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Open mapping theorem (functional analysis)

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Condition for a linear operator to be open

This article is about open mapping theorem infunctional analysis. For other results with the same name, seeOpen mapping theorem.

Infunctional analysis, theopen mapping theorem, also known as theBanach–Schauder theorem or theBanach theorem^[1] (named afterStefan Banach andJuliusz Schauder), is a fundamental result that states that if abounded orcontinuous linear operator betweenBanach spaces issurjective then it is anopen map.

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

Statement and proof

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Open mapping theorem—^[2]^[3] Let $T:E\to F$ be a surjective continuous linear map between Banach spaces (or more generallyFréchet spaces). Then $T {\displaystyle T}$ is an open mapping (that is, if $U\subset E$ is an open subset, then $T(U)$ is open).

The proof here uses theBaire category theorem, andcompleteness of both $E {\displaystyle E}$ and $F {\displaystyle F}$ is essential to the theorem. The statement of the theorem is no longer true if either space is assumed to be only anormed vector space; see§ Counterexample.

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 benearly open if, for each neighborhood $U {\displaystyle U}$ 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.

Lemma—^[4]^[5] A linear map $f:E\to F$ between normed spaces is nearly open if the image of $f {\displaystyle f}$ isnon-meager in $F {\displaystyle F}$ . (The continuity is not needed.)

Proof: Shrinking $U {\displaystyle U}$ , we can assume $U {\displaystyle U}$ 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 $y {\displaystyle y}$ ; that is, for some radius $r>0$ ,

B(y,r)\subset {\overline {f(nU)}}.

Then for any $v {\displaystyle v}$ in $F {\displaystyle F}$ 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 $E, F {\displaystyle E,F}$ are pre-Fréchet spaces.)

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

Lemma (Schauder)—^[6]^[7] Let $f:E\to F$ be a continuous linear map between normed spaces.

If $f {\displaystyle f}$ is nearly-open and if $E {\displaystyle E}$ is complete, then $f {\displaystyle f}$ is open and surjective.

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

B(0,\delta )\subset f(B(0,1))

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

Proof: Let $y {\displaystyle y}$ 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 $\epsilon >0$ and $z {\displaystyle z}$ in $F {\displaystyle F}$ , we can find an $x {\displaystyle x}$ with $\|x\|<\delta ^{-1}\|z\|$ and $z {\displaystyle z}$ 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 ^{-1}\|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 ^{-1}c_{1}

where we observed $\|x_{2}\|<\delta ^{-1}\|z\|<\delta ^{-1}c_{1}$ . Then so on. Thus, if $c:=\sum c_{n}<\infty$ , we found a sequence $x_{n}$ such that $x=\sum _{1}^{\infty }x_{n}$ converges and $f(x)=y$ . Also,

\|x\|\leq \sum _{1}^{\infty }\|x_{n}\|\leq \delta ^{-1}\|y\|+\delta ^{-1}c.

Since $\delta ^{-1}\|y\|<1$ , by making $c {\displaystyle c}$ small enough, we can achieve $\|x\|<1$ . $\square$ (Again the same proof is valid if $E, F {\displaystyle E,F}$ 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:

Corollary (Bounded inverse theorem)—^[8] 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 $T:E\to F$ factors as

T:E{\overset {p}{\to }}E/\operatorname {ker} T{\overset {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, $T {\displaystyle T}$ is open then.

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

Transpose formulation

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

Theorem—^[6]Let $X {\displaystyle X}$ and $Y {\displaystyle Y}$ be Banach spaces, let $B_{X}$ and $B_{Y}$ denote their open unit balls, and let $T:X\to Y$ be a bounded linear operator.If $\delta >0$ then among the following four statements we have $(1)\implies (2)\implies (3)\implies (4)$ (with the same $\delta$ )

$\delta \left\|y'\right\|\leq \left\|T'y'\right\|$ for all $y'\in Y'$ = continuous dual of $Y {\displaystyle Y}$ ;
$\delta B_{Y}\subset {\overline {T\left(B_{X}\right)}}$ ;
$\delta B_{Y}\subset {T\left(B_{X}\right)}$ ;
$T {\displaystyle T}$ is surjective.

Furthermore, if $T {\displaystyle T}$ is surjective then (1) holds for some $\delta >0.$

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

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

Quantative formulation

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Terence Tao gives the following quantitative formulation of the theorem:^[9]

Theorem—Let $T:E\to F$ be a bounded operator between Banach spaces. Then the following are equivalent:

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

The proof follows acycle 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 $B {\displaystyle B}$ 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 $E {\displaystyle E}$ is complete, $u=\sum _{0}^{\infty }u_{j}$ with $\|u_{j}\|\leq C\|f_{j}\|$ and $Tu_{j}=f_{j}$ is a required solution.

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

Counterexample

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The open mapping theorem may not hold for normed spaces that are not complete. A quickest way to see this is to note that theclosed graph theorem, a consequence of the open mapping theorem, fails without completeness. But here is a more concrete counterexample. Consider the spaceX ofsequencesx : N → R with only finitely many non-zero terms equipped with thesupremum norm. The mapT : X → X defined by

Tx=\left(x_{1},{\frac {x_{2}}{2}},{\frac {x_{3}}{3}},\dots \right)

is bounded, linear and invertible, butT⁻¹ is unbounded. This does not contradict the bounded inverse theorem sinceX is notcomplete, and thus is not a Banach space. To see that it's not complete, consider the sequence of sequencesx⁽ⁿ⁾ ∈ X given by

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

converges asn → ∞ to the sequencex^(∞) given by

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

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

The completion ofX is thespace $c_{0}$ of all sequences that converge to zero, which is a (closed) subspace of theℓ_p space ℓ_∞(N), which is the space of all bounded sequences. However, in this case, the mapT is not onto, and thus not a bijection. To see this, one need simply note that the sequence

x=\left(1,{\frac {1}{2}},{\frac {1}{3}},\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 $T {\displaystyle T}$ is also not onto in $l^{\infty }$ , for example $x=\left(1,1,1,\dots \right)$ is not in the range of $T {\displaystyle T}$ .

Consequences

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The open mapping theorem has several important consequences:

If $T:X\to Y$ is abijective continuous linear operator between the Banach spaces $X {\displaystyle X}$ and $Y, {\displaystyle Y,}$ then theinverse operator $T^{-1}:Y\to X$ is continuous as well (this is called thebounded inverse theorem).^[10]
If $T:X\to Y$ is a linear operator between the Banach spaces $X {\displaystyle X}$ and $Y, {\displaystyle Y,}$ and if for everysequence $\left(x_{n}\right)$ in $X {\displaystyle X}$ with $x_{n}\to 0$ and $Tx_{n}\to y$ it follows that $y=0,$ then $T {\displaystyle T}$ is continuous (theclosed graph theorem).^[11]
Given a bounded operator $T:E\to F$ between normed spaces, if the image of $T {\displaystyle T}$ isnon-meager and if $E {\displaystyle E}$ is complete, then $T {\displaystyle T}$ is open and surjective and $F {\displaystyle F}$ is complete (to see this, use the two lemmas in the proof of the theorem).^[12]
An exact sequence of Banach spaces (or more generally Fréchet spaces) istopologically exact.
Theclosed range theorem, which says an operator (under some assumption) has closed image if and only if its transpose has closed image (seeclosed 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:^[9]

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 theHahn–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 theBartle–Graves theorem.^[13]^[14]

Generalizations

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Local convexity of $X {\displaystyle X}$ or $Y {\displaystyle Y}$ is not essential to the proof, but completeness is: the theorem remains true in the case when $X {\displaystyle X}$ and $Y {\displaystyle Y}$ areF-spaces. Furthermore, the theorem can be combined with the Baire category theorem in the following manner:

Open mapping theorem for continuous maps^[12]^[15]—Let $A:X\to Y$ be acontinuous linear operator from a completepseudometrizable TVS $X {\displaystyle X}$ onto a Hausdorff TVS $Y . {\displaystyle Y.}$ If $\operatorname {Im} A$ isnonmeager in $Y {\displaystyle Y}$ then $A:X\to Y$ is a (surjective) open map and $Y {\displaystyle Y}$ is a complete pseudometrizable TVS.Moreover, if $X {\displaystyle X}$ is assumed to be hausdorff (i.e. aF-space), then $Y {\displaystyle Y}$ 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 $N {\displaystyle N}$ is thekernel of $A, {\displaystyle A,}$ then there is a canonical factorization of $A {\displaystyle A}$ in the form $X\to X/N{\overset {\alpha }{\to }}Y$ where $X/N$ is thequotient space (also an F-space) of $X {\displaystyle X}$ by theclosed subspace $N . {\displaystyle N.}$ The quotient mapping $X\to X/N$ is open, and the mapping $\alpha$ is anisomorphism oftopological vector spaces.^[16]

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

Theorem^[17]—Let $X {\displaystyle X}$ and $Y {\displaystyle Y}$ be twoF-spaces. Then every continuous linear map of $X {\displaystyle X}$ onto $Y {\displaystyle Y}$ is aTVS homomorphism,where a linear map $u:X\to Y$ is a topological vector space (TVS) homomorphism if the induced map ${\hat {u}}:X/\ker(u)\to Y$ 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^[15]—Let $A:X\to Y$ be a surjectivelinear map from acomplete pseudometrizable TVS $X {\displaystyle X}$ onto a TVS $Y {\displaystyle Y}$ and suppose that at least one of the following two conditions is satisfied:

$Y {\displaystyle Y}$ is aBaire space, or
$X {\displaystyle X}$ islocally convex and $Y {\displaystyle Y}$ is abarrelled space,

If $A {\displaystyle A}$ is aclosed linear operator then $A {\displaystyle A}$ is an open mapping.If $A {\displaystyle A}$ is acontinuous linear operator and $Y {\displaystyle Y}$ is Hausdorff then $A {\displaystyle A}$ is (a closed linear operator and thus also) an open mapping.

Nearly/Almost open linear maps

A linear map $A:X\to Y$ between two topological vector spaces (TVSs) is called anearly open map (or sometimes, analmost open map) if for every neighborhood $U {\displaystyle U}$ of the origin in the domain, the closure of its image $\operatorname {cl} A(U)$ is a neighborhood of the origin in $Y . {\displaystyle Y.}$ ^[18] 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 $Y, {\displaystyle Y,}$ ^[18] but for surjective maps these definitions are equivalent.A bijective linear map is nearly open if and only if its inverse is continuous.^[18]Every surjective linear map fromlocally convex TVS onto abarrelled TVS isnearly open.^[19] The same is true of every surjective linear map from a TVS onto aBaire TVS.^[19]

Open mapping theorem^[20]—If a closed surjective linear map from acomplete pseudometrizable TVS onto a Hausdorff TVS isnearly open then it is open.

Theorem^[21]—If $A:X\to Y$ is a continuous linear bijection from acomplete Pseudometrizable topological vector space (TVS) onto a Hausdorff TVS that is aBaire space, then $A:X\to Y$ is ahomeomorphism (and thus an isomorphism of TVSs).

Webbed spaces are a class oftopological vector spaces for which the open mapping theorem and theclosed graph theorem hold.

References

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^Trèves 2006, p. 166.
^Rudin 1973, Theorem 2.11.
^Vogt 2000, Theorem 1.6.
^Vogt 2000, Lemma 1.4.
^The first part of the proof ofRudin 1991, Theorem 2.11.
^^a ^bRudin 1991, Theorem 4.13.
^Vogt 2000, Lemma 1.5.
^Vogt 2000, Corollary 1.7.
^^a ^bTao, Terence (February 1, 2009)."245B, Notes 9: The Baire category theorem and its Banach space consequences".What's New.
^Rudin 1973, Corollary 2.12.
^Rudin 1973, Theorem 2.15.
^^a ^bRudin 1991, Theorem 2.11.
^Sarnowski, Jarek (October 31, 2020)."Can the inverse operator in Bartle-Graves theorem be linear?".MathOverflow.
^Borwein, J. M.; Dontchev, A. L. (2003). "On the Bartle–Graves theorem".Proceedings of the American Mathematical Society.131 (8):2553–2560.doi:10.1090/S0002-9939-03-07229-0.hdl:1959.13/940334.MR 1974655.
^^a ^bNarici & Beckenstein 2011, p. 468.
^Dieudonné 1970, 12.16.8.
^Trèves 2006, p. 170
^^a ^b ^cNarici & Beckenstein 2011, pp. 466.
^^a ^bNarici & Beckenstein 2011, pp. 467.
^Narici & Beckenstein 2011, pp. 466−468.
^Narici & Beckenstein 2011, p. 469.

Bibliography

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This article incorporates material from Proof of open mapping theorem onPlanetMath, which is licensed under theCreative Commons Attribution/Share-Alike License.

v t e Topological vector spaces (TVSs)
Basic concepts	Banach space Completeness Continuous linear operator Linear functional Fréchet space Linear map Locally convex space Metrizability Operator topologies Topological vector space Vector space
Main results	Anderson–Kadec Banach–Alaoglu Closed graph theorem F. Riesz's Hahn–Banach (hyperplane separation Vector-valued Hahn–Banach) Open mapping (Banach–Schauder) Bounded inverse Uniform boundedness (Banach–Steinhaus)
Maps	Bilinear operator form Linear map Almost open Bounded Continuous Closed Compact Densely defined Discontinuous Topological homomorphism Functional Linear Bilinear Sesquilinear Norm Seminorm Sublinear function Transpose
Types of sets	Absolutely convex/disk Absorbing/Radial Affine Balanced/Circled Banach disks Bounding points Bounded Complemented subspace Convex Convex cone(subset) Linear cone(subset) Extreme point Pre-compact/Totally bounded Prevalent/Shy Radial Radially convex/Star-shaped Symmetric
Set operations	Affine hull (Relative) Algebraic interior (core) Convex hull Linear span Minkowski addition Polar (Quasi) Relative interior
Types of TVSs	Asplund B-complete/Ptak Banach (Countably) Barrelled BK-space (Ultra-) Bornological Brauner Complete Convenient (DF)-space Distinguished F-space FK-AK space FK-space Fréchet tame Fréchet Grothendieck Hilbert Infrabarreled Interpolation space K-space LB-space LF-space Locally convex space Mackey (Pseudo)Metrizable Montel Quasibarrelled Quasi-complete Quasinormed (Polynomially Semi-) Reflexive Riesz Schwartz Semi-complete Smith Stereotype (B Strictly Uniformly) convex (Quasi-) Ultrabarrelled Uniformly smooth Webbed With the approximation property
Category

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Open mapping theorem (functional analysis)

Statement and proof

Transpose formulation

Quantative formulation

Counterexample

Consequences

Generalizations

See also

References

Bibliography

Further reading