Complemented subspace

In the branch of mathematics called functional analysis, when a topological vector space (TVS; e.g., a normed space or a Banach space) $X$ admits a direct sum decomposition $X ≅ Y \oplus Z$ (in the category of TVSs), the spaces $Y$ and $Z$ are called complements of each other. This happens if and only if the addition map $Y \times Z \to X$ , which is defined by $(y, z) \mapsto y + z$ , is a homeomorphism. Note that while this addition map is always continuous, it may fail to be a homeomorphism, which is why this definition is needed.

The concept of a complemented subspace in functional analysis should never be confused with that of a set complement in set theory, which is completely different.

Definition

Algebraic direct sum

If $X$ is a vector space and $M$ and $N$ are vector subspaces of $X$ then $X$ is the algebraic direct sum of $M$ and $N$ or the direct sum of $M$ and $N$ in the category of vector spaces if any of the following equivalent conditions are satisfied:

the canonical map $M \times N \to X$ defined by $(m, n) \mapsto m + n$ is a vector space isomorphism;[1]
the canonical map $M \times N \to X$ is bijective;
$M \cap N = { 0$ } and $M + N = X$ ; in this case $N$ is called an algebraic complementary or algebraic supplementary to $M$ in $X$ .

If $X$ is the algebraic direct sum of $M$ and $N$ and if $S : M \times N \to X$ is the canonical map defined by $(m, n) \mapsto m + n$ , then the inverse of the canonical map S can be written as $S -1 := (P M, P N) : X \to M \times N$ where $P M : X \to M$ and $P N : X \to N$ are called the canonical projections of $X$ onto $M$ and $N$ , respectively.[1] Note that $ker P M = N$ , $ker P N = M$ , and $x = P M (x) + P N (x)$ for all $x \in X$ .

Topological direct sum

If $X$ is the algebraic direct sum of $M$ and $N$ and if $X$ is also a topological vector space (TVS) then the canonical map $S : M \times N \to X$ defined by $(m, n) \mapsto m + n$ is necessarily continuous (since addition is continuous) but its inverse $S -1 : X \to M \times N$ may fail to be continuous; when it is continuous then $X$ is the direct sum of $M$ and $N$ in the category of TVSs. If $f$ is a discontinuous linear functional on a locally convex Hausdorff TVS $X$ , n is any element of $X$ such that $f (n) \neq 0$ , $N := span {n$ }, and $M := ker f$ , then $X$ is the algebraic direct sum of $M$ and $N$ but not the topological direct sum of $M$ and $N$ .

If $X$ is a topological vector space (TVS) and if $M$ and $N$ are vector subspaces of $X$ , then $X$ is the topological direct sum of $M$ and $N$ or the direct sum of $M$ and $N$ in the category of TVSs if any of the following equivalent conditions are satisfied:

the canonical map $S : M \times N \to X$ defined by $(m, n) \mapsto m + n$ is a TVS-isomorphism;
the canonical map $S : M \times N \to X$ is a bijective open map;
X is the algebraic direct sum of $M$ and $N$ (i.e. the canonical map $S : M \times N \to X$ is bijective) and the inverse of the canonical map, $S -1 : X \to M \times N$ , is continuous;
X is the algebraic direct sum of $M$ and $N$ and the canonical projections $P M : X \to M$ and $P N : X \to N$ are continuous (where $S -1 := (P M, P N$ );
X is the algebraic direct sum of $M$ and $N$ and at least one of the two canonical projections $P M : X \to M$ and $P N : X \to N$ is continuous;
X is the algebraic direct sum of $M$ and $N$ and the restriction to $N$ of the canonical quotient map $Q : X \to X / M$ (defined by $Q (x) := x + M$ ) is a TVS-isomorphism of $N$ onto $X / M$ ;[2]
The canonical vector space isomorphism $p : N \to X / M$ is bicontinuous (meaning $p$ and its inverse are continuous);
Note that this implies that any topological complement of $M$ in $X$ is TVS-isomorphic to $X / M$ .

in this case $N$ is called a (topological) complement, (topological) complementary, or (topological) supplementary to $M$ in $X$ and we say that $M$ (as well as $N$ ) is a (topologically) complemented subspace of $X$ .

Complemented subspaces

Let $X$ be a topological vector space and let $M$ be a vector subspace of $X$ . We say that $M$ is a complemented subspace of $X$ if any of the following equivalent conditions hold:

there exists a vector subspace $N$ of $X$ such that $X$ is the direct sum of $M$ and $N$ in the category of TVSs (or said differently, $M$ is topologically complemented in $X$ );
there exists a continuous linear map $P : X \to X$ with range $M$ such that $P \circ P = P$ ;
Such a map is called a continuous linear projection onto $M$ .
there exists a continuous linear projection $P : X \to X$ with range $M$ such that $X$ is the algebraic sum of $M$ and the null space of P.
for every TVS $Y$ , the restriction map $R : L(X; Y) \to L(M; Y)$ is surjective, where $R$ is defined by sending a continuous linear operator $u : X \to Y$ to the restriction $R(u):=u{\big \vert }_{M}:M\to Y$ .[2]

It can be shown that if $f : X \to Y$ is a continuous linear bijection between two TVSs, then the following conditions are equivalent:

the kernel of $f$ has a topological complement;
there exists a continuous linear map $g : Y \to X$ such that $f \circ g = Id Y$ , where $Id Y : Y \to Y$ is the identity map.[2]

For Banach spaces

For the special case of a Banach space, the following definition is equivalent to the one given above for TVSs.

Let $X$ be a Banach space and $Y$ a closed subspace of $X$ . Then $Y$ is called a complemented subspace of $X$ whenever there exists another closed subspace $Z$ of $X$ such that $X$ is isomorphic to the direct sum $Y \oplus Z$ ; in this case $Z$ is also a complemented subspace, and $Y$ and $Z$ are called complements of each other (in $X$ ).

Examples and sufficient conditions

If X and Y are TVS then X × { 0 } and { 0 } × Y are topological complements in X × Y (since the definition of the product topology implies that the addition map is open).
- Note if neither $X$ nor $Y$ are Hausdorff then neither $X \times { 0$ } nor ${ 0 } \times Y$ is a closed subspace of $X \times Y$ , but they are nevertheless topological complements.[note 1]
If X is a TVS then every vector subspace of X that is an algebraic complement of cl { 0 } is a topological complement of cl { 0 } (this is because cl { 0 } has the indiscrete topology).[3]
- Thus every TVS is the product of a Hausdorff TVS and a TVS with the indiscrete topology.[3]
Suppose $X$ is a Hausdorff locally convex TVS over the field $𝕂$ and $Y$ is a vector subspace of $X$ that is TVS-isomorphic to $𝕂 I$ for some set $I$ . Then $Y$ is a closed and complemented vector subspace of $X$ (see footnote for proof).[note 2]
Every finite-dimensional or finite-codimensional subspace of a Banach space (or TVS) is complemented.
If M is a proper maximal closed vector subspace of a TVS X and if N is any algebraic complement of M in X, then N is a topological complement of M in X.[4]
- In particular, if $L$ is a non-trivial continuous linear functional on $X$ , then the kernel of $L$ is topologically complemented in $X$ .

Non-complemented subspaces

Let $X$ be a non-normable Fréchet space on which there exists a continuous norm. Then $X$ contains a closed vector subspace that has no topological complement.[5]
If $X$ is a complete TVS and $M$ is a closed vector subspace of $X$ such that $X / M$ is not complete, then $H$ does not have a topological complement in $X$ .[5]

Properties

Every closed subspace of a Banach space (or TVS) $X$ is complemented if and only if $X$ is isomorphic to a Hilbert space.
Every topologically complemented vector subspace of a Hausdorff TVS is necessarily closed.[4]

The decomposition method

Theorem[6] — Let $X$ and $Y$ be TVSs such that $X = X \oplus X$ and $Y = Y \oplus Y$ . Suppose that $Y$ contains a complemented copy of $X$ and $X$ contains a complemented copy $Y$ . Then $X$ is TVS-isomorphic to $Y$ .

This theorem led to the Schroeder-Bernstein problem: If $X$ and $Y$ are Banach spaces and each is TVS-isomorphic to a complemented subspace of the other, then is $X$ TVS-isomorphic to $Y$ ? In 1996, Gowers provided a negative answer to this question.[6]

Classifying complemented subspaces of a Banach space

Consider a Banach space $X$ . What are the complemented subspaces of $X$ , up to isomorphism? Answering this question is one of the complemented subspace problems that remains open for a variety of important Banach spaces, most notably the space $L_{1}[0,1]$ (see below).

For some Banach spaces the question is closed. Most famously, if $1 \leq p \leq \infty$ then the only complemented subspaces of $\ell _{p}$ are isomorphic to $\ell _{p}$ , and the same goes for $c_{0}$ . Such spaces are called prime (when their only complemented subspaces are isomorphic to themselves). These aren't the only prime spaces, however.

The spaces $L_{p}[0,1]$ are not prime whenever $p\in (1,2)\cup (2,\infty )$ ; in fact, they admit uncountably many non-isomorphic complemented subspaces. The spaces $L_{2}[0,1]$ and $L_{\infty }[0,1]$ are isomorphic to $\ell _{2}$ and $\ell _{\infty }$ , respectively, so they are indeed prime.

The space $L_{1}[0,1]$ isn't prime due to the fact that it contains a complemented copy of $\ell _{1}$ , however no other complemented subspaces of $L_{1}[0,1]$ are currently known. It's undoubtedly one of the most interesting open problems in functional analysis whether $L_{1}[0,1]$ admits other complemented subspaces.

Indecomposable Banach spaces

An infinite-dimensional Banach space is called indecomposable whenever its only complemented subspaces are either finite-dimensional or finite-codimensional. Due to the fact that a finite-codimensional subspace of a Banach space $X$ is always isomorphic to $X$ , that makes indecomposable Banach spaces prime.

The most well-known example of indecomposable spaces are in fact hereditarily indecomposable, which means every infinite-dimensional subspace is also indecomposable. Such spaces are fairly nasty, and have been constructed specifically to defy the sort of behavior we typically desire in a Banach space.

Notes

Since $X$ and $Y$ aren't Hausdorff, neither is $X \times Y$ . Since the set ${ 0$ } isn't closed in $X \times Y$ , there is no requirement that topological complements in $X \times Y$ be closed.
Since $𝕂 I$ is a complete TVS, so is Y and since any complete subset of a Hausdorff TVS is closed, Y is a closed subset of $X$ . Let $f = (f i) i \in I : Y \to 𝕂 I$ be a TVS isomorphism, where each $f i : Y \to 𝕂$ is a continuous linear functional. By the Hahn–Banach theorem, we may extend each $f i$ to a continuous linear functional $F i : X \to 𝕂$ on $X$ . Let $F := (F i) i \in I : X \to 𝕂 I$ so $F$ is a continuous linear surjection such that its restriction to $Y$ is $F | Y = (F i | Y) i \in I = (f i) i \in I = f$ . It follows that if we define $P := f -1 \circ F : X \to Y$ then the restriction of this continuous linear map $P | Y : Y \to Y$ is the identity map on $Y$ so that $P$ is a continuous projection onto $Y$ (i.e. $P \circ P = P$ ). Thus $Y$ is complemented in $X$ and $X = Y \oplus ker P$ in the category of TVSs.

References

Schaefer & Wolff 1999, pp. 19-24.
Trèves 2006, p. 36.
Wilansky 2013, p. 63.
Narici & Beckenstein 2011, p. 97.
Schaefer & Wolff 1999, pp. 190-202.
Narici & Beckenstein 2011, pp. 100-101.

Bibliography

Narici, Lawrence; Beckenstein, Edward (2011). Topological Vector Spaces. Pure and applied mathematics (Second ed.). Boca Raton, FL: CRC Press. ISBN 978-1584888666. OCLC 144216834.
Schaefer, Helmut H.; Wolff, Manfred P. (1999). Topological Vector Spaces. GTM. 8 (Second ed.). New York, NY: Springer New York Imprint Springer. ISBN 978-1-4612-7155-0. OCLC 840278135.
Trèves, François (2006) [1967]. Topological Vector Spaces, Distributions and Kernels. Mineola, N.Y.: Dover Publications. ISBN 978-0-486-45352-1. OCLC 853623322.
Wilansky, Albert (2013). Modern Methods in Topological Vector Spaces. Mineola, New York: Dover Publications, Inc. ISBN 978-0-486-49353-4. OCLC 849801114.

This article is issued from Wikipedia. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.

[3] Since $X$ and $Y$ aren't Hausdorff, neither is $X \times Y$ . Since the set ${ 0$ } isn't closed in $X \times Y$ , there is no requirement that topological complements in $X \times Y$ be closed.

[5] Since $𝕂 I$ is a complete TVS, so is Y and since any complete subset of a Hausdorff TVS is closed, Y is a closed subset of $X$ . Let $f = (f i) i \in I : Y \to 𝕂 I$ be a TVS isomorphism, where each $f i : Y \to 𝕂$ is a continuous linear functional. By the Hahn–Banach theorem, we may extend each $f i$ to a continuous linear functional $F i : X \to 𝕂$ on $X$ . Let $F := (F i) i \in I : X \to 𝕂 I$ so $F$ is a continuous linear surjection such that its restriction to $Y$ is $F | Y = (F i | Y) i \in I = (f i) i \in I = f$ . It follows that if we define $P := f -1 \circ F : X \to Y$ then the restriction of this continuous linear map $P | Y : Y \to Y$ is the identity map on $Y$ so that $P$ is a continuous projection onto $Y$ (i.e. $P \circ P = P$ ). Thus $Y$ is complemented in $X$ and $X = Y \oplus ker P$ in the category of TVSs.

[FOOTNOTESchaeferWolff199919-24-1] Schaefer & Wolff 1999, pp. 19-24.

[FOOTNOTETrèves200636-2] Trèves 2006, p. 36.

[FOOTNOTEWilansky201363-4] Wilansky 2013, p. 63.

[FOOTNOTENariciBeckenstein201197-6] Narici & Beckenstein 2011, p. 97.

[FOOTNOTESchaeferWolff1999190-202-7] Schaefer & Wolff 1999, pp. 190-202.

[FOOTNOTENariciBeckenstein2011100-101-8] Narici & Beckenstein 2011, pp. 100-101.

Topological vector spaces (TVSs)
Basic concepts	Banach space Continuous linear operator Functionals Hilbert space Linear operators Locally convex space Homomorphism Topological vector space Vector space
Main results	Closed graph theorem F. Riesz's theorem Hahn–Banach (hyperplane separation Vector-valued Hahn–Banach) Open mapping (Banach–Schauder) (Bounded inverse) Uniform boundedness (Banach–Steinhaus)
Maps	Almost open Bilinear (form operator) and Sesquilinear forms Closed Compact operator Continuous and Discontinuous Linear maps Densely defined Homomorphism Functionals Norm Operator Seminorm Sublinear 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 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 (Ultra-) Bornological Brauner Complete (DF)-space Distinguished F-space Fréchet (tame Fréchet) Grothendieck Hilbert Infrabarreled Interpolation 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