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Open mapping theorem (functional analysis) - Wikipedia

  • ️Tue Mar 07 9939

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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 {\displaystyle T} from one Banach space to another has bounded inverse {\displaystyle T^{-1}}.

Statement and proof

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Here, {\displaystyle 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, {\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

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

Proof: The idea of 1. {\displaystyle \Rightarrow } 2. is to show: {\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 {\displaystyle T'} is injective and has closed image and then by the closed range theorem, that implies {\displaystyle T} has dense image and closed image, respectively; i.e., {\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]

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

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

4. {\displaystyle \Rightarrow } 3.: We can write {\displaystyle f=\sum _{0}^{\infty }f_{j}} with {\displaystyle f_{j}} in the dense subspace and the sum converging in norm. Then, since {\displaystyle E} is complete, {\displaystyle u=\sum _{0}^{\infty }u_{j}} with {\displaystyle \|u_{j}\|\leq C\|f_{j}\|} and {\displaystyle Tu_{j}=f_{j}} is a required solution. Finally, 3. {\displaystyle \Rightarrow } 2. is trivial. {\displaystyle \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

{\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

{\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

{\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 {\displaystyle c_{0}} 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

{\displaystyle x=\left(1,{\frac {1}{2}},{\frac {1}{3}},\dots \right),}

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

The open mapping theorem has several important consequences:

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]

Local convexity of {\displaystyle X} or {\displaystyle Y}  is not essential to the proof, but completeness is: the theorem remains true in the case when {\displaystyle X} and {\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 {\displaystyle A:X\to Y} be a continuous linear operator from a complete pseudometrizable TVS {\displaystyle X} onto a Hausdorff TVS {\displaystyle Y.} If {\displaystyle \operatorname {Im} A} is nonmeager in {\displaystyle Y} then {\displaystyle A:X\to Y} is a (surjective) open map and {\displaystyle Y} is a complete pseudometrizable TVS. Moreover, if {\displaystyle X} is assumed to be hausdorff (i.e. a F-space), then {\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 {\displaystyle N} is the kernel of {\displaystyle A,} then there is a canonical factorization of {\displaystyle A} in the form {\displaystyle X\to X/N{\overset {\alpha }{\to }}Y} where {\displaystyle X/N} is the quotient space (also an F-space) of {\displaystyle X} by the closed subspace {\displaystyle N.} The quotient mapping {\displaystyle X\to X/N} is open, and the mapping {\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 {\displaystyle X} and {\displaystyle Y} be two F-spaces. Then every continuous linear map of {\displaystyle X} onto {\displaystyle Y} is a TVS homomorphism, where a linear map {\displaystyle u:X\to Y} is a topological vector space (TVS) homomorphism if the induced map {\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:

Nearly/Almost open linear maps

A linear map {\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 {\displaystyle U} of the origin in the domain, the closure of its image {\displaystyle \operatorname {cl} A(U)} is a neighborhood of the origin in {\displaystyle Y.}[18] Many authors use a different definition of "nearly/almost open map" that requires that the closure of {\displaystyle A(U)} be a neighborhood of the origin in {\displaystyle A(X)} rather than in {\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]

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

  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.

This article incorporates material from Proof of open mapping theorem on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.