# Open mapping theorem (functional analysis)

Short description: Condition for a linear operator to be open

In functional analysis, the open mapping theorem, also known as the Banach–Schauder theorem or the Banach theorem[1] (named after Stefan Banach and Juliusz Schauder), is a fundamental result that states that if a bounded or 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 $\displaystyle{ X }$ and $\displaystyle{ Y }$ are Banach spaces and $\displaystyle{ A : X \to Y }$ is a surjective continuous linear operator, then $\displaystyle{ A }$ is an open map (that is, if $\displaystyle{ U }$ is an open set in $\displaystyle{ X, }$ then $\displaystyle{ A(U) }$ is open in $\displaystyle{ Y }$).

This proof uses the Baire category theorem, and completeness of both $\displaystyle{ X }$ and $\displaystyle{ Y }$ is essential to the theorem. The statement of the theorem is no longer true if either space is just assumed to be a normed vector space, but is true if $\displaystyle{ X }$ and $\displaystyle{ Y }$ are taken to be Fréchet spaces.

Proof

Suppose $\displaystyle{ A : X \to Y }$ is a surjective continuous linear operator. In order to prove that $\displaystyle{ A }$ is an open map, it is sufficient to show that $\displaystyle{ A }$ maps the open unit ball in $\displaystyle{ X }$ to a neighborhood of the origin of $\displaystyle{ Y. }$

Let $\displaystyle{ U = B_1^X(0), V = B_1^Y(0). }$ Then $\displaystyle{ X = \bigcup_{k \in \N} k U. }$

Since $\displaystyle{ A }$ is surjective: $\displaystyle{ Y = A(X) = A\left(\bigcup_{k \in \N} k U\right) = \bigcup_{k \in \N} A(kU). }$

But $\displaystyle{ Y }$ is Banach so by Baire's category theorem $\displaystyle{ \exists k \in \N: \qquad \left(\overline{A(kU)} \right)^{\circ} \neq \varnothing. }$

That is, we have $\displaystyle{ c \in Y }$ and $\displaystyle{ r \gt 0 }$ such that $\displaystyle{ B_r(c) \subseteq \left(\overline{A(kU)} \right)^\circ \subseteq \overline{A(kU)}. }$

Let $\displaystyle{ v \in V, }$ then $\displaystyle{ c, c + r v \in B_r(c) \subseteq \overline{A(kU)}. }$

By continuity of addition and linearity, the difference $\displaystyle{ r v }$ satisfies $\displaystyle{ r v \in \overline{A(kU)} + \overline{A(kU)} \subseteq \overline{A(kU) + A(kU)} \subseteq \overline{A(2kU)}, }$ and by linearity again, $\displaystyle{ V \subseteq \overline{A(LU)} }$

where we have set $\displaystyle{ L = 2 k / r. }$ It follows that for all $\displaystyle{ y \in Y }$ and all $\displaystyle{ \epsilon \gt 0, }$ there exists some $\displaystyle{ x \in X }$ such that $\displaystyle{ \qquad \|x\|_X \leq L \|y\|_Y \quad \text{and} \quad \|y - A x\|_Y \lt \epsilon. \qquad (1) }$

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

Let $\displaystyle{ y \in V. }$ By (1), there is some $\displaystyle{ x_1 }$ with $\displaystyle{ \left\|x_1\right\| \lt L }$ and $\displaystyle{ \left\|y - A x_1\right\| \lt 1/2. }$ Define a sequence $\displaystyle{ \left(x_n\right) }$ inductively as follows. Assume: $\displaystyle{ \|x_n\| \lt \frac{L}{2^{n-1}} \quad \text{and} \quad \left\|y - A\left(x_1 + x_2 + \cdots + x_n\right)\right\| \lt \frac{1}{2^n}. \qquad (2) }$

Then by (1) we can pick $\displaystyle{ x_{n+1} }$ so that: $\displaystyle{ \|x_{n+1}\| \lt \frac{L}{2^n} \quad \text{and} \quad \left\|y - A\left(x_1 + x_2 + \cdots + x_n\right) - A\left(x_{n+1}\right)\right\| \lt \frac{1}{2^{n+1}}, }$ so (2) is satisfied for $\displaystyle{ x_{n+1}. }$ Let $\displaystyle{ s_n = x_1 + x_2 + \cdots + x_n. }$

From the first inequality in (2), $\displaystyle{ \left(s_n\right) }$is a Cauchy sequence, and since $\displaystyle{ X }$ is complete, $\displaystyle{ s_n }$ converges to some $\displaystyle{ x \in X. }$ By (2), the sequence $\displaystyle{ A s_n }$ tends to $\displaystyle{ y }$ and so $\displaystyle{ Ax = y }$ by continuity of $\displaystyle{ A. }$ Also, $\displaystyle{ \|x\| = \lim_{n \to \infty} \|s_n\| \leq \sum_{n=1}^\infty \|x_n\| \lt 2 L. }$

This shows that $\displaystyle{ y }$ belongs to $\displaystyle{ A(2LU), }$ so $\displaystyle{ V \subseteq A(2LU) }$ as claimed. Thus the image $\displaystyle{ A(U) }$ of the unit ball in $\displaystyle{ X }$ contains the open ball $\displaystyle{ V / 2L }$ of $\displaystyle{ Y. }$ Hence, $\displaystyle{ A(U) }$ is a neighborhood of the origin in $\displaystyle{ Y, }$ and this concludes the proof.

### Related results

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

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. $\displaystyle{ \operatorname{Im} T = Y }$ (that is, $\displaystyle{ T }$ is surjective).

Furthermore, if $\displaystyle{ T }$ is surjective then (1) holds for some $\displaystyle{ \delta \gt 0 }$

### Consequences

The open mapping theorem has several important consequences:

• If $\displaystyle{ A : X \to Y }$ is a bijective continuous linear operator between the Banach spaces $\displaystyle{ X }$ and $\displaystyle{ Y, }$ then the inverse operator $\displaystyle{ A^{-1} : Y \to X }$ is continuous as well (this is called the bounded inverse theorem).[3]
• If $\displaystyle{ A : X \to Y }$ is a linear operator between the Banach spaces $\displaystyle{ X }$ and $\displaystyle{ Y, }$ and if for every sequence $\displaystyle{ \left(x_n\right) }$ in $\displaystyle{ X }$ with $\displaystyle{ x_n \to 0 }$ and $\displaystyle{ A x_n \to y }$ it follows that $\displaystyle{ y = 0, }$ then $\displaystyle{ A }$ is continuous (the closed graph theorem).[4]

## Generalizations

Local convexity of $\displaystyle{ X }$ or $\displaystyle{ Y }$  is not essential to the proof, but completeness is: the theorem remains true in the case when $\displaystyle{ X }$ and $\displaystyle{ Y }$ are F-spaces. Furthermore, the theorem can be combined with the Baire category theorem in the following manner:

Open mapping theorem for continuous maps[5][6] — Let $\displaystyle{ A : X \to Y }$ be a continuous linear operator from a complete pseudometrizable TVS $\displaystyle{ X }$ onto a Hausdorff TVS $\displaystyle{ Y. }$ If $\displaystyle{ \operatorname{Im} A }$ is nonmeager in $\displaystyle{ Y }$ then $\displaystyle{ A : X \to Y }$ is a (surjective) open map and $\displaystyle{ Y }$ is a complete pseudometrizable TVS. Moreover, if $\displaystyle{ X }$ is assumed to be hausdorff (i.e. a F-space), then $\displaystyle{ Y }$ is also an F-space.

Furthermore, in this latter case if $\displaystyle{ N }$ is the kernel of $\displaystyle{ A, }$ then there is a canonical factorization of $\displaystyle{ A }$ in the form $\displaystyle{ X \to X/N \overset{\alpha}{\to} Y }$ where $\displaystyle{ X / N }$ is the quotient space (also an F-space) of $\displaystyle{ X }$ by the closed subspace $\displaystyle{ N. }$ The quotient mapping $\displaystyle{ X \to X / N }$ is open, and the mapping $\displaystyle{ \alpha }$ is an isomorphism of topological vector spaces.[7]

An important special case of this theorem can also be stated as

Theorem[8] — Let $\displaystyle{ X }$ and $\displaystyle{ Y }$ be two F-spaces. Then every continuous linear map of $\displaystyle{ X }$ onto $\displaystyle{ Y }$ is a TVS homomorphism, where a linear map $\displaystyle{ u : X \to Y }$ 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.

On the other hand, a more general formulation, which implies the first, can be given:

Open mapping theorem[6] — Let $\displaystyle{ A : X \to Y }$ be a surjective linear map from a complete pseudometrizable TVS $\displaystyle{ X }$ onto a TVS $\displaystyle{ Y }$ and suppose that at least one of the following two conditions is satisfied:

1. $\displaystyle{ Y }$ is a Baire space, or
2. $\displaystyle{ X }$ is locally convex and $\displaystyle{ Y }$ is a barrelled space,

If $\displaystyle{ A }$ is a closed linear operator then $\displaystyle{ A }$ is an open mapping. If $\displaystyle{ A }$ is a continuous linear operator and $\displaystyle{ Y }$ is Hausdorff then $\displaystyle{ A }$ is (a closed linear operator and thus also) an open mapping.

Nearly/Almost open linear maps

A linear map $\displaystyle{ A : X \to Y }$ between two topological vector spaces (TVSs) is called a nearly open map (or sometimes, an almost open map) if for every neighborhood $\displaystyle{ U }$ of the origin in the domain, the closure of its image $\displaystyle{ \operatorname{cl} A(U) }$ is a neighborhood of the origin in $\displaystyle{ Y. }$[9] Many authors use a different definition of "nearly/almost open map" that requires that the closure of $\displaystyle{ A(U) }$ be a neighborhood of the origin in $\displaystyle{ A(X) }$ rather than in $\displaystyle{ Y, }$[9] but for surjective maps these definitions are equivalent. A bijective linear map is nearly open if and only if its inverse is continuous.[9] Every surjective linear map from locally convex TVS onto a barrelled TVS is nearly open.[10] The same is true of every surjective linear map from a TVS onto a Baire TVS.[10]

Open mapping theorem[11] — If a closed surjective linear map from a complete pseudometrizable TVS onto a Hausdorff TVS is nearly open then it is open.

### Consequences

Theorem[12] — If $\displaystyle{ A : X \to Y }$ is a continuous linear bijection from a complete Pseudometrizable topological vector space (TVS) onto a Hausdorff TVS that is a Baire space, then $\displaystyle{ A : X \to Y }$ is a homeomorphism (and thus an isomorphism of TVSs).

### Webbed spaces

Main page: Webbed space

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

## References

1. Trèves 2006, p. 166.
2. Rudin 1991, p. 100.
3. Rudin 1973, Corollary 2.12.
4. Rudin 1973, Theorem 2.15.
5. Rudin 1991, Theorem 2.11.
6. Narici & Beckenstein 2011, p. 468.
7. Dieudonné 1970, 12.16.8.
8. Trèves 2006, p. 170
9. Narici & Beckenstein 2011, pp. 466.
10. Narici & Beckenstein 2011, pp. 467.
11. Narici & Beckenstein 2011, pp. 466−468.
12. Narici & Beckenstein 2011, p. 469.