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 (in the category of TVSs), the spaces Y and Z are called complements of each other.

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 is satisfied:

  1. the canonical map M × N X defined by (m, n) ↦ m + n is a vector space isomorphism;[1]
  2. the canonical map M × N X is bijective;
  3. MN = { 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 × N X is the canonical map defined by (m, n) ↦ m + n, then the inverse of the canonical map S can be written as where PM : X M and PN : X N are called the canonical projections of X onto M and N, respectively.[1]

Topological direct sum

If X is the algebraic direct sum of M and N and if X is also a TVS then the canonical map S : M × N X defined by (m, n) ↦ m + n is necessarily continuous (since addition is continuous) but its inverse 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) ≠ 0, N := span { n }, and M := ker f, then X is the algebraic direct sum of M and N but not the direct sum of M and N in the category of TVSs.

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 is satisfied:

  1. the canonical map S : M × N X defined by (m, n) ↦ m + n is a TVS-isomorphism;
  2. the canonical map S : M × N X is a bijective open map;
  3. X is the algebraic direct sum of M and N (i.e. the canonical map S : M × N X is bijective) and the inverse of the canonical map, , is continuous;
  4. X is the algebraic direct sum of M and N and the canonical projections PM : X M and PN : X N are continuous (where );
  5. X is the algebraic direct sum of M and N and at least one of the two canonical projections PM : X M and PN : X N is continuous;
  6. X is the algebraic direct sum of M and N and the restriction to N of the canonical projection Q : X X/M (defined by Q(x) := x + M) is a TVS-isomorphism of N onto X/M;[2]
  7. The canonical vector space isomorphism p : N 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:

  1. there exists a vector subspace N of X such that X is the direct sum of M and N in the category of TVSs;
  2. there exists a continuous linear map P : X X with range M such that PP = P;
  3. there exists a continuous linear projection P : X X with range M such that X is the algebraic sum of M and the null space of P.
  4. for every TVS Y, the restriction map is surjective, where R is defined by sending a continuous linear operator u : X Y to the restriction .[2]

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

  1. the kernel of f has a topological complement;
  2. there exists a continuous linear map g : Y X such that fg = IdY, where IdY : Y 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 be a Banach space and a closed subspace of . Then is called a complemented subspace of whenever there exists another closed subspace of such that is isomorphic to the direct sum ; in this case is also a complemented subspace, and and are called complements of each other (in ).

Examples and sufficient conditions

  • 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.[3]
    • In particular, if L is a non-trivial continuous linear functional on X, then the kernel of L is topologically complemented in X.

Properties

  • Every closed subspace of a Banach space (or TVS) is complemented if and only if is isomorphic to a Hilbert space.
  • Every topologically complemented vector subspace of a Hausdorff TVS is necessarily closed. [3]

The decomposition method

Theorem:[4] Let X and Y be TVSs such that X = XX and Y = YY. Suppose that Y contains a complemented copy of X and X contains a complemented copy Y. Then X is TVS-isomorphic to Y.

This 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.[4]

Classifying complemented subspaces of a Banach space

Consider a Banach space . What are the complemented subspaces of , 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 (see below).

For some Banach spaces the question is closed. Most famously, if then the only complemented subspaces of are isomorphic to , and the same goes for . Such spaces are called prime (when their only complemented subspaces are isomorphic to themselves). These aren't the only prime spaces, however, as we see in the next section.

The spaces are not prime whenever ; in fact, they admit uncountably many non-isomorphic complemented subspaces. The spaces and are isomorphic to and , respectively, so they are indeed prime.

The space isn't prime due to the fact that it contains a complemented copy of , however no other complemented subspaces of are currently known. It's undoubtedly one of the most interesting open problems in functional analysis whether 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 is always isomorphic to , 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.

See also

References

  1. Schaefer 1999, pp. 19-24.
  2. Treves 2006, p. 36.
  3. Narici 2011, p. 97.
  4. Narici 2011, pp. 100-101.

Sources

  • Narici, Lawrence (2011). Topological vector spaces. Boca Raton, FL: CRC Press. ISBN 1-58488-866-0. OCLC 144216834.
  • Schaefer, Helmut H.; Wolff, Manfred P. (1999). Topological Vector Spaces. GTM. 8. New York, NY: Springer New York Imprint Springer. ISBN 978-1-4612-7155-0. OCLC 840278135.CS1 maint: ref=harv (link)
  • Trèves, François (2006). Topological vector spaces, distributions and kernels. Mineola, N.Y: Dover Publications. ISBN 978-0-486-45352-1. OCLC 853623322.CS1 maint: ref=harv (link)
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