# Open mapping theorem (functional analysis)

Jump to navigation Jump to search

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 ${\displaystyle U=B_{1}^{X}(0),V=B_{1}^{Y}(0)}$. Then

${\displaystyle X=\bigcup _{k\in \mathbb {N} }kU}$.

Since A is surjective:

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

But Y is Banach so by Baire's category theorem

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

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

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

Let vV, then

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

By continuity of addition and linearity, the difference rv satisfies

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

and by linearity again,

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

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

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

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:

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

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

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

so (2) is satisfied for xn+1. Let

${\displaystyle s_{n}=x_{1}+x_{2}+\cdots +x_{n}.}$

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,

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

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

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

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

## References

1. ^ Trèves (1967), 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.