Open mapping theorem (functional analysis)

In functional analysis, the open mapping theorem, also known as the Banach–Schauder theorem or the Banach theorem (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.

A special case is also called the bounded inverse theorem (also called inverse mapping theorem or Banach isomorphism theorem), which states that a bijective bounded linear operator $$T$$ from one Banach space to another has bounded inverse $$T^{-1}$$.

Statement and proof
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See for a more quantitative formulation of the theorem.

The proof here uses the Baire category theorem, and completeness of both $$E$$ and $$F$$ is essential to the theorem. The statement of the theorem is no longer true if either space is assumed to be only a normed vector space; see

The proof is based on the following lemmas, which are also somewhat of independent interest. A linear map $$f : E \to F$$ between topological vector spaces is said to be nearly open if, for each neighborhood $$U$$ of zero, the closure $$\overline{f(U)}$$ contains a neighborhood of zero. The next lemma may be thought of as a weak version of the open mapping theorem.

Proof: Shrinking $$U$$, we can assume $$U$$ is an open ball centered at zero. Then $$F = f(E) = f\left(\bigcup_{n \in \N} n U\right) = \bigcup_{n \in \N} f(nU)$$. Thus,
 * $$F = \bigcup_{n > 0} \overline{f(nU)}.$$

Since $$F$$ is non-meager, some $$\overline{f(nU)}$$ contains an interior point $$y$$; that is, for some radius $$r > 0$$,
 * $$B(y, r) \subset \overline{f(nU)}.$$

Then for any $$v$$ in $$F$$ with $$\|v\| < r$$, by linearity, convexity and $$(-1)U \subset U$$,
 * $$v = v - y + y \in \overline{f(-nU)} + \overline{f(nU)} \subset \overline{f(2nU)}$$,

which proves the lemma by dividing by $$2n$$.$$\square$$

(The same proof works if $$E, F$$ are pre-Fréchet spaces.)

The completeness on the domain then allows to upgrade nearly open to open.

Proof: Since $$f$$ is nearly open, for some $$\epsilon > 0$$, $$B(0, \epsilon) \subset \overline{f(B(0, 2^{-1}))}$$. Let $$y$$ be in $$B(0, \epsilon)$$. Then we can find an $$x_1$$ in $$E$$ such that
 * $$\|y - f(x_1) \| < \epsilon/2, \, \|x_1 \| < 2^{-1}.$$

Since $$B(0, \epsilon/2) \subset \overline{f(B(0, 2^{-2}))}$$ and $$y - f(x_1)$$ is in $$B(0, \epsilon/2)$$, we can then choose $$x_2$$ such that
 * $$\|y - f(x_1) - f(x_2)\| < \epsilon/4, \, \|x_2\| < 2^{-2}.$$

Then so on. Thus, we find a sequence $$x_n$$ such that $$x = \sum_1^{\infty} x_n$$ converges in $$B(0, 1)$$ and $$f(x) = y$$ by linearity and continuity. In other words, we have shown $$B(0, \epsilon) \subset f(B(0, 1))$$, which is enough to conclude the openness and surjectivity.$$\square$$

(Again the same proof is valid if $$E, F$$ are pre-Fréchet spaces, using distance functions in place of norms.)

Proof of the theorem: By Baire's category theorem, the first lemma applies. Then the conclusion of the theorem follows from the second lemma. $$\square$$

In general, a continuous bijection between topological spaces is not necessarily a homeomorphism. The open mapping theorem, when it applies, implies the bijectivity is enough:

Even though the above bounded inverse theorem is a special case of the open mapping theorem, the open mapping theorem in turns follows from that. Indeed, a surjective linear operator $$T : E \to F$$ factors as
 * $$T : E \overset{p}\to E/\operatorname{ker} T \overset{T_0}\to F.$$

Here, $$T_0$$ is bijective and thus is a homeomorphism by the bounded inverse theorem; in particular, it is an open mapping. As a quotient map for topological groups is open, $$T$$ is open then.

Because the open mapping theorem and the bounded inverse theorem are essentially the same result, they are often simply called Banach's theorem.

Related results
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See also:.

Counterexample
The open mapping theorem may not hold for normed spaces that are not complete. A quickest way to see this is to note that the closed graph theorem, a consequence of the open mapping theorem, fails without completeness. But here is a more concrete counterexample. Consider the space X of sequences x : N &rarr; R with only finitely many non-zero terms equipped with the supremum norm. The map T : X &rarr; X defined by


 * $$T x = \left( x_{1}, \frac{x_{2}}{2}, \frac{x_{3}}{3}, \dots \right)$$

is bounded, linear and invertible, but T&minus;1 is unbounded. This does not contradict the bounded inverse theorem since X is not complete, and thus is not a Banach space. To see that it's not complete, consider the sequence of sequences x(n) &isin; X given by


 * $$x^{(n)} = \left( 1, \frac1{2}, \dots, \frac1{n}, 0, 0, \dots \right)$$

converges as n &rarr; &infin; to the sequence x(&infin;) given by


 * $$x^{(\infty)} = \left( 1, \frac1{2}, \dots, \frac1{n}, \dots \right),$$

which has all its terms non-zero, and so does not lie in X.

The completion of X is the space $c_0$ of all sequences that converge to zero, which is a (closed) subspace of the ℓp space ℓ&infin;(N), which is the space of all bounded sequences. However, in this case, the map T is not onto, and thus not a bijection. To see this, one need simply note that the sequence


 * $$x = \left( 1, \frac12, \frac13, \dots \right),$$

is an element of $$c_0$$, but is not in the range of $$T:c_0\to c_0$$.

Consequences
The open mapping theorem has several important consequences:
 * If $$A : X \to Y$$ is a bijective continuous linear operator between the Banach spaces $$X$$ and $$Y,$$ then the inverse operator $$A^{-1} : Y \to X$$ is continuous as well (this is called the bounded inverse theorem).
 * If $$A : X \to Y$$ is a linear operator between the Banach spaces $$X$$ and $$Y,$$ and if for every sequence $$\left(x_n\right)$$ in $$X$$ with $$x_n \to 0$$ and $$A x_n \to y$$ it follows that $$y = 0,$$ then $$A$$ is continuous (the closed graph theorem).
 * An exact sequence of Banach spaces (or more generally Fréchet spaces) is topologically exact.
 * The closed range theorem, which says an operator (under some assumption) has closed image if and only if its transpose has closed image (see closed range theorem).

The open mapping theorem does not imply that a continuous surjective linear operator admits a continuous linear section. What we have is: In particular, the above applies to an operator between Hilbert spaces or an operator with finite-dimensional kernel (by the Hahn–Banach theorem).
 * A surjective continuous linear operator admits a continuous linear section if and only if the kernel is topologically complemented.

Quantative version
A T. Tao’s blog post gives the following quantitative formulation of the theorem:

Proof: 2. $$\Rightarrow$$ 1. is the usual open mapping theorem.

1. $$\Rightarrow$$ 4.: For some $$r > 0$$, we have $$B(0, 2) \subset T(B(0, r))$$ where $$B$$ means an open ball. Then $$\frac{f}{\|f\|} = T \left(\frac{u}{\|f\|} \right)$$ for some $$\frac{u}{\|f\|}$$ in $$B(0, r)$$. That is, $$Tu = f$$ with $$\|u\| < r\|f\|$$.

4. $$\Rightarrow$$ 3.: We can write $$f = \sum_0^{\infty} f_j$$ with $$f_j$$ in the dense subspace and the sum converging in norm. Then, since $$E$$ is complete, $$u = \sum_0^{\infty} u_j$$ with $$\|u_j\| \le C \|f_j\|$$ and $$Tu_j = f_j$$ is a required solution. Finally, 3. $$\Rightarrow$$ 2. is trivial. $$\square$$

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:

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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 $$\alpha$$ is an isomorphism of topological vector spaces.

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

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

Nearly/Almost open linear maps

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

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