Open mapping theorem (functional analysis)

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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 [math]\displaystyle{ X }[/math] and [math]\displaystyle{ Y }[/math] are Banach spaces and [math]\displaystyle{ A : X \to Y }[/math] is a surjective continuous linear operator, then [math]\displaystyle{ A }[/math] is an open map (that is, if [math]\displaystyle{ U }[/math] is an open set in [math]\displaystyle{ X, }[/math] then [math]\displaystyle{ A(U) }[/math] is open in [math]\displaystyle{ Y }[/math]).

This proof uses the Baire category theorem, and completeness of both [math]\displaystyle{ X }[/math] and [math]\displaystyle{ Y }[/math] 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 [math]\displaystyle{ X }[/math] and [math]\displaystyle{ Y }[/math] are taken to be Fréchet spaces.

Proof

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

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

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

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

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

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

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

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

Our next goal is to show that [math]\displaystyle{ V \subseteq A(2LU). }[/math]

Let [math]\displaystyle{ y \in V. }[/math] By (1), there is some [math]\displaystyle{ x_1 }[/math] with [math]\displaystyle{ \left\|x_1\right\| \lt L }[/math] and [math]\displaystyle{ \left\|y - A x_1\right\| \lt 1/2. }[/math] Define a sequence [math]\displaystyle{ \left(x_n\right) }[/math] inductively as follows. Assume: [math]\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) }[/math]

Then by (1) we can pick [math]\displaystyle{ x_{n+1} }[/math] so that: [math]\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}}, }[/math] so (2) is satisfied for [math]\displaystyle{ x_{n+1}. }[/math] Let [math]\displaystyle{ s_n = x_1 + x_2 + \cdots + x_n. }[/math]

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

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

Related results

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

  1. [math]\displaystyle{ \left\|T^* y^*\right\| \geq \delta \left\|y^*\right\| }[/math] for all [math]\displaystyle{ y^* \in Y^* }[/math];
  2. [math]\displaystyle{ \overline{T\left(B_X\right)} \supseteq \delta B_Y }[/math];
  3. [math]\displaystyle{ {T\left(B_X\right)} \supseteq \delta B_Y }[/math];
  4. [math]\displaystyle{ \operatorname{Im} T = Y }[/math] (that is, [math]\displaystyle{ T }[/math] is surjective).

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

Consequences

The open mapping theorem has several important consequences:

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

Generalizations

Local convexity of [math]\displaystyle{ X }[/math] or [math]\displaystyle{ Y }[/math]  is not essential to the proof, but completeness is: the theorem remains true in the case when [math]\displaystyle{ X }[/math] and [math]\displaystyle{ Y }[/math] 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 [math]\displaystyle{ A : X \to Y }[/math] be a continuous linear operator from a complete pseudometrizable TVS [math]\displaystyle{ X }[/math] onto a Hausdorff TVS [math]\displaystyle{ Y. }[/math] If [math]\displaystyle{ \operatorname{Im} A }[/math] is nonmeager in [math]\displaystyle{ Y }[/math] then [math]\displaystyle{ A : X \to Y }[/math] is a (surjective) open map and [math]\displaystyle{ Y }[/math] is a complete pseudometrizable TVS. Moreover, if [math]\displaystyle{ X }[/math] is assumed to be hausdorff (i.e. a F-space), then [math]\displaystyle{ Y }[/math] is also an F-space.

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

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

Theorem[8] — Let [math]\displaystyle{ X }[/math] and [math]\displaystyle{ Y }[/math] be two F-spaces. Then every continuous linear map of [math]\displaystyle{ X }[/math] onto [math]\displaystyle{ Y }[/math] is a TVS homomorphism, where a linear map [math]\displaystyle{ u : X \to Y }[/math] is a topological vector space (TVS) homomorphism if the induced map [math]\displaystyle{ \hat{u} : X / \ker(u) \to Y }[/math] 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 [math]\displaystyle{ A : X \to Y }[/math] be a surjective linear map from a complete pseudometrizable TVS [math]\displaystyle{ X }[/math] onto a TVS [math]\displaystyle{ Y }[/math] and suppose that at least one of the following two conditions is satisfied:

  1. [math]\displaystyle{ Y }[/math] is a Baire space, or
  2. [math]\displaystyle{ X }[/math] is locally convex and [math]\displaystyle{ Y }[/math] is a barrelled space,

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

Nearly/Almost open linear maps

A linear map [math]\displaystyle{ A : X \to Y }[/math] between two topological vector spaces (TVSs) is called a nearly open map (or sometimes, an almost open map) if for every neighborhood [math]\displaystyle{ U }[/math] of the origin in the domain, the closure of its image [math]\displaystyle{ \operatorname{cl} A(U) }[/math] is a neighborhood of the origin in [math]\displaystyle{ Y. }[/math][9] Many authors use a different definition of "nearly/almost open map" that requires that the closure of [math]\displaystyle{ A(U) }[/math] be a neighborhood of the origin in [math]\displaystyle{ A(X) }[/math] rather than in [math]\displaystyle{ Y, }[/math][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 [math]\displaystyle{ A : X \to Y }[/math] is a continuous linear bijection from a complete Pseudometrizable topological vector space (TVS) onto a Hausdorff TVS that is a Baire space, then [math]\displaystyle{ A : X \to Y }[/math] 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.

See also

References

Bibliography