open mapping theorem (functional analysis)

{{Math theorem|name=Open mapping theorem|math_statement={{harvnb|Rudin|1973|loc=Theorem 2.11.}}{{harvnb|Vogt|2000|loc=Theorem 1.6.}} Let $T:\; E\; \backslash to\; F$ be a surjective continuous linear map between Banach spaces (or more generally Fréchet spaces). Then $T$ is an open mapping (that is, if $U\; \backslash subset\; E$ is an open subset, then $T(U)$ is open).}}

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 {{section link||Counterexample}}.

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

{{math_theorem|name=Lemma|{{harvnb|Vogt|2000|loc=Lemma 1.4.}}The first part of the proof of {{harvnb|Rudin|1991|loc=Theorem 2.11.}} A linear map $f\; :\; E\; \backslash to\; F$ between normed spaces is nearly open if the image of $f$ is non-meager in $F$. (The continuity is not needed.)}}

Proof: Shrinking $U$, we can assume $U$ is an open ball centered at zero. We have $f(E)\; =\; f\backslash left(\backslash bigcup\_\{n\; \backslash in\; \backslash N\}\; n\; U\backslash right)\; =\; \backslash bigcup\_\{n\; \backslash in\; \backslash N\}\; f(nU)$. Thus, some $\backslash overline\{f(nU)\}$ contains an interior point $y$; that is, for some radius $r\; >\; 0$,

:$B(y,\; r)\; \backslash subset\; \backslash overline\{f(nU)\}.$

Then for any $v$ in $F$ with , by linearity, convexity and $(-1)U\; \backslash subset\; U$,

:$v\; =\; v\; -\; y\; +\; y\; \backslash in\; \backslash overline\{f(-nU)\}\; +\; \backslash overline\{f(nU)\}\; \backslash subset\; \backslash overline\{f(2nU)\}$,

which proves the lemma by dividing by $2n$.$\backslash square$ (The same proof works if $E,\; F$ are pre-Fréchet spaces.)

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

{{math_theorem|name=Lemma|note=Schauder|math_statement={{harvnb|Vogt|2000|loc=Lemma 1.5.}} Let $f:\; E\; \backslash to\; F$ be a continuous linear map between normed spaces.

If $f$ is nearly-open and if $E$ is complete, then $f$ is open and surjective.

More precisely, if $B(0,\; \backslash delta)\; \backslash subset\; \backslash overline\{f(B(0,\; 1))\}$ for some $\backslash delta\; >\; 0$ and if $E$ is complete, then

:$B(0,\; \backslash delta)\; \backslash subset\; f(B(0,\; 1))$

where $B(x,\; r)$ is an open ball with radius $r$ and center $x$.}}

Proof: Let $y$ be in $B(0,\; \backslash delta)$ and $c\_n\; >\; 0$ some sequence. We have: $\backslash overline\{B(0,\; \backslash delta)\}\; \backslash subset\; \backslash overline\{f(B(0,1))\}$. Thus, for each $\backslash epsilon\; >\; 0$ and $z$ in $F$, we can find an $x$ with and $z$ in $B(f(x),\; \backslash epsilon)$. Thus, taking $z\; =\; y$, we find an $x\_1$ such that

:

Applying the same argument with $z\; =\; y\; -\; f(x\_1)$, we then find an $x\_2$ such that

:

where we observed . Then so on. Thus, if , we found a sequence $x\_n$ such that $x\; =\; \backslash sum\_1^\{\backslash infty\}\; x\_n$ converges and $f(x)\; =\; y$. Also,

:$\backslash |x\backslash |\; \backslash le\; \backslash sum\_1^\{\backslash infty\}\; \backslash |x\_n\backslash |\; \backslash le\; \backslash delta^\{-1\}\; \backslash |y\backslash |\; +\; \backslash delta^\{-1\}\; c.$

Since , by making $c$ small enough, we can achieve . $\backslash square$ (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. $\backslash square$

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:

{{math_theorem|name=Corollary|note=Bounded inverse theorem|math_statement={{harvnb|Vogt|2000|loc=Corollary 1.7.}} A continuous bijective linear operator between Banach spaces (or Fréchet spaces) has continuous inverse. That is, the inverse operator is continuous.}}

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 $T\; :\; E\; \backslash to\; F$ factors as

:$T\; :\; E\; \backslash overset\{p\}\backslash to\; E/\backslash operatorname\{ker\}\; T\; \backslash overset\{T\_0\}\backslash to\; F.$

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.

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

{{Math theorem|name=Theorem|math_statement={{harvnb|Rudin|1991|loc=Theorem 4.13.}}

Let $X$ and $Y$ be Banach spaces, let $B\_X$ and $B\_Y$ denote their open unit balls, and let $T\; :\; X\; \backslash to\; Y$ be a bounded linear operator.

If $\backslash delta\; >\; 0$ then among the following four statements we have $(1)\; \backslash implies\; (2)\; \backslash implies\; (3)\; \backslash implies\; (4)$ (with the same $\backslash delta$)

1. $\backslash delta\; \backslash left\backslash |y\text{'}\backslash right\backslash |\; \backslash le\; \backslash left\backslash |T\text{'}\; y\text{'}\backslash right\backslash |$ for all $y\text{'}\; \backslash in\; Y\text{'}$ = continuous dual of $Y$;
2. $\backslash delta\; B\_Y\; \backslash subset\; \backslash overline\{T\backslash left(B\_X\backslash right)\}$;
3. $\backslash delta\; B\_Y\; \backslash subset\; \{T\backslash left(B\_X\backslash right)\}$;
4. $T$ is surjective.

Furthermore, if $T$ is surjective then (1) holds for some $\backslash delta\; >\; 0.$

}}

Proof: The idea of 1. $\backslash Rightarrow$ 2. is to show: $y\; \backslash notin\; \backslash overline\{T(B\_X)\}\; \backslash Rightarrow\; \backslash |y\backslash |\; >\; \backslash delta,$ and that follows from the Hahn–Banach theorem. 2. $\backslash Rightarrow$ 3. is exactly the second lemma in {{section link||Statement and proof}}. Finally, 3. $\backslash Rightarrow$ 4. is trivial and 4. $\backslash Rightarrow$ 1. easily follows from the open mapping theorem. $\backslash square$

Alternatively, 1. implies that $T\text{'}$ 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.

Terence Tao gives the following quantitative formulation of the theorem:{{cite web|url=https://terrytao.wordpress.com/2009/02/01/245b-notes-9-the-baire-category-theorem-and-its-banach-space-consequences/|first=Terence|last=Tao|author-link=Terence Tao|title=245B, Notes 9: The Baire category theorem and its Banach space consequences|work=What's New|date=February 1, 2009}}

{{math_theorem|math_statement=Let $T:\; E\; \backslash to\; F$ be a bounded operator between Banach spaces. Then the following are equivalent:

1. $T$ is open.
2. $T$ is surjective.
3. There exists a constant $C\; >\; 0$ such that, for each $f$ in $F$, the equation $Tu\; =\; f$ has a solution $u$ with $\backslash |u\backslash |\; \backslash le\; C\; \backslash |f\backslash |$.
4. 3. holds for $f$ in some dense subspace of $F$.}}

The proof follows a cycle of implications $1\backslash Rightarrow\; 4\backslash Rightarrow\; 3\backslash Rightarrow\; 2\backslash Rightarrow\; 1$. Here $2\; \backslash Rightarrow\; 1$ is the usual open mapping theorem.

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

$4\; \backslash Rightarrow\; 3$: We can write $f\; =\; \backslash sum\_0^\{\backslash infty\}\; f\_j$ with $f\_j$ in the dense subspace and the sum converging in norm. Then, since $E$ is complete, $u\; =\; \backslash sum\_0^\{\backslash infty\}\; u\_j$ with $\backslash |u\_j\backslash |\; \backslash le\; C\; \backslash |f\_j\backslash |$ and $Tu\_j\; =\; f\_j$ is a required solution.

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

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

:$T\; x\; =\; \backslash left(\; x\_\{1\},\; \backslash frac\{x\_\{2\}\}\{2\},\; \backslash frac\{x\_\{3\}\}\{3\},\; \backslash dots\; \backslash right)$

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

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

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

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

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

:$x\; =\; \backslash left(\; 1,\; \backslash frac12,\; \backslash frac13,\; \backslash dots\; \backslash right),$

is an element of $c\_0$, but is not in the range of $T:c\_0\backslash to\; c\_0$. Same reasoning applies to show $T$ is also not onto in $l^\backslash infty$, for example $x\; =\; \backslash left(\; 1,\; 1,\; 1,\; \backslash dots\; \backslash right)$ is not in the range of $T$.

The open mapping theorem has several important consequences:

• If $T\; :\; X\; \backslash to\; Y$ is a bijective continuous linear operator between the Banach spaces $X$ and $Y,$ then the inverse operator $T^\{-1\}\; :\; Y\; \backslash to\; X$ is continuous as well (this is called the bounded inverse theorem).{{sfn|Rudin|1973|loc=Corollary 2.12}}
• If $T\; :\; X\; \backslash to\; Y$ is a linear operator between the Banach spaces $X$ and $Y,$ and if for every sequence $\backslash left(x\_n\backslash right)$ in $X$ with $x\_n\; \backslash to\; 0$ and $T\; x\_n\; \backslash to\; y$ it follows that $y\; =\; 0,$ then $T$ is continuous (the closed graph theorem).{{sfn|Rudin|1973|loc=Theorem 2.15}}
• Given a bounded operator $T\; :\; E\; \backslash to\; F$ 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).{{harvnb|Rudin|1991|loc=Theorem 2.11.}}
• 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.{{cite web|url=https://mathoverflow.net/questions/375326/can-the-inverse-operator-in-bartle-graves-theorem-be-linear|work=MathOverflow|title=Can the inverse operator in Bartle-Graves theorem be linear?|date=October 31, 2020|first=Jarek|last=Sarnowski}}{{cite journal

| last1 = Borwein | first1 = J. M. | author1-link = Jonathan Borwein

| last2 = Dontchev | first2 = A. L.

| doi = 10.1090/S0002-9939-03-07229-0

| issue = 8

| journal = Proceedings of the American Mathematical Society

| mr = 1974655

| pages = 2553–2560

| title = On the Bartle–Graves theorem

| volume = 131

| year = 2003| hdl = 1959.13/940334

| hdl-access = free

}}

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:

{{Math theorem|name=Open mapping theorem for continuous maps{{sfn|Narici|Beckenstein|2011|p=468}}|note=|math_statement=

Let $A\; :\; X\; \backslash to\; Y$ be a continuous linear operator from a complete pseudometrizable TVS $X$ onto a Hausdorff TVS $Y.$

If $\backslash operatorname\{Im\}\; A$ is nonmeager in $Y$ then $A\; :\; X\; \backslash to\; Y$ is a (surjective) open map and $Y$ is a complete pseudometrizable TVS.

Moreover, if $X$ is assumed to be hausdorff (i.e. a F-space), then $Y$ is also an F-space.

}}

(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

$$X\; \backslash to\; X/N\; \backslash overset\{\backslash alpha\}\{\backslash to\}\; Y$$

where $X\; /\; N$ is the quotient space (also an F-space) of $X$ by the closed subspace $N.$

The quotient mapping $X\; \backslash to\; X\; /\; N$ is open, and the mapping $\backslash alpha$ is an isomorphism of topological vector spaces.{{sfn|Dieudonné|1970|loc=12.16.8}}

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

{{Math theorem|name=Theorem{{harvnb|Trèves|2006|p=170}}|math_statement=

Let $X$ and $Y$ be two F-spaces. Then every continuous linear map of $X$ onto $Y$ is a TVS homomorphism,

where a linear map $u\; :\; X\; \backslash to\; Y$ is a topological vector space (TVS) homomorphism if the induced map $\backslash hat\{u\}\; :\; X\; /\; \backslash ker(u)\; \backslash to\; Y$ is a TVS-isomorphism onto its image.

}}

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

{{Math theorem|name=Open mapping theorem{{sfn|Narici|Beckenstein|2011|p=468}}|note=|math_statement=

Let $A\; :\; X\; \backslash to\; Y$ be a surjective linear map from a complete pseudometrizable TVS $X$ onto a TVS $Y$ and suppose that at least one of the following two conditions is satisfied:

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

If $A$ is a closed linear operator then $A$ is an open mapping.

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

}}

Nearly/Almost open linear maps

A linear map $A\; :\; X\; \backslash to\; Y$ between two topological vector spaces (TVSs) is called a {{em|{{visible anchor|nearly open map}}}} (or sometimes, an {{em|almost open map}}) if for every neighborhood $U$ of the origin in the domain, the closure of its image $\backslash operatorname\{cl\}\; A(U)$ is a neighborhood of the origin in $Y.${{sfn|Narici|Beckenstein|2011|pp=466}} 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 $Y,${{sfn|Narici|Beckenstein|2011|pp=466}} but for surjective maps these definitions are equivalent.

A bijective linear map is nearly open if and only if its inverse is continuous.{{sfn|Narici|Beckenstein|2011|pp=466}}

Every surjective linear map from locally convex TVS onto a barrelled TVS is nearly open.{{sfn|Narici|Beckenstein|2011|pp=467}} The same is true of every surjective linear map from a TVS onto a Baire TVS.{{sfn|Narici|Beckenstein|2011|pp=467}}

{{Math theorem|name=Open mapping theorem{{sfn|Narici|Beckenstein|2011|pp=466−468}}|note=|math_statement=

If a closed surjective linear map from a complete pseudometrizable TVS onto a Hausdorff TVS is nearly open then it is open.

}}

{{Math theorem|name=Theorem{{sfn|Narici|Beckenstein|2011|p=469}}|math_statement=

If $A\; :\; X\; \backslash to\; Y$ is a continuous linear bijection from a complete Pseudometrizable topological vector space (TVS) onto a Hausdorff TVS that is a Baire space, then $A\; :\; X\; \backslash to\; Y$ is a homeomorphism (and thus an isomorphism of TVSs).

}}

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

• {{annotated link|Closed graph}}
• {{annotated link|Closed graph theorem}}
• {{annotated link|Closed graph theorem (functional analysis)}}
• {{annotated link|Open mapping theorem (complex analysis)}}
• {{annotated link|Surjection of Fréchet spaces}}
• {{annotated link|Ursescu theorem}}
• {{annotated link|Webbed space}}

{{reflist}}

{{reflist|group=note}}

• {{Adasch Topological Vector Spaces}}
• {{Banach Théorie des Opérations Linéaires}}
• {{Berberian Lectures in Functional Analysis and Operator Theory}}
• {{Bourbaki Topological Vector Spaces}}
• {{Conway A Course in Functional Analysis}}
• {{citation|last=Dieudonné|first=Jean|authorlink=Jean Dieudonné|title=Treatise on Analysis, Volume II|publisher=Academic Press|year=1970}}
• {{Edwards Functional Analysis Theory and Applications}}
• {{Grothendieck Topological Vector Spaces}}
• {{Jarchow Locally Convex Spaces}}
• {{Köthe Topological Vector Spaces I}}
• {{Narici Beckenstein Topological Vector Spaces|edition=2}}
• {{Robertson Topological Vector Spaces}}
• {{Rudin Walter Functional Analysis|edition=1}}
• {{Rudin Walter Functional Analysis|edition=2}}
• {{Schaefer Wolff Topological Vector Spaces|edition=2}}
• {{Swartz An Introduction to Functional Analysis}}
• {{Trèves François Topological vector spaces, distributions and kernels}}
• {{cite web|first=Dietmar|last=Vogt|url=https://www2.math.uni-wuppertal.de/~vogt/vorlesungen/fs.pdf|title=Lectures on Fréchet spaces|year=2000|publisher=Bergische Universität Wuppertal}}
• {{Wilansky Modern Methods in Topological Vector Spaces}}

{{PlanetMath attribution|id=8537|title=Proof of open mapping theorem}}

• {{cite web|url=https://mathoverflow.net/questions/383316/when-is-a-complex-of-banach-spaces-exact-as-condensed-abelian-groups|work=MathOverflow|title=When is a complex of Banach spaces exact as condensed abelian groups?|date=February 6, 2021}}

{{Functional analysis}}

{{Topological vector spaces}}

Category:Articles containing proofs

Category:Theorems in functional analysis