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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