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A FUNCTIONAL ANALYSIS POINT OF VIEW ON

ARZELA-ASCOLI THEOREM

GABRIEL NAGY

Abstract. We discuss the Arzela-Ascoli pre-compactness Theorem from thepoint of view of Functional Analysis, using compactness in and its dual.

The Arzela-Ascoli Theorem is a very important technical result, used in manybranches of mathematics. Aside from its numerous applications to Partial Differ-ential Equations, the Arzela-Ascoli Theorem is also used as a tool in obtaining

Functional Analysis results, such as the compactness for duals of compact oper-ators, as presented for example [1]. The purpose of this note is to offer a newperspective on the Arzela-Ascoli Theorem based on a functional analytic proof.

The theorem of Arzela and Ascoli deals with (relative) compactness in the Ba-nach space C(K) of complex valued continuous functions on a compact Hausdorffspace K. One helpful characterization of pre-compactness for sets in a completemetric space is the following well-known criterion, which we state without proof.

Proposition 1. Let(Y, d) be a complete metric space. For a subsetM Y, thefollowing are equivalent:

(i) M is relatively compact inY, i.e. its closureM inYis compact;(ii) M contains no infinite subsetsT, satisfying

inf

d(x, y) : x, y T, x=y

> 0.

Proposition 2. LetX be a normed vector space, letS X be a compact subset,and letBbe the unit ball inX the (topological) dual ofX equipped with thew-topology. If we consider the Banach algebraA= C(S), equipped with the uniformtopology, then the restriction map : (B, w)

S

(A, . )is continuous.In particular, the set(B) is compact inA.

Proof. To prove contnuity, we start with a net () in B, that converges tosome Bin the w-topology, and let us show that the net

S

converges

to S

uniformly on S. Fix some >0, and (use compactness ofS) choose pointss1, . . . , sn S, such that

() for everys S, there existsk {1, . . . , n}, withs sk< /3.

Using the conditionw

, there exists , such that

(1) |(sk) (sk)|< /3, , k {1, . . . , n}.

Now we are done, since if we start with some arbitrary s S, and we choosek {1, . . . , n}, such that s sk< /3, then using (1) we get

|(s) (s)| |(s) (sk)| + |(sk) (sk)| + |(sk) (s)|

s sk + |(sk) (sk)| + s sk

2s sk + |(sk) (sk)| , .

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2 GABRIEL NAGY

Having proven the continuity of , the second assertion follows from AlaoglusTheorem (which states that Bis compact in the w-topology).

Theorem (Arzela-Ascoli). LetKbe a compact Hausdorff space, and letM C(K)be a set which is

pointwise bounded: sup

|f(p)| : f M

< , p K; equicontinuous: for everyp K and >0, there exists a neighborhood

Np, ofp inK, such that: supfM |f(q) f(p)| , q Np,.

ThenM is relatively compact inC(K) in the uniform topology.

Proof. The main observation is that, using pointwise boundedness, any set T Mgives rise to a map

T :K p

f(p)fT

(T).

(Here (T) denotes the Banach space of all bounded functions from T to C.),Furthermore, by equicontinuity the map T is in fact continuous, when

(T) is

equipped with the norm topology. Since Kis compact, so is the setST = T(K)

(T).

We are going to argue by contradiction (see Proposition 1), assuming the existenceof some >0, and of an infinite set T M, such that

(2) f g> , f, g T, f=g.

Consider now the unit ball B in the dual space (T), and use Proposition 2 toconclude that the set

T =

ST

: B

C(ST)

is compact in C(ST) in the norm topology. Consider now the coordinate mapsef :

(T) C, f T, and their restrictionsf=efST

C(ST), which satisfy

(3) f

T(p)

= f(p), p K, f T.Now, if we start withf , g T, f=g, then using (2) there exists p K, such that|f(p) g(p)|> , so by (3) we also get

f

T(p)

g

T(p)

> .

This way we have shown that

f gC(ST) > , f, g T, f=g,

and therefore the set T which contains all the fs cannot be compact in thenorm topology.

References

[1] W. Rudin,Functional Analysis, McGraw-Hill, Springer-Verlag, New York, 1973

Department of Mathematics, Kansas State university, Manhattan KS 66506, U.S.A.

E-mail address: nagy@math.ksu.edu