Open mapping theorem (functional analysis)

Condition for a linear operator to be open

In functional analysis, the open mapping theorem, also known as the Banach–Schauder theorem or the Banach theorem[1] (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 {\displaystyle T} from one Banach space to another has bounded inverse T 1 {\displaystyle T^{-1}} .

Statement and proof

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

The proof here uses the Baire category theorem, and completeness 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 a normed vector space; see § Counterexample.

The proof is based on the following lemmas, which are also somewhat of independent interest. A linear map f : E F {\displaystyle f:E\to F} between topological vector spaces is said to be nearly open if, for each neighborhood U {\displaystyle U} of zero, the closure f ( U ) ¯ {\displaystyle {\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 F {\displaystyle f:E\to F} between normed spaces is nearly open if the image of f {\displaystyle f} is non-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 ( n N n U ) = n N f ( n U ) {\displaystyle f(E)=f\left(\bigcup _{n\in \mathbb {N} }nU\right)=\bigcup _{n\in \mathbb {N} }f(nU)} . Thus, some f ( n U ) ¯ {\displaystyle {\overline {f(nU)}}} contains an interior point y {\displaystyle y} ; that is, for some radius r > 0 {\displaystyle r>0} ,

B ( y , r ) f ( n U ) ¯ . {\displaystyle B(y,r)\subset {\overline {f(nU)}}.}

Then for any v {\displaystyle v} in F {\displaystyle F} with v < r {\displaystyle \|v\|<r} , by linearity, convexity and ( 1 ) U U {\displaystyle (-1)U\subset U} ,

v = v y + y f ( n U ) ¯ + f ( n U ) ¯ f ( 2 n U ) ¯ {\displaystyle v=v-y+y\in {\overline {f(-nU)}}+{\overline {f(nU)}}\subset {\overline {f(2nU)}}} ,

which proves the lemma by dividing by 2 n {\displaystyle 2n} . {\displaystyle \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 F {\displaystyle 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 , δ ) f ( B ( 0 , 1 ) ) ¯ {\displaystyle B(0,\delta )\subset {\overline {f(B(0,1))}}} for some δ > 0 {\displaystyle \delta >0} and if E {\displaystyle E} is complete, then

B ( 0 , δ ) f ( B ( 0 , 1 ) ) {\displaystyle B(0,\delta )\subset f(B(0,1))}

where B ( x , r ) {\displaystyle 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 , δ ) {\displaystyle B(0,\delta )} and c n > 0 {\displaystyle c_{n}>0} some sequence. We have: B ( 0 , δ ) ¯ f ( B ( 0 , 1 ) ) ¯ {\displaystyle {\overline {B(0,\delta )}}\subset {\overline {f(B(0,1))}}} . Thus, for each ϵ > 0 {\displaystyle \epsilon >0} and z {\displaystyle z} in F {\displaystyle F} , we can find an x {\displaystyle x} with x < δ 1 z {\displaystyle \|x\|<\delta ^{-1}\|z\|} and z {\displaystyle z} in B ( f ( x ) , ϵ ) {\displaystyle B(f(x),\epsilon )} . Thus, taking z = y {\displaystyle z=y} , we find an x 1 {\displaystyle x_{1}} such that

y f ( x 1 ) < c 1 , x 1 < δ 1 y . {\displaystyle \|y-f(x_{1})\|<c_{1},\,\|x_{1}\|<\delta ^{-1}\|y\|.}

Applying the same argument with z = y f ( x 1 ) {\displaystyle z=y-f(x_{1})} , we then find an x 2 {\displaystyle x_{2}} such that

y f ( x 1 ) f ( x 2 ) < c 2 , x 2 < δ 1 c 1 {\displaystyle \|y-f(x_{1})-f(x_{2})\|<c_{2},\,\|x_{2}\|<\delta ^{-1}c_{1}}

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

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

Since δ 1 y < 1 {\displaystyle \delta ^{-1}\|y\|<1} , by making c {\displaystyle c} small enough, we can achieve x < 1 {\displaystyle \|x\|<1} . {\displaystyle \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. {\displaystyle \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 turns follows from that. Indeed, a surjective linear operator T : E F {\displaystyle T:E\to F} factors as

T : E p E / ker T T 0 F . {\displaystyle T:E{\overset {p}{\to }}E/\operatorname {ker} T{\overset {T_{0}}{\to }}F.}

Here, T 0 {\displaystyle 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 {\displaystyle 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.

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

  1. δ y T y {\displaystyle \delta \left\|y'\right\|\leq \left\|T'y'\right\|} for all y Y {\displaystyle y'\in Y'} = continuous dual of Y {\displaystyle Y} ;
  2. δ B Y T ( B X ) ¯ {\displaystyle \delta B_{Y}\subset {\overline {T\left(B_{X}\right)}}} ;
  3. δ B Y T ( B X ) {\displaystyle \delta B_{Y}\subset {T\left(B_{X}\right)}} ;
  4. T {\displaystyle T} is surjective.

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

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

Alternatively, 1. implies that T {\displaystyle T^{*}} is injective and has closed image and then by the closed 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

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

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

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

Proof: 2. {\displaystyle \Rightarrow } 1. is the usual open mapping theorem.

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

4. {\displaystyle \Rightarrow } 3.: We can write f = 0 f j {\displaystyle f=\sum _{0}^{\infty }f_{j}} with f j {\displaystyle f_{j}} in the dense subspace and the sum converging in norm. Then, since E {\displaystyle E} is complete, u = 0 u j {\displaystyle u=\sum _{0}^{\infty }u_{j}} with u j C f j {\displaystyle \|u_{j}\|\leq C\|f_{j}\|} and T u j = f j {\displaystyle Tu_{j}=f_{j}} is a required solution. Finally, 3. {\displaystyle \Rightarrow } 2. is trivial. {\displaystyle \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 = ( x 1 , x 2 2 , x 3 3 , ) {\displaystyle Tx=\left(x_{1},{\frac {x_{2}}{2}},{\frac {x_{3}}{3}},\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 ( n ) = ( 1 , 1 2 , , 1 n , 0 , 0 , ) {\displaystyle x^{(n)}=\left(1,{\frac {1}{2}},\dots ,{\frac {1}{n}},0,0,\dots \right)}

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

x ( ) = ( 1 , 1 2 , , 1 n , ) , {\displaystyle 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 in X.

The completion of X is the space c 0 {\displaystyle c_{0}} of all sequences that converge to zero, which is a (closed) subspace of the ℓp 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 , 1 2 , 1 3 , ) , {\displaystyle x=\left(1,{\frac {1}{2}},{\frac {1}{3}},\dots \right),}

is an element of c 0 {\displaystyle c_{0}} , but is not in the range of T : c 0 c 0 {\displaystyle T:c_{0}\to c_{0}} . Same reasoning applies to show T {\displaystyle T} is also not onto in l {\displaystyle l^{\infty }} , for example x = ( 1 , 1 , 1 , ) {\displaystyle x=\left(1,1,1,\dots \right)} is not in the range of T {\displaystyle T} .

Consequences

The open mapping theorem has several important consequences:

  • If T : X Y {\displaystyle T:X\to Y} is a bijective continuous linear operator between the Banach spaces X {\displaystyle X} and Y , {\displaystyle Y,} then the inverse operator T 1 : Y X {\displaystyle T^{-1}:Y\to X} is continuous as well (this is called the bounded inverse theorem).[10]
  • If T : X Y {\displaystyle T:X\to Y} is a linear operator between the Banach spaces X {\displaystyle X} and Y , {\displaystyle Y,} and if for every sequence ( x n ) {\displaystyle \left(x_{n}\right)} in X {\displaystyle X} with x n 0 {\displaystyle x_{n}\to 0} and T x n y {\displaystyle Tx_{n}\to y} it follows that y = 0 , {\displaystyle y=0,} then T {\displaystyle T} is continuous (the closed graph theorem).[11]
  • Given a bounded operator T : E F {\displaystyle T:E\to F} between normed spaces, if the image of T {\displaystyle T} is non-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) 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:[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 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.[13][14]

Generalizations

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} are F-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 Y {\displaystyle A:X\to Y} be a continuous linear operator from a complete pseudometrizable TVS X {\displaystyle X} onto a Hausdorff TVS Y . {\displaystyle Y.} If Im A {\displaystyle \operatorname {Im} A} is nonmeager in Y {\displaystyle Y} then A : X Y {\displaystyle 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. a F-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 the kernel of A , {\displaystyle A,} then there is a canonical factorization of A {\displaystyle A} in the form X X / N α Y {\displaystyle X\to X/N{\overset {\alpha }{\to }}Y} where X / N {\displaystyle X/N} is the quotient space (also an F-space) of X {\displaystyle X} by the closed subspace N . {\displaystyle N.} The quotient mapping X X / N {\displaystyle X\to X/N} is open, and the mapping α {\displaystyle \alpha } is an isomorphism of topological 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 two F-spaces. Then every continuous linear map of X {\displaystyle X} onto Y {\displaystyle Y} is a TVS homomorphism, where a linear map u : X Y {\displaystyle u:X\to Y} is a topological vector space (TVS) homomorphism if the induced map u ^ : X / ker ( u ) Y {\displaystyle {\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 Y {\displaystyle A:X\to Y} be a surjective linear map from a complete pseudometrizable TVS X {\displaystyle X} onto a TVS Y {\displaystyle Y} and suppose that at least one of the following two conditions is satisfied:

  1. Y {\displaystyle Y} is a Baire space, or
  2. X {\displaystyle X} is locally convex and Y {\displaystyle Y} is a barrelled space,

If A {\displaystyle A} is a closed linear operator then A {\displaystyle A} is an open mapping. If A {\displaystyle A} is a continuous 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 Y {\displaystyle A:X\to Y} between two topological vector spaces (TVSs) is called a nearly open map (or sometimes, an almost open map) if for every neighborhood U {\displaystyle U} of the origin in the domain, the closure of its image cl A ( U ) {\displaystyle \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 ) {\displaystyle A(U)} be a neighborhood of the origin in A ( X ) {\displaystyle 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 from locally convex TVS onto a barrelled TVS is nearly open.[19] The same is true of every surjective linear map from a TVS onto a Baire TVS.[19]

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

Theorem[21] — If A : X Y {\displaystyle A:X\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 Y {\displaystyle A:X\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.

See also

  • Almost open linear map – Map that satisfies a condition similar to that of being an open map.Pages displaying short descriptions of redirect targets
  • Bounded inverse theorem – Condition for a linear operator to be openPages displaying short descriptions of redirect targets
  • Closed graph – Graph of a map closed in the product spacePages displaying short descriptions of redirect targets
  • Closed graph theorem – Theorem relating continuity to graphs
  • Closed graph theorem (functional analysis) – Theorems connecting continuity to closure of graphs
  • Open mapping theorem (complex analysis) – Theorem that holomorphic functions on complex domains are open mapsPages displaying wikidata descriptions as a fallback
  • Surjection of Fréchet spaces – Characterization of surjectivity
  • Ursescu theorem – Generalization of closed graph, open mapping, and uniform boundedness theorem
  • Webbed space – Space where open mapping and closed graph theorems hold

References

  1. ^ Trèves 2006, p. 166.
  2. ^ Rudin 1973, Theorem 2.11.
  3. ^ Vogt 2000, Theorem 1.6.
  4. ^ Vogt 2000, Lemma 1.4.
  5. ^ The first part of the proof of Rudin 1991, Theorem 2.11.
  6. ^ a b Rudin 1991, Theorem 4.13.
  7. ^ Vogt 2000, Lemma 1.5.
  8. ^ Vogt 2000, Corollary 1.7.
  9. ^ a b Tao, Terence (February 1, 2009). "245B, Notes 9: The Baire category theorem and its Banach space consequences". What's New.
  10. ^ Rudin 1973, Corollary 2.12.
  11. ^ Rudin 1973, Theorem 2.15.
  12. ^ a b Rudin 1991, Theorem 2.11.
  13. ^ Sarnowski, Jarek (October 31, 2020). "Can the inverse operator in Bartle-Graves theorem be linear?". MathOverflow.
  14. ^ 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. MR 1974655.
  15. ^ a b Narici & Beckenstein 2011, p. 468.
  16. ^ Dieudonné 1970, 12.16.8.
  17. ^ Trèves 2006, p. 170
  18. ^ a b c Narici & Beckenstein 2011, pp. 466.
  19. ^ a b Narici & Beckenstein 2011, pp. 467.
  20. ^ Narici & Beckenstein 2011, pp. 466−468.
  21. ^ Narici & Beckenstein 2011, p. 469.

Bibliography

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  • Banach, Stefan (1932). Théorie des Opérations Linéaires [Theory of Linear Operations] (PDF). Monografie Matematyczne (in French). Vol. 1. Warszawa: Subwencji Funduszu Kultury Narodowej. Zbl 0005.20901. Archived from the original (PDF) on 2014-01-11. Retrieved 2020-07-11.
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This article incorporates material from Proof of open mapping theorem on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.

Further reading

  • "When is a complex of Banach spaces exact as condensed abelian groups?". MathOverflow. February 6, 2021.
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