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Open mapping theorem (functional analysis)

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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 from one Banach space to another has bounded inverse .

Statement and proof

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Open mapping theorem — [2][3] Let be a surjective continuous linear map between Banach spaces (or more generally Fréchet spaces). Then is an open mapping (that is, if is an open subset, then is open).

The proof here uses the Baire category theorem, and completeness of both and 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 between topological vector spaces is said to be nearly open if, for each neighborhood of zero, the closure 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 between normed spaces is nearly open if the image of is non-meager in . (The continuity is not needed.)

Proof: Shrinking , we can assume is an open ball centered at zero. We have . Thus, some contains an interior point ; that is, for some radius ,

Then for any in with , by linearity, convexity and ,

,

which proves the lemma by dividing by . (The same proof works if are pre-Fréchet spaces.)

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

Lemma (Schauder) — [6][7] Let be a continuous linear map between normed spaces.

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

More precisely, if for some and if is complete, then

where is an open ball with radius and center .

Proof: Let be in and some sequence. We have: . Thus, for each and in , we can find an with and in . Thus, taking , we find an such that

Applying the same argument with , we then find an such that

where we observed . Then so on. Thus, if , we found a sequence such that converges and . Also,

Since , by making small enough, we can achieve . (Again the same proof is valid if 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:

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 factors as

Here, 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, 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

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

Theorem — [6] Let and be Banach spaces, let and denote their open unit balls, and let be a bounded linear operator. If then among the following four statements we have (with the same )

  1. for all = continuous dual of ;
  2. ;
  3. ;
  4. is surjective.

Furthermore, if is surjective then (1) holds for some

Proof: The idea of 1. 2. is to show: and that follows from the Hahn–Banach theorem. 2. 3. is exactly the second lemma in § Statement and proof. Finally, 3. 4. is trivial and 4. 1. easily follows from the open mapping theorem.

Alternatively, 1. implies that is injective and has closed image and then by the closed range theorem, that implies has dense image and closed image, respectively; i.e., 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 be a bounded operator between Banach spaces. Then the following are equivalent:

  1. is open.
  2. is surjective.
  3. There exists a constant such that, for each in , the equation has a solution with .
  4. 3. holds for in some dense subspace of .

Proof: 2. 1. is the usual open mapping theorem.

1. 4.: For some , we have where means an open ball. Then for some in . That is, with .

4. 3.: We can write with in the dense subspace and the sum converging in norm. Then, since is complete, with and is a required solution. Finally, 3. 2. is trivial.

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 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

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

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

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

The completion of X is the space of all sequences that converge to zero, which is a (closed) subspace of the p space(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

is an element of , but is not in the range of . Same reasoning applies to show is also not onto in , for example is not in the range of .

Consequences

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The open mapping theorem has several important consequences:

  • If is a bijective continuous linear operator between the Banach spaces and then the inverse operator is continuous as well (this is called the bounded inverse theorem).[10]
  • If is a linear operator between the Banach spaces and and if for every sequence in with and it follows that then is continuous (the closed graph theorem).[11]
  • Given a bounded operator between normed spaces, if the image of is non-meager and if is complete, then is open and surjective and 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

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Local convexity of or  is not essential to the proof, but completeness is: the theorem remains true in the case when and 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 be a continuous linear operator from a complete pseudometrizable TVS onto a Hausdorff TVS If is nonmeager in then is a (surjective) open map and is a complete pseudometrizable TVS. Moreover, if is assumed to be hausdorff (i.e. a F-space), then 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 is the kernel of then there is a canonical factorization of in the form where is the quotient space (also an F-space) of by the closed subspace The quotient mapping is open, and the mapping is an isomorphism of topological vector spaces.[16]

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

Theorem[17] — Let and be two F-spaces. Then every continuous linear map of onto is a TVS homomorphism, where a linear map is a topological vector space (TVS) homomorphism if the induced map 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 be a surjective linear map from a complete pseudometrizable TVS onto a TVS and suppose that at least one of the following two conditions is satisfied:

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

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

Nearly/Almost open linear maps

A linear map between two topological vector spaces (TVSs) is called a nearly open map (or sometimes, an almost open map) if for every neighborhood of the origin in the domain, the closure of its image is a neighborhood of the origin in [18] Many authors use a different definition of "nearly/almost open map" that requires that the closure of be a neighborhood of the origin in rather than in [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 is a continuous linear bijection from a complete Pseudometrizable topological vector space (TVS) onto a Hausdorff TVS that is a Baire space, then 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

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References

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  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. hdl:1959.13/940334. 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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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

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