∈N has no convergent subse-quence which is a contradiction to T being compact. Hence there exists an n ∈ N suchthat N(An−1) =N(An). Set nλ := n. Then for m> n, x ∈N(Am):

Am−n x ∈N(An) =N(An−1)

(since An(Am−n x) = Am x = 0). Then An−1Am−n x = 0, so Am−1 x = 0 and x ∈ N(Am−1).So N(Am) = N(Am−1). By induction it follows N(Am) = N(An) = N(An−1) for all m ¾ n.Hence nλ <∞ and since λ ∈ σ(T ) \ 0 ⊆ σp(T ) we have

N(A) =N(λI − T ) =N(T −λI) 6= 0

so nλ ¾ 1.

(3) Let again A := λI − T and λ 6= 0, since λ ∈ σ(T ) \ 0, so by Theorem 1.3

A= λ(I − T/λ)

is a Fredholm operator. We claim that N(Anλ) ⊕ R(Anλ) ⊆ X , i.e. the two subspacesform a direct sum. Their intersection is 0. True because if x ∈ N(Anλ) ∩R(Anλ) thenAnλ x = 0 and x = Anλ y for some y ∈ X . Then A2nλ y = AnλAnλ x = Anλ x = 0. Soy ∈N(A2nλ) =N(Anλ) (by (2)), so x = Anλ y = 0, i.e. N(Anλ)∩R(Anλ) = 0. Note that5

Anλ = (λI − T )nλ = λnλ I +nλ∑

k=1

nλk

λnλ−k(−T )k

Note that if B, C ∈ B(X ), B or C compact, then BC is compact6, so (−T )k is compact.Hence

nλ∑

k=1

nλk

λnλ−k(−T )k ()

is in K(X ). So Anλ = λnλ(I − T ), T ∈ K(X ), i.e. by Theorem 1.3, Anλ is Fredholm.

5Binomial Formula (B+ C)n =∑n

k=0

nk

BkCn−k if BC = CB, B, C ∈B(X )6By exercise sheet 1

16

NB. Note that there is nothing particular about nλ here. This would have worked withevery n ∈ N.

Therefore, by 1.3 (4) and 1.3 (1) we have

co dimR(N nλ)¶ dimN(Anλ)<∞

so it follows that X = N(Anλ)⊕R(Anλ). Note that T commutes with A (AT = TA) and soTAnλ = AnλT , therefore both N(Anλ) and R(Anλ) are T -invariant.

(4) From (3) we have that R(Anλ) =R((λI−T )nλ) is T -invariant. Let Tλ be the restriction ofT to R(Anλ). Then Tλ is still compact and by Theorem 1.3 (2) R(Anλ) is a closed subspaceof X , since Anλ is Fredholm (see above). Hence, R(Anλ) is a Banach space. Also we have7

N(λI − Tλ) =N(A)∩R(Anλ) = 0

So λI − Tλ is injective and by 1.3 (3) it follows that λI − Tλ is onto. So we have

R(λI − Tλ) =R(Anλ)

So λ ∈ ρ(Tλ).

It remains to prove σ(Tλ) \ λ = σ(T ) \ λ. Let µ ∈ C \ λ. Recall that N(Anλ) is alsoT -invariant, hence so is under µI − T (so was R(Anλ) – already used). We claim: µI − Tis injective on N(Anλ). Assume x ∈N(µI − T ), then (µI − T )x = 0, i.e. (λ−µ)x = Ax . IfAm x = 0 for some m¾ 1, then

(λ−µ)Am−1 x = Am−1(λ−µ)x = Am−1(Ax) = Am x = 0

Hence (λ 6= µ), Am−1 x = 0. By induction we get that x = A0 x = 0. So N(µI − T ) ∩N(Am) = 0 for all m¾ 1. In particular, for m= nλ: N(µI − T )∩N(Anλ) = 0. So

(µI − T )|N(Anλ )

is injective. But N(Anλ) is finite dimensional (by 1.3), so µI − T is a bijection on N(Anλ).Hence, wheter or not µI − T is invertible (with bounded inverse...) on X is equivalent towhether or not it is invertible on R(Anλ). That is µ ∈ ρ(T )⇐⇒ µ ∈ ρ(Tλ).

Summing up: By splitting off (off X ) the finite dimensional characteristic subspace N(Anλ)belonging to the eigenvalue λ, we get a “rest operator” Tλ (on R(Anλ)) for which σ(Tλ) =σ(T ) \ λ.

(5) Let λ,µ ∈ σ(T ) \ 0, λ 6= µ and let Aλ := λI − T , Aµ := µI − T . Let x ∈ N(Anµµ ) ⊆

X = N(Anλλ)⊕R(Anλ

λ). Then there exist unique elements z ∈ N(Anλ

λ), y ∈ R(Anλ

λ) such

that x = z + y . Both N(Anλλ) and R(Anλ

λ) are T -invariant, so also Aµ-invariant, hence

Anµµ -invariant. So, since x ∈N(A

nµµ )

0= Anµµ x = A

nµµ z

∈N(Anλλ)

+ Anµµ y

∈R(Anλλ)

so, by X =N(Anλλ)⊕R(Anλ

λ) we get

Anµµ z = 0= Anλ

µ y

7Since N(Anλ)∩R(Anλ) = 0 and N(A)⊆N(Anλ).

17

Recall that Aµ is a bijection on N(Anλλ), so A

nµµ is also a bijection on N(Anλ

λ). Hence z = 0.

It follows that x = y ∈R(Anλλ). So

N(Anµµ )⊆R(Anλ

λ) (∗)

Now Eµ : X −→ X is the projection on N(Anµµ ) (ditto for λ). So

R(Eµ) =N(Anµµ ) R(Eλ) =N(Anλ

λ)

So (∗) says: R(Eµ)⊆N(Eλ). Interchanching the roles of λ and µ, we get R(Eλ)⊆N(Eµ)hence (check!) EµEλ = δλµEλ for µ,λ ∈ σ(T ) \ 0.

Theorem 1.5. Let X be a normed space and E an n-dimensional subspace with basis e1, . . . , en.Let Y ⊆ X be a closed subspace with Y ∩ E = 0. Then

(1) There exists e′1, . . . , e′n ∈ X ′ with e′j = 0 on Y and e′i(e j) = δi j for i, j = 1, . . . , n

(2) There exists a projection P ∈B(X ) on E, with Y ⊆N(P).

Proof. (1) Let Yj := spanek | k 6= j ⊕ Y . Then, by Lemma 1.6, Yj ⊆ X is a closed subspace.Also e j /∈ Yj . Hence, by Lemma 1.7, there is e′j ∈ X ′ with e′j ≡ 0 on Yj and e′j(e j) = 1. Inparticular e′j ≡ 0 on Y and e′j(ek) = 0 for all k 6= j, i.e. e′j(ek) = δ jk.

(2) Let

P x :=n∑

j=1

e′j(x)e j ∈ E

P continuous because it has finite rank and identically 0 on Y .

Lemma 1.6. Let X be a normed space. For all finite-dimensional subspaces E ⊆ X and Y ⊆ Xclosed with Y ∩ E = 0, then the direct sum Y ⊕ E ⊆ X is also a closed subspace.

Proof. Let X be the space X with the equivalence relation

x1 ∼ x2 :⇐⇒ x1− x2 ∈ Y

Then X with ‖x‖X := dist(x , Y ) is a normed vector space (since Y is closed). Since Y ∩E = 0the dimension of E, as a subspace of X is the same as its dimension as a subspace of X . Let

(yn)n ⊆ Y and (zn)n ⊆ E with yk + zkk→∞−−−→ x ∈ X . Then

‖zk − x‖X = dist(zk − x , Y )k→∞−−−→ 0

since ‖(zk− x)+ yk‖k→∞−−−→ 0 and yk ∈ Y . Hence (zk)k ⊆ E is a Cauchy sequence in (X ,‖ · ‖X ).

Since E is complete (it is a finite dimensional subspace of X ), so there is z ∈ E such that

‖zk − z‖Xk→∞−−−→ 0. I.e. x = z ∈ X by uniqueness of limits, i.e. x − z =: y ∈ Y by the definition

of X . That is, x = z+ y ∈ E ⊕ Y .

Lemma 1.7. Let X be a normed space and Y & X a closed proper subspace and x0 /∈ Y . Thenthere is x ′ ∈ X ′ such that x ′ ≡ 0 on Y , ‖x ′‖= 1 and x ′(x0) = dist(x0, Y ).

18

Proof. On Y0 := Y ⊕ spanx0, define y ′0(y + αx) := αdist(x0, Y ) for y ∈ Y , α ∈ K. Theny ′0 : Y0 −→K is linear and y ′ ≡ 0 on Y . For y ∈ Y , 0 6= α ∈K, then

dist(x0, Y )¶

x0−−y

α

Hence

|y ′0(y +αx0)|¶ |α|

x0−−y

α

=

αx0+ y

so y ′0 ∈ Y ′ with ‖y ′0‖ ¶ 1. To prove ‖y ′0‖ = 1. Since Y is closed, we have dist(x0, Y ) > 0, sofor every ε > 0 there is some yε ∈ Y such that

‖x0− yε‖¶ (1+ ε)dist(x0, Y )

Then

y ′0(x0− yε) = dist(x0, Y )¾1

1+ ε‖x0− yε‖

Now x0− yε 6= 0, so ‖y ′0‖¾1

1+εε→0−−→ 1, so ‖y ′0‖= 1.

1.2 Consequences of the spectral Theorem for compact operators

Proposition 1.8 (Fredholm Alternative). Let X be a Banach space. For T ∈ K(X ) and λ 6= 0one has that either the equation T x −λx = y has a unique solution x for any y ∈ X or that theequation T x −λx = 0 has non-trivial solutions.

NB. In this case “Either – or” is mutually-exclusive (not “∨”).

Proof. Follows from σ(T ) \ 0 ⊆ σp(T ).

NB. Compare with finite-dimensional case!

Proposition 1.9. Let X be a Banach space and T ∈ K(X ) and λ ∈ σ(T )\0. Then the resolventmap

ρ(T ) 3 µ 7−→ Rµ(T ) = (T −µI)−1

has an isolated pole of order nλ at λ (see 1.4 (2)). That is, the map

ρ(T ) 3 µ 7−→ (µ−λ)nλRµ(T ) ∈B(X )

can be continued at the point λ to an analytic map (i.e. with a convergent power series represen-tation, with values in B(X )) and its value at µ= λ is different from the 0-operator.

Proof. See e.g. Alt, “Lineare Funktionalanalysis”, 6. Auflage, p. 399.

1.3 The finite-dimensional case

Proposition 1.10 (finite-dimensional case). Let X be a finite-dimensional C-vector space, andT : X −→ X linear. Then there exist pairwise different λ1, . . . ,λm ∈ C, 1 ¶ m ¶ dim X such thatσ(T ) = σp(T ) = λ1, . . . ,λm and orders nλ j

(of λ j) with the properties in 1.4 (2) - (5). Thatis,

X =N((T −λ1 I)nλ1 )⊕ · · · ⊕N((T −λm I)nλm )

19

Proof. Equip X with your favorite norm, then X is Banach and T ∈ K(X ). So Tµ = T −µI forany µ ∈ C is also compact since I ∈ K(X ). Now apply 1.4 on (for example) T0 and T1.

Proposition 1.11 (Jordan Normal Form). Let X be a finite-dimensional C-vector space andT ∈ K(X ), λ ∈ σp(T ). For A= T −λI one has

(1) For n= 1, . . . , nλ there exist subspaces En ⊆N(An) \N(An−1) such that

N(Anλ) =nλ⊕

k=1

Nk Nk :=k−1⊕

`=0

A`(Ek)

(2) The subspaces Nk (k = 1, . . . , nλ) are T-invariant and dk = dim A`(Ek) is independent of` ∈ 0, . . . , k− 1

(3) If ek, j | j = 1, . . . , dk ⊆ Ek are a basis for Ek then

A`ek, j | 0¶ ` < k ¶ nλ j, 1¶ j ¶ dk

is a basis for N(Anλ) and with

x =∑

k, j,`

αk, j,`A`ek, j y =

∑

k, j,`

βk, j,`A`ek, j

the equation T x = y is equivalent to

λ −1 0 · · · · · · 0

0 λ −1 0 · · ·...

.... . . . . . . . . . . .

......

. . . . . . . . . . . . 0...

. . . . . . . . . λ −10 · · · · · · · · · 0 λ

αk, j,0............

αk, j,k−1

=

βk, j,0............

βk, j,k−1

(that is, the matrix representing T in this basis has Jordan Normal Form).

Proof. If E is a subspace withN(An−1)⊕ E ⊆N(An)

then N(An−`−1)⊕ A`(E) ⊆ N(An−`) for all 0 ¶ ` < n and A` is injective on E, since if x ∈ Ewith A`x = 0 then An−1 x = 0 since ` ¶ n− 1, hence x ∈ N(An−1) ∩ E = 0. Now using thisinductively, choose En for n= nλ, nλ− 1, . . . , 2, 1 such that

N(An) =N(An−1)⊕nλ−n⊕

`=0

A`(En+`)

from which the claim follows.

Theorem 1.12 (Schauder). T ∈ K(X , Y ) if and only if T ′ ∈ K(Y ′, Y ′)

Proof. See [FA1] Theorem 5.30 (b)

20

1.4 Spectral Theory for normal compact operators on Hilbert spaces

The last part of this chapter on spectral theory of compact operators is to investigate the caseof compact, normal operators on a Hilbert space.

We shall see: The Riesz-Schauder decomposition (Theorem 1.4)

X =N((T −λI)nλ)⊕R((T −λI)nλ)

in this case results in an orthonormal decomposition of the whole space X (H when Hilbert)in eigenspaces. This is the “diagonalisation” in the infinite-dimensional case.

Throughout let H be a Hilbert space.

Definition 1.13. T ∈B(H) is called a normal operator iff T T ∗ = T ∗T (i.e. [T, T ∗] = T T ∗ −T ∗T = 0 ∈B(H)).

Lemma 1.14. Let T ∈B(H). T is normal iff ‖T x‖= ‖T ∗x‖ for every x ∈ H.

Proof. (⇒) This is done by the computation

‖T x‖2 = ⟨T x , T x⟩=

x , T ∗T x

=

x , T T ∗x

=

T ∗x , T ∗x

= ‖T ∗x‖2

(⇐) From “polarisation”

1

4

‖x + y‖2−‖x − y‖2

=Re

x , y

∀x , y ∈ H

follows Re

T x , T y

=Re

T ∗x , T ∗ y

. If K= C, then replace y by iy to get

0=

T x , T y

−

T ∗x , T ∗ y

=

T ∗T x − T T ∗x , y

=

(T ∗T − T T ∗)x , y

Theorem 1.15. Let X be a Banach space and T ∈ B(X ). Then σ(T ) is a compact, non-emptysubset of C (if X 6= 0!) and

supλ∈σ(T )

|λ|= limm→∞

‖T m‖1/m ¶ ‖T‖

We call r(T ) := supλ∈σ(T ) the spectral radius of T .

NB. The sup is in fact a max and σ(T )⊆ B‖T‖(0).

We shall need some complex analysis for the proof.

Theorem (Cauchy’s Integral Theorem). Let Ω ⊆ C open and simply connected and assumef : Ω−→ C is analytic. Let γ: [a, b]−→ Ω by any closed path, i.e. γ(a) = γ(b). Then

∮

γ

f (z)dz = 0

The same holds when f : Ω −→ X where X is a C-Banach space (!!). Only the C-linear struc-ture and completeness are needed! Obviously we need to generalize the Riemann-integral tofunctions g : [a, b]−→ X – “easy” – using “middle sums”.

21

Proof. By Proposition 0.3, σ(T ) ⊆ C is open, so σ(T ) ⊆ C is closed. By Proposition 0.12, theoperator I − T/λ is invertible in B(X ) if ‖T‖< |λ| and in this case

Rλ(T ) =−1

λ(I − T/λ)−1 =−

1

λ

∞∑

n=0

T

λ

n

=−∞∑

n=0

1

λn+1 T n

Hence r(T ) ¶ ‖T‖ (since |λ| > ‖T‖ =⇒ λ ∈ ρ(T )). So σ(T ) is bounded and so (by Heine-Borel) σ(T ) is compact8). Note that

T m−λm I = (T −λI)pm(T ) = pm(T )(T −λI) pm(T ) =m−1∑

i=1

λm−1−i T i (!)

Hence if λ ∈ σ(T ) then9 λm ∈ σ(T m). So |λm| ¶ ‖T m‖ (just proved!) and |λ| ¶ ‖T m‖1/m forall m ∈ N. Hence r(T )¶ lim infm→∞ ‖T m‖1/m. We now show also

r(T )!¾ limsup

m→∞‖T m‖1/m

Then it follows that (‖T m‖1/m)m∈N is convergent, and r(T ) = limm→∞ ‖T m‖1/m. Recall thatProposition 0.3 says that ρ(T )⊆ C is open and that the map λ 7−→ Rλ(T ) is complex analytic,i.e. for all λ0 ∈ C\Br(0) (since σ(T )⊆ Br(0) we have ρ(T )⊇ C\Br(0) with r ≡ r(T )), thereexist c j(λ0) ∈B(X ), δ(λ0)> 0 such that

Rλ(T ) =∞∑

j=0

c j(λ0)(λ−λ0)j ∀λ ∈ Br(λ0)

Hence, as a consequence of the Cauchy Integral Theorem: For and 0¶ j ∈ N, and s > r(T ) = rthe (curve) integral

1

2πi

∮

∂ Bs(0)

λ jRλ(T )dλ

is independent of s. To see this we “deform the curve” and use Cauchy Integral Theorem (notdone here). Hence

1

2πi

∮

∂ Bs(0)

λ jRλ(T )dλ=−1

2πi

∮

∂ Bs(0)

∞∑

n=0

λ j−n−1T n

!

dλ

We curve ∂ Bs(0) as γ: [0,2π)−→ C, θ 7−→ seiθ with γ′(θ) = iseiθ

=−1

2π

∞∑

n=0

s j−n

2π∫

0

eiθ( j−n)dθ

T n

since the convergence is uniform

=−1

2π

∞∑

n=0

s j−n2πδ j,nT n =−T j

8This holds for any bounded operator – not only compact operators!9This relates σ( f (T )) and σ(T ) with f (x) = xm!

22

Hence, for any j ¾ 0 and s > r(T )

‖T j‖=1

2π

∮

∂ Bs(0)λ jRλ(T )dλ

=1

2π

∫ 2π

0

(seiθ ) jRseiθ (T )iseiθdθ

¶1

2π

2π∫

0

(seiθ ) jRseiθ (T )iseiθ

dθ ¶ s j+1 sup|λ|=s‖Rλ(T )‖ ·

1

1

2π

2π∫

0

dθ

= s j+1 · sup|λ|=s‖Rλ(T )‖

So for any s > r(T ) and any sequence j→∞ we have

‖T j‖1/j ¶ s

s · sup|λ|=s‖Rλ(T )‖

!1/j

j→∞−−→ s

Hence,limsup

j→∞‖T j‖1/j ¶ s ∀s > r(T )

So lim sup j→∞ ‖T j‖1/j ¶ r(T ). Assume that σ(T ) = ;, then for j = 0 and s 0

‖I‖= ‖T0‖¶ s sup|λ|=1‖Rλ(T )‖

s→0−−→ 0

i.e. ‖I‖= 0, so X = 0.

Proposition 1.16. Let H be a C-Hilbert space, H 6= 0 and T ∈B(H) be normal. Then

supλ∈σ(T )

|λ|= ‖T‖

Proof. Let T 6= 0. Since limm→∞ ‖T m‖1/m ¶ ‖T‖ it suffices to prove that

‖T m‖¾ ‖T‖m

for m¾ 0 (then in fact ‖T m‖= ‖T‖m). For m= 0, 1 this is clear. For m¾ 1 and x ∈ H:

‖T m x‖2 =

T m x , T m x

=¬

T ∗T m x , T m−1 x¶

¶ ‖T ∗T m x‖ · ‖T m−1 x‖

and since ‖T ∗ y‖= ‖T y‖ for (in particular) y = T m x , by Lemma 1.14, we have

¶ ‖T m+1‖ · ‖T m−1‖ · ‖x‖2

Hence ‖T m‖2 ¶ ‖T m+1‖ · ‖T m−1‖. If we assume, by induction ‖T m‖ ¾ ‖T‖m (so, in fact,equality!), then

‖T m+1‖¾‖T m‖2

‖T‖m−1 = ‖T‖2m−(m−1) = ‖T‖m+1

Definition 1.17. Let T ∈B(H) be self adjoint. T is called positive semi-definite, T ¾ 0, iff

⟨x , T x⟩¾ 0 ∀x ∈ H

23

Proposition 1.18. Let H be a C-Hilbert space, T ∈B(H).

(i) If T ∗ = T, thenσ(T )⊆ [−‖T‖,‖T‖]⊆ R

If even T ∈ K(H), then −‖T‖ or ‖T‖ is an eigenvalue of T (or both).

(ii) If T = T ∗ and T is positive semi-definite, then

σ(T )⊆ [0,‖T‖]⊆ R

If also T ∈ K(H) then ‖T‖ is an eigenvalue.

Proof. Let U := ⟨x , T x⟩ | ‖x‖= 1, x ∈ H ⊆ C. Since

⟨x , T x⟩= ⟨T x , x⟩= ⟨x , T x⟩

we have U ⊆ R. Let λ ∈ U c . We have to show that T −λI is invertible.

• Let 0 6= x ∈ ker(T −λI). Then for y ∈ H

0= ⟨y, (T −λI)x⟩= ⟨y, T x⟩ −λ⟨y, x⟩

Inserting y := ‖x‖−2 x yields λ= ⟨x , T x⟩, hence λ ∈ U . So T −λI is injective.

• Let A := T − λI . Since U is closed, we have dist(λ, U) ¾ ε > 0. Then R(A) is closed.True because, for x ∈ H with ‖x‖ = 1, we have ε ¶ |⟨x , T x⟩ − λ| and so for every0 6= x ∈ H:

⟨x , x⟩ε ¶ |⟨x , T x⟩ −λ⟨x , x⟩|= |⟨x , (T −λ)x⟩|

Let (T −λ)xn be a Cauchy-sequence in R(A) converging to z ∈ H. Then

‖xn− xm‖2 =

xn− xm, xn− xm

¶1

ε|

xn− xm, (T −λ(xn− xm)

¶1

ε‖xn− xm‖ · ‖(T −λ)(xn− xm)‖

So

‖xn− xm‖¶1

ε‖(T −λ)(xn− xm)‖

Hence, (xn)n ⊆ H is a Cauchy-sequence and therefore xn→ x for some x ∈ H and sinceA is continuous, we have Ax = z, so z ∈R(A).

• Now assume for contradiction that A is not surjective. Then there exists some non-zeroy ∈R(A)⊥ and for all x ∈R(A) it follows that

0= ⟨Ax , y⟩= ⟨x , Ay⟩

Inserting x = Ay gives Ay = 0. Hence A is not injective .

• Summing up, we have U ⊆ R and if λ ∈ U c , then A= T −λI is invertible. Hence

σ(T )⊆h

infx∈H,‖x‖=1

⟨x , T x⟩ , supx∈H,‖x‖=1

⟨x , T x⟩i

⊆ R (∗)

24

By 1.15, we have σ(T )⊆ B‖T‖(0)⊆ C, hence σ(T )⊆ [−‖T‖,‖T‖]. In fact

supλ∈σ(T )

|λ|= r(T ) = ‖T‖

by 1.16. If T ∈ K(H), then σ(T ) \ 0 consists of isolated eigenvalues, so

supλ∈σ(T )

|λ|= maxλ∈σ(T )

|λ|= λ0|

where λ0 ∈ σp(T ), hence λ0 =±‖T‖. Again by (∗)

σ(T )⊆h

0, supx∈H,‖x‖=1

⟨x , T x⟩i

⊆ R

So σ(T ) ⊆ [0,‖T‖]. If T ∈ K(H), we also have that ‖T‖ is an eigenvalue (all non-zero-eigenvalues are non-negative, so λ0 = ‖T‖ above.

Note (to 1.18 (2)). In particular: If T ∈ K(H), T ∗ = T and T is positive semi-definite, then

λ0 := supλ∈σ(T )

λ= supx∈H,‖x‖=1

⟨x , T x⟩=

x0, T x0

= λ0‖x0‖2 = λ0

I.e. maximising ⟨x , T x⟩ under the constraint ‖x‖ = 1 gives the largest eigenvalue, i.e λ0 ¾⟨x , T x⟩, ∀x ∈ H, ‖x‖= 1. This we can use to compute a lower bound to λ0 by choosing somex and compute. Afterwards we repeat this, for λ1 ¶ λ0 (next eigenvalue) by restricting T tospanx0⊥ =: H. H is invariant, since for x ∈ H:

T x , x0

=

x , T x0

= λ0

x , x0

= 0

So T x ∈ H. This can be used to compute the λ’s and their eigenspaces.

Recall the following (Problem 6, Sheet 2, T : `2 −→ `2 translated to general setting)

Example 1.19. Let H be a C-Hilbert space, e j j∈N , N ⊆ N an orthonormal system andλkk∈N ⊆K with |λk|¶ r <∞ for k ∈ N . Then

T x :=∑

j∈N

λ j⟨e j , x⟩e j

defines a normal, bounded operator. Also T is compact iff10 limk→∞λk = 0.

Proof. Let |N | = ∞ and N = n1, n2, . . . be an increasing enumeration of N ⊆ N. Let T be

compact. We know that en j+ 0 as j→∞. Hence, since T is compact, we have ‖Ten j

‖j→∞−−→ 0

and

0k→∞←−−− ⟨Tenk

, Tenk⟩=

*

∑

i∈N

λi⟨ei , enk⟩ei ,∑

j∈N

λ j⟨e j , enk⟩e j

+

=∑

i, j∈N

λiλ j ⟨enk, ei⟩

︸ ︷︷ ︸

=δi,nk

⟨e j , enk⟩

︸ ︷︷ ︸

=δ j,nk

⟨ei , e j⟩= |λnk|2

So λk→ 0. On the other hand if λkk→∞−−−→ 0 let Tk x :=

∑kj=1λn j

⟨en j, x⟩en j

, then

‖T x − Tk x‖=

∞∑

j=k+1

λn j⟨en j

, x⟩en j

¶∞∑

j=k+1

|λn j| · |⟨en j

, x⟩| · ‖en j‖¶ ‖x‖ ·

∞∑

j=k+1

|λn j| k→∞−−−→ 0

So T is a limit of finite-rank operators.

The following Theorem shows that every compact normal operator has this form.10If N is finite, T is compact.

25

1.4.1 The Spectral Theorem for normal compact operators on a Hilbert space

Theorem 1.20 (Spectral Theorem for normal compact operators on a Hilbert space). Let Hbe a C-Hilbert space, T ∈ K(H), T 6= 0 and T ∗T = T T ∗. Then

(1) There exists an orthonormal system e j j∈N with N ⊆ N and λk ∈ C \ 0 such that

Tek = λkek k ∈ N

and σ(T ) \ 0 = λk | k ∈ N, i.e. the λk ’s are the eigenvalues of T different from 0 witheigenvectors ek.

Note. Here (with this notation) λk ’s for different k’s may be equal.

If N is infinite, then λkk→∞−−−→ 0.

(2) For all k ∈ N we have nλk= 1.

(3) H =N(T )⊕ spanek | k ∈ N

Note. This is the orthogonal direct sum: N(T )⊥ spanek | k ∈ N

(4) We haveT x =

∑

k∈N

λk⟨ek, x⟩ek ∀x ∈ H

“T is diagonal in ekk∈N ”

Proof. Apply first the spectral Theorem for compact operators 1.4 on T : σ(T ) \ 0 consistsof eigenvalues λk, k ∈ N ⊆ N with λk → 0 if N is infinite11. Moreover, the space Nk :=N(T − λk I) is finite dimensional (see 1.4). Let N0 :=N(T ), λ0 := 0. Note that

N(T − λk I) =N

T ∗− λk I

k ∈ N ∪ 0 (∆)

which follows from ‖T x‖ = ‖T ∗x‖ for x ∈ H by 1.14. Claim: Nk ⊥ N` for k,` ∈ N ∪ 0,k 6= `. True, because

λ`⟨xk, x`⟩= ⟨xk, T x`⟩(∆)= ⟨λk xk, x`⟩= λk⟨xk, x`⟩

Hence ⟨xk, x`⟩= 0 and

H =⊕

k∈N∪0

Nk =: W ()

True because let y ∈ Y := W⊥. By the projection Theorem we have H = Y ⊕W (W closed).By (∆), we have for x ∈ Nk, k ∈ N ∪ 0:

⟨x , T y⟩= ⟨T ∗x , y⟩= ⟨λk x , y⟩= λk⟨x , y⟩= 0

Hence T y ⊥ Nk for k ∈ N ∪ 0, so, since T is continuous, T y ∈ Y . Hence, Y is a T -invariantclosed subspace of H. We look at T0 := T |Y : Y −→ Y . Then T0 ∈ K(Y ) and T0 is normal. Inthe case Y 6= 0, then since

maxλ∈σ(T )

|λ| != ‖T0‖

11In this ennumeration, the λk ’s are pairwise different.

26

there exists λ ∈ σ(T0) such that |λ| = ‖T0‖. If T0 6= 0 then ‖T0‖ 6= 0, so λ 6= 0 and λ is by 1.4(on T0) an eigenvalue for T0 : Y −→ Y . I.e. there is some non-zero u ∈ Y such that T0u= λu,i.e. Tu = λu, so λ is an eigenvalue for T . So u ∈ Nk ∩ Y 6= 0 for some k ∈ N . This leads to⟨u, u⟩ = 0 since Y ⊥ Nk which is a contradiction to u 6= 0. Hence T0 = 0, so Y ⊆ N(T ), butY ∩ N0 = 0 so Y = 0 .

Let Ek for k ∈ N ∪ 0 be the orthogonal projection on Nk (Problem 5 on Sheet 2), then, from()

x =∑

k∈N∪0

Ek x ∀x ∈ H

So

T x =∑

k∈N∪0

T (Ek x) =∑

k∈N

λkEk x (Ek x ∈ Nk) ()

Now the representation of T follows. Let dk := dim Nk <∞ and choose an orthonormal basis:ek1, . . . , ekdk

. Then

Ek x =dk∑

j=1

⟨ek j , x⟩ek j

From the representation () of T follows that

T x =∑

k∈N

λk

dk∑

i=1

⟨ek j , x⟩ek j

≡∑

k

λk⟨ek, x⟩ek

(re-number the λk ’s and ek j ’s). From the representation of T also follows that

Nk =N(T − λk I) =N((T − λk I)2)

True because for x ∈N((T − λk I)2):

0= (T − λk I)2 x()=

∑

j∈N∪0

(λk −λ j)2E j x (E2

j = E j)

So E j x = 0 for j 6= k, hence x = Ek x ∈ Nk.

Recall that T ∗T = T T ∗ and T ∈ K(H). Let Ek be the orthogonal projection on the eigenspaceN(T − λk I) = Nk, belonging to the eigenvalue λk (λk 6= λ`, k 6= `), that is

Ek x =dk∑

j=1

⟨ek j , x⟩ek j

The spectral Theorem says

T x =∞∑

k=1

λkEk x ∀x ∈ H

I.e. the series on the right converges pointwise to T x at any x ∈ H. In fact, we have more:

27

Corollary 1.21 (Spectral Theorem for normal, compact operators – Projection version). LetH be a C-Hilbert space, T ∈ K(H), T 6= 0 and and T ∗T = T T ∗, then

T =∞∑

k=1

λkEk

with convergence in operator norm on B(H).

Proof. The calculation with the notation as above gives

T −M∑

k=1

λkEk

=

∑

k>M

λkEk

and the operator∑

k>M λkEk is normal and compact, so its norm is its biggest eigenvalue:

= supk>M|λk|

M→∞−−−→ 0

Next, we shall use the spectral Theorem to compute square roots of positive semi-definiteoperators12.

Theorem 1.22. Let T ∈ K(H), T ∗ = T, T ¾ 0. Then there exists a unique, self-adjoint operatorS ¾ 0 with S2 = T. We write S := T 1/2 =

pT.

Proof. Write from the Spectral Theorem 1.20

T x =∑

k

λk⟨ek, x⟩ek

Since T ¾ 0, we have λk ¾ 0 1.18. Let

Sx :=∑

k

p

λk⟨ek, x⟩ek

Then, since (λk → 0, k→∞) =⇒ (p

λk → 0, k→∞) it follows from Example 1.19 that S iscompact, that S is self-adjoint and positive semi-definite. Also we have S2 x = T x , x ∈ H. Let’sprove uniqueness. Assume R ∈ K(H) satisfies R ¾ 0, R2 = T , R = R∗. Write, as in Corollary1.21

R=∞∑

k=1

νkFk

(convergence in operator norm, νk eigenvalues of R, Fk orthogonal projection on the eigenspaces).Then

T =∞∑

k=1

ν2k Fk

since FkF` = δk`Fk and νk ¾ 0 (since R ¾ 0, 1.18). Hence, the ν2k are the eigenvalues of T ,

and the Fk ’s are the eigenprojections of T on the eigenspaces. Hence, since the convergenceis pointwise unconditionally, we have R= S.

12I.e. f (T ), for f (x) =p

x .

28

Let T : H1 −→ H2 be a compact operator between two Hilbert spaces, then

T ∗T : H1 −→ H1

is positive semi-definite, self-adjoint and compact. Its unique square root is denoted by |T | :=(T ∗T )1/2.

Theorem 1.23 (Polar decomposition). For T ∈ K(H1, H2) there exists an operator U ∈B(H1, H2)with

T = U |T |

so that U |(N(U))⊥ is an isometry and hence a unitary operator between N(U)⊥ and R(U). U isuniquely determined by the demand N(U) =N(T ).

Proof. |T | is selfadjoint (all square roots are, by Definition). So

|T |x

2=¬

x , |T |2 x¶

=

x , T ∗T x

= ⟨T x , T x⟩= ‖T x‖2 (∆)

Hence: U(|T |x) := T x defines an isometric operator from R(|T |) to R(T ), which can by(uniquely) extended13 to

U : R(|T |)−→R(T )

(same notation: U). Let U x = 0 for x ∈ (R(|T |))⊥ = N(|T |) (since N(S) = (R(S∗))⊥,N(S∗) = R(S)⊥). This proves the existence of U (such that T = U(|T |)). The uniquenessfollows from N(|T |) =N(T ) (see (∆).

Remarks. (1) An operator with the properties of U is called a partial isometry.

(2) T = U |T | reminds of λ= eit |λ| (“U∗U = UU∗ = 1”). However, in general,

|S+ T |¶ |S|+ |T |

does not hold (for example). Here A¶ B :⇐⇒ B− A¾ 0.

Theorem 1.24 (Singular Value Decomposition – SVD). For any compact T ∈ K(H1, H2) thereexist orthonormal systems eii∈N ⊆ H1, f j j∈N ⊆ H2 and numbers

s1 ¾ s2 ¾ s3 ¾ · · ·¾ 0

with skk→∞−−−→ 0 such that

T x =∑

k∈N

sk⟨ek, x⟩ fk ∀x ∈ H1

The numbers s2k are the eigenvalues of T ∗T (counted with multiplicity). The sk ’s are called the

singular values of T .

Proof. Write T = U |T | as above, and use the Spectral Theorem for normal compact operators1.20 on |T |:

|T |x =∑

k∈N

sk⟨ek, x⟩ek

Then the sk ’s and the ek ’s have the properties in the Theorem. Let fk := U(ek) ∈ H2. SinceU is an isometry on R(|T |), which, by the polarization identity, means, it conserves the scalarproduct, it maps ekk∈N into an orthonormal system fkk∈N .

13This is Theorem 2.32 in [FA1].

29

Remarks. (1) There is no condition on “normal” or “self-adjoint” (applies when H1 = H2 = Hand T not normal).

(2) SVD is important, also on matrices in numerics (which are compact operators).

Outlook. (1) Compact operators occur (and were first studied) as integral operators:

(Tk f )(x) =

∫

Ω

k(x , y) f (y)dy f ∈ X

for various kinds/types of k, and various choices of X . For instance

• X =C ([0,1]), k continuous, ‖ · ‖X = ‖ · ‖∞• X = L2([0, 1]), k ∈ L2([0, 1]× [0, 1])

But, also weakly singular, or strongly singular kernels k for which Tk might not be com-pact ; Harmonic Analysis.

(2) Compact operators occur as embeddings: Certain embeddings of one space X into anotherspace Y (ι : X −→ Y , x 7−→ x , injective; different norms on X and Y ) are not onlycontinuous/bounded but also compact14.

(3) Both (1) and (2) are important in the study of partial differential equations (hence, ofquantum mechanics), for example in so called “boundary value problems”. For Exampleconsider the Dirichlet Problem: For a bounded domain Ω ⊆ Rn, ϕ ∈ C (∂Ω), find u ∈C 2(Ω)∩C (Ω) such that ∆u= 0 on Ω and u|∂Ω = ϕ.

(4) One can study compact operators according to the properties of their singular values: Letthe p’th Schatten-class be defined by

Sp :=

T ∈ K(H) : ‖T‖p :=∑

n∈Nsn(T )

p1/p

<∞

(with sn(T ) the singular values of T). Then ‖ · ‖p is a norm and (Sp,‖ · ‖p) is a Banachspace. These spaces share many properties with `p (for example (Sp)′ ∼= Sq with p, qHölder conjugate). Of special interest are S1, which allows to define the trace, and S2,the so-called Hilbert operators. In fact, S2 becomes a Hilbert space (just like `2). Inparticular: Th ∈ S2(L2[0, 1]) if h ∈ L2([0,1]× [0,1]) (see (1) above). For more on this,see “Trace Ideals” by B. Simon.

(5) For compact operators between Banach space, there is a notion of „nuclear operators“:T ∈B(X , Y ) is called nuclear iff

T x =∞∑

n=1

x ′n(x)yn

for x ∈ X with x ′nn ⊆ X ′, ynn ⊆ Y with∑

n‖x ′n‖ · ‖yn‖<∞

See books by Werner and Alt.

14For example, exercise 20 in FA1 Müller 2014: Let X := x ∈ `1 :∑

j∈N j|x j |<∞ and define ‖x‖∗ :=∑

j∈N j|x j |for x ∈ X . Then (X ,‖·‖∗) is a Banach space, whereas (X ,‖·‖1) is not. Moreover J : (X ,‖·‖∗)−→ (X ,‖·‖1), x 7−→ xis a continuous, compact operator.

30

2 Spectral Theory for bounded, self-adjoint operators on a Hilbertspace

In the previous chapter we completely characterized normal, compact operators on a Hilbertspace H – and used this to give a representation for all compact operators (see 1.24). In thischapter, we shall look at more general operators, namely bounded ones – but we shall restrictourselves to self-adjointness. As before, all Hilbert spaces will be over C and separable.

Motivation. If T : Cn −→ Cn is normal, then it can be diagonalised: T is unitarily equi-valent to a diagonal matrix D: U T U−1 = D (i.e. U is unitary). The diagonal elements areof course the eigenvalues of T (with multiplicities). So (U T U−1 x)i = λi x i (with λ1, . . . ,λnthe eigenvalues of T). So D can be thought of as a multiplication operator. Remember theanalogy1

(T f )(x) =

∫

Ω

k(x , y) f (y)dy (Ax)i =n∑

j=1

a(i, j)x j

and the multplication operator (Problems 4,16)

(Mm f )(x) := m(x) f (x)

for example with m ∈ L∞([0, 1]), f ∈ L2([0, 1]). One can then define the matrix f (T ) as

(U f (T )U−1 x)i = f (λi)x i (∗∗)

(undo the U ’s yourself to get f (T )!) as long as λi belongs to the domain of f . We can alsodo this for compact operators – although then we have to think about convergence (as well:think about whether f is well-defined, i.e. defined on σ(T )). We have also already seen howto define f (T ) for T ∈B(X ) if f is analytic. Another way of looking at (∗∗) is:

T =∑

j

µ j E j

with µi 6= µk, i 6= k the different eigenvalues and the E j ’s the orthogonal projection on thecorresponding eigenspaces (see also 1.21!).

We shall elaborate on these different aspects and start by defining f (T ) for continuous func-tions f and T ∈B(H), T self-adjoint.

2.1 The Continuous Functional Calculus

Theorem 2.1 (Continuous Functional Calculus). Let T ∈ B(H), T ∗ = T. Then there exists aunique map

Φ: C (σ(T ))−→B(H)

f 7−→ Φ( f ) =: f (T )

such that:

(a) Φ(t) = T and Φ(1) = 1H where t : σ(T )−→ C, t 7−→ t, and 1: σ(T )−→ C, t 7−→ 1.

(b) Φ is an “involution”, i.e. Φ( f ) = Φ( f )∗, and Φ is an algebra homomorphism.

1One can think of x ∈ Cn as x : 1, . . . , n −→ C, i 7−→ x(i)≡ x i

31

(c) Φ is continuous – i.e. bounded – from (C (σ(T )),‖ · ‖∞) to (B(H),‖ · ‖). In fact: ‖Φ( f )‖=‖ f ‖∞ (hence Φ is isometric and therefore injective).

The map Φ is called the Continuous Functional Calculus of T . We write f (T ) := Φ( f ) forf ∈ C (σ(T )), i.e. t 7−→ T , 1 7−→ 1H . “Any ‘formula’ that holds for continuous functions,holds for the Functional Calculus”. This can be thought of as “ex is the unique solution tof ′ = f , f (0) = 1. On the other hand it satisfies ex =

∑∞n=0

xn

n!”. Hence, every function that

satisfies either of the properties is necessarily the exponential.

We shall, in both the proofs of uniqueness and existence use the following two theorems:

Theorem 2.2 (Weierstraß’ Approximation Theorem). Let f ∈ C ([a, b]), ε > 0. Then thereexists a polynomial p = pε, such that ‖ f − p‖∞ < ε.

In other words2 P ([a, b]) is dense in (C ([a, b]),‖ · ‖∞).

Proof. Exercise on Problem sheet/later (?) or see [FA1] Corollary 1.45.

Theorem 2.3 (Tietze-Urysohn for metric spaces). Let M be a metric space, and A ⊆ M aclosed subset. For all continuous functions f : A −→ [c, d] there exists a continuous extensionF : M −→ [c, d] (i.e. F is continuous and F |A = f ). In particular: ∀t0 /∈ A there is ϕ : M −→[0,1] continuous with ϕ|A = 0, ϕ(t0) = 1.

Proof. It is enough to look at c = 0, d = 1 (map to [a, b] linearly). Let

F(t) :=

f (t) t ∈ A

inf f (s)·d(s,t) | s∈Ainfd(s,t) | s∈A t /∈ A

and note that F(t) ∈ [0,1] for each t ∈ M (since f (t) ∈ [0, 1] for every t ∈ A). One easily seesthat F is continuous. For t0 /∈ A define ϕ0 : A0 := A∪ t0 −→ [0,1] by ϕ0|A = 0, ϕ0(t0) = 1,and extend to ϕ as above (ϕ0 is continuous on A, since t0 /∈ A, A closed!).

Note. Tietze-Urysohn holds in more generality in topological spaces – but not in all – only inso-called normal spaces. In particular, it holds in compact Hausdorff spaces.

Proof of Theorem 2.1. Theorem 2.1 will mean that Φ is unique if it is unique on P (since it islinear, and continuous/bounded, it has a unique extension). Recall that σ(T )⊆ R is compactby 1.15, so there are a, b ∈ R such that σ(T ) ⊆ [a, b]. We show uniqueness first. SinceΦ is linear, it is enough that Φ(tn), n ¾ 0 is unique and by multiplication property (b) thisreduces to Φ(1),Φ(t) – and their calues are explicitely determined in the theorem (Φ(t) = T ,Φ(1) = 1H). Now we turn over to the existence. Let p ∈ P ([a, b]), then if p : t 7−→

∑nk=0 ak tk,

ak ∈ C we define

Φ0(p) :=n∑

k=0

akT k ∈B(H)

Clearly, (a) holds and so does (b) for Φ0. Again by 1.15, σ(T )⊆ C is closed. Let f ∈ C (σ(T )),i.e. f : σ(T )−→ C is continuous. By 2.3 (Tietze-Urysohn), there exists

f : B‖T‖(0)−→ C

2Let P ([a, b]) be the subalgebra of C ([a, b]) of all polynomials on [a, b].

32

continuous with f |σ(T ) = f . In other words, for f ∈ C (σ(T )), there exists f : [a, b] −→ Rcontinuous fσ(T ) = f . Then, by 2.2 there is, for all ε > 0, a polynomial p such that

supt∈[a,b]

|p(t)− f (t)|= ‖p− f ‖∞ < ε

Then (since σ(T ) ⊆ [a, b]) obviously supt∈σ(T ) |p(t)− f (t)| < ε. I.e. (by abuse of notation!)‖p − f ‖∞,σ(T ) < ε. It now remains to prove that Φ0 : P −→ B(H) is bounded/continuous.Then by density of P in C (σ(T )), by [FA1] Theorem 2.32, we will have a unique extensionof Φ0 to Φ with the desired properties ((a) is automatic and (b) follows from “passing to thelimit”, (c) follows by extension). To see, for example, the involution in (b), let f ∈ C (σ(T )),then choose a sequence pnn∈N of polynomials, such that ‖ f − pn‖∞

n→∞−−−→ 0. Then

Φ( f ) = Φ

limn→∞

pn

(1)= Φ

limn→∞

pn

(2)= lim

n→∞Φ(pn)

(3)= lim

n→∞Φ0(pn)

(4)= lim

n→∞Φ0(pn)

∗ (5)=

limn→∞

Φ0(pn)∗ (3)=

limn→∞

Φ(pn)∗ (2)= Φ

limn→∞

pn

∗

= Φ( f )∗

where

(1) is by the continuity of C−→ C, z 7−→ z.

(2) is by the continuity of Φ (which is still to be proven).

(3) is by Φ(p) = Φ0(p) for every p ∈ P (σ(T )).

(4) is by the involutive property of Φ0.

(5) is by the continuity of B(H)−→B(H), T 7−→ T ∗.

To prove Φ0 : P (σ(T ))−→B(H) is bounded, we shall in fact prove that

‖Φ0(p)‖B(H) = ‖p‖∞ = supλ∈σ(T )

|p(λ)|

For this, we need the following3:

σ(Φ0(p)) = σ(p(T )) = p(σ(T ))

which holds for T ∈ B(X ) on a Banach space X . Then, by the involutive property, Φ0(p) isnormal and since4 ‖S‖2 = ‖S∗S‖ for S normal, we have

‖Φ0(p)‖2 = ‖Φ0(p)∗Φ0(p)‖= ‖Φ0(pp)‖

= sup|λ| | λ ∈ σ(Φ0(pp))

and since, Φ( f f ) is self-adjoint, by 1.16, its norm equals its spectral radius

= sup|(pp)(λ) | λ ∈ σ(T )

= sup|p(λ)|2 | λ ∈ σ(T )

=

sup|p(λ)| | λ ∈ σ(T )2 = ‖p‖2∞

3This is Problem 17 on Sheet 54By the proof of 1.16.

33

2.1.1 Properties of the Continuous Functional Calculus

We collect some more properties of the Continuous Functional Calculus for T :

Theorem 2.4. Let T ∈ B(H), T ∗ = T and f 7−→ f (T ) the Continuous Functional Calculus onC (σ(T )). Then

(a) ‖ f (T )‖B(H) = ‖ f ‖∞

(b) If f ¾ 0 on σ(T ), then f (T )¾ 0 (i.e. f (T ) is positive semi-definite).

(c) If T x = λx, then f (T )x = f (λ)x

(d) (Spectral Mapping Theorem): σ( f (T )) = f (σ(T )) = f (λ) | λ ∈ σ(T ) ⊆ C.

(e) A := f (T ) | f ∈ C (σ(T )) ⊆ B(H) is a commutative subalgebra of operators and forevery f ∈ C (σ(T )): f (T ) is normal and f (T ) is self-adjoint iff f = f ( f is real-valued).

Proof. (a) Follows from the proof of Theorem 2.1, since the statement is true for polynomialsand these are dense.

(b) Let f ¾ 0 on σ(T ) (i.e. f (x) ¾ 0 ∀x ∈ σ(T )). Let f = g2 with g ∈ C (σ(T )), g ¾ 0 (i.e.g =

p

f on σ(T )). Then for x ∈ H:

f (T )x , x

=

g(T )x , g(T )∗x

=

g(T )x , g(T )x

=

g(T )x , g(T )x

= ‖g(T )x‖2 ¾ 0

(by the properties of the Continuous Functional Calculus (b) in Theorem 2.1, in particu-lar).

(c) Is obvious for polynomials, for general f ∈ C (σ(T ) it follows by a usual limit-argument:Choose a sequence pnn∈N of polynomials such that pn

n→∞−−−→ f . Then pn(T )→ f (T ) inB(H) by Theorem 2.1 (c). In particular

‖ f (T )x − f (λ)x‖= ‖ f (T )x − pn(T )x + pn(T )x − pn(λ)x + pn(λ)x − f (λ)x‖¶ ‖ f (T )− pn(T )‖ · ‖x‖+ ‖pn(T )x − pn(λ)x‖

︸ ︷︷ ︸

=0

+|pn(λ)− f (λ)| · ‖x‖

¶ 2‖pn− f ‖∞ · ‖x‖n→∞−−−→ 0

(d) By problem 17 on sheet 5 the statement holds for polynomials. Let f ∈ C (σ(T )). Assumefirst that µ /∈ f (σ(T )) then µ 6= f (t) for any t ∈ σ(T ). So ( f (t)−µ)−1 =: g(t), t ∈ σ(T )is well-defined, g ∈ C (σ(T )) and

g(t)( f (t)−µ) = ( f (t)−µ)g(t) = 1 ∀t ∈ σ(T )

So by the Continuous Functional Calculus

g(T )( f (T )−µ) = ( f (T )−µ)g(T ) = 1H

Hence, µ ∈ ρ( f (T )). This shows σ( f (T )) ⊆ f (σ(T )). On the other hand, let µ = f (λ)for some λ ∈ σ(T ). Let n ∈ N, choose a polynomial pn with ‖ f − pn‖∞ ¶

1n, then also

| f (λ)− pn(λ)|¶1

n

34

since this holds for any λ ∈ σ(T ). Again by the Continuous Functional Calculus

‖ f (T )− pn(T )‖¶1

n

As noted, λ ∈ σ(T ) implies pn(λ) ∈ σ(pn(T )) (Problem 17). Hence there exists a Weyl-sequence (xn)n ⊆ H with ‖xn‖= 1 and ‖(pn(T )− pn(λ))(xn)‖¶

1n. Hence

‖( f (T )−µ)xn‖= ‖( f (T )− f (λ))xn‖= ‖( f (t)− pn(T ) + pn(T )− pn(λ) + pn(λ)− f (λ))xn‖¶ ‖( f (T )− pn(T ))xn‖+ ‖(pn(T )− pn(λ))xn‖+ ‖(pn(λ)− f (λ))xn‖¶ ‖ f (t)− pn(T )‖ · ‖xn‖+ ‖(pn(T )− pn(λ))xn‖+ |pn(λ)− f (λ)| · ‖xn‖

¶1

n+

1

n+

1

n=

3

n

Hence there exists a sequence (xn)n ⊆ H, ‖xn‖= 1 for every n such that

‖( f (T )−µ)xn‖n→∞−−−→ 0

So (xn)n is a Weyl-sequence for f (T ) at µ ∈ C, hence f (λ) = µ ∈ σ( f (T )) and f (σ(T ))⊆σ( f (T )).

(e) We have

f (T ) f (T )∗ = f (T ) f (T ) = ( f f )(T ) = ( f f )(T ) = f (T ) f (T ) = f (T )∗ f (T )

so f (T ) is normal. Further we have

g(T ) f (T ) = (g f )(T ) = ( f g)(T ) = f (T )g(T )

soA is commutative and f (T ) = f (T )∗ iff f = f on σ(T ). The remaining properties areclear.

2.2 Outlook on the Measurable Functional Calculus

Outlook (Functional Calculus). Holomorphic Functional Calculus: Cauchy’s Integral Formula.Let f : Ω−→ C holomorphic/analytic with Ω⊆ C open and simply connected, Γ a rectifiable,closed curve in Ω. Then

f (T ) =1

2πi

∮

Γ

f (ξ)ξ− z

dξ

for any z inside Γ. This construction (due to Riesz – Dunford) gives a Holomorphic FunctionalCalculus: For T ∈B(X ) (X is C-Banach - for example C-Hilbert). T is not necessarily neithernormal nor self-adjoint, D ⊆ C open set containing the spectrum of T (σ(T )⊆ D), f : D −→ Cholomorphic and Γ = γimi=1 a collection of (Jordan) curves in D, such that σ(T ) lies insideof Γ and each γi to be positively oriented. Then let

f (T ) :=1

2πi

∮

Γ

f (ξ)(ξI − T )−1dξ

Then f 7−→ f (T ) is an involutive, linear algebra homomorphism from H (σ(T )) – the set ofholomorphic functions – to B(X ) which is continuous with respect to what is called compact

35

convergence (uniform convergence on all compacts). Note that the Continuous FunctionalCalculus now allows to “compute”/define f (T ) ∈ B(H) for T bounded, self-adjoint and fcontinuous (on σ(T )). However, there is nothing on diagonalisation. Assume now that T ∈K(H), T ∗ = T and let

T =∞∑

k=0

µkEk

be the spectral decomposition of T from the Spectral Theorem for normal, compact operators1.20 + 1.21. Here µ0 = 0 ∈ σ(T ) and E0 is the orthogonal projection on N(T ), µk∞k=1 arethe sequence of (distinct and real) eigenvalues of T , µk 6= 0, k 6= 0. It then follows

C (σ(T ))−→B(H)

f 7−→∞∑

k=0

f (µk)Ek

satisfies (a),(b),(c) in Theorem 2.1. Hence, this map is the Continuous Functional Calculusof T . That is, the diagonalisation of T ∈ K(H), T ∗ = T , allowed already to define the Con-tinuous Functional Calculus of T . On the other hand (in this case: T ∈ K(H), T ∗ = T) theContinuous Functional Calculus allows to “re-find” (compute/define) the projections Ek (thespectral projection of T). Recall σ(T ) \ 0 consists of isolated points. Hence for j ¾ 1:

f j : σ(T )−→ R f j(t) :=

(

1 t = µ j

0 otherwise

then f j is continuous on σ(T ) and (µ0 = 0!)

f j(T ) =∞∑

k=1

f j(µk)Ek = E j j ¾ 1

Retoric question: How to get E0?? Problem: If we imagine there is, also for general bounded(self-adjoint) operators, something like “spectral projections”, one could hope to “get them”by using, as above, a continuous function f with values in 0,1 ( f 2

j (t) = f j(t), f j(t) = f j(t))and define an operator f (T ) (to be a projection). BUT: f has to be continuous (on σ(T )),for us to use Theorem 2.1) but in general “ f ∈ C (σ(T ))” and “ f (σ(T )) ⊆ 0,1” will notbe compatible. This is one (major!) motivation for the following: Extending the ContinuousFunctional Calculus, to define f (T ) for bounded, measurable functions on σ(T ) (i.e. Borel-measurable).

For this we shall need a certain knowledge on Lebesgue integration theory – not only for theLebesgue integral/measure on Rn, but in general...

2.3 A recall on Measure Theory and Integration Theory

Definitions/results/recall (Lebesgue Integration Theory). Let T be any set. A family ofsubsets Σ⊆ 2T – the power-set of T – is called σ-algebra iff

(a) ; ∈ Σ

(b) If E ∈ Σ, then T \ E ∈ Σ

(c) If En ∈ Σ for n ∈ N, then⋃∞

n=1 Ei ∈ Σ

36

For any family Γ⊆ 2T , there exists a smallest σ-algebra Σ≡ Σ(T ), containing Γ. The smallestσ-algebra (in Rn or any topological space T) containing all open sets is called the Borel-(σ)-algebra and its elements are called Borel-sets. A family of subsets ∆ ⊆ 2T is called aDynkin-system († 14.11.2014) iff

(a) T ∈∆

(b) If E, F ∈∆ with E ⊆ F , then F \ E ∈∆.

(c) If En ∈∆ for n ∈ N with Ei ∩ E j for i 6= j then⋃∞

i=1 Ei ∈∆.

For a family Γ⊆ 2T there exists a smallest Dynkin-system∆≡∆(Γ) containing Γ. Note that ifΣ is a σ-algebra, then Σ is a Dynkin system. A family Γ⊆ 2T is called stable unter intersectioniff whenever A, B ∈ Γ then also A∩ B ∈ Γ.

Proposition 2.5. If Γ⊆ 2T is stable under intersection then Σ(Γ) = ∆(Γ).

A function f : T −→ R is called Borel-measurable (“measurable” from now on) iff f −1([a, b])⊆T is a Borel-set for all a, b ∈ R. A function f : T −→ C is measurable iff Re f and Im f aremeasurable. A function of the form f =

∑ni=1αiχEi

with αi ∈ C and

χEi(t) =

(

1 t ∈ Ei

0 t /∈ Ei

for Ei ∈ Σ is called a stepfunction and is measurable.

Proposition 2.6. Let f , fn, g : T −→K (K either R or C) be measurable. Then

(a) Then f + g, f · g, fg

(if g 6= 0), α f , | f |, max f , g, min f , g (if K = R), supn fn, infn fn,lim infn fn, limsupn fn (if K= R) (and hence limn fn) are all measurable functions.

(b) There exists a sequence ϕnn∈N of step functions such that f (t) = limn→∞ϕn(t) for allt ∈ T. If f ¾ 0, the ϕn’s can be chosen such that ϕ1(t)¶ ϕ2(t)¶ · · ·¶ f (t) for all t ∈ T.

(c) If f is measurable and bounded, then there exists a sequence of step functions that convergesuniformly to f on T (not only pointwise, as in (b)).

Proof. No proof. See e.g. Elstrodt (Maß & Integrationstheorie), Bauer, Rudin (“Real & Com-plex Analysis”).

If T is a set, Σ is a σ-algebra on T , then µ: Σ −→ R (or C) is called a signed (or complex)measure iff for every sequence Aii∈N ⊆ Σ, Ai ∩ A j = ;, i 6= j we have

µ

∞⋃

i=1

Ai

=∞∑

i=1

µ(Ai)

M (T,Σ) is the space of all complex measures on Σ. It is a C-vector space. If T is a topologicalor metric space, we let Σ be the Borel-algebra. The variation of a measure µ is the positivemeasure |µ| defined by

|µ|(A) := sup n∑

i=1

|µ(Ai)|

∃n ∈ N :n⋃

i=1

Ai = A, Ai ∩ A j = ;, i 6= j

Then (!) |µ|(T )<∞. The variation norm of µ, µ ∈M (T,Σ) is ‖µ‖ := |µ|(T ).

37

Proposition 2.7. (M (T,Σ),‖ · ‖) is a Banach space.

Proof. No proof.

Theorem 2.8 (Riesz/Riesz-Markov-Kakutani Representation Theorem, “big Riesz”). Let K bea compact topological space. Then C (K)′ – the dual of C (K) with ‖ · ‖∞ – is isometricallyisomorphic toM (K), C (K)′ ∼=M (K), via the map

T :M (K)−→C (K)′

µ 7−→ Tµ

given by Tµ( f ) :=∫

Kf dµ for f ∈ C (K).

Proof. See e.g. Elstrodt (Maß & Integrationstheorie), Bauer, Rudin (“Real & Complex Analy-sis”).

2.4 Statements aboutMb(D)

Let D ⊆ C be compact. LetMb(D) be the vector space of bounded, borel-measurable functionson D.

Lemma 2.9. (a) (Mb(D),‖ · ‖∞) is a Banach space.

(b) Assume U ⊆Mb(D) satisfies

(1) C (D)⊆ U

(2) Whenever fnn∈N ⊆ U, supn∈N ‖ fn‖∞ <∞ (“ fn is uniformly bounded”) and for everyt ∈ D the limit

f (t) := limn→∞

fn(t)

exists, then f ∈ U.

Then U =Mb(D).

Proof. (a) Follows from the fact that the space of all bounded functions on D, with the supre-mum norm, is a Banach space. Let X be any set,

B(X ,C) := f : X −→ C : ∃c > 0 : | f (x)|¶ c ∀x ∈ X

with ‖ f ‖∞ := supx∈X | f (x)|. Then (B(X ,C),‖ · ‖∞) is Banach andMb(D) ⊆ B(D,C) andthe pointwise limit of measurable functions is a measurable function. HenceMb(D) is aclosed subspace of B(D,C).

(b) LetF := S ⊆Mb(D) :C (D)⊆ S and S satiesfies (2) in Lemma 2.9 (b)

and let V :=⋂

S∈F S. Note: C (D) ⊆ V (and V ⊆ Mb(D)). We claim that V is a vectorspace. At least 0 ∈ C (D)⊆ V . For f0 ∈ V , define

Vf0 = g ∈Mb(D) : f0+ g ∈ V

Note:

(i) For f0 ∈ C (D), we have C (D) ⊆ Vf0 , since if f0 ∈ C (D) and if g ∈ C (D) ⊆Mb(D)then f0+ g ∈ C (D)⊆ V . So g ∈ Vf0 for f0 ∈ C (D).

38

(ii) Vf0 satisfies (2) in Lemma 2.9 (b). True because assume fnn∈N ⊆ Vf0 with ‖ fn‖∞ ¶C for n ∈ N and f (t) := limn→∞ fn(t) exists for x ∈ D. Since fn ∈ Vf0 for all n ∈ Nwe have f0+ fn ∈ V for all n ∈ N. Let hn := f0+ fn, then

‖hn‖∞ ¶ ‖ fn‖∞+ ‖ f0‖∞ ¶ C + ‖ f0‖ ≡ C ∀n ∈ N

and h(t) := limn→∞ hn(t) = f0(t) + limn→∞ fn(t) exists for all t ∈ D and hence,since hnn ⊆ V and V satisfies (2) in Lemma 2.9 (b), we have h ∈ V (by definitionof V ). Hence h = f0 + f ∈ V , so f ∈ Vf0 . So Vf0 satisfies (2) in Lemma 2.9 (b) asclaimed.

From (i) and (ii) follows by definition of V that for f0 ∈ C (D) we have V ⊆ Vf0 (sinceC (D)⊆ Vf0 and Vf0 satisfies (2) in Lemma 2.9 (b)). In other words: If f0 ∈ C (D), g ∈ V ,then f0 + g ∈ V . Let now g0 ∈ V . The above shows that f + g0 ∈ V for every f ∈ C (D)so C (D) ⊆ Vg0

for every g0 ∈ V . Again, Vg0satisfies (2) in Lemma 2.9 (b)5. Hence, by

definition of V , we have V ⊆ Vg0. So g0 ∈ V , g ∈ V implies g ∈ Vg0

so g + g0 ∈ V .Similarly α ∈ C, g ∈ V implies αg ∈ V . So V is a vector space. Aim is now to proveV =Mb(D). We know that V is a vector space. Claim:

(1) All step-functions belong to V .

(2) As a consequence, we obtain V = Mb(D), since the step-functions lie dense inMb(D). True because let f ∈Mb(D). Then, by Proposition 2.6 (c) there exists a se-quence fnn∈N of step-functions such that ‖ fn− f ‖∞

n→∞−−−→ 0, i.e. supn∈N ‖ fn‖∞ <∞,fn(t)

n→∞−−−→ f (t) for all t ∈ D, hence by definition of V , f ∈ V , so Mb(D) = V .

To see that all step-functions belong to V let Σ be the Borel-σ-algebra. If we prove thatfor E ∈ Σ, χE ∈ V then it follows (since V is a vector space) that all stepfunctions are inV . Let ∆ := E ∈ Σ : χE ∈ V (we hope ∆≡ Σ). We claim that ∆ is a Dynkin-system, i.e.:

(a) D ∈∆(b) If E, F ∈∆ and E ⊆ F , then F \ E ∈∆(c) If En ∈∆ for n ∈ N and Ei ∩ E j = ;, i 6= j then E :=

⋃∞i=1 Ei ∈∆.

Proof:

(a) χD ≡ 1 (on D) and 1 ∈ C (D)⊆ V , i.e. χD ∈ V , so D ∈∆.

(b) χF\E = χF − χE (evaluate at x and check cases) and so E, F ∈ ∆ implies χE ,χF ∈ V ,further χF\E = χF −χE ∈ V (since V is a vector space). Hence F \ E ∈∆.

(c) If Ei ∈ ∆, i ∈ N are pairwise disjoint, then χEi∈ V , ‖χEi

‖∞ ¶ 1 for i ∈ N andχE =

∑∞i=1χEi

(pointwise limit for all x ∈ D), so χE ∈ V by (2) in Lemma 2.9 (b).

Hence∆ is a Dynkin-system. Claim: T ⊆∆ (all open sets belong to∆, T is the topology).Given this, since T is stable under intersection, Proposition 2.5 gives that Σ = Σ(T ) =∆(T ). But Σ(T ) = Σ and T ⊆ ∆, ∆ is a Dynkin-system, so Σ = Σ(T ) = ∆(T ) ⊆ ∆,hence, by ∆ ⊆ Σ we have ∆ = Σ, so that E ∈ Σ implies χE ∈ V which we want to prove.Proof of T ⊆ ∆. For any E ∈ T (i.e. E open), by the below, there exists a sequence fnn∈N ⊆ C (D) of continuous functions, 0 ¶ fn ¶ 1, such that fn(t)

n→∞−−−→ χE(t) for allt ∈ D. Then, since supn∈N ‖ fn‖∞ ¶ 1<∞ it follows, by definition of V , that χE ∈ V , sinceC (D)⊆ V , so E ∈∆. To construct such a sequence fnn∈N we use Tietze-Urysohn in the

5The proof of (ii) above for Vf0 did not use the fact that f ∈ C (D).

39

version: For K , H closed, K ∩ H = ;, there is a continuous function g such that g|H = 0,g|K = 1, 0 ¶ g ¶ 1. Use Fn := y : d(y, Ec) ¾ 1

n. Since E is open, Ec is closed, Fn is

closed and Ec ∩ Fn = ; for each n ∈ N.

Motivation. (1) Extension of the Functional Calculus to measurable bounded functions

Φ:Mb(σ(T ))−→B(H) f 7−→ Φ( f ) =: f (T )

if T = T ∗ ∈B(H) given.

(2) Introduce projection valued measures (pvm) E : Σ−→B(H), A 7−→ EA and define∫

f dEfor bouned measurable functions.

(3) Connection between (1) and (2):

• If T = T ∗ is given then EA := χA(T ) defines a projection valued measure.

• If E is a projection valued measure, then∫

λdEλ =: T = T ∗.

These are inverse operations.

2.5 The Measurable Functional Calculus

First we would like to establish the extension of the Functional Calculus to Mb(σ(T )), solet T = T ∗ ∈ B(H), let x , y ∈ H and f ∈ C (σ(T )). We define `x ,y( f ) := ⟨x , f (T )y⟩ then`x ,y : C (σ(T )) −→ C is a bounded linear functional since |`x ,y( f )| ¶ ‖ f ‖∞‖x‖ · ‖y‖ (re-call ‖ f (T )‖ = ‖Φ( f )‖ = ‖ f ‖∞). By the R.M.K-Theorem 2.8, there exists a unique complexmeasure µx ,y such that

`x ,y( f ) =

∫

σ(T )

f dµx ,y

which satisfies |µx ,y |(σ(T )) = ‖µx ,y‖ = ‖`x ,y‖ ¶ Cx ,y := ‖x‖ · ‖y‖. (x , y) 7−→ µx ,y issesquilinear and bounded.

Theorem 2.10 (Measurable Functional Calculus). Let T ∈B(H), T = T ∗. Then there exists aunique map

Φ:Mb(σ(T ))−→B(H)

such that

(a) Φ(t) = T, Φ(1) = 1H

(b) Φ is an involutive, linear algebra homomorphism

(c) Φ is continuous

(d) If fnn∈N ⊆Mb(σ(T )) with supn∈N ‖ fn‖∞ <∞ and fn(t)−→ f (t) then

¬

Φ( fn)x , y¶ n→∞−−−→

¬

Φ( f )x , y¶

∀x , y ∈ H

hence Φ( fn) converges weakly to Φ( f ). In fact, the convergence is also strongly, i.e.

Φ( fn)xn→∞−−−→ Φ( f )x ∀x ∈ H

The proof uses

40

Theorem 2.11 (Lax-Milgram-Theorem). Let H be a complex Hilbert space, B : H × H −→ C asesquilinear form. Then

(a) the following statements are equivalent

(i) B is continuous

(ii) B is partially continuous, i.e. x 7−→ B(x , y) is continuous for y ∈ H and y 7−→ B(x , y)is continuous for x ∈ H.

(iii) There exists M ¾ 0 such that |B(x , y)|¶ M‖x‖ · ‖y‖ for every x , y ∈ H.

(b) If B is contiuous with |B(x , y)| ¶ M‖x‖ · ‖y‖ for x , y ∈ H then there exists a uniqueA∈B(H) such that B(x , y) =

Ax , y

for x , y ∈ H with ‖A‖¶ M.

Proof. FA1.

Proof of Theorem 2.10. For uniqueness we assume that there exists a Φ:Mb(σ(T ))−→B(H)satisfying (a) - (d). Use Lemma 2.9, so let

U := f ∈Mb(σ(T )) | Φ( f ) uniquely given by (a) - (d)

(i) If (a)-(c) hold and f ∈ C (σ(T )) then Φ( f ) = Φ( f ) is unique by the Continuous Func-tional Calculus.

(ii) Let fnn∈N ⊆ U , supn ‖ fn‖ < ∞, fn(t) −→ f (t) for every t ∈ σ(T ), then by (d):Φ( f ) = limn→∞ Φ( fn).

For the existence let f ∈ Mb(σ(T )), µx ,y as defined above. Then (x , y) 7−→∫

σ(T )f dµx ,y is

sesquilinear and bounded. Boundedness comes from

∫

σ(T )f dµx ,y

¶ ‖ f ‖∞‖µx ,y‖¶ ‖ f ‖∞‖x‖ · ‖y‖

By Lax-Milgram 2.11 there exists a unique operator Φ(g) ∈B(H) such that∫

σ(T )

f dµx ,y =¬

Φ( f )x , y¶

∀x , y ∈ H

This defines a map Φ:M (σ(T ))−→B(H).

(a) By the Continuous Functional Calculus: Φ(t) = Φ(t) = T , Φ(1) = Φ(1) = 1H

(b) To show multiplicativity use Lemma 2.9 twice. First let g ∈ C (σ(T )) be fixed and set

Ug := f ∈Mb(σ(T )) | Φ( f g) = Φ( f )Φ(g)

(i) Clearly C (σ(T ))⊆ Ug because Φ is multiplicative and

Φ( f g) = Φ( f g) = Φ( f )Φ(g) = Φ( f )Φ(g)

(ii) Let fnn ⊆ Ug such that supn ‖ fn‖ < ∞ and fn(t)n→∞−−−→ f (t) for every t ∈ σ(T ).

Then, since ‖g‖∞ < ∞, also fn gn satisfies the conditions in (d) and hence forevery x , y ∈ H we have:¬

Φ( f g)x , y¶ n→∞←−−−

¬

Φ( fn g)x , y¶

=¬

Φ( fn)(Φ(g)x), y¶ n→∞−−−→

¬

Φ( f )Φ(g)x , y¶

Hence Φ( f )Φ(g) = Φ( f g).

41

So Ug =Mb(σ(T )). Now fix h ∈Mb(σ(T )) and let

Vh := f ∈Mb(σ(T )) | Φ( f h) = Φ( f ) · Φ(h)

(i) By the above, C (σ(T ))⊆ Vh

(ii) Let fnn ⊆ Vh, supn ‖ fn‖∞ <∞, fn(t) −→ f (t) for all t ∈ σ(T ). Exactly as above(here, bounedeness of h is given) we show that Φ( f )Φ(h) = Φ( f h). Hence f ∈ Uh

So Uh = Mb(σ(T )). Just as the above we show that Φ( f )∗ = Φ( f ) (use the strongconvergence in (d)).

(c) By Lax-Milgram ‖Φ( f )‖¶ ‖ f ‖∞, so Φ is continuous.

(d) Let fnn ⊆ Mb(σ(T )) with supn ‖ f ‖ < ∞ and fn(t)n→∞−−−→ f (t) for every t ∈ σ(T )

and some f , then since |µx ,y |(σ(T )) < ∞ and | fn(t)| ¶ ‖ fn‖∞ ¶ supn ‖ fn‖∞ which isintegrable on σ(T ), we can use the Dominated Convergence Theorem to show:

¬

Φ( fn)x , y¶

=

∫

σ(T )

fn dµx ,yn→∞−−−→DCT

∫

σ(T )

f dµx ,y =¬

Φ( f )x , y¶

For the strong convergence let fnn ⊆Mb(σ(T )) with supn ‖ fn‖<∞, fn(t)−→ f (t) forall t ∈ σ(T ). By the above Φ( fn)x

n→∞−−−+ Φ( f )x for all x ∈ H and

‖Φ( fn)x‖2 =¬

x , Φ( fn)∗Φ( fn)x

¶

=D

x , Φ( f n)Φ( fn)xE

=D

x ,Φ( f n fn)xE

n→∞−−−→D

x , Φ( f f )xE

= ‖Φ( f )x‖2

because f f also satisfies the assumptions in (d). Hence6,we have Φ( fn)x −→ Φ( f )x forall x ∈ H.

2.5.1 Projection Valued Measures

Lemma 2.12. Let T ∈B(H), T ∗ = T. Then

(a) If A ⊆ σ(T ) is a Borel set in R, set EA := Φ(χA) =: χA(T ). Then EA is an orthogonalprojection.

(b) E; = 0, Eσ(T ) = 1H

(c) If An ⊆ σ(T ) are disjoint Borel sets, A :=⋃

n∈N An and x ∈ H. Then

∞∑

n=1

EAnx = EAx

(d) If A, B ⊆ σ(T ) are Borel sets, then EAEB = EA∩B.

Proof. χA ∈Mb(σ(T )) if A is Borel. Hence

(a) E2A = χA(T )2 = Φ(χA)Φ(χA) = Φ(χ2

A) = Φ(χA) = EA and since χA = χA, we have E∗A = EA.So EA is an orthogonal projection.

6If zn + z, ‖zn‖ −→ ‖z‖, then zn −→ z, because ‖zn − z‖= ‖zn‖2 + 2Re⟨zn, z⟩+ ‖z‖2 n→∞−−→ 0.

42

(b) E; = Φ(χ;) = Φ(0) = 0 and Eσ(T ) = Φ(χσ(T )) = Φ(1) = 1H .

(c) Let fk :=∑k

n=1χAn. Then, since the An are pairwise disjoint, this means ‖ fk‖∞ ¶ 1 for all

k, and fk(t)k→∞−−−→

∑∞n=1χAn

(t) = χA(t) for all t ∈ σ(T ) and by 2.10 (d) we have

k∑

n=1

EAkx = Φ( fk)x

k→∞−−−→ Φ(χA)x = EAx

(d) We have EAEB = Φ(χA)Φ(χB) = Φ(χAχB) = Φ(χA∩B) = EA∩B.

Definition 2.13 (spectral measure/projection valued measure). Let Σ be the Borel σ-Algebraon R. A map

E : Σ−→B(H)

A 7−→ EA

is called a spectral measure (or projection valued measure) iff EA is an orthogonal projectionand

(a) E; = 0, ER = 1H

(b) Whenever An ∈ Σ, n ∈ N are pairwise disjoint, then∑

n∈N EAnx = E⋃

n∈N Anx for every

x ∈ H.

E is said to have compact support iff there exists a compact set K ⊆ R such that EK = 1H .

Note. If A, B ∈ Σ then EAEB = EB EA = EA∩B.

Proof. • If A∩ B = ;, then EA+ EB = EA∪B = E2A∪B = (EA+ EB)2 = EA+ EB + EAEB + EB EA.

From this followsEAEB = EAEAEB =−EAEB EA = EB EA

Hence EAEB = 0.

• If A⊆ B, then B = A∪ B \ A and

EAEB = EA(EA+ EB\A) = EA

Hence EB EA = EA.

• For A, B ∈ Σ arbitrary we write B = (A∩ B)∪ B \ A and then

EAEB = EA(EA∩B + EB\A) = EA∩B = EB EA

In the following we intend to define an integration with respect to such a projection valuedmeasure E for every f ∈Mb(R). In the first step, we look at f = χA, A∈ Σ and define

∫

χA dE := EA

Secondly, for f =∑n

i=1αiχAi, αi ∈ C, Ai ∈ Σ pairwise disjoint, we define7

∫

f dE :=m∑

i=1

αi EAi

7As in measure theory we can show that this definition is independent of the choice of (αi , Ai)mi=1.

43

In the third step let f ∈Mb(R). By proposition 2.6 there exists a sequence of step functions fnn with ‖ fn− f ‖∞

n→∞−−−→ 0. We wish to have∫

f dE := limn→∞

∫

fn dE

Lemma 2.14. If f is a step function, then

∫

f dE

¶ ‖ f ‖∞

Proof. Let f =∑n

i=1αiχAi, Ai ∈ Σ pairwise disjoint, αi ∈ C. Let x ∈ H. Then

∫

f dE

!

x

2

=

n∑

i=1

αi EAix

2

=n∑

i, j=1

αiα j

D

EA jx , EAi

xE

whereD

EA jx , EAi

xE

=D

x , EA jEAi

xE

=D

x , EAi∩A jxE

= δi j‖EAix‖2, then

=n∑

i=1

|αi|2‖EAix‖2

¶ supi=1,...,n

|αi|2 ·n∑

i=1

‖EAix‖2 = ‖ f ‖2∞‖E

⋃

n∈N Anx‖2

¶ ‖ f ‖2∞‖x‖2

Back in step 3 we know that∫

fn dE

n

⊆B(H)

is a Cauchy-sequence because by Lemma 2.14

∫

fn dE −∫

fm dE

¶ ‖ fn− fm‖∞

Hence we define∫

f dE := limn→∞

∫

fn dE

If fn is another sequence with ‖ fn− f ‖∞n→∞−−−→ 0 then

∫

fn dE −∫

fn dE

¶ ‖ fn− fn‖∞ ¶ ‖ fn− f ‖∞+ ‖ fn− f ‖∞n→∞−−−→ 0

so the limit is independent of the choice of the fn.

Definition (Definition of∫

f dE for E with compact support). If E has compact support, i.e.if there is a compact set K ⊆ R, such that EK = 1H , then for f : K −→ C bounded andmeasurable, set

∫

f dE :=

∫

K

f dE :=

∫

χK f dE

(independent of the choice of K! Check!).

44

Theorem 2.15. Let E be a spectral measure on R and f ∈ Mb(R). Then∫

f dE ∈ B(H) and

the map∫

:Mb(R)−→B(H), f 7−→∫

f dE is linear and continuous. Moreover

∫

f dE

¶ ‖ f ‖∞ ∀ f ∈Mb(R)

and if f is real-valued then∫

f dE is self-adjoint. If there is K ⊆ R compact such that EK = 1H ,then it is enough that f is (defined and) bounded on K.

Note. • If E has compact support, then the monomial λ 7−→ λ on R is bounded on eachcompact set. Hence T :=

∫

λdEλ ∈B(H) is self-adjoint.

• On the other hand, if T = T ∗ ∈B(H) then by Lemma 2.12 ET : Σ−→B(H),

ETA := χA∩σ(T )(T )

defines a spectral measure ET with compact support, since

ETσ(T ) = χσ(T )(T ) = Φ(χσ(T )) = Φ(1) = 1H

We will see that these are inverse operations.

2.5.2 Characterization of the Measurable Functional Calculus

Theorem 2.16. Let E be a spectal measure on R with compact support and T :=∫

λdEλ. Thenthe map

ψ:Mb(σ(T ))−→B(H)

f 7−→∫

σ(T )

f dE

is the (unique) Measurable Functional Calculus of T , i.e. ψ= Φ and

f (T ) =

∫

σ(T )

f dE

In particular for all A⊆ σ(T ) Borel, we have8 χA(T ) =∫

σ(T )χA dE.

Definition 2.17. A measure on the Borel-σ-algebra is called a Borel-measure. A positiveBorel-measure µ is called regular iff

(a) µ(C)<∞ for every compact set C .

(b) For every A∈ Σ:

µ(A) = supµ(C) | C ⊆ A, C compact= infµ(O) | A⊆ O, O open

A signed/complex measure is regular iff its variation (measure) is regular.

Theorem 2.18. Let X be one of the following:

8The left hand side is the spectal measure from Lemma 2.12: ETA and the right hand side is EA.

45

(a) a compact metric space

(b) a complete, separable space

(c) an open subset of Rn

Then every finite Borel measure on X (|µ(X )|<∞) is regular. Also the Lebesgue measure on RN

is regular.

Proof. No proof here. See Rudin “Real and Complex Analysis”, Elstrodt etc. (favorite integra-tion theory book).

Proof of Theorem 2.16. We only need to show that ψ satisfies (a) - (d) of Theorem 2.10, sincefrom Theorem 2.15 it follows that

∫

σ(T )f dE =

∫

χσ(T ) f dE ∈B(H).

(a) Ψ(1) = 1H is equivalent to proving that Eσ(T ) = 1, since Ψ(1) = Ψ(χσ(T )) = Eσ(T ), andfrom this follows that Ψ(t) =

∫

σ(T )λdEλ = T (from Theorem 2.15). So it remains to

show Eσ(T ) = 1H . E has, by assumption, compact support K ⊆ R. Choose an interval withK ⊆ (a, b], then

E(a,b] = EK = 1H ()

Let µ ∈ ρ(T ). Claim: There exists a neighborhood U of µ such that EU = 0. Note that ifµ /∈ (a, b] then such a U trivially exists by (). Hence assume that µ ∈ (a, b]. Note thatT −µI and hence µI − T is invertible. It follows (from 0.14) that there exists δ > 0 suchthat if ‖S− (µ− T )‖¶ δ then S is also invertible and

‖S−1‖¶ C := ‖(µ− T )−1‖+ 1

We can assume that δ = b−aN

for some N ∈ N (make δ smaller! – choose N large), and

δ < 1C

(ditto). Let then ak := a + k · δ for k = 0,1, . . . , N (so aN = b) and consider thestep function

f :=N∑

k=1

akχ(ak−1,ak]

By 2.13, with

Ek := E(ak−1,ak] =

∫

χ(ak−1,ak]dE

we have

T −N∑

k=1

akEk

=

∫

λdEλ−∫

f (λ)dEλ

¶ ‖λ− f ‖∞ ¶maxk|ak−1− ak|¶ δ

Note (ak−1, ak]∩ (a j−1, a j] = ;, k 6= j. So

(a, b] =N⋃

k=1

(ak−1, ak]

can be written as a disjoint union. Hence (E is a spectral measure):

1H = E(a,b] = E⋃Nk=1(ak−1,ak]

=N∑

k=1

E(ak−1,ak] ≡N∑

k=1

Ek

46

Hence

(µ− T )−N∑

k=1

(µ− ak)Ek

=

T −N∑

k=1

akEk

¶ δ

Therefore S :=∑N

k=1(µ− ak)Ek is invertible, and

N∑

k=1

(µ− ak)Ek

!−1

¶ C

On the other hand, since EkE j = δk j Ek,

N∑

k=1

(µ− ak)Ek

!−1

= sup(µ− ak)−1 | Ek 6= 0¶ C

So |µ− ak| <1C

implies Ek = 0. Now, choosing N large (such that δ < 1C

– see earlier),

there exists some ak0such that |µ− ak0

| < 1C

, since µ ∈ (a, b]. Hence EU = 0 for someneighbourhood U of µ. Let now K ⊆ ρ(T ) be compact. Then, by the above, for each µ ∈ Kthere is a neighbourhood Uµ of µ with EUµ = 0. Since K is compact there exist µ1, . . . ,µm

with K ⊆⋃m

i=1 Uµi. Then EK = 0 (since A⊆ B, EB = 0 ⇒ EA = 0,

∑mi=1 EUµ =

∑

0 = 0).

This means for every K ⊆ ρ(T ), K compact: EK = 0. By 2.18, it follows (!) that for allex ∈ H, the positive measure

Σ(ρ(T ))−→ R+A 7−→

x , EAx

is regular (ρ(T )⊆ R is open) and EK = 0 for every K ⊆ ρ(T ), K compact. So ⟨x , Eρ(T )x⟩=0 for every x ∈ H. Hence, since Eρ(T ) is self-adjoint,

‖Eρ(T )‖= supx∈H,‖x‖¶1

|⟨x , Eρ(T )x⟩|

Eρ(T ) = 0. Since R = σ(T ) ∪ ρ(T ), ρ(T ) ∩ σ(T ) = ; and ER = 1H (def. of spectralmeasure). It follows that Eσ(T ) = 1H .

(b) ψ is linear. ψ is multplicative since

• Let A, B ⊆ σ(T ) Borel, then

ψ(χAχB) =ψ(χA∩B) = EA∩B = EAEB =ψ(χA)ψ(χB)

• By linearity ψ( f g) =ψ( f )ψ(g) is clear for step functions f , g.

• Let f , g ∈ Mb(σ(T )) and pick fn, gn which converge uniformly to f , g respectively.Then fn gn converges uniformly to f g and

ψ( f g) = limn→∞

ψ( fn gn) =ψ( f )ψ(g)

Similarly we get ψ( f )∗ =ψ( f ).

(c) ψ is continuous by Theorem 2.15.

47

(d) Note that for a spectral measure E and x , y ∈ H, the map

νx ,y : Σ−→ CA 7−→

x , EA y

is a complex measure, and that

x ,Ψ( f )y

=

∫

f dνx ,y (∗)

To see (∗): This is trivial for f = χA – then for stepfunctions by linearity, and for f ∈Mb(σ(T )) by approximation (use Lemma 2.14). Then (d) follows from the DominatedConvergence Theorem: Let fnn∈N ⊆Mb(σ(T )) with supn∈N ‖ fn‖ ¶ C <∞ and f (t) :=limn→∞ fn(t) exists for every t ∈ σ(T ). Then

x ,Ψ( fn)y

=

∫

fn dνx ,yn→∞−−−→

∫

f dνx ,y =

x ,Ψ( f )y

because C is an integrable dominant on the compact set σ(T ).

2.6 Spectral Theorem for self-adjoint bounded operators

Recall that if A⊆ σ(T ) is a measurable (Borel) set, then

χA(T ) = Ψ(χA) =

∫

χA dE = EA

so the spectral measure associated to T : A 7−→ χA∩σ(T )(T ) and the spectral measure used todefine T : T :=

∫

λdE and A 7−→ EA coincide.

Reversely:

Theorem 2.19 (Spectral Theorem for self-adjoint bounded operators). Let T ∈B(H) be self-adjoint. Then there exists a unique spectral measure with compact support in R, such that

T =

∫

σ(T )

λdEλ (∗)

The map f 7−→ f (T ) =∫

f (λ)dEλ (for f ∈ Mb(σ(T )) defines the Measurable FunctionalCalculus. f (T ) is given by

x , f (T )y

=

∫

σ(T )

f (λ)d⟨x , Eλ y⟩ x , y ∈ H

Remarks. (1) (∗) above is the generalisation of “T =∑∞

i=0µi Ei” for compact (in particularmatrices) operators. Hence T is “built” of the orthogonal projections EA (weighted withλ) but in a “continuous” way.

(2) The symbol d⟨x , Eλ y⟩means integration with respect to the Borel measure A 7−→ ⟨x , EA y⟩(x , y ∈ H fixed). In the proof of 2.10, it was called µx ,y .

48

Proof. Let T ∈ B(H) self-adjoint and let E be its associated spectral measure E : A 7−→χA∩σ(T )(T ). Define S :=

∫

λdEλ. Then S ∈ B(H) and S∗ = S by Theorem 2.15. We needto prove that S = T (the rest of the statements in 2.19 are just repetition). Firstly, E hascompact support: σ(T ) ⊆ R ⊆ C is compact and χσ(T )(T ) = Eσ(T ) = 1H (Lemma 2.12 (b)).Let, for g ∈ Mb(σ(T )), g(T ) denote the Measurable Functional Calculus of T (from 2.10)on g and Ψ(g) the Measurable Functional Calculus of S (from 2.16). Let t be the functiont : σ(T )−→ R, t 7−→ t and for ε > 0 choose a step function f on σ(T ) such that ‖ f − t‖∞ ¶ ε.According to 2.10:

‖T − f (T )‖¶ ‖t − f ‖∞ ¶ ε

(T = Φ(T )) and according to 2.15 (used on the spectral measure associated to T , and S :=∫

λdEλ)

‖S−Ψ( f )‖=

∫

σ(T )(t − f )dE

¶ ‖t − f ‖∞ ¶ ε

Finally with f :=∑n

i=1αiχAiand since, by definition, EAi

= χAi(T ), we have

f (T )−Ψ( f ) =n∑

i=1

αiχAi(T )−

n∑

i=1

αi EAi= 0

So‖T − S‖¶ ‖T − f (T )‖+ ‖ f (T )−Ψ( f )‖+ ‖Ψ( f )− S‖¶ ε+ 0+ ε = 2ε

and since ε was arbitrary, it follows that ‖T − S‖= 0, so T = S.

Remark. Let T ∈ B(H), T ∗ = T and E its spectral measure. Let S ∈ B(H) be an arbitraryoperator (not necessarily self-adjoint). Then: Σ 3 A 7−→ ⟨x , SEA y⟩ = ⟨S∗x , EA y⟩ ∈ C forx , y ∈ H fixed is also a (complex) measure.

Corollary 2.20 (Fuglede’s Theorem). Let T ∈B(H), T ∗ = T and E be its spectral measure. LetS ∈B(H). Then [S, T] = 0 if and only if [S, EA] = 0 for every A∈ Σ (i.e. S comutes with all ofT ’s spectral projection9.

Proof. We have [S, T] = 0⇐⇒ [S, T n] = 0 for any n ¾ 0. This is equivalent to ⟨x , ST n y⟩ =⟨x , T nS y⟩ for every n¾ 0 and x , y ∈ H. Note that

x , ST n y

=

S∗x , T n y

=

∫

λn d⟨S∗x , Eλ y⟩=∫

λn d⟨x , SEλ y⟩

and similarly, ⟨x , T nS y⟩=∫

λd⟨x , EλS y⟩. Hence [S, T] = 0 is equivalent to

∫

λn d⟨x , SEλ y⟩=∫

λn d⟨x , EλS y⟩ ∀n¾ 0, ∀x , y ∈ H

That is, thinking of the (complex) measure d⟨x , SEλ y⟩ and d⟨x , EλS y⟩ as bounded linearfunctionals on C (σ(T )), these coincide on fn(λ) = λn for all n ¾ 0. Hence, by linearity,on all polynomials. By density of the polynomials (Weierstraß Approximation Theorem) on

9One can see this as „S leaves all eigenspaces of T invariant“.

49

C (σ(T )). Hence, by Riesz-Markov-Kakutani Representation Theorem 2.8, the two measuresare equal for all x , y ∈ H iff [S, T] = 0. Now, for any A∈ Σ and any x , y ∈ H:

x , SEA y

=

∫

χA d⟨x , SEλ y⟩=∫

χA d⟨x , EλS y⟩=

x , EAS y

Hence SEA = EAS for every A∈ Σ⇐⇒ [S, T] = 0.

Examples 2.21. (a) Let H be finite-dimensional (H ∼= Cn). Let T be a self-adjoint matrix –assume T has m (m¶ n) pairwise different eigenvalues µ1, . . . ,µm. Then

T =m∑

i=1

µi Eµi ()

when the Eµi’s are the orthogonal projections onto the eigenspace N(T − µi I) corre-sponding to µi . For A∈ Σ, the spectral measure of T is

EA =∑

j|µ j∈A

Eµ j

and Theorem 2.21 reduces to () above.

(b) Let H be arbitrary (but separable!) and T = T ∗, T ∈ K(H). Then by 1.21 (SpectralTheorem for compact operators, projection version):

T =∞∑

i=1

µi Eµi

(converging in operator norm), with µi and Eµ being as in (a). As above, the spectralmeasure is, for A∈ Σ

EA =∑

j|µ j∈A

Eµi

by only pointwise convergent in the sense that EAx =∑

j|µ j∈A(Eµix) and not in opera-tor norm, in general.

(c) Let H = L2([0,1]). Let(T x)(t) := t · x(t)

for x ∈ L2([0, 1]). Then T is self-adjoint (and bounded), σ(T ) = σc(T ) = [0, 1](σp(T ) = ; by Exercise 16 (ii) on sheet 4). For A∈ Σ, let

EAx := χA∩[0,1] · x

i.e. (EAx)(t) = χA∩[0,1](t) · x(t), t ∈ [0,1]. Then E is the spectral measure of T (i.e.T =

∫

λdEλ (Details: Exercise!)

Theorem 2.22. Let T ∈B(H), T ∗ = T and let E be its spectral measure.

(a) λ ∈ ρ(T ) iff there is U open with λ ∈ U and EU = 0.

(b) λ is an eigenvalue of T (λ ∈ σp(T )) iff Eλ 6= 0. In this case, Eλ is the orthogonalprojection on the eigenspace N(T −λI) associated to λ.

50

(c) The isolated points λ of σ(T ) are eigenvalues (i.e. if there is U ⊆ R open, such that U ∩σ(T ) = λ, then λ is an eigenvalue)10

Proof. (a) By construction of E, Eρ(T ) = 0, since Eσ(T ) = 1H . Since ρ(T ) is open, this proves(⇒). On the other hand, assume U 3 λ, U open, EU = 0. Let

f (t) =

(

1λ−t

t /∈ U

0 otherwise

Then f is measurable and bounded onσ(T ). Same for g(t) = λ−t (measurable, boundedon σ(T )). Then, by the Measurable Functional Calculus

f (T )(λ− T ) = f (T )g(T ) = ( f g)(T ) = χU c (T ) = EU c = 1H

by definition of the spectral measure E of T and since EU = 0. Similarly (λ−T ) f (T ) = 1H .Hence λ ∈ ρ(T ).

(b) It is enough to prove R(Eλ) =N(T −λ). Let x ∈R(Eλ) then Eλx = x . Therefore

y, (λ− T )x

=¬

y, (λ− T )Eλx¶

=

∫

(λ− t)χλ(t)d⟨y, Et x⟩= 0 ∀y ∈ H

Hence (λ− T )x = 0, so λ ∈ σp(T ). Reversely let x ∈ N(T − λI), i.e. T x = λx for somex 6= 0. Then for f ∈ C (σ(T )) we have f (T )x = f (λ)x (by 2.4 (c)) and therefore (by 2.9(b), applied on the set, where this equality holds), this also holds for f ∈Mb(σ(T )). Inparticular, for f = χλ (λ Borel!) Eλx = χλ(λ)x = 1x = x . So x ∈R(Eλ).

(c) Let U be open, such that U ∩ σ(T ) = λ. Since U \ λ ⊆ ρ(T ) we have EU\λ = 0(recall A⊆ B =⇒ EA ¶ EB). If Eλ = 0 then EU = 0 (EA∪B = EA+ EB if A∩ B = ;). Whichby (a) implies λ ∈ ρ(T ), hence Eλ 6= 0. So by (b), λ is an eigenvalue.

Corollary 2.23. Let T ∈ B(H), T ∗ = T, E its spectral measure. Then σ(T ) is the smallestcompact set such that Eσ(T ) = 1H .

Proof. Assume K ⊆ R, λ /∈ K , EK = 1. Then ER\K = 0 and λ ∈ R \ K , R \ K open. Hence by2.22 (a), λ ∈ ρ(T ) (i.e. λ ∈ R \σ(T )). So σ(T )⊆ K .

Recall that one way (among others!) of thinking of the Spectral Theorem for symmetric self-adjoint matrices is T : Cn −→ Cn, T = T ∗, then T is unitarily equivalent to a diagonal matrix,i.e. there is U : Cn −→ Cn unitary (U∗U = UU∗ = En or U∗ = U−1) and D = diag(λ1, . . . ,λn)(λi ’s counted with multiplicities) such that

(U T U−1 x)i = λi x i i = 1, . . . , n

So, D can be thought of as a multiplication operator. We would like to think of Cn as a spaceof functions x : 1, . . . , n −→ C, j 7−→ x( j) = x j and λ: 1, . . . , n −→ C, j 7−→ λ( j) = λ j and

λx : 1, . . . , n −→ C, j 7−→ (λx)( j) = (λx) j := λ( j)x( j) = λ j x j

Here is the analogon (a bit more complicated...) for infinite dimensions.

10The set λ is open in the relative topology of σ(T ).

51

Theorem 2.24. Let T ∈B(H), T = T ∗, E its spectral measure and assume T has a cyclic vectorx0, i.e. there exists x0 ∈ H such that H = spanT n x0 | n¾ 0. Let dµ be the finite positivemeasure11 d⟨x0, Eλx0⟩. Then there exists a unitary operator U : H −→ L2(dµ) such that

(U T U−1ϕ)(t) = tϕ(t) µ-a.e.

Here,

L2(µ) =

¨

f : σ(T )−→ C

f is µ-measurable and

∫

| f |2 dµ <∞«

Proof. Let ϕ ∈ C (σ(T )), then∫

σ(T )

|ϕ(t)|2 dµ(t) =

∫

σ(T )

ϕ(t)ϕ(t)d⟨x0, Et x0⟩=∫

σ(T )

ϕ(t)ϕ(t)d⟨x0, Et x0⟩

= ⟨x0,ϕ(T )∗ϕ(T )x0⟩= ‖ϕ(T )x0‖2HHence, the linear map

V : C (σ(T ))−→ H

ϕ 7−→ ϕ(T )x0

is isometric with respect to the L2(µ)-norm (on C (σ(T )). Since, according to the propositionbelow,C (σ(T )) is dense in L2(µ)we get (via the BLT-Theorem, [FA1] Theorem 2.32) a uniqueextension

V : L2(µ)−→ H

which is linear and an isometry. Note that T n x0 ∈ V (σ(T )) for all n¾ 0 (since f ∈ C (σ(T )),f (t) = tn), i.e. T n x0 ∈ V (σ(T )) ⊆ V (σ(T )) ⊆ H for all n ¾ 0 and since by assumption “x0 isa cyclic vector”

spanT n x0 | n¾ 0= H

we get that the range of V is dense in H, hence V is surjective12. Therefore V is unitary(V V−1 = 1H , V−1V = 1L2(µ)). Also, for ϕ ∈ C (σ(T )):

T (V (ϕ)) = T (V (ϕ)) = T (ϕ(T )x0) = (T ϕ(T ))x0 = V (t ·ϕ)

Hence (V−1T V )(ϕ) = t ·ϕ for every ϕ ∈ C (σ(T )) and hence, by density, for ϕ ∈ L2(µ). Nowlet U := V−1 then the statement of the theorem follows.

We needed:

Proposition. If µ is a finite regular Borel measure on a compact metric/topological space X ,then C (X )⊆ Lp(µ) for 1¶ p <∞ is dense.

More generally, one has:

Theorem 2.25 (Multiplication operator-version of the Spectral Theorem for bounded opera-tors). Every bounded, self-adjoint operator is unitarily equivalent to a multiplication operator.More precisely: For and T ∈B(H), T ∗ = T there exists a measure space13 (Ω,Σ,µ), a boundedmeasurable function f : Ω −→ R, and a unitary operator U : H −→ L2(µ) (UU−1 = 1L2(µ),U−1U = 1H) such that

(U T U−1)ϕ = f ϕ

i.e. (U T U−1ϕ)(t) = f (t)ϕ(t) for µ-a.e. t ∈ Ω.11Was ist der Unterschied zwischen dµ und µ???12The range of an isometry is always a closed subset!13If H is separable, this is σ-finite. This is the only case we will treat here.

52

We do the proof only for H separable. Notation: We write H = ⊕2Hi iff for every x ∈ H, x =∑

i<N x i , x i ∈ Hi (unique decomposition) and ‖x‖2 =∑

i<N ‖x i‖2 (here N ∈ N or N = ∞).We identify x with the possibly finite sequence x ii<N (i.e. the Hi are pairwise orthogonalsubspaces of H such that their linear span is dense). We need:

Lemma 2.26. Let H be a separable Hilbert space, T ∈ B(H), and T = T ∗. Then there exists adecomposition H =⊕2Hi such that T (Hi)⊆ Hi and the restricted operator Ti := T |Hi

: Hi −→ Hihas a cyclic vector x i ∈ Hi for all i.

Proof of Theorem 2.25. By 2.24 there exist unitary operators Ui : Hi −→ L2(µi) and boundedmeasurable functions fi : σ(Ti) −→ R such that (Ui TiU

−1i )(ϕi) = fi ·ϕi , µi-a.e. Let Ω be the

the disjoint union of the sets σ(Ti)⊆ R. One way of thinking of this is:

Ω =⋃

i<N

(σ(Ti)× i)⊆ R2

LetΣ = A⊆ Ω | π1(A∩σ(Ti)× i) is Borel (in R) ∀i

and let µ: Σ−→ [0,∞] be given by

µ(A) =∑

i<N

µ(A∩σ(Ti))

Recall that dµi is the measure d⟨x i , E iλ

x i⟩, x i the cyclic vector of Hi and E i the spectral mea-sure of Ti . Then Σ is a σ-algebra, and µ is a (σ-finite) measure. Let f (t) = fi(t) if t ∈ σ(Ti)(rather f (t, i) = fi(t), (t, i) ∈ Ω). We write f = fii<N , similarly ϕ = ϕii<N for ϕ ∈ L2(µ).Let U : H −→ L2(µ) be defined by

U(x ii<N ) = Ui x ii<N

Then U is unitary, and (U T U−1)ϕ = f ·ϕ.

Proof of Lemma 2.26. We will use Zorn’s Lemma (Axiom 4.3 in [FA1]). Let H be the familyof finite or countable families Hi of pairwise orthogonal closed subspaces Hi ⊆ H withT (Hi)⊆ Hi , and Hi = spanT n x i | n¾ 0 for some x i ∈ Hi , i.e.

H =

Hii<N

N ∈ N or N =∞, Hi ⊆ H closed linear subspacesHi ⊥ H j , i 6= j, T (Hi)⊆ Hi ∀i∀i ∃x i ∈ Hi , Hi = spanT n x i | n 6= 0

NoteH 6= ;! This is since

0

∈H . Also,H is partially ordered by inclusion:

Hii<N K j j<M :⇐⇒ Hii<N ⊆ K j j<M

LetK be a chain inH , i.e. a totally ordered subset (∀h, k ∈K : h k or k h). Write k ∈Kas k = Hik | i < Nk. Let h0 :=

⋃

k∈K k = Hik | i < Nk, k ∈ K . Then h0 ∈ H . Since H isseparable, h0 can at most contain countably many (different) Hik ’s, or we could construct anuncountable orthonormal basis in H (That rest of conditions for “h0 ∈H ”are trivially satisfiedsince K is totally ordered). Note that h0 is an upper bound for K (k h0 for all k ∈ K ).Hence, by Zorn’s Lemma H contains at least one maximal element h (h maximal: If h ∈ Hsuch that h h, then h= h). Let

U := span⋃

K∈h

K

53

(K ⊆ H subspace, h = K). Assume for contradiction that U 6= H, then there would existx ∈ U⊥ \ 0 (U is closed). Let V := spanT n x | n 6= 0, then T (V ) ⊆ V by construction andx is a cyclic vector for the restriction T |V : V −→ V . Hence h h∪ V (trivial: h∪ V ∈ H ,since T (V ) ⊆ V and V ⊥ U). Since h was maximal, h = h∪ V hence V ∈ H but x ∈ V ,so x ∈ U by definition of U . This implies x = 0 . Hence U = H and we have the disireddecomposition of H given by h.

Remark. The Spectral Theorem can be generalised to normal operators T (T ∗T = T T ∗) usingthat for S1 =

T+T ∗

2, S2 =

T−T ∗

2i, we have T = S1+ iS2, S1, S2 self-adjoint and [S1, S2] = 0 (since

T is normal, see Problem 30). We shall not do this here in detail.

54

3 Unbounded operators

3.1 Motivation

In particular we shall study symmetric operators and quadratic forms. In [FA1] one briefly dis-cusses that not all operators are bounded. Typical examples are differential operators (partialdifferential equations, quantum mechanics). We do have that d

dxis bounded as an operator

from the Banach space (C 1([0, 1]),‖·‖C 1) to (C 0([0,1]),‖·‖C 0), with ‖ f ‖C 1 = ‖ f ‖∞+‖ f ′‖∞,‖ f ‖C0

= ‖ f ‖∞. T = ddx

and T : C 1 −→ C0. Then

‖T f ‖C 0 = ‖ f ′‖∞ ¶ ‖¶ ‖ f ‖∞+ ‖ f ′‖∞ = ‖ f ‖C 1

But this choise of spaces is – for many practical purposes – not suitable. Here enough motiva-tion: They are not Hilbert spaces. However, we need an operator on a Hilbert space into itself,to talk about spectral theory! But (C 1([0,1]),‖ · ‖2), ‖ f ‖22 =

∫

| f |2 with (C 1([0, 1]),‖ · ‖H ′),with ‖ f ‖2H ′ =

∫

| f ′|2+| f |2 are not Hilbert spaces. (L2([0, 1]),‖·‖2) is a Hilbert space, but howto differentiate any general L2-function?? We could dicuss this in distribution theory and/orweak differentiation. But we can already see that this does not give bounded operators: Let

en(x) = einx , x ∈ [0, 1] then ‖en‖22 =∫ 1

01dt = 1 but

‖Ten‖22 = n2

1∫

0

1 dt = n2

Hence @C > 0 : ‖Ten‖2 ¶ C‖en‖ for all n! So we need to study unbounded operators. Recallthat such maps are NOT continuous! I.e. xn

n→∞−−−→ x needs not imply T xn −→ T x in general!We know that f ′ = a f iff f (t) = eta f0 and u′ = Ax iff u(t) = etAu0 with eA =

∑∞n=0

An

n!.

But, if we like to solve the heat equation, ut = ∆u we need to make (if possible) senseof u(t) = et∆u0, e∆ =

∑∞n=0

∆n

n!. We will need to study operators on subspaces of Hilbert

spaces (motivated by, for example, C 1[0, 1] ⊆ L2[0,1] above). The “theory” of unboundedoperators is less “complete” than that of bounded ones, but still gives results for a large class ofinteresting (concrete!) operators from partial differential equations and quantum mechanics.

3.2 Unbounded operators and their adjoint

Let H be a separable, complex Hilbert space.

Definition 3.1. (a) An operator (operator in a Hilbert space H) T : D(T ) −→ H is a linearmap whose domain (operator domain) D(T ) ⊆ H is a linear subspace1 of H. T is said tobe densely defined (in H) iff D(T ) = H.

(b) An operator S : D(S) −→ H is called an extension of the operator T : D(T ) −→ H iffD(T ) ⊆ D(S) and Sx = T x for every x ∈ D(T ). We write T ⊂ S. Two operators S and Tare equal iff S ⊂ T and T ⊂ S (we write T = S).

(c) An operator T : D(T ) −→ H is called symmetric (Hermitian) iff ⟨y, T x⟩ = ⟨T y, x⟩ for allx , y ∈ D(T ).

1In general this is not a closed subspace!

55

Note. (1) If H = L2[0, 1] and T = i ddx

in

D(T ) = x ∈ C 1[0, 1] | x(0) = x(1) = 0 ⊆ H

and S = i ddx

inD(S) = x ∈ C 1[0,1] | x(0) = x(1)& D(T )⊆ H

Then both S and T are symmetric and T ⊂ S.

(2) Recall the Hellinger-Toeplitz-Theorem: If T : H −→ H is a linear map, such that ⟨y, T x⟩=⟨T y, x⟩ for all x , y ∈ H. Then T is continuous, i.e. T ∈B(H) (and so T ∗ = T).

We now define the adjoint of a densely defined operator: Let T : D(T ) −→ H be a denselydefined operator in H and let

D(T ∗) = y ∈ H | x 7−→ ⟨y, T x⟩ is continuous on D(T )

Then D(T ∗) ⊆ H is a linear subspace. For y ∈ D(T ∗) fixed we know that x 7−→ ⟨y, T x⟩is continuous on a dense subspace of H (namely, D(T )), hence has a unique extension(BLT-Theorem, [FA1] Theorem 2.32) to a bounded linear functional on H, which, by Riesz’Representation Theorem can be written uniquely as x 7−→ ⟨z, x⟩ for some z ∈ H. Hence⟨z, x⟩= ⟨y, T x⟩ for all x ∈ D(T ) and y ∈ D(T ∗). We let T ∗ y := z (z is uniquely given, so well-defined). Claim: The map T ∗ : D(T ∗) −→ H, y 7−→ T ∗ y is linear. Clearly ⟨T ∗ y, x⟩ = ⟨y, T x⟩for every x ∈ D(T ), y ∈ D(T ∗).

Definition 3.2. (a) The operator T ∗ defined above is the adjoint (operator) of T .

(b) If T = T ∗ then T is called self-adjoint.

Note. (1) If T ∈B(H), then this definition coincides with the old one.

(2) T = T ∗ in particular demands that D(T ) = D(T ∗) (recall 3.1 (b)).

(3) Recall that T is symmetric iff ⟨y, T x⟩ = ⟨T y, x⟩ for all x , y ∈ D(T ). If T = T ∗, thenD(T ) = D(T ∗) and T x = T ∗x for all x ∈ D(T ) = D(T ∗), hence ⟨T y, x⟩ = ⟨T ∗ y, x⟩ =⟨y, T x⟩ for every x ∈ D(T ); y ∈ D(T ∗) = D(T ) so

T = T ∗ =⇒ T is symmetric

In general, the Spectral Theorem holds only for self-adjoint operators (not for symmetricones). We shall study the question about when an operator is self-adjoint in more detail. Infact:

Lemma 3.3. Let T : D(T ) −→ H densely defined and symmetric. Then T ⊂ T ∗. In particular,D(T ∗)⊆ H is dense, so (T ∗)∗ =: T ∗∗ is well-defined.

Proof. Let x , y ∈ D(T ). Then T symmetric implies ⟨T y, x⟩ = ⟨y, T x⟩ and hence the linearfunctional x 7−→ ⟨y, T x⟩ is given by the scalar product with a fixed vector (namely T y).Hence y ∈ D(T ∗). Further we have that for x , y ∈ D(T )⊆ D(T ∗):

⟨T ∗ y, x⟩= ⟨y, T x⟩= ⟨T y, x⟩

So ⟨(T ∗ − T )y, x⟩ = 0 for every x , y ∈ D(T ). By denseness, this implies T ∗ y = T y fory ∈ D(T ).

Remark 3.4. If T is not symmetric, then it may happen that D(T ∗) = 0.

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3.3 Closed operators

For the study of self-adjointness, the concept of a closed operator is important (see also [FA1],Definition 4.14).

Definition 3.5. Let X , Y be normed spaces, D ⊆ X a linear subspace, T : D −→ Y . Then T iscalled closed (a closed operator) iff: Whenever xnn∈N ⊆ D, xn

n→∞−−−→ x ∈ X which satisfiesT xn

n→∞−−−→ y ∈ Y , then x ∈ D and T x = y .

LetΓ(T ) = gr(T ) = (x , T x) ∈ X × Y | x ∈ D ⊆ D× Y ⊆ X × Y

be the graph of T . Then T is closed iff Γ(T ) is a closed subspace in X ⊕2 Y (i.e. with norm‖(u, v)‖X⊕2Y =

p

‖u‖2X + ‖v‖2Y ).

Note. If D = D(T ) = X , T linear, consider the following three statements:

(a) xnn→∞−−−→ x

(b) T xnn converges in Y : ∃y ∈ Y : T xnn→∞−−−→ y

(c) T x = y .

T is continuous (bounded) iff (a) implies (b) and (c). Now T is closed iff (a) and (b) imply(c)

Recall [FA1], Theorem 4.16: “Closed Graph Theorem”.

Theorem 3.6. Let X , Y be Banach spaces, T : X −→ Y closed and linear. Then T is bounded (i.e.T ∈B(X , Y )).

Proposition 3.7. Let T : D(T )−→ H be densely defined.

(a) T ∗ is closed.

(b) If T ∗ is densely defined (for example if T is symmetric) then T ⊂ T ∗∗.

(c) Assume T ∗ is densely defined, S : D(S)−→ H is closed with T ⊂ S. Then T ∗∗ ⊂ S.

As an immediate consequence we get

Corollary 3.8. Let T : D(T )−→ H be densely defined.

(a) T is symmetric⇐⇒ T ⊂ T ∗. In this case, T ⊂ T ∗∗ ⊂ T ∗ = T ∗∗∗. Hence T ∗∗ is also symmetric.

(b) T is closed and symmetric⇐⇒ T = T ∗∗ ⊂ T ∗.

(c) T = T ∗ if and only if T = T ∗∗ = T ∗.

Proof. Exercise!

Remark. As a consequence from (a): T = T ∗ implies T is closed. Hence non-closed operatorscannot be self-adjoint. (c) means that T ∗∗ is called the closure T of T : The smallest closedextension.

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Proof of 3.7. (a) We need to prove: If yn ⊆ D(T ∗) and ynn→∞−−−→ y ∈ H, T ∗ yn

n→∞−−−→ z ∈ Hthen y ∈ D(T ∗) and z = T ∗ y . Now, for all x ∈ D(T ):

⟨y, T x⟩= limn→∞⟨yn, T x⟩= lim

n→∞⟨T ∗ yn, x⟩= ⟨z, x⟩

because x ∈ D(T ), yn ∈ D(T ∗) for all n. Hence x 7−→ ⟨y, T x⟩ = ⟨z, x⟩ is continuous andbounded (on D(T )). So y ∈ D(T ∗) and ⟨y, T x⟩= ⟨z, x⟩ shows T ∗ y = z.

(b) Assume T ∗ is densely defined (i.e. D(T ∗) = H). Let x ∈ D(T ) and y ∈ D(T ∗), then

⟨y, T x⟩= ⟨T ∗ y, x⟩

Hence y 7−→ ⟨x , T ∗ y⟩ = ⟨T ∗ y, x⟩ is continuous on D(T ∗) (it equals ⟨y, T x⟩). So x ∈D((T ∗)∗) = D(T ∗∗) and

⟨x , T ∗ y⟩= ⟨T ∗∗x , y⟩ ∀y ∈ D(T ∗)

Since x ∈ D(T ), this implies ⟨T x , y⟩ = ⟨x , T ∗ y⟩ = ⟨T ∗∗x , y⟩ for all y ∈ D(T ∗) and sinceD(T ∗) ⊆ H is dense, it follows that ⟨T x , y⟩ = ⟨T ∗∗x , y⟩ for all y ∈ H, hence T x = T ∗∗xfor all x ∈ D(T ), that is, T ⊂ T ∗∗.

(c) The claim follows from Γ(T ) = Γ(T ∗∗) (closure with respect to the norm on H ⊕2 H).Γ(T ) ⊆ Γ(T ∗∗) follows from (a) and (b): By (b) T ⊂ T ∗∗, i.e. Γ(T ) ⊆ Γ(T ∗∗) and by(a) T ∗∗ = (T ∗)∗ is a closed operator so Γ(T ) ⊆ Γ(T ∗∗) = Γ(T ∗∗). For Γ(T ) ⊇ Γ(T ∗∗)it is enough to prove that Γ(T )⊥ ⊆ Γ(T ∗∗)⊥ where “⊥” means orthogonal complementwith respect to the scalar product on H × H, ⟨(u, v), (x , y)⟩H×H = ⟨u, x⟩H + ⟨v, y⟩H . Let(u, v) ∈ Γ(T )⊥ i.e.

0= ⟨(u, v), (x , T x)⟩= ⟨u, x⟩+ ⟨v, T x⟩ ∀x ∈ D(T )

Hence ⟨v, T x⟩ = −⟨u, x⟩ for all x ∈ D(T ), so v ∈ D(T ∗) (x 7−→ ⟨v, T x⟩ is bounded) andT ∗v =−u. Let now (z, T ∗∗z) ∈ Γ(T ∗∗), then

⟨(u, v), (z, T ∗∗z)⟩= ⟨u, z⟩+ ⟨v, T ∗∗z⟩= ⟨u, z⟩+ ⟨T ∗v, z⟩= ⟨u+ T ∗v, z⟩= 0

so (u, v) ∈ Γ(T ∗∗)⊥.

3.4 Essential self-adjointness

Note. For T : D(T )−→ H densely defined and symmetric:

T ⊂ T ∗∗ ⊂ T ∗ = T ∗∗∗ (∆)

If T is self-adjoint T = T ∗∗ = T ∗. “In between” we have:

Definition 3.9. If T : D(T ) −→ H is densely defined and symmetric. T is called essentiallyself-adjoint iff T = T ∗∗ is self-adjoint.

Since T = T ∗∗ by the remark after 3.7, we have T is essentially self-adjoint ⇐⇒ T ⊂ T ∗∗

(since T ∗∗ = T ∗∗∗, see (∆) above).

Remark. If T is symmetric, but not self-adjoint then T ⊂ T ∗∗ but T 6= T ∗.

Note (Exercise!). If T ⊂ S, then T ∗ ⊃ S∗.

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Hence: The domain of T is too small – enlarging it (i.e. finding a symmetric extension) willdiminish the domain of the adjoint. The “goal” is to find an extension S such that the domains“meet”/“fit”, i.e. S = S∗. T may or may not have a self-adjoint extension! It may have morethan one! An essentially self-adjoint operator has exactly one self-adjoint extension (namelyT ∗∗). We continue to search for criterions for self-adjointness.

Lemma 3.10. Let T : D(T )−→ H be densely defined.

(a) N(T ∗∓ i) =R(T ± i)⊥. Hence, N(T ∗∓ i) = 0 iff R(T ± i)⊆ H is dense.

(b) Assume, T is symmetric and closed. Then R(T ± i)⊆ H is closed.

Proof. (a) Note that (T + i)∗ = (T + iI)∗!= T ∗ − iI = T ∗ − i (!! Use the definition, the sum

T + iI is defined on D(T )).

“⊇” Let y ∈ R(T + i)⊥. Then ⟨y, (T + i)z⟩ = 0 for every z ∈ D(T ). So y 7−→ ⟨y, Tz⟩is is continuous (it equals −⟨y, iz⟩. Hence y ∈ D(T ∗) and 0 = ⟨(T ∗ − i)y, z⟩ for allz ∈ D(T ). So (T ∗ − i)y = 0, since D(T ) is dense, i.e. y ∈ N(T ∗ − i). The proof issimilar for “−”.

“⊆” Read the above backwards!

(b) We have ⟨x , T x⟩ ∈ R for every x ∈ D(T ) since T is symmetric. Hence for x ∈ D(T )

‖(T + i)x‖2 = ‖T x‖2+ ‖x‖2+ ⟨ix , T x⟩+ ⟨ix , T x⟩

= ‖T x‖2+ ‖x‖2+ 2Re(⟨ix , T x⟩)

= ‖T x‖2+ ‖x‖2+ 2Re(−i⟨x , T x⟩) = ‖T x‖2+ ‖x‖2 ¾ ‖x‖2

Hence (T + i)−1 : R(T + i)−→ D(T ) exists and is bounded (put y = (T + i)x above). Letxnn ⊆ D(T ) such that (T + i)xn

n→∞−−−→ y ∈ R(T + i). Then (T + i)xnn ⊆ R(T + i) isCauchy (since convergent) and since (T + i)−1 is bounded, it follows that xnn is Cauchy(xn = (T + i)−1 yn = (T + i)−1(T + i)xn). Hence there exists x ∈ H such that xn → x (HHilbert). Therefore: T xn = (T + i)xn− ixn

n→∞−−−→ y− ix . So, since xnn ⊆ D(T ), xn→ x ,T xnn converges and T xn

n→∞−−−→ y − ix . Since T is closed, it follows that x ∈ D(T ) andT xn

n→∞−−−→ T x , i.e. y− ix = T x , that is y = (T + i)x ∈R(T + i). Hence R(T + i) is closed.Again, similarly for T − i.

Theorem 3.11. Let T : D(T )−→ H be densely defined and symmetric. The following statementsare equivalent:

(i) T = T ∗ (T is self-adjoint)

(ii) T is closed and N(T ∗± i) = 0

(iii) R(T ± i) = H

Proof. (i)⇒ (ii) By 3.7 (a), T = T ∗ is closed. Let x ∈ N(T ∗ + i), i.e. (T ∗ + i)x = 0. Recallthat for a symmetric operator S (⟨z, Sz⟩ ∈ R for all z ∈ D(S)) we have ‖(S+i)z‖2 ¾ ‖z‖2,in particular, since T ∗ is symmetric, x = 0 (since (T ∗ + i)x = 0). So N(T ∗ + i) = 0(similarly for T ∗− i).

(ii)⇒ (iii) By 3.10 (b), R(T+ i)⊆ H is closed (since T is closed), and by 3.10 (a), it is dense(since N(T ∗± i) = 0).

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(iii)⇒ (i) We have T ⊂ T ∗, hence we need to prove that D(T ∗) ⊆ D(T ). Then if x ∈ D(T ∗)we have x ∈ D(T ) and T ⊂ T ∗, T x = T ∗x . Let y ∈ D(T ∗), then (T ∗−i)y ∈ H =R(T−i),so let x ∈ D(T ) with (T ∗− i)y = (T − i)x . Since T ⊂ T ∗ this gives

(T ∗− i)y = (T ∗− i)x

i.e. (T ∗ − i)(y − x) = 0. But R(T + i) = H means R(T + i) is dense in H, so 3.10 (a)gives N(T ∗− i) = 0. So y − x = 0, i.e. x = y ∈ D(T ).

Corollary 3.12. Let T : D(T )−→ H be densely defined and symmetric. The following statementsare equivalent:

(i) T is essentially self-adjoint (i.e. T ∗∗ = T is self-adjoint)

(ii) N(T ∗± i) = 0

(iii) R(T ± i)⊆ H are dense

Proof. (i)⇒ (ii) (i) implies T ∗∗ = (T ∗∗)∗ so by Corollary 3.8, T ∗∗ = T ∗∗∗ = T ∗. By 3.11(used on T ∗∗!) it follows that N(T ∗± i) =N((T ∗∗)∗± i) = 0. This is (ii).

(ii)⇒ (i) By Corollary 3.8 (a) we have T ∗∗∗ = T ∗, so N((T ∗∗)∗ ± i) = 0 and T ∗∗ is closedby Proposition 3.7. So, by 3.11 (on T ∗) T ∗∗∗ = T ∗∗ hence T is essentially self-adjoint.

(ii)⇔ (iii) Is in 3.10 (a).

3.5 Existence of self-adjoint extensions

Definition 3.13. We call dimN(T ∗±i) the deficiency indices of T (here, this means the Hilbertspace dimension).

We have the following, which answers, when a symmetric operator has self-adjoint extensions:

Theorem 3.14. Let T : D(T ) −→ H be densely defined and symmetric. T has a self-adjointextension2 if and only if dimN(T ∗+ i) = dimN(T ∗− i).

Note. Recall 3.11: If T is self-adjoint, then dimN(T ∗ + i) = dimN(T ∗ − i) = 0. Also T isessentially self-adjoint iff dimN(T ∗+ i) = dimN(T ∗− i) = 0 (the extension is T ∗∗).

Proof. Let’s firstly prove (⇐). Assume dimN(T ∗+ i) = dimN(T ∗− i). From the proof of 3.10(b) we saw that: ‖(T + i)x‖ = ‖(T − i)x‖ for every x ∈ D(T ) (because ⟨x , T x⟩ ∈ R since Tsymmetric). So the map

U : R(T − i)−→R(T + i)

(T − i)x 7−→ (T + i)x

is well-defined, linear, isometric and surjective. Note that by 3.10 (a) R(T±i)⊥ =N(T ∗∓i). Sothe orthogonal complements of R(T + i) and R(T − i) have the same Hilbert space dimension.Hence, we can extend the map U to a unitary operator V : H −→ H (take an orthonormalbasis for R(T − i)⊥ and R(T + i)⊥ and map the first to the second). Claim: V − I is injective.

2I.e. there exists S : D(S)−→ H such that S = S∗ and T ⊂ S.

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Assume (V − I)y = 0, i.e. V y = y . Then V ∗ y = y (V is unitary V V ∗ = V ∗V = I , so V y = yimplies V ∗V = V ∗ y so y = V ∗ y). Let now x ∈ D(T ), then

2i⟨x , y⟩= ⟨y, 2ix⟩= ⟨y, (T + i)x − (T − i)x⟩= ⟨y, (V − I)(T − i)x⟩= ⟨V ∗ y − y, (T − i)x⟩= 0

So y ⊥ D(T ) which is dense in H. So y = 0. This means we an define the operator

S : R(V − I)−→ H

Vz− z 7−→ i(Vz+ z)

In other words S = i(V+ I)(V− I)−1 on R(V− I). V− I is injective so (V− I)−1 is well-definedon R(V − I). The goal is to show that S is a self-adjoint extension of T . First: For x ∈ D(T )we have (see above) (V − I)(T − i)x = 2ix , i.e. x ∈R(V − I) = D(S). So D(T )⊆ D(S) and

Sx = i(V − I)(V − I)−1

1

2i(V − I)(T − i)x

=1

2(V + I)(T − i)x =

1

2V (T + i)x +

1

2(T − i)x

=1

2(T + i)x +

1

2(T − i)x = T x

So T ⊂ S. Secondly we prove that S is symmetric. Let x ∈ D(S), i.e. x = V y − y for somey ∈ H, then

⟨x , Sx⟩= ⟨(V − I)y, i(V + I)y⟩= i

⟨V y, V y⟩+ ⟨V y, y⟩ − ⟨y, V y⟩ − ⟨y, y⟩

= i

⟨V y, y⟩ − ⟨y, V y⟩

=−2Im⟨V y, y⟩ ∈ R

Hence (!), S is symmetric (use polarisation!). To prove S is self-adjoint, we use Theorem3.11. We need to prove R(S ± i) = H. Note that (compute!) S − i = 2i(V − I)−1 andS+ i= 2iV (V − I)−1 (on D(S)) and so for all z ∈ H:

z = (S− i)(V − I) z

2i

= (S+ i)(V − I)V ∗ z

2i

So z ∈R(S− i), z ∈R(S+ i). So S∗ = S and T ⊂ S. For (⇒) assume now S (not the one above– any S) is a self-adjoint extension of T : T ⊂ S, S = S∗. Let V = (S+ i)(S− i)−1 on D(V ) = H(possible because, by 3.11, R(S− i) = H) and U = (T + i)(T − i)−1 on D(U) =R(T − i)⊆ H.Since T ⊂ S, we have U ⊂ V (!) and V is unitary: By ‖(S ± i)x‖ ¾ ‖x‖ (S symmetric), henceV is injective and (again 3.11), R(S + i) = H, so V is surjective. By construction, U maps thesubspace R(T − i) onto the subspace R(T + i). Since V is unitary and U ⊂ V , V maps theorthogonal complement of R(T− i) to the orthogonal complement of R(T+ i) (use ProjectionTheorem). But (by Lemma 3.10), N(T ∗ ± i) = R(T ± i)⊥, so V maps N(T ∗ + i) unitarily toN(T ∗− i). Hence dimN(T ∗− i) = dimN(T ∗+ i).

Note. Re-cap: T symmetric implied ‖(T + i)x‖ = ‖(T − i)x‖ for all x ∈ D(T ) and makedsure that U : R(T − i) −→ R(T + i), (T − i)x 7−→ (T + i)x is well-defined. This we extendedto a unitary V : H −→ H via the assumption and R(T + i)⊥ = N(T + i). Then we definedS : R(V − I) −→ H, Vz − z 7−→ i(Vz + z). S∗ = S, T ⊂ S is algebraic manipulation – payingattention on domains.

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3.6 The spectrum of an unbounded operator

Definition 3.15. Let T : D(T )−→ H be densely defined

(a) The resolvent set of T is

ρ(T ) := λ ∈ C | T −λI : D(T )−→ H is bijective and (T −λ)−1 ∈B(H)

(b) R: ρ(T )−→B(H), λ 7−→ Rλ = (T −λI)−1 is the resolvent map.

(c) σ(T ) := C \ρ(T ) is the spectrum of T .

(d) If 0 6= x ∈ H and λ ∈ C are such that T x = λx , then λ is called eigenvalue of T and x aneigenvector of T (this automatically implies x ∈ D(T )!).

Remarks. (1) (T −λ)−1 : H −→ D(T )⊆ H, so think of as (T −λ)−1 : H −→ H, but then it isnot necessarily onto!

(2) If T is a closed operator and T − λ is bijective for some λ ∈ C, then it is automatic that(T −λ)−1 is bounded (Open Mapping Theorem!)

(3) However, if T is not closed, then σ(T ) = C (Exercise!). So it only makes sense to studyspectral theory of closed operators!

(4) Obviously, λ ∈ σ(T ) if λ is an eigenvalue.

Proposition 3.16. Let T : D(T )−→ H be densely defined. Then

(a) ρ(T )⊆ C is open, hence σ(T )⊆ C is closed.

(b) The resolvent map λ 7−→ Rλ = (T −λ)−1 ∈B(H) is analytic and

Rµ− Rλ = (µ−λ)RµRλ

Proof. As (!) in the bounded case (modulo domains!). See 0.3.

Note. (1) If T is not closed, then (a) is trivial.

(2) σ(T ) = ; is possible! (but not if T ∈B(H)).

(3) σ(T ) need not be compact (since not necessarily bounded).

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4 Spectral Theory for unbounded self-adjoint operators

For the spectum of a self-adjoint (unbounded) operator we first prove:

Proposition 4.1. Let T : D(T )−→ H be densely defined, T = T ∗. Then σ(T )⊆ R.

Proof. Let z = λ+ iµ ∈ C \R, i.e. µ 6= 0, µ,λ ∈ R and let

S :=T

µ−λ

µ

on D(S) = D(T ). Then (!) S = S∗ (since T = T ∗ exercise!) in particular (S symmetric),

‖(T − z)x‖2 = ‖µ(S− i)x‖2 = µ2‖(S− i)x‖2 ¾ µ2‖x‖2

Hence (T−z)−1 : R(T−z)−→ D(T ) exists where R(T−z) =R(S− i) and (by the calculationabove) is bounded. Since S = S∗ we have by Theorem 3.11 that R(S − i) = H, i.e. (T −z)−1 : H −→ D(T ) is bounded, so z ∈ ρ(T ).

4.1 Spetral Theorem – Multiplication operator version

Theorem 4.2 (Spectral Theorem for self-adjoint operators – Multiplication operator version).Let T : D(T ) −→ H be self-adjoint. Then there exists a measure space (Ω,Σ,µ) (if H is separa-ble, it is σ-finite), a measurable function f : Ω −→ R (not necessarily bounded) and a unitaryoperator U : H −→ L2(µ) such that

(a) x ∈ D(T )⇐⇒ f · U x ∈ L2(µ)

(b) U T U∗ϕ = f ·ϕ =: M f (ϕ) for every ϕ ∈ D(M f ) = ϕ ∈ L2(µ) : f ϕ ∈ L2(µ).

Remark. From a later exercise, we will know that the operator M f is self-adjoint (and theexercise will give the “spectral theory” of such an operator – see also Problem 16, 39 for B(H).Theorem 4.2 says that any self-adjoint operator is of this form (i.e. is unitarily equivalent tosuch an operator).

For the proof, we know from the previous theorem that σ(T ) ⊆ R, hence R := (T + i)−1 ∈B(H) is well-defined. We aim to prove RR∗ = R∗R (R is normal) and then use the SpectralTheorem, operator version, 2.25, for normal operators (see discussion after proof of 2.24 atthe end of chapter 2). A reformulation of 2.25 reads:

Theorem (Spectral Theorem for normal bounded operators). For any normal bounded opera-tor S there exists a unique (complex) measure G (with compact support) on the Borel-σ-algebraof C with

S =

∫

σ(S)

z dGz

The formula

g(S) =

∫

σ(S)

g(z)dGz

defines the uniquely determined measurable functional calculus. Every normal bounded opera-tor is unitarily equivalent to a multiplication operator defined by multiplication by a boundedmeasurable (complex valued) function.

63

Proof of Theorem 4.2. We prove that R= (T+i)−1 (well-defined by Proposition 4.1) is normal.Let z1, z2 ∈ H. Since T is self-adjoint, Theorem 3.11 gives that T+ i, T− i are surjective, hencethere exist x , y ∈ D(T ) such that z1 = (T + i)x , z2 = (T − i)y . Then

⟨z2, Rz1⟩= ⟨(T − i)y, x⟩= ⟨(T ∗− i)y, x⟩= ⟨y, (T + i)x⟩= ⟨(T − i)−1z2, z1⟩

I.e. R∗ = [(T + i)−1]∗ = (T − i)−1 =: Ri , R= (T + i)−1 = R−i. Hence by 3.16 (b)

RR∗ = R−iRi =−1

2i(R−i− Ri) =

1

2i(Ri− R−i) = RiR−i = R∗R

Therefore, by the previous reformulation of 2.25, we have URU∗ = Mg (multiplication ope-rator by g) for a bounded measurable function g : Ω −→ C, and a unitary operator U : H −→L2(µ). We need to construct a representation of T out of this. Note that

1

τ+ i= γ⇐⇒ τ=

1

γ− i ∀τ,γ ∈ C, τ 6=−i, γ 6= 0

We shall use this on T and R. Let

f (ω) =1

g(ω)− i w ∈ Ω (∗)

Note that R is injective (but maybe not surjective!) so Mg is injective, hence ω ∈ Ω : g(ω) =0 is a set of measure 0 (0 is not an eigenvalue – otherwise could construct an L2-functionϕ, “supported” where g = 0, such that Mgϕ = 0). Hence, f in (∗) is well-defined for almostevery ω ∈ Ω (this will be the f in the theorem). I.e. f (ω) = 1

g(ω) − i µ-almost everywhere.

(a) Let x ∈ D(T ). Since R: H −→ H has range equal to D(T ), we have x = Ry for somey ∈ H, hence

U x = URy = gU y (∆)

(URU∗ = Mg , so UR = Mg U). Also, f · U x = f · g · U y . Now, f · g = 1− ig for almostevery ω ∈ Ω and g ∈ L∞(µ), so f · g is bounded. Also U y ∈ L2(µ), so f · g · U y ∈ L2(µ),so f · U x ∈ L2(µ). On the other hand, assume, f · U x ∈ L2(µ) (for some x ∈ H). Thenalso iU x ∈ L2(µ), so ( f + i)U x ∈ L2(µ) and U : H −→ L2(µ) is bijective so there is y ∈ H:U y = ( f + i)U x , hence

Mg(U y) = g · U y = g( f + i)U x = 1 · U x

for almost every ω (using (∗)). So

x = (U∗Mg U)y = Ry = (T + i)−1 y ∈ D(T )

(b) By (a), D(M f ) = U(D(T )) ⊆ L2(µ), and, as above, for x ∈ D(T ) there is y ∈ H withx = Ry , so (T + i)x = y , i.e. T x = y − ix . Hence, by (∆)

U T x = U y − iU x =1

gU x − iU x =

1

g− i

U x = f · U x = M f U x

So U T U∗ϕ = f ·ϕ = M f (ϕ) for every ϕ ∈ D(M f ) (ϕ = U x). Since T is symmetric, alsoM f is symmetric and hence f is real-valued (see later exercise). So f : Ω−→ R.

64

4.2 Spectral Theorem – Spectral measure version

We continue with the spectral measure version of the Spectral Theorem.

Theorem 4.3 (Spectral decomposition of unbounded self-adjoint operators). Let T : D(T )−→H be self-adjoint. Then there exists a unique spectral measure E such that

⟨y, T x⟩=∫

Rλd⟨y, Eλx⟩ ∀x ∈ D(T ), y ∈ H

If h: R−→ R is measurable and

Dh =

¨

x ∈ H :

∫

R|h(λ)|2 d⟨x , Eλx⟩<∞

«

then

⟨y, h(T )x⟩=∫

R

h(y)d⟨y, Eλx⟩ x ∈ Dh, y ∈ H

defines a self-adjoint operator h(T ): Dh −→ H (Dh ⊆ H).

Proof. By 4.2, there exist a measure space (Ω,Σ,µ), a measurable function f : Ω −→ R, aunitary U : H −→ L2(µ) such that U T U∗ϕ = f · ϕ = M f (ϕ) µ-almost everywhere for ϕ ∈D(M f ) = ϕ ∈ L2(µ) : f · ϕ ∈ L2(µ). Let h ∈ Mb(R) (i.e. h: R −→ C bounded andmeasurable). Then define the operator h(M f ) := Mh f , i.e

(h(M f )ϕ)(w) = (Mh f ϕ)(w) = (h f )(w) ·ϕ(w) = h( f (w)) ·ϕ(w) µ-a.e.

Since h is bounded, the function h f is bounded, hence Mh f is a bounded (multiplication)operator. Such operators are always normal. Hence h(M f ) ∈B(L2(µ)), h(M f ) normal. Claim:The map

Mb(R)−→B(L2(µ))

h 7−→ h(M f )()

is linear, bounded/continuous and multiplicative. For linearity let h, g ∈Mb(R), α ∈ C:

[((αh+ g)M f )ϕ](ω) = [M(αh+g) f ](ω) = ((αh+ g) f )(ω) ·ϕ(ω)= (αh+ g)( f (ω)) ·ϕ(ω)= α · h(( f (ω)) ·ϕ(ω) + g( f (ω) ·ϕ(ω)= [αh(M f )ϕ](ω) + [g(M f )ϕ](ω) µ-a.e.

For boundedness let ϕ ∈ L2(µ):

‖h(M f )ϕ‖2L2(µ) =

∫

Ω

|h( f (ω))ϕ(ω)|2 dµ(ω)¶ ‖h‖2∞

∫

Ω

|ϕ(ω)|2 dµ(ω) = ‖h‖2∞‖ϕ‖2L2(µ)

when ‖h‖∞ = ess supt∈R |h(t)|. So ‖h(M f )‖B(L2(µ)) ¶ ‖h‖∞. Hence the operator in () isbounded with norm ¶ 1 (i.e. an element of B((Mb(R),‖ · ‖∞);B(L2(µ)),‖ · ‖op). For multi-plicativity let h, g ∈Mb(R). For any ϕ ∈ L2(µ):

[h(M f )g(M f )ϕ](ω) = (h f )(ω) · (g f )(ω)ϕ(ω) = h( f (ω))g( f (ω))ϕ(ω)

= (h g)( f (ω))ϕ(ω)

= [(h · g) f ](ω) ·ϕ(ω) = [(h · g)(M f )ϕ](ω) µ-a.e.

65

i.e. [h·g](M f ) = h(M f )·g(M f ). Hence, in particular, for any Borel-set A⊆ R the characteristicfunction χA is measurable and bounded (even if A is unbounded set). So χA ∈Mb(R) and wedefine FA := χA(M f ) = MχA f . Note that

(χA f )(ω) = χA( f (ω)) =

(

1 f (ω) ∈ A

0 f (ω) /∈ A

)

= χ f −1(A)(ω)

Claim: A 7−→ FA is a spectral measure (on L2(µ))1. Clearly, F∗A = FA = F2A (so, orthogonal

projection).

F∗A = (MχA f )∗ = MχA f = MχA f = FA

andF2

A = (χA(M f ))2 = χA(M f ) ·χA(M f ) = (χ

2A)(M f ) = χA(M f ) = FA

It remains to prove that it is a spectral measure (see Definition 2.13)

(a) We have F; = Mχ; f , but χ; f = 0 so F; = M0 =OL2(µ) and FR = MχR f and χR f = 1,i.e. FR = M1 = 1L2(µ)

(b) Let A1, A2, . . . be pairwise disjoint Borel-sets and ϕ ∈ L2(µ). Then:

∞∑

n=1

(FAnϕ)(ω) =

∞∑

n=1

(MχAn f ϕ)(ω) =∞∑

n=1

(χAn f )(ω)ϕ(ω)

Now

(χAn f )(ω) =

(

1 f (ω) ∈ An

0 f (ω) /∈ An

and An ∩Am = ;, n 6= m. I.e. for almost all ω: Either (χAn f )(ω) = 0 for all n or there is

n0 ∈ N such that (χAn f )(ω) = δn,n0

for all n ∈ N. In both cases, also

(χ⋃∞n=1 An

f )(ω) =

(

0 f (ω) /∈⋃∞

n=1 An

1 ∃! n0 : f (ω) ∈ An0

I.e. for almost all ω

∞∑

n=1

(χAn f )(ω) = [χ⋃∞

n=1 An f ](ω)

This means

∞∑

n=1

(χAn f )(ω) ·ϕ(ω) = (χ⋃∞

n=1 An f )(ω) ·ϕ(ω) = Mχ⋃∞

n=1 An fϕ)(ω)

= (F⋃∞n=1 An

ϕ)(ω)

I.e.∑∞

n=1 FAn(ϕ) = F⋃∞

n=1 An(ϕ). Hence A 7−→ FA is a spectral measure on L2(µ).

1However, in general, it has not got compact support.

66

It follows (using Theorem 2.15 and a density argument) that

h(M f ) =

∫

R

h(λ)dFλ =

∫

hdF h ∈ Mb(R)

and that this uniquely determines F (i.e. there is only one spectral measure such that thisholds). Now: Let EA := U∗FAU ∈ B(H) (U : H −→ L2(µ), FA : L2(µ) −→ L2(µ)). ThenA 7−→ EA is a spectral measure (on H). Clearly

E2A = (U

∗FAU)(U∗FAU) = U∗F2A U = U∗FAU = EA

and E∗A = (U∗FAU)∗ = U∗F∗AU∗∗ = U∗FAU = EA (since FA orthogonal projection and U is

bounded). Also E; = U∗F;U = U∗OL2(µ)U =OH and similarly ER = 1H (since FR = 1L2(µ) andU unitary). Finally for A1, A2, . . . disjoint Borel-sets (in R), x ∈ H:

∞∑

n=1

EAn(x) =

∞∑

n=1

U∗FAn(U x) = U∗

∞∑

n=1

FAn(U x)

!

= U∗F⋃∞n=1 An

(U x)

= (U∗F⋃∞n=1 An

U)(x) = E⋃∞n=1 An

x

Note that as for A 7−→ FA (called F), the spectral measure E (i.e. A 7−→ EA) does not in generalhave compact support. We can now, for h ∈Mb(R) define

h(T ) :=

∫

R

h(λ)dEλ =

∫

hdE

(using Theorem 2.15). The map

Mb(R)−→B(H)

h 7−→ h(T )()

is linear and bounded/continuous. So

‖h(T )‖B(H) =

∫

hdE

B(H)

¶ ‖h‖∞

This fact is useful and not stated explicitly in the theorem! Hence, this defines h(T ) forh ∈Mb(R). Note h(T ) = U∗h(M f )U with h(M f ) as before (this is true for h= χA, since then

h(T ) = χA(T ) =

∫

χA dE = EA = U∗FAU = U∗χA(M f )U = U∗h(M f )U

It then follows by linearity for simple functions h=∑N

i=1αiχAiand then for general bounded,

measurable functions by a limiting argument – use 2.6 (c) (Exercise!)). The map () hasthe properties of a functional calculus (by 2.15). Recall: T : D(T ) −→ H, T = T ∗, U : H −→L2(µ), (Ω,Σ,µ), U∗T U = M f , f : Ω −→ R, measurable, D(M f ) = ϕ ∈ L2(µ) : f ϕ ∈ L2(µ).For h ∈Mb(R): h(M f ) := Mh f ∈B(L2(µ)):

Mb(R)−→B(L2(µ))

h 7−→ h(M f )

67

also has the properties of a functional calculus. Let FA := χA(M f ) = Mχ f−1(A), A Borel, then

A 7−→ FA is a spectral measure (on L2(µ)) and

h(M f ) =

∫

R

h(λ)dFλ

Let EA := U∗FAU . Then A 7−→ EA is a spectral measure (on H), and for h ∈Mb(R)

h(T ) :=

∫

R

h(λ)dEλ = U∗h(M f )U

Let now h: R −→ R be measurable, but not necessarily bounded. For x ∈ H, d⟨x , Eλx⟩ is apositive measure, as earlier (given by A 7−→ ⟨x , EAx⟩ = ⟨x , E2

A x⟩ = ⟨EAx , EAx⟩ = ‖EAx‖2 ¾ 0,A Borel). Let

Dh :=

(

x ∈ H

∫

R

|h(λ)|2 d⟨x , Eλx⟩<∞

)

Claims:

(1) Dh ⊆ H is dense.

(2) The integral∫

R

h(λ)d⟨y, Eλx⟩

exists for x ∈ Dh, y ∈ H (Here, d⟨y, Eλx⟩ is a complex measure).

(3) Therefore there exists an element h(T )x ∈ H such that

⟨y, h(T )x⟩=∫

R

h(λ)d⟨y, Eλx⟩ y ∈ H

(i.e. this defines the map Dh 3 x 7−→ h(T )x ∈ H).

We shall “transport” things to the “concrete” setting on L2(µ) via the unitary U : H −→ L2(µ).To see this: Let x , y ∈ H, write ϕ = U x , ψ= U y . Then

⟨y, EAx⟩H = ⟨U∗ψ, EAU∗ϕ⟩H = ⟨ψ, U EAU∗ϕ⟩L2(µ)

= ⟨ψ, FAϕ⟩L2(µ) = ⟨ψ, Mχ f−1(A)ϕ⟩L2(µ) =

∫

Ω

χ f −1(A)ψϕ dµ=

∫

f −1(A)

ψϕ dµ

(EA = U∗FAU , U EAU∗ = FA). Let

ν(B) :=

∫

B

ψϕ dµ B ∈ Σ

then

⟨y, EAx⟩H =∫

f −1(A)

ψϕ dµ= ν( f −1(A))

68

That is, the measure d⟨y, Eλx⟩ (say, λx ,y) is the pushforward f∗ν of ν via f (“Bildmaß”)2.Hence by the Transformation Theorem (simple in this case!)

∫

R

g(λ)d⟨y, Eλx⟩=∫

Ω

g f dν =

∫

Ω

(g f )ψϕ dµ (?)

for g : R−→ R integrable. Using (?), we get that x ∈ Dh iff ϕ = U x stafisfies∫

Ω

|h f |2|ϕ|2 dµ=

∫

|h|2 f ·ϕϕ dµ=

∫

R

|h(λ)|2 d⟨x , Eλx⟩<∞

By Exercise 46 (i), the subset (in L2(µ)) of such ϕ’s is dense in L2(µ), hence (since U isunitary, Dh ⊆ H is also dense. Note that for h: R−→ C, x ∈ Dh, y ∈ H, we have

∫

h(λ)d⟨y, Eλx⟩

¶∫

R

|h(λ)|d|⟨y, Eλx⟩| (∆)

where d|⟨y, Eλx⟩| is the variation (measure) of the (complex) measure d⟨y, Eλx⟩ (i.e. the“smallest” positive measure |ρ| such that |ρ(A)| ¶ |ρ|(A), A measurable. Clearly, d|⟨y, EAx⟩|is a positive measure, so this is the variation of ⟨y, EAx⟩. Using the analogue of (?) ond|⟨y, Eλx⟩|, we get

∫

R

|h(λ)|d|⟨y, Eλx⟩|=∫

Ω

|h f | · |ϕψ|dµ¶∫

Ω|h f |2|ϕ|2 dµ

1/2∫

Ω|ψ|2 dµ

1/2

=

∫

Ω|h f |2|ϕ|2 dµ

1/2

‖ψ‖L2(µ)

=

∫

Ω|h f |2|ϕ|2 dµ

1/2

· ‖y‖H

since ψ= U y , U unitary. Hence, since x ∈ Dh⇐⇒ (h f )ϕ ∈ L2(µ), we have by (∆) that

∫

Rh(λ)d⟨y, Eλx⟩

¶ C · ‖y‖H

so that y 7−→∫

R h(λ)d⟨y, Eλx⟩ is an (anti-)linear bounded functional on H. By Riesz-Fischer(Riesz Representation Theorem) this functional is given by the scalar product of a uniqueelement z ∈ H. Denote this element h(T )x := z ∈ H. Then, for all x ∈ Dh, y ∈ H:

∫

R

h(λ)d⟨y, Eλx⟩= ⟨y, h(T )x⟩H ()

This defines a map (by the uniqueness of z):

h(T ): Dh −→ H

x 7−→ h(T )x

2I.e. ( f∗ν)(A) = ν( f −1(A)); ν measure on (Ω,Σ), f : Ω−→ R.

69

This is clearly linear (since the EA’s are linear) and we write

h(T ) =

∫

R

h(λ)dEλ ()

Note, however, that () only holds in the sense of ()!!! Let now h: R −→ R be thefunction h(t) = t. Let, as before, x ∈ Dh ≡ Dt , y ∈ H, and write ϕ = U x , ψ = U y . Note that(h f )(ω) = f (ω), ω ∈ Ω. Hence

∫

R

h(λ)d⟨y, Eλx⟩=∫

R

λd⟨y, Eλx⟩=∫

Ω

(h f )ψϕ dµ=

∫

Ω

fψϕ dµ=

∫

Ω

ψ(M f ϕ)dµ

= ⟨ψ, M f ϕ⟩L2(µ) = ⟨U y, M f U x⟩L2(µ) = ⟨y, U∗M f U⟩H = ⟨y, T x⟩H

So Dt = D(T ), and T =∫

RλdEλ (in the sense that ⟨y, T x⟩H =∫

Rλd⟨y, Eλx⟩ for all x ∈ D(T ),y ∈ H – “sense of quadratic forms”). Similar of course,

⟨y, h(T )x⟩H = ⟨y, U∗Mh f U x⟩H ∀x ∈ Dh, y ∈ H

That h(T ) is self-adjoint if h: R −→ R follows from this last formula and from Exercise 46(ii).

4.3 Consequences of the Spectral Theorems

Within the proof, we have seen:

Corollary 4.4. Let T : D(T )−→ H, T = T ∗. Then there is a unique map

Φ:Mb(R)−→B(H)

h 7−→ h(T )

such that

(i) Φ is an involutive algebra homomorphism

(ii) Φ is continuous/bounded. In fact ‖h(T )‖B(H) ¶ ‖h‖∞.

(iii) If hnn∈N ⊆Mb(R), supn ‖hn‖∞ <∞ and hn(t)n→∞−−−→ h(t) for all t ∈ R then

Φ(hn)xn→∞−−−→ Φ(h)x ∀x ∈ H

i.e. hn(T )xn→∞−−−→ h(T )x for all x ∈ H.

We have

h(T ) =

∫

R

h(λ)dEλ ∈B(H)

Noteworthy is the following corollary:

Corollary 4.5. Let T : D(T )−→ H be self-adjiont.

(a) If x ∈ Dh (h: R−→ R measurable), then

‖h(T )x‖2H =∫

R

|h(λ)|2 d⟨x , Eλx⟩

70

(b) If hnn∈N ⊆Mb(R) (measurable and bounded) with

(i) hn(t)n→∞−−−→ t for all t ∈ R.

(ii) |hn(t)|¶ |t| for all t ∈ R, n ∈ N.

then, for any x ∈ D(T )hn(T )x

n→∞−−−→ T x

(i.e. hn(T )→ T strongly).

Proof. (a) Assume first that h: R−→ R is such that h ∈Mb(R). Then

‖h(T )x‖2H = ⟨h(T )x , h(T )x⟩H = ⟨x , h(T )∗h(T )x⟩H = ⟨x , h(T )h(T )x⟩H

= ⟨x , (hh)(T )x⟩H = ⟨x , |h|2(T )x⟩=∫

R

|h(λ)|2 d⟨x , Eλx⟩

all by Corollary 4.4. Assume now that h: R −→ R is general (measurable but not neces-sarily bounded). Note that above we proved that

∫

Rh(λ)d⟨y, Eλx⟩

¶ C‖y‖H

with C =

∫

Ω|h f |2|ϕ|2 dµ

1/2, ϕ = U x , U : H −→ L2(µ) and so that

y 7−→∫

R

h(λ)d⟨y, Eλx⟩

is a bounded (anti-)linear functional, given by the scalarproduct with h(T )x , and of normat most C – i.e.

‖h(T )x‖2H ¶∫

Ω

|h f |2|ϕ|2 dµ=

∫

R

|h(λ)|2 d⟨x , Eλx⟩ (∗∆)

Let now, for h unbounded,

hn(t) =

(

h(t) |h(t)|< n

0 |h(t)|¾ n

Then hn ∈Mb(R) and Dh−hn= Dh and for x ∈ Dh, by (∗∆)

‖h(T )x − hn(T )x‖H = ‖(h− hn)(T )x‖H ¶∫

R

|(h− hn)(λ)|2 d⟨x , Eλx⟩ n→∞−−−→ 0

by Lebegue’s Dominated Convergence Theorem (a dominant is 4|h(λ)|2 ∈ L1(d⟨x , Eλx⟩)).Now,

limn→∞

∫

R

|hn(λ)|2 d⟨x , Eλx⟩=∫

R

|h(λ)|2 d⟨x , Eλx⟩

71

(again, by Dominated Convergence) and∫

R

|hn(t)|2 d⟨x , Eλx⟩= ‖hn(T )x‖2H

since hn ∈Mb(R). It follows that

‖h(T )x‖2H =∫

R

|h(λ)|2 d⟨x , Eλx⟩

(b) The proof is similar to the above, again using Dominated Convergence (see conditions onthe functions hnn∈N.

4.4 Outlook on other approaches to the Spectral Theorem (for bounded/unboundedself-adjoint operators

The proofs we have given follow D. Werner, “Funktionalanalysis”, Springer Lehrbuch (Online!7. Auflage 2011). Other approaches are3

One defines and studies projection valued measures (PVM), spectral measures respectively,i.e. integration with respect to them, and proves that this implies a measurable functionalcalculus (i.e. ΦE :Mb(R) −→B(H), f 7−→

∫

f (λ)dEλ. BUT one gets back in a different way(i.e. given a self-adjoint operator, how to get hold of (or construct) the spectral measure).Namely E ; self-adjoint operator: T :=

∫

RλdEλ – how to “invert” the map: T self-adjointoperator ; E??

(Other) Strategy: The resolvent should be:

RT (z) =

∫

R

(λ− z)−1 dEλ

Define

Fψ(z) := ⟨ψ, RT (z)ψ⟩H =∫

R

1

λ− zd⟨ψ, Eλψ⟩

For a Borel measure µ, the map

Fµ(z) :=

∫

R

1

λ− zdµ(λ)

is known as the Borel transform of µ. It turns out (!) that4 Fµ : H −→ H and it is holomor-phic/complex analytic. Such functions are called Herglotz, or Nevanlinna functions. One can(!) reconstruct the measure µ from Fµ(z) by the Stieltje’s inversion formula

µ(λ) = limδ0

limε0

1

π

λ+δ∫

−∞

Im(Fµ(t + iε))dt (∗)

3After G. Teschl: “Math. Methodes in Quantum Methods”, AMS 2009 (online, for example).4Set H := z ∈ C | Im(z)> 0.

72

Conversely. If Fψ(z) is a Herglotz function, satisfying

|Fµ(z)|¶M

Im(z)

then it is the Borel-transform of a unique measure µψ (given by (∗)), satisfying µψ(R) ¶ M .So: One lets Fψ(z) = ⟨ψ, RT (z)ψ⟩ and prove (!) that this is holomorphic/analytic (for z ∈ρ(T )) and satisfies

|Fψ(z)|¶‖ψ‖2

Im(z)

That Fψ : H −→ H follows from (!) the first resolvent formula/identity. Then one gets the(unique) measure µψ by the Stieltje’s Inversion Formula. By polarization, one can then getthe complex measure µφ,ψ (which will be d⟨ψ, Eλφ⟩). One then defines operators via thequadratic form:

sA(φ,ψ) =

∫

χA(λ)dµφ,ψ(λ) A⊆ R Borel

One proves (!) the existence of corresponding projections EA such that

sA(φ,ψ) = ⟨φ, EAψ⟩=∫

R

χA(λ)dµφ,ψ(λ)

Then one proves (!) that A 7→ EA is a spectral measure and that the corresponding self-adjointoperator

∫

RλdEλ is the original T .

This procedure needs a lot (!) of Complex Analysis.

Yet another approach (After J. Weidmann, “Lineare Operatoren in Hilberträumen”, Teubner,Reed & Simon I, FA). “Resolution of the identity” (“Spektralschar”).

E(λ) := E(−∞,λ] λ ∈ R

(i) E(λ) are orthogonal projections.

(ii) E(λ1)¶ E(λ2) for λ1 ¶ λ2

(iii) limλnλ Eλn= E(λ) strongly5 (strong right continuity).

(iv) limλ→−∞ E(λ) =OH (strongly) and limλ→∞ E(λ) = 1H (strongly).

For a resolution of the identity, there is a unique spectral measure EA such that (∗) holds6. Ifof compact support: There exist m, M ∈ R such that E(λ) = OH for λ < m and E(λ) = 1H forλ > M . One defines, for f : R−→ C continuous and x ∈ H: Want to write

∫

R

f (λ)d⟨x , E(λ)x⟩

5This means the application of an x ∈ H yields the equation.6Define distribution functions µψ(λ) = ⟨ψ, E(λ)ψ⟩ then there is a unique Borel measure such that can recon-

struct EA.

73

as Riemann-Stieltje’s integral as limit of the Riemann-sumsn−1∑

i=0

f (ξi)

⟨x , E(t i+1)x⟩ − ⟨x , E(t i)x⟩

Spectral Theorem: Unique correspondence between self-adjoint bounded operators and spec-tral resolutions with compact support.

The Riemann-Stieltjes integral of a real-valued function f with respect to the real function g(g = ⟨x , E(λ)x⟩) is written

∫ b

a

f (x)dg(x) (∗)

and is the limit of approximating sumsn−1∑

i=0

f (ci)(g(x i+1)− g(x i)) ci ∈ [x i , x i+1], x0 = a, xn = b

If g is everywhere differentiable and g ′ is continuous, thenb∫

a

f (x)dg(x) =

b∫

a

f (x)g ′(x)dx

(∗) exists for example if f is continuous and g is of so-called bounded variation (BV) – exam-ple: g = g1− g2, g1, g2 both monotone7.

4.5 Another outlook: Spectral theory of bounded operatos; Generalisation toBanach algebras

4.5.1 Banach algebras setup (B(X ))

Definition. A complex algebra is a complex vector space A with a multiplication such thatlinear structure and the multiplication “interact well”:

(i) x(y x) = (x y)z

(ii) (x + y)z = xz+ yz, x(y + z) = x y + xz

(iii) α(x y) = (αx)y = x(αy) for every x , y, z ∈ A, α ∈ C

A Banach algebra is a complex algebra A with a norm ‖·‖, making A a Banach space (complete)and satisfying

(iv) ‖x y‖¶ ‖x‖ · ‖y‖ for all x , y ∈ A.

and with a unit element e ∈ A such that

(v) ex = xe = x for all x ∈ A

(vi) ‖e‖= 1.

Note. (1) We’ve not assumed that A is commutative with respect to multiplication.

(2) If A has no unit, then there is a standard way of “adding a unit”: Making, in a canonicalway, a new algebra with a unit (see next time).

(3) (iv) implies that multiplication is continuous.7For more: See Stroock, “Essentials of integration theory for analysis”, GTM 262, Springer (online!).

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4.5.2 Adding a unit to a unit-less algebra

Let A be a complex Banach algebra without a unit, let A1 = A× C, and define the algebraicoperations on A1 by

(i) (a,α) + (b,β) := (a+ b,α+ β)

(ii) β(a,α) = (βa,βα)

(iii) (a,α) · (b,β) = (ab+αb+ βa,αβ)

Define ‖(a,α)‖ := ‖a‖+ |α|. Then A1 with this norm, and the algebraic operations, defined in(i) - (iii) is a Banach algebra with unit (0,1), and a 7−→ (a, 0) is an isometric isomorphism ofA into A1.

Examples. (a) Let K be a non-empty compact Hausdorff space and C (K) the set of all com-plex values continuous functions on K . This clearly is a complex vector space. Let( f g)(x) := f (x)g(x), x ∈ K and the constant function 1 is a unit. The norm ‖ f ‖C (K) :=supx∈K | f (x)|. C (K) is commutative! Hence C (K) is a commutative Banach algebra.

(b) If K is finite, say, K = 1, . . . , n, then, C(K) ∼= Cn with coordinatewise multiplication(n= 1: C).

(c) Let X be a Banach space. Then B(X ) is a Banach algebra with operator norm, unit1X . If dim X = n < ∞, then B(X ) ∼=Mn(C) ≡ Cn×n. If dim X > 1 then B(X ) is non-commutative.

(d) Let K ⊆ Cn, non-empty compact and A⊆C (K) as:

A= f ∈ C (K) | f : K −→ C holomorpic

Then A is a Banach algebra (in the norm of C (K)). If K = D ⊆ C (closed unit disk) thenA is called the disc algebra.

(e) L1(Rn) with convolution as multiplication

( f ∗ g)(x) :=

∫

Rn

f (y)g(x − y)dy

is almost a Banach algebra. It has no unit! One can make an algebra with unit (as above)– or, concretely, enlarge L1(Rn) to the algebra (!) of all complex Borel measures µ on Rn

of the form dµ= f dλn+αδ, where f ∈ L1(Rn), λn is the Lebesgue measure of Rn, α ∈ Cand δ is the Dirac measure at 0 ∈ Rn (δ is the missing unit!)

(f) LetM (Rn) be the algebra (!) of all complex (finite) Borel measures on Rn, with convo-lution as multiplication and with total variation as norm. This is a commutative Banachalgebra, with unit δ, containing the algebra in (e) as a closed (!) subalgebra.

(g) Let (X ,Ω,µ) be a σ-finite measure space, and A= L∞(X ,Ω,µ), then A is a commutative(Abelian) Banach algebra with unit (with all operations defined pointwise).

(h) Let X be a Banach space, and let A= K(X ) be the space of compact operators. Then K(X )is a Banach algebra – this has no unit if dim X =∞ (and if dim X <∞ then K(X ) =B(X )).Note that A is an ideal of B(X ) – in particular, it is a subalgebra of B(X ).

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(i) The construction in (e) (L1(Rn) with convolution) can be generalised, as soon asone has a measure space with a group structure: Let G be a σ-compact locallygroup, with m as the (right) Haar-measure on G (translation invariant). For f , g ∈L1(G, dm), let

( f ∗ g)(x) =

∫

G

f (x y−1)g(y)dm(y)

Then (!) f ∗ g ∈ L1(G, dm) and L1(G) Abelian iff G is Abelian, and L1(G) has a unitiff G is discrete.

If A is a Banach algebra, and there is a map ∗: A−→ A such that for x , y ∈ A, λ ∈ C:

(i) (x∗)∗ = x

(ii) (x y)∗ = y∗x∗

(iii) (x + y)∗ = x∗+ y∗

(iv) (λx)∗ = λx∗

(an involution), then A is called a ∗-Banach algebra (“star-Banach algebra”). If the norm on Asatisfies the ∗-identity

‖x∗x‖= ‖x∗‖ · ‖x‖

(“¶” is always true) then A is called C∗-star algebra (“C-star algebra”). This is a very strongcondition: It turns out that there can only be one norm which can satisfy this, given the alge-braic structure8. Example B(H).

Spectral Theory of Banach algebras: Rudin, “Functional Analysis”.

Definition. (i) An element x ∈ A is invertible iff ∃x−1 ∈ A such that x−1 x = x x−1 = e

(ii) Let G = G(A) be the set of all invertible elements in A.

(iii) For x ∈ A, the spectrum σ(x) is the set of all λ ∈ C such that λe− x is not invertible (inthe algebra A). The complement of the resolvent set ρ(x).

(iv) The spectral radius of x is

r(x) := sup|λ| | λ ∈ σ(x)

Theorem. (0) The inverse x−1 is unique.

(1) G(A)⊆ A is an open subset and x 7−→ x−1 is a homeomorphism of G(A) onto G(A).

(2) σ(x)⊆ C is compact and non-empty for every x ∈ A.

(3) r(x) = limn→∞ ‖xn‖1/m = infn¾1 ‖xn‖1/n

Parts follow from:

Proposition. Let A be a Banach algebra, x ∈ A, ‖x‖< 1 Then

(i) e− x is invertible8Bratteli & Robinson “Operator algebras and Quantum Statistical Mechanics”, Murphy “C∗-algebra and Opera-

tor Theory”

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(ii) ‖(e− x)−1− e− x‖¶ ‖x‖2

1−‖x‖

Theorem (Gel’fand-Mazur). Let A be a Banach algebra satisfying: For every x ∈ A \ 0: x isinvertible (i.e. G(A) = A). Then A∼= C

Note. If B ⊆ A is a subalgebra, and if x ∈ B, then x may have no inverse as element in B buthave an inverse as element in A. In particular, σ(x) depends on if “computed” relative to B orrelative to A – However, by (3) above r(x) does not depend on this (norm of x in A and B arethe same).

4.5.3 Functional Calculus

As for B(X ), if x ∈ A, then p(x) ∈ A makes sense for p : C−→ C polynomial, p(t) =∑n

i=0αi ti ,

αi ∈ C. We set p(x) ∈ A equal to p(x) = α0e + α1 x + · · ·αn xn. Also if f (t) =∑∞

i=0αi ti is

entire (t ∈ C) then f (x) :=∑∞

i=0αi xi (x0 = e) makes sense ( f (x) ∈ A). If f (t9 = 1

α−tthen

f (x9= (αe− x)−1 makes sense in A iff α /∈ σ(x). So if f : Ω−→ C holomorpic on an open setΩ containing σ(x) (Ω ⊇ σ(x)) and we put a curve Γ around σ(x) (in Ω) then (by Cauchy’sFormula)

f (t) =1

2πi

∮

Γ(z− t)−1 f (z)dz ∀t ∈ σ(x)

Defining integrals of Banach space values functions as earlier described (!) one gets

Lemma. We have

1

2πi

∮

Γ(α− z)n(ze− x)−1 dz = (αe− x)n ∈ A

One then defines (as seen earlier)

Definition. Let A be a Banach with unit, let a ∈ A and let Ω ⊆ C be open with σ(a) ⊆ Ω. LetΓ be a contour surrounding σ(a) in Ω Let f : Ω−→ C be holomorpic. Then we define

f (a) :=1

2πi

∮

Γf (z)(ze− a)−1 dz

( f (a) is, in fact, “ f (a)”). This turns out to be independent of Γ and of Ω (i.e. if g : Ω −→ C,σ(a) ⊆ Ω, Γ contour surrounding σ(a) in Ω, and g ≡ f on Ω1 ∩Ω1 then f (a) = g(a) in A).Let

Hol(a) = f analytic in a neighborhood of σ(a)= f | ∃Ω⊆ C open, Ω⊇ σ(a), f : Ω−→ C holomorphic

Hol(a) is an algebra (However, Hol(a) is not a Banach algebra in general).

Theorem (The Riesz Funcional Calculus). Let A be a Banach algebra with a unit, a ∈ A (fixed).

(a) Φ: Hol(a)−→ A, f 7−→ f (a) is an algebra homomorphism.

(b) In particular, if f (z) = 1, then f (a) = e (the unit) f (z) = z, then f (a) = a.

(c) If f (z) =∑∞

n=0αnzn has radius of convergence > r(a) (spectral radius) then f ∈ Hol(A)and f (a) =

∑∞k=0αkak.

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(d) Let fn∞n=1 be holomorphic on G ⊆ C (G open, G ⊇ σ(a)) and assume fn(z) → f (z) oncompact subsets of G, then ‖ fn(a)− f (a)‖A→ 0 as n→∞.

The map Hol(a) −→ A, f 7−→ f (a) is unique: Any map satisfying (a), (b) and (d) equals Φ.More generally

Definition. Let, for Ω⊆ C open, H(Ω) be the algebra of all (complex) holomorphic (complexanalytic) functions in Ω (Ω now fixed). Let

AΩ = x ∈ A | σ(x)⊆ Ω ⊆ A

Then (!) AΩ ⊆ A is open (in the norm-topology on A). Let H(AΩ) be the set of all A-valuedfunctions f with domain AΩ ( f : AΩ −→ A, f ∈ H(AΩ)) which arises from f ∈ Hol(Ω) be theformula

f (x) =1

2πi

∮

Γf (z)(ze− x)−1 dz

(where Γ is a contour that surrounds σ(x) in Ω).

Theorem. Then H(AΩ) is a complex algebra and the map

H(Ω)−→ H(AΩ)

f 7−→ f

is an algebra homomorphism which is continuous in the following sense. If fnn ⊆ H(Ω), fn→ funiformly on compact subsets of Ω, then f (x) = limn→∞ fn(x) for every x ∈ AΩ. If u(z) = z andv(z) = 1 in Ω, then u(x) = x, v(x) = e for every x ∈ AΩ.

Theorem. Let x ∈ AΩ, f ∈ Hol(Ω)

(a) f (a) ∈ A is invertible iff f (z) 6= 0 for all z ∈ σ(x).

(b) σ( f (x)) = f (σ(x)) (as sets) (“Spectral Mapping Theorem”).

4.6 Outlook on the Functional Calculus and Spectral decompositions: Usefull-ness?!?

For self-adjoint operators on the Hilbert space T (bounded or unbounded), then the functionalcalculus allows to define f (T ) for certain functions f . Then spectral decomposition (T =∫

λdEλ) gives a “formula” – a representation of f (T ):

f (T ) =

∫

f (λ)dEλ (∗)

with E the spectral measure (or “resolution of the identity” – see earlier) of T . I.e. (∗)gives f (T ) as an integral, built out of simpler functions of T – namely spectral projectionsE (EA = χA(T ), A Borel). However, often (!) one does not know (sufficiently explicitly)EA = χA(T ) (if we not know the spectrum, how to find the spectral projections??). Of course,we know that T is unitarily equivalent fo a multiplication operator – and for multiplicationoperators, we know “everything” – however: often (!) we don’t know neither the unitary,measure (L2(µ), nor “ f ” (as in M f ). Hence (∗) and “unitarly equivalent to moultiplicationoperator” is sometimes usefull in practice – but often not! However, the Functional Calculuscan be sued to get (many!) other representations of f (T ): “Every formula involving f (t)gives a representation of f (T ) via the Spectral Theorem” (whether usefull or not is anotherquestion...!).

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Example 4.6 (to see how use in practice, what type of questions can be addressed, and how).Problem 38, Sheet 10. Let I ⊆ R be an interval, f : I −→ R, continuous is called operatormonotone (on I) iff A ¶ B (A, B ∈ B(H), σ(A),σ(B) ⊆ I) implies f (A) ¶ f (B) (Interest:If A, B have just eigenvalues, then A ¶ B implies λJ (A) ¶ λ j(B) by min-max-principle (UseProblem 27). Answer: operator monotone functions. In exercise: Prove that fα(t) =

t1+αt

,α ¾ 0 are operator monotone on R+ (i.e. 0 ¶ A¶ B). Then (!) prove gα(x) := xα, α ∈ [0, 1]is operator monotone on R+ (0 ¶ A ¶ B). How (?!): By using the Spectral Theorem and a“good/usefull” formula (not Tα =

∫

λα dEλ)

tα =sin(απ)π

∫ ∞

0

t

t +λdλ

λ1−α α ∈ (0,1)

I.e.

Tα =sin(απ)π

∫ ∞

0

T

T +λdλ

λ1−α α ∈ (0,1)

=sin(απ)π

∞∫

0

f1/λ

dλ

λ2−α

“nicefrac building blocks”.

“Technique”: For f (T ) (for f for which we want to know “something”) find a good/usefullformula (often, integral representation) that writes f as an integral (or otherwiese) of nicefunctinos of T (“nice” in the sense that we can answer the question (maybe partially) for therebuilding blocks).

Example 4.7. Cauchy:

f (T ) =1

2πi

∮

γ

(z− T )−1 f (z)dz

Idea: Often (!) can say (enough...) on resolvents (for all z ∈ γ).

Example 4.8. We have 1x=∫∞

0e−x tdt (∗) i.e.

(T − z)−1 =1

T − z=

∞∫

0

e−(T−z)t dt =

∞∫

0

e−tT ezt dt

(T − z ¾ 0). I.e. if have “information” on t 7−→ e−tT (T ¾ 0) then we can get info on theresolvent.

Note. Just (!) like U(t) = e−itT , ϕ(t) = U(t)ϕ0, ϕ0 ∈ H, solves idϕdt= Tϕ, ϕ|t=0 = ϕ0,

i.e. i∂tψ = Hψ, time-dependend Schrödinger equation, V (t) = e−tT , u(t) = V (t)u0 solves∂tu+Tu= 0, u|t=0 = u0 the heat equation for T (V (t) is a semi-group in t) (i.e. T =−∆¾ 0).Note that (∗) is a special case of the Laplace-Transform (L f )(x) =

∫∞0

e−x t f (t)dt).

Example 4.9 (Fouriertransform (a little more later maybe)). Let f : R −→ R (ex f ∈ L1(R),or f ∈ S (R)

f (ω) :=1p

2π

∞∫

−∞

e−itω f (t)dt

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Then (!)

f (t) =1p

2π

∞∫

−∞

eitω f (ω)dω

Hence (!)

f (T ) =1p

2π

∞∫

−∞

eiTω(ω)dω=1p

2πe−itT f (−t)dt

=1p

2π

∞∫

−∞

U(t) f (−t)dt

Idea: Since U(t) = e−itT , V (t) = e−tT solve a (two) PDE (“time dependent Schrödingerequation”/“heat equation”) can, via this, sometimes get (partial) info on U(t), V (t), that canbe used above – Info that maybe cannot get for EA = χA(T ).

Example 4.10 (Mellin transform).

T s =i

2π

∫

Γ

zs(T − z)−1dz

Res < 0 and Γ some (!) curve in C (not closed!) (“complex power of operators”).

f (T ) =1

2πi

σ+i∞∫

σ−i∞

F(s)T−s ds

where

F(s) =

∞∫

0

ts−t f (t)dt

is the Mellin-transform of f . Then Study (T − z)−1 ; T−s −→ f (T ).

Example 4.11. Other usefull (!!) integrals:

px =−

1

π

∞∫

0

1

x + t−

1

t

ptdt =⇒

pT =−

1

π

∞∫

0

(T + t)−1−1

t

ptdt T ¾ 0

1

xa = cs

∞∫

0

∞∫

0

1

x + t

dt

ts

For x > 0 and s ∈ (0,1) so

1p

T= (p

T )−1 =1

π

∞∫

0

(T + t)−1 dtp

t=

1

π

∞∫

0

R−t(T )dtp

t

Typical (!) examples (!!) of (unbounded) operators are

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4.7 (Partial) Differential Operators

(from PDE, and Quantum Mechanics (QM)).

(P(x ,∂ )u)(x) =∑

|α|¶m

aα(x)∂αx u(x)

u: Ω−→ R (or C), Ω⊆ Rn open and a multi-index α. Either

(1) Ω& Rn (typically, Ω bounded) + boundary conditions (or, initial value)

P(x ,∂ )u= 0 in Ω

u= f on ∂Ω

unbounded operators in L2(µ) (Questions of self-adjointness, spectrum etc (Ex: Pb 4, 16,39, 44, 45) (multiplication operator! m = 0). Problem 51: P(x ,∂ ) = −i d

dx(n = 1) (+

boundary conditions) or Pb 54: P(x ,∂ ) =− d2

dx2 (+ boundard conditions)

Or

(2) Ω = Rn, i.e. P(x ,∂ ) on/in L2(Rn) (ex. Pb 46 (multiplication operator m = 0!). Pb 48Schrödinger operator (QM!) P(x ,∂ ) =−∆+ V (x).

−∆ + V (x) is a Huge topic in QM! Asider: Fouriertransform: S (Rn) ⊆ L2(Rn) Schwartz-functions.

S (Rn) = f : Rn −→ C | f ∈ C∞, xβ∂ αx f ∈ L∞(Rn)∀α,β

(here, xα := xα11 · · · x

αnn , x ∈ Rn, α ∈ Nn

0) For f ∈ S (Rn)

f (ξ) = (F f )(ξ) =1

(2π)n/2

∫

Rn

e−i x ·ξ f (x)dx (∗)

Then (!) F : S (Rn)−→S (Rn), and

⟨F f ,F g⟩L2(Rn) = ⟨ f , g⟩L2(Rn) (∗∗)

So F is L2-bounded on S (Rn) (dense in L2, C∞0 (Rn) ⊆ S (Rn) ⊆ L2(Rn)). Hence there

exists a unique F : L2(Rn)−→ L2(Rn), f 7−→F f bounded & linear.

NB. No longer given by (∗)!

Note (∗∗) shows that F is unitary on L2(Rn): FF ∗ =F ∗F = 1L2(Rn). In fact

(F ∗ f )(x) =1

(2π)n/2

∫

Rn

eix p f (p)dp ∀ f ∈ S (Rn)

I.e.

f (x) =1

(2π)n/2

∫

Rn

eix p f (p)dp

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Note.

(∇x f )(x)!=

1

(2π)n/2

∫

Rn

(ip)eix p f (p)dp

I.e. ∇x =F ∗(ip)F (ip ≡ Mip, so∇x is unitarily eqivalent fo a multiplication operator (namelyMip on some L2-space (namely, L2(Rn)). So −i∇x = F ∗(p)F = F ∗(Mp)F (hence we knowthe spectrum! know spectral decomposition (spectral projections)). Also −∆ = F ∗(p2)F(p2 = pp = |p|2). I.e. σ(−∆) = [0,∞)⊆ R and D(−∆) = f | p2 f ∈ L2= H2(Rn)

More generally P(∂ ) =∑

|α|¶m aα∂αx (constant coefficients: aα ∈ C, indep of x) then P(∂ ) =

F ∗P(ip)F , P(ip) ≡ MP(ip). Here P(ip) =∑

|α|¶m aα(ip)α (polynomial in p of order m. So“Fouriertransform diagonalises PDO’s with constant coefficients” (on L2(Rn)). Of course (!compute): [−∆, V (x)) = (−∆)V − V (−∆) 6= 0 So cannot simultaneously “diagonalise” −∆and V (x). So Question:

(0) Self-adjointness of −∆+ V ?? (& domain questions; V can be also unbounded (Pb 46),also Pb 48).

(1) Spectral questions (when self-adjoint) – For example, “perturbation theory”. “If V is nottoo “big”/“bad” then properties of −∆ are conserved when adding V (ex. D(−∆+ V ) =D(−∆) (domains of operators). Not true for spectrum in general, but: One splits thespectrum in various parts/types (like: σp,σc ,σr (if s.a.)). There are other! ways of“splitting the spectum” – some of these (typically, σc) are stable under perturbation.Since σ(−∆) = [0,∞) (“continuous spectrum”) (no eigenvalues!) σ(−∆− 1

|x |) containseigenvalues and continuous spectrum (Hydrogen atom).

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References

[FA1] Peter Müller, funcional analysis, lecture course (2013). The script by can be found onhttps://www.dropbox.com/s/dnceea2tob4ubz6/functional%20analysis.pdf?dl=0.

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