# 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.

## Classical (Banach space) form

Open mapping theorem for Banach spaces (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).

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.

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 cY 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 for all yY and all 𝜀 > 0, there exists some xX such that

${\displaystyle \qquad \|x\|_{X}\leq L\|y\|_{Y}\quad {\text{and}}\quad \|y-Ax\|_{Y}<\epsilon .\qquad (1)}$

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:

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

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

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

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 the origin in Y, and this concludes the proof.

Theorem [1]   Let X and Y be Banach spaces, let BX and BY denote their open unit balls, and let T : XY be a bounded linear operator. If δ > 0 then among the following four statements we have ${\displaystyle (1)\implies (2)\implies (3)\implies (4)}$ (with the same δ)

1. ${\displaystyle \left\|T^{*}y^{*}\right\|\geq \delta \left\|y^{*}\right\|}$ for all ${\displaystyle y^{*}\in Y^{*}}$;
2. ${\displaystyle {\overline {T\left(B_{X}\right)}}\supseteq \delta B_{Y}}$;
3. ${\displaystyle {T\left(B_{X}\right)}\supseteq \delta B_{Y}}$;
4. Im T = Y (i.e. T is surjective).

Furthermore, if T is surjective then (1) holds for some δ > 0

### Consequences

The open mapping theorem has several important consequences:

• If A : XY is a bijective continuous linear operator between the Banach spaces X and Y, then the inverse operator A−1 : YX is continuous as well (this is called the bounded inverse theorem). [2]
• If A : XY is a linear operator between the Banach spaces X and Y, and if for every sequence (xn) in X with xn → 0 and Axny it follows that y = 0, then A is continuous (the closed graph theorem). [3]

## 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:

Theorem ((Rudin 1991, 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. [4]

Open mapping theorem ( [5] )  If A : XY is a surjective closed linear operator from an complete pseudometrizable TVS X into a topological vector space Y and if at least one of the following conditions is satisfied:

1. Y is a Baire space, or
2. X is locally convex and Y is a barrelled space,

either A(X) is a meager set in Y, or A(X) = Y. then A is an open mapping.

Open mapping theorem for continuous maps ( [5] )  Let A : XY be a continuous linear operator from an complete pseudometrizable TVS X into a Hausdorff topological vector space Y. If Im A is nonmeager in Y then A : XY is a surjective open map and Y is a complete pseudometrizable TVS.

The open mapping theorem can also be stated as

Theorem [6]   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.

### Consequences

Theorem [7]   If A : XY is a continuous linear bijection from a complete Pseudometrizable topological vector space (TVS) onto a Hausdorff TVS that is a Baire space, then A : XY is a homeomorphism (and thus an isomorphism of TVSs).

### Webbed spaces

Webbed spaces are a class of topological vector spaces for which the open mapping theorem and the closed graph theorem hold.

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## References

1. Rudin 1991, p. 100.
2. Rudin 1973, Corollary 2.12.
3. Rudin 1973, Theorem 2.15.
4. Dieudonné 1970, 12.16.8.
5. Narici & Beckenstein 2011, p. 468.
6. Trèves 2006, p. 170
7. Narici & Beckenstein 2011, p. 469.