# Open mapping theorem (functional analysis)

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).

One proof uses Baire's 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

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

That is, we have c in Y and r > 0 such that

$B_{r}(c)\subseteq \left({\overline {A(kU)}}\right)^{\circ }\subseteq {\overline {A(kU)}}.$

Let vV, then

$c,c+rv\in B_{r}(c)\subseteq {\overline {A(kU)}}.$

By continuity of addition and linearity, the difference rv satisfies

$rv\in {\overline {A(kU)}}+{\overline {A(kU)}}\subseteq {\overline {A(kU)+A(kU)}}\subseteq {\overline {A(2kU)}},$

and by linearity again,

$V\subseteq {\overline {A\left(LU\right)}}.$

where we have set L=2k/r. It follows that

$\forall y\in Y,\forall \varepsilon >0,\exists x\in X:\qquad \|x\|_{X}\leq L\|y\|_{Y}\quad {\text{and}}\quad \|y-Ax\|_{Y}<\varepsilon .\qquad (1)$

Our next goal is to show that VA(2LU).

Let yV. By (1), there is some x1 with ||x1|| < L and ||yAx1|| < 1/2. Define a sequence {xn} inductively as follows. Assume:

$\|x_{n}\|<{\frac {L}{2^{n-1}}}\quad {\text{and}}\quad \left\|y-A(x_{1}+x_{2}+\cdots +x_{n})\right\|<{\frac {1}{2^{n}}}.\qquad (2)$

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

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

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,

$\|x\|=\lim _{n\to \infty }\|s_{n}\|\leq \sum _{n=1}^{\infty }\|x_{n}\|<2L.$

This shows that y belongs to A(2LU), so VA(2LU) as claimed. Thus the image A(U) of the unit ball in X contains the open ball V/2L of 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

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