In the branch of mathematics called functional analysis, a complemented subspace of a topological vector space [math]\displaystyle{ X, }[/math] is a vector subspace [math]\displaystyle{ M }[/math] for which there exists some other vector subspace [math]\displaystyle{ N }[/math] of [math]\displaystyle{ X, }[/math] called its (topological) complement in [math]\displaystyle{ X }[/math], such that [math]\displaystyle{ X }[/math] is the direct sum [math]\displaystyle{ M \oplus N }[/math] in the category of topological vector spaces. Formally, topological direct sums strengthen the algebraic direct sum by requiring certain maps be continuous; the result retains many nice properties from the operation of direct sum in finite-dimensional vector spaces. Every finite-dimensional subspace of a Banach space is complemented, but other subspaces may not. In general, classifying all complemented subspaces is a difficult problem, which has been solved only for some well-known Banach spaces.
The concept of a complemented subspace is analogous to, but distinct from, that of a set complement. The set-theoretic complement of a vector subspace is never a complementary subspace.
If [math]\displaystyle{ X }[/math] is a vector space and [math]\displaystyle{ M }[/math] and [math]\displaystyle{ N }[/math] are vector subspaces of [math]\displaystyle{ X }[/math] then there is a well-defined addition map [math]\displaystyle{ \begin{alignat}{4} S :\;&& M \times N &&\;\to \;& X \\ && (m, n) &&\;\mapsto\;& m + n \\ \end{alignat} }[/math] The map [math]\displaystyle{ S }[/math] is a morphism in the category of vector spaces — that is to say, linear.
The vector space [math]\displaystyle{ X }[/math] is said to be the algebraic direct sum (or direct sum in the category of vector spaces) [math]\displaystyle{ M\oplus N }[/math] when any of the following equivalent conditions are satisfied:
When these conditions hold, the inverse [math]\displaystyle{ S^{-1} : X \to M \times N }[/math] is well-defined and can be written in terms of coordinates as[math]\displaystyle{ S^{-1} = \left(P_M, P_N\right)\text{.} }[/math] The first coordinate [math]\displaystyle{ P_M : X \to M }[/math] is called the canonical projection of [math]\displaystyle{ X }[/math] onto [math]\displaystyle{ M }[/math]; likewise the second coordinate is the canonical projection onto [math]\displaystyle{ N. }[/math][4]
Equivalently, [math]\displaystyle{ P_M(x) }[/math] and [math]\displaystyle{ P_N(x) }[/math] are the unique vectors in [math]\displaystyle{ M }[/math] and [math]\displaystyle{ N, }[/math] respectively, that satisfy [math]\displaystyle{ x = P_M(x) + P_N(x)\text{.} }[/math] As maps, [math]\displaystyle{ P_M + P_N = \operatorname{Id}_X, \qquad \ker P_M = N, \qquad \text{ and } \qquad \ker P_N = M }[/math] where [math]\displaystyle{ \operatorname{Id}_X }[/math] denotes the identity map on [math]\displaystyle{ X }[/math].[2]
Suppose that the vector space [math]\displaystyle{ X }[/math] is the algebraic direct sum of [math]\displaystyle{ M\oplus N }[/math]. In the category of vector spaces, finite products and coproducts coincide: algebraically, [math]\displaystyle{ M \oplus N }[/math] and [math]\displaystyle{ M \times N }[/math] are indistinguishable. Given a problem involving elements of [math]\displaystyle{ X }[/math], one can break the elements down into their components in [math]\displaystyle{ M }[/math] and [math]\displaystyle{ N }[/math], because the projection maps defined above act as inverses to the natural inclusion of [math]\displaystyle{ M }[/math] and [math]\displaystyle{ N }[/math] into [math]\displaystyle{ X }[/math]. Then one can solve the problem in the vector subspaces and recombine to form an element of [math]\displaystyle{ X }[/math].
In the category of topological vector spaces, that algebraic decomposition becomes less useful. The definition of a topological vector space requires the addition map [math]\displaystyle{ S }[/math] to be continuous; its inverse [math]\displaystyle{ S^{-1} : X \to M \times N }[/math] may not be.[1] The categorical definition of direct sum, however, requires [math]\displaystyle{ P_M }[/math] and [math]\displaystyle{ P_N }[/math] to be morphisms — that is, continuous linear maps.
The space [math]\displaystyle{ X }[/math] is the topological direct sum of [math]\displaystyle{ M }[/math] and [math]\displaystyle{ N }[/math] if (and only if) any of the following equivalent conditions hold:
The topological direct sum is also written [math]\displaystyle{ X = M \oplus N }[/math]; whether the sum is in the topological or algebraic sense is usually clarified through context.
Every topological direct sum is an algebraic direct sum [math]\displaystyle{ X = M \oplus N }[/math]; the converse is not guaranteed. Even if both [math]\displaystyle{ M }[/math] and [math]\displaystyle{ N }[/math] are closed in [math]\displaystyle{ X }[/math], [math]\displaystyle{ S^{-1} }[/math] may still fail to be continuous. [math]\displaystyle{ N }[/math] is a (topological) complement or supplement to [math]\displaystyle{ M }[/math] if it avoids that pathology — that is, if, topologically, [math]\displaystyle{ X = M \oplus N }[/math]. (Then [math]\displaystyle{ M }[/math] is likewise complementary to [math]\displaystyle{ N }[/math].)[1] Condition 1(d) above implies that any topological complement of [math]\displaystyle{ M }[/math] is isomorphic, as a topological vector space, to the quotient vector space [math]\displaystyle{ X / M }[/math].
[math]\displaystyle{ M }[/math] is called complemented if it has a topological complement [math]\displaystyle{ N }[/math] (and uncomplemented if not). The choice of [math]\displaystyle{ N }[/math] can matter quite strongly: every complemented vector subspace [math]\displaystyle{ M }[/math] has algebraic complements that do not complement [math]\displaystyle{ M }[/math] topologically.
Because a linear map between two normed (or Banach) spaces is bounded if and only if it is continuous, the definition in the categories of normed (resp. Banach) spaces is the same as in topological vector spaces.
The vector subspace [math]\displaystyle{ M }[/math] is complemented in [math]\displaystyle{ X }[/math] if and only if any of the following holds:[1]
If in addition [math]\displaystyle{ X }[/math] is Banach, then an equivalent condition is
For any two topological vector spaces [math]\displaystyle{ X }[/math] and [math]\displaystyle{ Y }[/math], the subspaces [math]\displaystyle{ X \times \{0\} }[/math] and [math]\displaystyle{ \{0\} \times Y }[/math] are topological complements in [math]\displaystyle{ X \times Y }[/math].
Every algebraic complement of [math]\displaystyle{ \overline{\{0\}} }[/math], the closure of [math]\displaystyle{ 0 }[/math], is also a topological complement. This is because [math]\displaystyle{ \overline{\{0\}} }[/math] has the indiscrete topology, and so the algebraic projection is continuous.[6]
If [math]\displaystyle{ X=M\oplus N }[/math] and [math]\displaystyle{ A:X\to Y }[/math] is surjective, then [math]\displaystyle{ Y=AM\oplus AN }[/math].[2]
Suppose [math]\displaystyle{ X }[/math] is Hausdorff and locally convex and [math]\displaystyle{ Y }[/math] a free topological vector subspace: for some set [math]\displaystyle{ I }[/math], we have [math]\displaystyle{ Y\cong\mathbb{K}^I }[/math] (as a t.v.s.). Then [math]\displaystyle{ Y }[/math] is a closed and complemented vector subspace of [math]\displaystyle{ X }[/math].[proof 1] In particular, any finite-dimensional subspace of [math]\displaystyle{ X }[/math] is complemented.[7]
In arbitrary topological vector spaces, a finite-dimensional vector subspace [math]\displaystyle{ Y }[/math] is topologically complemented if and only if for every non-zero [math]\displaystyle{ y\in Y }[/math], there exists a continuous linear functional on [math]\displaystyle{ X }[/math] that separates [math]\displaystyle{ y }[/math] from [math]\displaystyle{ 0 }[/math].[1] For an example in which this fails, see § Fréchet spaces.
Not all finite-codimensional vector subspaces of a TVS are closed, but those that are, do have complements.[7][8]
In a Hilbert space, the orthogonal complement [math]\displaystyle{ M^{\bot} }[/math] of any closed vector subspace [math]\displaystyle{ M }[/math] is always a topological complement of [math]\displaystyle{ M }[/math]. This property characterizes Hilbert spaces within the class of Banach spaces: every infinite dimensional, non-Hilbert Banach space contains a closed uncomplemented subspace.[3]
Let [math]\displaystyle{ X }[/math] be a Fréchet space over the field [math]\displaystyle{ \mathbb{K} }[/math]. Then the following are equivalent:[9]
A complemented (vector) subspace of a Hausdorff space [math]\displaystyle{ X }[/math] is necessarily a closed subset of [math]\displaystyle{ X }[/math], as is its complement.[1][proof 2]
From the existence of Hamel bases, every infinite-dimensional Banach space contains unclosed linear subspaces.[proof 3] Since any complemented subspace is closed, none of those subspaces is complemented.
Likewise, if [math]\displaystyle{ X }[/math] is a complete TVS and [math]\displaystyle{ X / M }[/math] is not complete, then [math]\displaystyle{ M }[/math] has no topological complement in [math]\displaystyle{ X. }[/math][10]
If [math]\displaystyle{ A : X \to Y }[/math] is a continuous linear surjection, then the following conditions are equivalent:
Topological vector spaces admit the following Cantor-Schröder-Bernstein–type theorem:
The "self-splitting" assumptions that [math]\displaystyle{ X = X \oplus X }[/math] and [math]\displaystyle{ Y = Y \oplus Y }[/math] cannot be removed: Tim Gowers showed in 1996 that there exist non-isomorphic Banach spaces [math]\displaystyle{ X }[/math] and [math]\displaystyle{ Y }[/math], each complemented in the other.[11]
Understanding the complemented subspaces of an arbitrary Banach space [math]\displaystyle{ X }[/math] up to isomorphism is a classical problem that has motivated much work in basis theory, particularly the development of absolutely summing operators. The problem remains open for a variety of important Banach spaces, most notably the space [math]\displaystyle{ L_1[0,1] }[/math].[12]
For some Banach spaces the question is closed. Most famously, if [math]\displaystyle{ 1 \leq p \leq \infty }[/math] then the only complemented subspaces of [math]\displaystyle{ \ell_p }[/math] are isomorphic to [math]\displaystyle{ \ell_p, }[/math] and the same goes for [math]\displaystyle{ c_0. }[/math] Such spaces are called prime (when their only infinite-dimensional complemented subspaces are isomorphic to the original). These are not the only prime spaces, however.[12]
The spaces [math]\displaystyle{ L_p[0,1] }[/math] are not prime whenever [math]\displaystyle{ p \in (1, 2) \cup (2, \infty); }[/math] in fact, they admit uncountably many non-isomorphic complemented subspaces.[12]
The spaces [math]\displaystyle{ L_2[0,1] }[/math] and [math]\displaystyle{ L_{\infty}[0,1] }[/math] are isomorphic to [math]\displaystyle{ \ell_2 }[/math] and [math]\displaystyle{ \ell_{\infty}, }[/math] respectively, so they are indeed prime.[12]
The space [math]\displaystyle{ L_1[0,1] }[/math] is not prime, because it contains a complemented copy of [math]\displaystyle{ \ell_1 }[/math]. No other complemented subspaces of [math]\displaystyle{ L_1[0,1] }[/math] are currently known.[12]
An infinite-dimensional Banach space is called indecomposable whenever its only complemented subspaces are either finite-dimensional or -codimensional. Because a finite-codimensional subspace of a Banach space [math]\displaystyle{ X }[/math] is always isomorphic to [math]\displaystyle{ X, }[/math] indecomposable Banach spaces are prime.
The most well-known example of indecomposable spaces are in fact hereditarily indecomposable, which means every infinite-dimensional subspace is also indecomposable.[13]
Let [math]\displaystyle{ f = \left(f_i\right)_{i \in I} : Y \to \mathbb{K}^I }[/math] be a TVS-isomorphism; each [math]\displaystyle{ f_i : Y \to \mathbb{K} }[/math] is a continuous linear functional. By the Hahn–Banach theorem, we may extend each [math]\displaystyle{ f_i }[/math] to a continuous linear functional [math]\displaystyle{ F_i : X \to \mathbb{K} }[/math] on [math]\displaystyle{ X. }[/math] The joint map [math]\displaystyle{ F : X \to \mathbb{K}^I }[/math] is a continuous linear surjection whose restriction to [math]\displaystyle{ Y }[/math] is [math]\displaystyle{ f }[/math]. The composition [math]\displaystyle{ P = f^{-1} \circ F : X \to Y }[/math] is then a continuous continuous projection onto [math]\displaystyle{ Y }[/math].
Original source: https://en.wikipedia.org/wiki/Complemented subspace.
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