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 : 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).

One 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 (the closed graph theorem). (Rudin 1973, Theorem 2.15)

Proof

Suppose A : X → Y 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 v ∈ V, 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\|_{X}<\varepsilon .\qquad (1)

Our next goal is to show that $V \subseteq A (2 LU)$ .

Let $y \in V$ . By (1), there is some $x 1$ with $|| x 1 || < L$ and $|| y - Ax 1 || < 1/2$ . Define a sequence {x_n} 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 $x n +1$ 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 $x n +1$ . 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 $x \in X$ . By (2), the sequence $As n$ 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 $A (2 LU)$ , so $V \subseteq A (2 LU)$ as claimed. Thus the image A(U) of the unit ball in X contains the open ball $V /2 LU$ 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 $A : X \to Y$ 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 $X \to X / N$ 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 $u : X \to Y$ 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.

References

↑ Trèves (1995), p. 170

Rudin, Walter (1973), Functional Analysis, McGraw-Hill, ISBN 0-07-054236-8
Dieudonné, Jean (1970), Treatise on Analysis, Volume II, Academic Press
Trèves, François (1995), Topological Vector Spaces, Distributions and Kernels, Academic Press, Inc., pp. 166, 170, ISBN 0-486-45352-9

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Open mapping theorem (functional analysis)

Consequences

Proof

Generalizations

See also

References