Bi-Continuous

7/30/2019 Bi-Continuous

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Bi-Continuous Semigroups, Generators and

Resolvents

Saw Marlar Aung

Abstract

In this paper, some properties of bicontinuous semi group, generators and re-

solvents are studied and the relation between semi group operators and -topology

is discussed. Then the fact that generator of a bicontinuous semi group is bi-

closed and the domain is bi dense is proved.

Key words and phrases: multiplication operator, sequence spaces

1 Bi-Continuous Semigroups

Definition 1.1 An operator family

{T(t) : t 0} L(X)

is called bi-equicontinuous if for every -bounded sequence (xn)nN X which is-convergent to x X we have

limn

(T(t)(xn x)) = 0

uniformly for all t 0. It is called locally bi-equicontinuous if for every t0 0 the

subset {T(t) : 0 t t0} is bi-equicontinuous.

Definition 1.2. An operator family

{T(t) : t 0} L(X)

is called a bi-continuous semigroup (with respect to and of type ) if the following

conditions hold.

Dr, Lecturer, Department of Mathematics, University of Magway

2010 Mathematics Subject Classification: 47B33, 47B99


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1. T(0) = Id and T(t + s) = T(t)T(s) for all s, t 0.

2. The operators T(t) are exponentially bounded, that is,

T(t)L(X) M ewt

for all t 0 and some constants M 1 and w R .

3. (T(t))t0 is strongly -continuous, i.e.,the map

R+ t T(t)x X

is -continuous for each x X.

4. (T(t))t0 is locally bi-equicontinuous.

Proposition 1.3 Let (T(t))t0 be a bi-continuous semigroup of type w on X. Thenthe following properties hold.

1. For every a 0 and there exists the operator Ra() : X X defined as

Ra()x := a

0

etT(t)x dt (1.1)

for all x X. The integral has to be understood as a Riemann integral

(sometimes denoted by

a0

etT(t)x dt ).

2. The rescaled semigroup (etT(t)t0 is globally bi-equicontinuous for every > w.

Proof. Assertion (1) is immediately hold. Indeed, for fixed C and a 0,

x() := eT(t)()x

is a uniformly continuous X-valued function on the interval [0, a] for all x X.Therefore, the Riemann sums S(x(), ) defined as

S(x(), ) :=n

k=1

x(tk)(tk tk1),

: 0 = t0 t1 t1 t

n tn = a,


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from a Cauchy net. Taking

S(x(), n) :=a

n

x(a

k

n), n = 1, 2,

We obtain an equivalent -bounded -Cauchy sequence. Since (X, ) is sequentiallycomplete on -bounded sets, S(x(), n) converges, and hence

x() = eT()x

is Riemann integrable for allx X .

To prove property (2), let > w, > 0, p Pr, and (xn)nN X be a -bounded sequence which is -convergent to x X. Then there exists t0 0 suchthat

supt>t0

p(etT(t)(xn x)) supt>t0

etT(t)(xn x)

supt>t0

e(w)tM(xn + x)

2

for all n N.

Then there exists n0 N such that

sup0tt0

p(etT(t)(xn x))

2

for all n n0. Therefore,

supt0

p(etT(t)(xn x)) sup0tt0

p(etT(t)(xn x))

+supt>t0

p(etT(t)(xn x))

2

+ supt>t0

etT(t)(xn x)

for all n n0.


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2 Generators and Resolvents

Definition 2.1. Let X be a Banach space, C, and consider operators J() L(X) for each . The family {J() : } is called a pseudoresolvent if

J() J() = ( )J()J()

holds for all , .

Definition 2.2. The generator A : D(A) X X of a bi-continuous semigroup(T(t))t0 on X is the unique operator on X such that its resolvent R(, A) is

R(, A)x =

0

etT(t)xdt

for all 0 and x X.

Lemma 2.3. Let (T(t))t0 be a bi-continuous semigroup on X. Then the followingproperties hold.

1. Let C and a 0. Then Ra() L(X) and

R() := lima

Ra()

for w0 exists with respect to the operator norm and satisfies the estimate

R()L(X) M

Re w

for all w, w > w0, and some constant M 1.

2. For every x X we have

limw


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Therefore assertion (1) holds.

(2) Let x X, p Pr, and > 0. There exists > 0 such that 0 t impliesp(T(t)x x) < . Thus, we have

p(R()x x) = p

0

etT(t)xdt

0

etxdt

p

0

et(T(t)x x)dt

+ p

et(T(t)x x)dt

= T1 + T2.

For the terms T2 e obtain with as above that

T2

sup

sup

et |(T(t)x x)dt,|

x

M

e() + e

,

which converges to zero as tends to infinity. For the term T1 we obtain

T1

0

etp(T(t)x x)dt

0

etdt = ,

which concludes the proof.

Lemma 2.4. Let (T(t))t>0 be a bi-continuous semigroup on X and (Ar, D(Ar)) of abi-continuous semigroup is defined by

Arx = limt0

T(t)x x

t.

Then the property holds. If x D(Ar), then T(t)x D(Ar) for all t 0, T(t)x iscontinuously differentiable in t with respect to the topology , and

d

dtT(t)x = AT(t)x = T(t)Ax

for all t 0.

Proof. Ifx D(Ar), then for t 0 we have


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T(t)Arx = limt0

T(t + h)x T(t)xh

= limt0

(T(h) Id)T(t)xh

which shows that T(t)x D(Ar) and the right derivatived+

dtT(t)x exists. Thus we

haved+

dtT(t)x = ArT(t)x = T(t)Arx.

Let now (X, ). Then

d

dtT(t)x, =

d+

dtT(t)x,

= T(t)Arx, ,

which implies the continuity in t ofd+

dtT(t)x, . Therefore T(t), is differentiable

in t andd

dtT(t)x, = T(t)Arx, .

Since (T(t))t0 is bi-continuous, the integral

t0

T(s)Arxds exists in X, and we obtain

T(t)x x, =

t0

d

dsT(s)x, ds

=

t

0

T(s)Arx, ds

=

t0

T(s)Arxds,

Hence,

T(t)x x =

t0

T(s)Arxds

for all t 0, and T(t)x is differentiable in t with

d

dt T(t)x = T(t)Arx.

Theorem 2.5. Let (T(t))t0 be a bi-continuous semigroup on X with generator(A, D(A)) and define (Ar, D(Ar)) as Lemma (2.4). Then A = Ar.

Proof. We show first that A Ar. For x X and A0 we have

T(h) Id

hR(, A)x =

(eh 1)

h

0

etT(t)xdt eh

h

h0

etT(t)xdt,


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which converges to R(, A)x x = AR(, A)x as h 0. Thus A Ar.

On the other hand, for x D(Ar) we define y := (Ar)x.Then we have

Ar

0

etT(t)xdt =

0

etT(t)Arxdt.

Therefore, we obtain

R(, A)y = ( Ar)

0

etT(t)xdt = ( Ar)R(, A)x = x,

and hence Ar A.

Proposition 2.6. Let {J() : } be a pseudoresolvent on a Banach space X.Then J()J() = J()J(), kerJ() = kerJ() and rgJ() = rgJ() holdfor all , .

Moreover, the following assertions are equivalent.

1. There exists a closed operator (A, D(A)) such that (A) and J() =R(, A) for all .

2. kerJ() = {0} for some .

Definition 2.7. A subset M X is called bi-dense if for every x X there exists a -bounded sequence (xn)nN M which is -convergent to x.

Proposition 2.8. The family of operators (R())0 is a resolvent.

Proof. Let = 0 . We assume without loss of generality that Re > Re andobtain

R()x R()x

=

0

e()tdtR()x

0

e()t

( )etT(t)xdt

=

0

e()t0

esT(s)xdsdt

e()t

0

esT(s)xds

t=t=0

0

e()t0

esT(s)xdsdt

=

0

e()t0

esT(s)xdsdt

=

0

et0

esT(s)T(t)xdsdt

= R()R()x


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for all x X.

Therefore, (R())0 is a pseudoresolvent . Then we obtain the injectivity ofthe operators R(). In fact, for x kerR() and an unbounded sequence (n)nN 0 R+, we have x kerR(n) for all n N and lim

nnR(n)x = 0. Since

is Hausdorff, it follows x = 0.This completes the proof.

We observe that the map 0 R() L(X) is holomorphic and

dk

dkR()x = (1)kk!R()k+1x (2.1)

for all x X, k N, and 0. The above observations allow the definition of the

generator of a bi-continuous semigroup.

Proposition 2.9. Let (A, D(A)) be the generator of a bi-continuous semigroup (T(t))t0of type on X. Then we have

dk

dkR(, A)x = (1)k

0

tketT(t)xdt (2.2)

for all x X, k N and 0. In particular, there exists for each > 0 a constantM 1 such that

R(, A)k M(Re )k

for all k N and .

Proof. Let x X, , > 0, and = { (X, ) : X 1}. Since the

space (X, ) is norming for (X, ), for every we haveR()x R()x +0

tetT(t)xdt

sup

0

et et

+ tet

|T(t), |dt Mx

0et et

+ tet etdt,

which converges to zero as tends to as a consequence of Lebesgues dominated

convergence theorem. Via induction we obtain the desire equality. Also, we get

R()kx sup

1

(n 1)!

0

tk1|etT(t)x, |dt

M

(n 1)!x

0

tk1e(Re)tdt

=M

(Re )kx


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for all x X and .

Proposition 2.10. Let (A, D(A)) be the generator of a bi-continuous semigroup(T(t))t0 of type on X. Then the following properties hold.

(a) The generator (A, D(A)) is bi-closed.

(b) The domain ofA is bi-dense in X.

(c) Let D D(A) be a bi-dense subset in X. Then R(, A)D, , is bi-densein D(A).

Proof. Assertion (a) follows directly from Proposition (2.4).

(b) Let x X. By Lemma 1.7 the sequence (xn)nN D(A) defined as

xn =

nR(n, A)x if n > ,

0 else,

is -bounded and -convergent to x.To prove (c), let x D(A), , > 0, and p P. There exists z X suchthat R(, A)z = x. Since D is bi-dense in X, there exists a -bounded sequence(yn)nN D and n0 N such that

p(yn z)

for all n n0. Further, the sequence (R(, A)yn)nN is -bounded, and, by thebi-continuity of(T(t))t0, we obtain that there exists n0 n0 such that

p(R(, A)yn x) = p(R(, A)(yn z))

0

e()tp(etT(t)(yn z))dt

for all n n0.

References

[1] Engel, K.L., Nagel and R., A Short Course on Operator Semi Groups, Springer

Science and Business Media, 2006.

[2] Nickel, G., A Semi Group Approach to Dynamic Boundary Value Problems, Semi

Group Forum 69, pp. 159189, 2004.

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