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 $T^$.

Statement and proof

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 $\overline$ 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. We have $f(E) = f\left(\bigcup_ n U\right) = \bigcup_ f(nU)$. Thus, some $\overline$ contains an interior point $y$; that is, for some radius $r > 0$,
:$B(y, r) \subset \overline.$
Then for any $v$ in $F$ with $\, v\, < r$, by linearity, convexity and $(-1)U \subset U$,
:$v = v - y + y \in \overline + \overline \subset \overline$,
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: Let $y$ be in $B(0, \delta)$ and $c_n > 0$ some sequence. We have: $\overline \subset \overline$. Thus, for each $\epsilon > 0$ and $z$ in $F$, we can find an $x$ with $\, x\, < \delta^\, z\,$ and $z$ in $B(f(x), \epsilon)$. Thus, taking $z = y$, we find an $x_1$ such that
:$\, y - f(x_1) \, < c_1, \, \, x_1 \, < \delta^ \, y\, .$
Applying the same argument with $z = y - f(x_1)$, we then find an $x_2$ such that
:$\, y - f(x_1) - f(x_2)\, < c_2, \, \, x_2\, < \delta^c_1$
where we observed $\, x_2\, < \delta^ \, z\, < \delta^ c_1$. Then so on. Thus, if $c := \sum c_n < \infty$, we found a sequence $x_n$ such that $x = \sum_1^ x_n$ converges and $f(x) = y$. Also,
:$\, x\, \le \sum_1^ \, x_n\, \le \delta^ \, y\, + \delta^ c.$
Since $\delta^ \, y\, < 1$, by making $c$ small enough, we can achieve $\, x\, < 1$. $\square$ (Again the same proof is valid if $E, F$ are pre-Fréchet spaces.)

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 turn follows from that. Indeed, a surjective continuous linear operator $T : E \to F$ factors as
:$T : E \overset\to E/\operatorname T \overset\to F.$
Here, $T_0$ is continuous and 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.

Transpose formulation

Here is a formulation of the open mapping theorem in terms of the transpose of an operator.

Proof: The idea of 1. $\Rightarrow$ 2. is to show: $y \notin \overline \Rightarrow \, y\, > \delta,$ and that follows from the Hahn–Banach theorem. 2. $\Rightarrow$ 3. is exactly the second lemma in . Finally, 3. $\Rightarrow$ 4. is trivial and 4. $\Rightarrow$ 1. easily follows from the open mapping theorem. $\square$

Alternatively, 1. implies that $T'$ is injective and has closed image and then by the closed range theorem, that implies $T$ has dense image and closed image, respectively; i.e., $T$ is surjective. Hence, the above result is a variant of a special case of the closed range theorem.

Quantative formulation

Terence Tao gives the following quantitative formulation of the theorem:

The proof follows a cycle of implications $1\Rightarrow 4\Rightarrow 3\Rightarrow 2\Rightarrow 1$. Here $2 \Rightarrow 1$ 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 $\frac = T \left(\frac \right)$ for some $\frac$ in $B(0, r)$. That is, $Tu = f$ with $\, u\, < r\, f\,$.

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

Finally, $3 \Rightarrow 2$ is trivial. $\square$

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 → R with only finitely many non-zero terms equipped with the supremum norm. The map ''T'' : ''X'' → ''X'' defined by

:$T x = \left( x_, \frac, \frac, \dots \right)$

is bounded, linear and invertible, but ''T''⁻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'') ∈ ''X'' given by

:$x^ = \left( 1, \frac1, \dots, \frac1, 0, 0, \dots \right)$

converges as ''n'' → ∞ to the sequence ''x''^(∞) given by

:$x^ = \left( 1, \frac1, \dots, \frac1, \dots \right),$

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 ℓ_''p'' space ℓ_∞(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 $T:c_0\to c_0$. Same reasoning applies to show $T$ is also not onto in $l^\infty$, for example $x = \left( 1, 1, 1, \dots \right)$ is not in the range of $T$.

Consequences

The open mapping theorem has several important consequences:
* If $T : X \to Y$ is a bijective continuous linear operator between the Banach spaces $X$ and $Y,$ then the inverse operator $T^ : Y \to X$ is continuous as well (this is called the bounded inverse theorem).
* If $T : X \to Y$ 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 $T x_n \to y$ it follows that $y = 0,$ then $T$ is continuous (the closed graph theorem).
*Given a bounded operator $T : E \to F$ between normed spaces, if the image of $T$ is non-meager and if $E$ is complete, then $T$ is open and surjective and $F$ is complete (to see this, use the two lemmas in the proof of the 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#Sketch of proof).

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

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:

(The proof is essentially the same as the Banach or Fréchet cases; we modify the proof slightly to avoid the use of convexity,)

Furthermore, in this latter case if $N$ is the kernel of $A,$ then there is a canonical factorization of $A$ in the form
$X \to X/N \overset Y$
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 $\operatorname A(U)$ 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.

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

See also

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References

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Further reading

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{{Topological vector spaces

Articles containing proofs
Theorems in functional analysis