Open mapping theorem (functional analysis): Difference between revisions
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==Proof== |
==Proof== |
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One has to prove that if <math>\underset{=}{A} \in \{\underset{=}{A} \in B(X,Y) : im(\underset{=}{A})=Y\}</math> <math>\underset{=}{A |
One has to prove that if <math>\underset{=}{A} \in \{\underset{=}{A} \in B(X,Y) : im(\underset{=}{A})=Y\}</math>, <math>\underset{=}{A}</math> is an open map. It suffices to show that ''A'' maps the open [[unit ball]] in ''X'' to a neighborhood of the origin of ''Y''. |
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Let <math>U=B_1^X(\underline{0})</math>, <math>V=B_1^Y(\underline{0})</math>. Then <math>X=\bigcup_{k\in\mathbb{N}}kU</math>. |
Let <math>U=B_1^X(\underline{0})</math>, <math>V=B_1^Y(\underline{0})</math>. Then <math>X=\bigcup_{k\in\mathbb{N}}kU</math>. |
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Well <math>im(\underset{=}{A})=Y</math> so |
Well <math>im(\underset{=}{A})=Y</math> so |
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:<math>Y=A(X)=A\Bigl(\bigcup_{k \in \mathbb{N}} kU\Bigr) = \bigcup_{k \in \mathbb{N}} A(kU).</math> |
:<math>Y=\underset{=}{A}(X)=\underset{=}{A}\Bigl(\bigcup_{k \in \mathbb{N}} kU\Bigr) = \bigcup_{k \in \mathbb{N}} \underset{=}{A}(kU).</math> |
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But <math>Y</math> is Banach so by [[Baire's category theorem]] <math>\exists k \in \mathbb{N}</math> <math>(\overline{\underset{=}{A}(kU)})^\circ \neq \varnothing</math>; let <math>\underline{c} \in (\overline{\underset{=}{A}(kU)})^\circ</math>. |
But <math>Y</math> is Banach so by [[Baire's category theorem]] <math>\exists k \in \mathbb{N}</math> <math>(\overline{\underset{=}{A}(kU)})^\circ \neq \varnothing</math>; let <math>\underline{c} \in (\overline{\underset{=}{A}(kU)})^\circ</math>. |
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Further <math>(\overline{\underset{=}{A}(kU)})^{\circ\circ}=(\overline{\underset{=}{A}(kU)})^\circ </math> so <math>\exists r>0</math> <math>B_r(\underline{c}) \subseteq (\overline{\underset{=}{A}(kU)})^\circ </math> |
Further <math>(\overline{\underset{=}{A}(kU)})^{\circ\circ}=(\overline{\underset{=}{A}(kU)})^\circ </math> so <math>\exists r>0</math> <math>B_r(\underline{c}) \subseteq (\overline{\underset{=}{A}(kU)})^\circ </math>. |
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If ''v'' ∈ ''V'', then {{nowrap|''c'' + ''r'' ''v''}} and ''c'' are in ''B''(''c'', ''r''), hence are [[limit points]] of {{nowrap|''A''(''k'' ''U'')}}. By continuity of addition, their difference ''rv'' is a limit point of {{nowrap|''A''(''k'' ''U'') − ''A''(''k'' ''U'') ⊂}} {{nowrap|''A''(2''k'' ''U'')}}. By linearity of ''A'', this implies that any ''v'' ∈ ''V'' is in the closure of {{nowrap|''A''(''δ'' <sup>−1</sup> ''U'')}}, where ''δ'' = {{nowrap|''r'' / (2''k'')}}. It follows that for any ''y'' ∈ ''Y'' and {{nowrap|any ''ε'' > 0}}, there is an ''x'' ∈ ''X'' with: |
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:<math>\ ||x||< \delta^{-1} ||y|| </math> {{pad|2.5em}}and{{pad|2.5em}} <math> ||y - Ax||< \varepsilon. \quad (1) </math> |
:<math>\ ||x||< \delta^{-1} ||y|| </math> {{pad|2.5em}}and{{pad|2.5em}} <math> ||y - Ax||< \varepsilon. \quad (1) </math> |
Revision as of 02:04, 2 June 2013
In functional analysis, the open mapping theorem, also known as the Banach–Schauder theorem (named after Stefan Banach and Juliusz Schauder), is a fundamental result which states that if a continuous linear operator between Banach spaces is surjective then it is an open map. More precisely, (Rudin 1973, Theorem 2.11):
- If X and Y are Banach spaces and A : X → Y is a surjective continuous linear operator, then A is an open map (i.e. if U is an open set in X, then A(U) is open in Y).
The proof uses the Baire category theorem, and completeness of both X and Y is essential to the theorem. The statement of the theorem is no longer true if either space is just assumed to be a normed space, but is true if X and Y are taken to be Fréchet spaces.
Consequences
The open mapping theorem has several important consequences:
- If A : X → Y is a bijective continuous linear operator between the Banach spaces X and Y, then the inverse operator A−1 : Y → X is continuous as well (this is called the bounded inverse theorem). (Rudin 1973, Corollary 2.12)
- If A : X → Y is a linear operator between the Banach spaces X and Y, and if for every sequence (xn) in X with xn → 0 and Axn → y it follows that y = 0, then A is continuous (Closed graph theorem). (Rudin 1973, Theorem 2.15)
Proof
One has to prove that if , is an open map. It suffices to show that A maps the open unit ball in X to a neighborhood of the origin of Y.
Let , . Then .
Well so
But is Banach so by Baire's category theorem ; let .
Further so .
If v ∈ V, then c + r v and c are in B(c, r), hence are limit points of A(k U). By continuity of addition, their difference rv is a limit point of A(k U) − A(k U) ⊂ A(2k U). By linearity of A, this implies that any v ∈ V is in the closure of A(δ −1 U), where δ = r / (2k). It follows that for any y ∈ Y and any ε > 0, there is an x ∈ X with:
- and
Fix y ∈ δ V (where δ V means the ball V stretched by a factor of δ, rather than the boundary of V). By (1), there is some x 1 with ||x 1|| < 1 and ||y − A x 1|| < δ / 2. Define a sequence {xn} inductively as follows. Assume:
- and
by (1) we can pick x n +1 so that:
- and
so (2) is satisfied for x n +1. Let
From the first inequality in (2), {sn} is a Cauchy sequence, and since X is complete, sn converges to some x ∈ X. By (2), the sequence A sn tends to y, and so A x = y by continuity of A. Also,
This shows that every y ∈ δ V belongs to A(2 U), or equivalently, that the image A(U) of the unit ball in X contains the open ball (δ / 2) V in Y. Hence, A(U) is a neighborhood of 0 in Y, and this concludes the proof.
Generalizations
Local convexity of X or Y is not essential to the proof, but completeness is: the theorem remains true in the case when X and Y are F-spaces. Furthermore, the theorem can be combined with the Baire category theorem in the following manner (Rudin, Theorem 2.11) :
- Let X be a F-space and Y a topological vector space. If A : X → Y is a continuous linear operator, then either A(X) is a meager set in Y, or A(X) = Y. In the latter case, A is an open mapping and Y is also an F-space.
Furthermore, in this latter case if N is the kernel of A, then there is a canonical factorization of A in the form
where X / N is the quotient space (also an F-space) of X by the closed subspace N. The quotient mapping X → X / N is open, and the mapping α is an isomorphism of topological vector spaces (Dieudonné, 12.16.8) .
References
- Rudin, Walter (1973), Functional Analysis, McGraw-Hill, ISBN 0-07-054236-8
- Dieudonné, Jean (1970), Treatise on Analysis, Volume II, Academic Press
This article incorporates material from Proof of open mapping theorem on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.