3. The Banach-Steinhaus theorem

Transcription

1 Institut for Matematiske Fag Det Naturvidenskabelige Fakultet Aarhus Universitet September 2005

2 We read in W. Rudin: Functional Analysis

3 We read in W. Rudin: Functional Analysis Covering Chapter 2, from 2.3 up to Theorem 2.7.

4 Equicontinuity Let X and Y be topological vector spaces. A collection Γ of linear maps from X to Y is equi-continuous when the following holds: For every open neighborhood W of 0 in Y there is an open neighborhood V of 0 in Y such that L(V) W for all L Γ.

5 Equicontinuity Let X and Y be topological vector spaces. A collection Γ of linear maps from X to Y is equi-continuous when the following holds: For every open neighborhood W of 0 in Y there is an open neighborhood V of 0 in Y such that L(V) W for all L Γ. This should be compared with the (familiar?) definition of equi-continuity for families F of functions f : R R: F is equi-continuous when: t R, ǫ > 0 δ > 0 : t s < δ f (t) f (s) < ǫ f F.

6 Equicontinuity Theorem Let X and Y be topological vector spaces, Γ an equi-continuous family of linear maps from X to Y and E a bounded subset of X. Then Y has a bounded subset F such that L(E) F for all L Γ.

7 Equicontinuity Theorem Let X and Y be topological vector spaces, Γ an equi-continuous family of linear maps from X to Y and E a bounded subset of X. Then Y has a bounded subset F such that L(E) F for all L Γ. Let F = L Γ L(E). We must show that F is bouned. Let W be an open neighborhood of 0 in Y.

8 Equicontinuity Theorem Let X and Y be topological vector spaces, Γ an equi-continuous family of linear maps from X to Y and E a bounded subset of X. Then Y has a bounded subset F such that L(E) F for all L Γ. Let F = L Γ L(E). We must show that F is bouned. Let W be an open neighborhood of 0 in Y. Since Γ is equi-continuous there is a neighborhood V of 0 in X such that L(V) W for all L Γ.

9 Equicontinuity Theorem Let X and Y be topological vector spaces, Γ an equi-continuous family of linear maps from X to Y and E a bounded subset of X. Then Y has a bounded subset F such that L(E) F for all L Γ. Let F = L Γ L(E). We must show that F is bouned. Let W be an open neighborhood of 0 in Y. Since Γ is equi-continuous there is a neighborhood V of 0 in X such that L(V) W for all L Γ. Since E is bounded, E tv for all sufficiently large t. Then F = L Γ L(E) L Γ L(tV) t ( L Γ L(V)) tw for all sufficiently large t.

10 Banach-Steinhaus, first version Theorem Let X and Y be topological vector spaces, Γ a collection of continuous linear maps from X to Y and B the set of elements in X with the property that is bounded in Y. Γ(x) = {L(x) : L Γ}

11 Banach-Steinhaus, first version Theorem Let X and Y be topological vector spaces, Γ a collection of continuous linear maps from X to Y and B the set of elements in X with the property that is bounded in Y. Γ(x) = {L(x) : L Γ} If B is of the second category in X, then B = X and Γ is equi-continuous.

12 Banach-Steinhaus, first version Theorem Let X and Y be topological vector spaces, Γ a collection of continuous linear maps from X to Y and B the set of elements in X with the property that is bounded in Y. Γ(x) = {L(x) : L Γ} If B is of the second category in X, then B = X and Γ is equi-continuous. Let W be a neighborhood of 0 in Y and let U be an open symmetric neighborhood of 0 in Y such that U + U W.

13 Banach-Steinhaus, first version Set E = L Γ L 1 ( U ). If x B, then Γ(x) nu for some n N, so that x ne. Hence B ne. n=1

14 Banach-Steinhaus, first version Set E = L Γ L 1 ( U ). If x B, then Γ(x) nu for some n N, so that x ne. Hence B ne. n=1 Since each ne is closed and B of second category by assumption, there is an n such that ne has non-empty interior. It follows that E has an interior point x.

15 Banach-Steinhaus, first version Set E = L Γ L 1 ( U ). If x B, then Γ(x) nu for some n N, so that x ne. Hence B ne. n=1 Since each ne is closed and B of second category by assumption, there is an n such that ne has non-empty interior. It follows that E has an interior point x. Then x E contains a non-empty open neighborhood V of 0 in X, and L(V) L(x) L(E) U U = U + U W for all L Γ. It follows that Γ is equi-continuous.

16 Banach-Steinhaus, first version Set E = L Γ L 1 ( U ). If x B, then Γ(x) nu for some n N, so that x ne. Hence B ne. n=1 Since each ne is closed and B of second category by assumption, there is an n such that ne has non-empty interior. It follows that E has an interior point x. Then x E contains a non-empty open neighborhood V of 0 in X, and L(V) L(x) L(E) U U = U + U W for all L Γ. It follows that Γ is equi-continuous. That B = X follows from Theorem 1.

17 Banach-Steinhaus, Banach space version Theorem Let X and Y be Banach spaces (real or complex) and Γ a collection of continuous linear maps from X to Y. Assume that sup { L(x) : L Γ} < for all x X. It follows that sup L Γ L <.

18 Banach-Steinhaus, Banach space version Theorem Let X and Y be Banach spaces (real or complex) and Γ a collection of continuous linear maps from X to Y. Assume that sup { L(x) : L Γ} < for all x X. It follows that sup L Γ L <. We claim that Γ(x) = {L(x) : L Γ} is bounded in Y for all x. To see this, let W be an open neighborhood of 0 in Y.

19 Banach-Steinhaus, Banach space version Theorem Let X and Y be Banach spaces (real or complex) and Γ a collection of continuous linear maps from X to Y. Assume that sup { L(x) : L Γ} < for all x X. It follows that sup L Γ L <. We claim that Γ(x) = {L(x) : L Γ} is bounded in Y for all x. To see this, let W be an open neighborhood of 0 in Y. There is then a δ > 0 such that {y Y : y < δ} W. Let x X. When t > δ 1 sup{ L(x) : L Γ} we have that L ( t 1 x ) = t 1 L(x) < δ.

20 Banach-Steinhaus, Banach space version Theorem Let X and Y be Banach spaces (real or complex) and Γ a collection of continuous linear maps from X to Y. Assume that sup { L(x) : L Γ} < for all x X. It follows that sup L Γ L <. We claim that Γ(x) = {L(x) : L Γ} is bounded in Y for all x. To see this, let W be an open neighborhood of 0 in Y. There is then a δ > 0 such that {y Y : y < δ} W. Let x X. When t > δ 1 sup{ L(x) : L Γ} we have that ( L t 1 x ) = t 1 L(x) < δ. Hence Γ(x) = L Γ {L(x)} = L Γ t { L ( t 1 x )} tw, proving the claim.

21 Banach-Steinhaus, Banach space version By the Baire category theorem X is of the second category, so we conclude from the Banach-Steinhaus theorem that Γ is equi-continuous. There is therefore an open neighborhood V of zero in X such that for all L Γ. L(V) {y Y : y 1}

22 Banach-Steinhaus, Banach space version By the Baire category theorem X is of the second category, so we conclude from the Banach-Steinhaus theorem that Γ is equi-continuous. There is therefore an open neighborhood V of zero in X such that L(V) {y Y : y 1} for all L Γ. Choose ǫ > 0 such that {x X : x ǫ} V. Then L(x) ǫ 1 for all L Γ and all x X, x 1.

23 Banach-Steinhaus, Banach space version By the Baire category theorem X is of the second category, so we conclude from the Banach-Steinhaus theorem that Γ is equi-continuous. There is therefore an open neighborhood V of zero in X such that L(V) {y Y : y 1} for all L Γ. Choose ǫ > 0 such that {x X : x ǫ} V. Then L(x) ǫ 1 for all L Γ and all x X, x 1. Hence L ǫ 1 for all L Γ.