# Open mapping theorem (functional analysis)

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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 : XY 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:

## Proof

One has to prove that if $\underset{=}{A} \in \{\underset{=}{A} \in B(X,Y) : im(\underset{=}{A})=Y\}$, $\underset{=}{A}$ 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 $U=B_1^X(\underline{0})$, $V=B_1^Y(\underline{0})$. Then $X=\bigcup_{k\in\mathbb{N}}kU$.

Well $im(\underset{=}{A})=Y$ so

$Y=\underset{=}{A}(X)=\underset{=}{A}\Bigl(\bigcup_{k \in \mathbb{N}} kU\Bigr) = \bigcup_{k \in \mathbb{N}} \underset{=}{A}(kU).$

But $Y$ is Banach so by Baire's category theorem $\exists k \in \mathbb{N}$ $(\overline{\underset{=}{A}(kU)})^\circ \neq \varnothing$; let $\underline{c} \in (\overline{\underset{=}{A}(kU)})^\circ$.

Further $(\overline{\underset{=}{A}(kU)})^{\circ\circ}=(\overline{\underset{=}{A}(kU)})^\circ$ so $\exists r>0$ $B_r(\underline{c}) \subseteq (\overline{\underset{=}{A}(kU)})^\circ$.

Let $\underline{v}\in V$; then $\underline{c},\underline{c}+r\underline{v}\in B_r(\underline{c}) \subseteq (\overline{\underset{=}{A}(kU)})^\circ \subseteq \overline{\underset{=}{A}(kU)}$. By continuity of $+$, their difference $r\underline{v}\in\overline{\underset{=}{A}(kU)-\underset{=}{A}(kU)}\subseteq\overline{\underset{=}{A}(2kU)}$. And $\underset{=}{A}$ is linear so $V\subseteq\overline{\underset{=}{A}(\frac{2kU}{r})}$. It follows that $\forall \underline{y}\in Y \forall\epsilon >0\exists \underline{x}\in X$:

$\ \|\underline{x}\|_X< \frac{2k\|\underline{y}\|_Y}{r}$  and  $\|\underline{y} - \underset{=}{A}\underline{x}\|_X< \varepsilon. \quad (1)$

Fix yδV (where δV means the ball V stretched by a factor of δ, rather than the boundary of V). By (1), there is some x1 with ||x1|| < 1 and ||yAx1|| < δ / 2. Define a sequence {xn} inductively as follows. Assume:

$\ ||x_{n}||< 2^{-(n-1)}$  and  $||y - A(x_1+x_2+ \cdots +x_n)|| < \delta \, 2^{-n} \, ; \quad (2)$

by (1) we can pick xn +1 so that:

$\ ||x_{n+1}||< 2^{-n}$  and  $||y - A(x_1+x_2+ \cdots +x_n) - A(x_{n+1})|| < \delta \, 2^{-(n+1)},$

so (2) is satisfied for xn +1. Let

$\ s_n=x_1+x_2+ \cdots + x_n.$

From the first inequality in (2), {sn} is a Cauchy sequence, and since X is complete, sn converges to some xX. By (2), the sequence Asn tends to y, and so Ax = y by continuity of A. Also,

$||x||=\lim_{n \rightarrow \infty} ||s_n|| \leq \sum_{n=1}^\infty ||x_n|| < 2.$

This shows that every yδV 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 : XY 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

$X\to X/N \overset{\alpha}{\to} Y$

where X / N is the quotient space (also an F-space) of X by the closed subspace N. The quotient mapping XX / N is open, and the mapping α is an isomorphism of topological vector spaces (Dieudonné, 12.16.8).

The open mapping theorem can also be stated as[1]

Let X and Y be two F-spaces. Then every continuous linear map of X onto Y is a TVS homomorphism.

where a a linear map $u : X \to Y$ is a topological vector space homomorphism if the induced map $\hat{u} : X/\ker(u) \to Y$ is a TVS-isomorphism onto its image.

## References

1. ^ Trèves (1995), p. 170

This article incorporates material from Proof of open mapping theorem on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.