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):

• If X and Y are Banach spaces and A : X → Y 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:

• If A : X → Y is a bijective continuous linear operator between the Banach spaces X and Y, then the inverse operator A⁻¹ : Y → X is continuous as well (this is called the bounded inverse theorem). (Rudin 1973, Corollary 2.12)
• If A : X → Y is a linear operator between the Banach spaces X and Y, and if for every sequence (x_n) in X with x_n → 0 and Ax_n → y it follows that y = 0, then A is continuous (Closed graph theorem). (Rudin 1973, Theorem 2.15)

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

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

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

Thus, there is an open ball B(c, r) in Y, with center c and radius r > 0, contained in the closure of A(kU). If v ∈ V, then c + r v and c are in B(c, r), hence are limit points of A(k U). By continuity of addition, their difference rv is a limit point of A(k U) − A(k U) ⊂ A(2k U). By linearity of A, this implies that any v ∈ V is in the closure of A(δ ⁻¹ U), where δ = r / (2k). It follows that for any y ∈ Y and any ε > 0, there is an x ∈ X with:

$\ ||x||<\delta ^{-1}||y||$  and  $||y-Ax||<\varepsilon .\quad (1)$

Fix y ∈ δ V (where δ V means the ball V stretched by a factor of δ, rather than the boundary of V). By (1), there is some x ₁ with ||x ₁|| < 1 and ||y − A x ₁|| < δ / 2. Define a sequence {x_n} 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 x _n +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 x _n +1. Let

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

From the first inequality in (2), {s_n} is a Cauchy sequence, and since X is complete, s_n converges to some x ∈ X. By (2), the sequence A s_n tends to y, and so A x = 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 y ∈ δ V 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 : X → Y 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 X → X / N is open, and the mapping α is an isomorphism of topological vector spaces (Dieudonné, 12.16.8).

References

• Rudin, Walter (1973), Functional Analysis, McGraw-Hill, ISBN 0-07-054236-8
• Dieudonné, Jean (1970), Treatise on Analysis, Volume II, Academic Press

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