# Open mapping theorem (functional analysis)

(Redirected from Banach–Schauder theorem)

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):

Open Mapping Theorem. 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

Suppose A : XY is a surjective continuous linear operator. In order to prove that A is an open map, it is sufficient to show that A maps the open unit ball in X to a neighborhood of the origin of Y.

Let $U=B_1^X(0), V=B_1^Y(0)$. Then $X=\bigcup_{k\in\mathbb{N}}kU$.

Since A is surjective:

$Y=A(X)=A\left(\bigcup_{k \in \mathbb{N}} kU\right) = \bigcup_{k \in \mathbb{N}} A(kU).$

But Y is Banach so by Baire's category theorem

$\exists k \in \mathbb{N}: \qquad \left (\overline{A(kU)} \right )^\circ \neq \varnothing.$

Let

$c \in \left (\overline{A(kU)} \right )^\circ.$

Since:

$\left (\overline{A(kU)} \right )^{\circ\circ}= \left (\overline{A(kU)} \right )^\circ$

we have:

$\exists r>0: \qquad B_r(c) \subseteq \left (\overline{A(kU)} \right )^\circ.$

Let vV, then

$c, c+rv \in B_r(c)\subseteq \left (\overline{A(kU)} \right )^\circ \subseteq \overline{A(kU)}.$

By continuity of addition, the difference

$rv\in\overline{A(kU)-A(kU)}\subseteq \overline{A(2kU)}.$

And A is linear so

$V\subseteq \overline{A\left (\tfrac{2k}{r}U \right )}.$

It follows that

$\forall y \in Y, \forall \varepsilon > 0, \exists x \in X: \qquad \|x\|_X< \tfrac{2k}{r} \|y\|_Y \quad \text{and} \quad \|y - Ax\|_X< \varepsilon. \qquad (1)$

Fix yδV, by (1), there is some x1 with ||x1|| < 1 and ||yAx1|| < δ/2. Define a sequence {xn} inductively as follows. Assume:

$\|x_n\|< 2^{-(n-1)} \quad \text{and} \quad \left\|y - A(x_1+x_2+ \cdots +x_n) \right \| < \delta 2^{-n} \qquad (2)$

by (1) we can pick xn+1 so that:

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

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,

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

This shows that every yδV belongs to A(2U), or equivalently, that the image A(U) of the unit ball in X contains the open ball δ/2VY. 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 linear map u : XY 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