Open mapping theorem (functional analysis)

Template:Short description Template:CS1 config Template:About In functional analysis, the open mapping theorem, also known as the Banach–Schauder theorem or the Banach theoremTemplate:Sfn (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

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 Template:Section link.

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. We have $f(E)=f\left(\bigcup _{n\in \mathbb{N}}nU\right)=\bigcup _{n\in \mathbb{N}}f(nU)$. Thus, 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 $\Vert v\Vert <r$, 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 $2n$.$◻$ (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: $\overline{B(0,\delta )}\subset \overline{f(B(0,1))}$. Thus, for each $ϵ>0$ and $z$ in $F$, we can find an $x$ with $\Vert x\Vert <{\delta }^{-1}\Vert z\Vert$ and $z$ in $B(f(x),ϵ)$. Thus, taking $z=y$, we find an $x_1$ such that

$\Vert y-f(x_1)\Vert <c_1,\Vert x_1\Vert <{\delta }^{-1}\Vert y\Vert .$

Applying the same argument with $z=y-f(x_1)$, we then find an $x_2$ such that

$\Vert y-f(x_1)-f(x_2)\Vert <c_2,\Vert x_2\Vert <{\delta }^{-1}{c}_1$

where we observed $\Vert x_2\Vert <{\delta }^{-1}\Vert z\Vert <{\delta }^{-1}{c}_1$. Then so on. Thus, if $c:=\sum c_n<\mathrm{\infty }$, we found a sequence $x_n$ such that $x=\sum _1^{\mathrm{\infty }}{x}_n$ converges and $f(x)=y$. Also,

$\Vert x\Vert \le \sum _1^{\mathrm{\infty }}\Vert x_n\Vert \le {\delta }^{-1}\Vert y\Vert +{\delta }^{-1}c.$

Since ${\delta }^{-1}\Vert y\Vert <1$, by making $c$ small enough, we can achieve $\Vert x\Vert <1$. $◻$ (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. $◻$

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 continuous linear operator $T:E\to F$ factors as

$T:E\stackrel{p}{\to }E/\ker T\stackrel{T_0}{\to }F.$

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: $y\notin \overline{T(B_X)}\Rightarrow \Vert y\Vert >\delta ,$ and that follows from the Hahn–Banach theorem. 2. $\Rightarrow$ 3. is exactly the second lemma in Template:Section link. Finally, 3. $\Rightarrow$ 4. is trivial and 4. $\Rightarrow$ 1. easily follows from the open mapping theorem. $◻$

Alternatively, 1. implies that $T^{\prime }$ 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:

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 $\frac{f}{\Vert f\Vert }=T\left(\frac{u}{\Vert f\Vert }\right)$ for some $\frac{u}{\Vert f\Vert }$ in $B(0,r)$. That is, $Tu=f$ with $\Vert u\Vert <r\Vert f\Vert$.

4. $\Rightarrow$ 3.: We can write $f=\sum _0^{\mathrm{\infty }}{f}_j$ with $f_j$ in the dense subspace and the sum converging in norm. Then, since $E$ is complete, $u=\sum _0^{\mathrm{\infty }}{u}_j$ with $\Vert u_j\Vert \le C\Vert f_j\Vert$ and $T{u}_j=f_j$ is a required solution. Finally, 3. $\Rightarrow$ 2. is trivial. $◻$

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

$Tx=(x_1,\frac{x_2}{2},\frac{x_3}{3},\dots )$

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

$x^{(n)}=(1,\frac{1}{2},\dots ,\frac{1}{n},0,0,\dots )$

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

$x^{(\mathrm{\infty })}=(1,\frac{1}{2},\dots ,\frac{1}{n},\dots ),$

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

$x=(1,\frac{1}{2},\frac{1}{3},\dots ),$

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 $l^{\mathrm{\infty }}$, for example $x=(1,1,1,\dots )$ 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 $T^{-1}:Y\to X$ is continuous as well (this is called the bounded inverse theorem).Template:Sfn
• 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).Template:Sfn
• 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 $$X\to X/N\stackrel{\alpha }{\to }Y$$ 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.Template:Sfn

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

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

Nearly/Almost open linear maps

A linear map $A:X\to Y$ between two topological vector spaces (TVSs) is called a Template:Em (or sometimes, an Template:Em) if for every neighborhood $U$ of the origin in the domain, the closure of its image $\mathrm{cl}A(U)$ is a neighborhood of the origin in $Y.$Template:Sfn 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 $Y,$Template:Sfn but for surjective maps these definitions are equivalent. A bijective linear map is nearly open if and only if its inverse is continuous.Template:Sfn Every surjective linear map from locally convex TVS onto a barrelled TVS is nearly open.Template:Sfn The same is true of every surjective linear map from a TVS onto a Baire TVS.Template:Sfn

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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Bibliography

• Template:Adasch Topological Vector Spaces
• Template:Banach Théorie des Opérations Linéaires
• Template:Berberian Lectures in Functional Analysis and Operator Theory
• Template:Bourbaki Topological Vector Spaces
• Template:Conway A Course in Functional Analysis
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• Template:Edwards Functional Analysis Theory and Applications
• Template:Grothendieck Topological Vector Spaces
• Template:Jarchow Locally Convex Spaces
• Template:Köthe Topological Vector Spaces I
• Template:Narici Beckenstein Topological Vector Spaces
• Template:Robertson Topological Vector Spaces
• Template:Rudin Walter Functional Analysis
• Template:Rudin Walter Functional Analysis
• Template:Schaefer Wolff Topological Vector Spaces
• Template:Swartz An Introduction to Functional Analysis
• Template:Trèves François Topological vector spaces, distributions and kernels
• Template:Cite web
• Template:Wilansky Modern Methods in Topological Vector Spaces

Further reading

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