Complemented subspace

In the branch of mathematics called functional analysis, a complemented subspace of a topological vector space ${\displaystyle X,}$ is a

${\displaystyle M}$ for which there exists some other vector subspace ${\displaystyle N}$ of ${\displaystyle X,}$ called its (topological) complement in ${\displaystyle X}$, such that ${\displaystyle X}$ is the direct sum ${\displaystyle M\oplus N}$ 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.

Preliminaries: definitions and notation

If ${\displaystyle X}$ is a vector space and ${\displaystyle M}$ and ${\displaystyle N}$ are

of ${\displaystyle X}$ then there is a well-defined addition map The map ${\displaystyle S}$ is a .

Algebraic direct sum

The vector space ${\displaystyle X}$ is said to be the algebraic direct sum (or direct sum in the category of vector spaces) ${\displaystyle M\oplus N}$ when any of the following equivalent conditions are satisfied:

1. The addition map ${\displaystyle S:M\times N\to X}$ is a
   vector space isomorphism.^[1][2]
2. The addition map is bijective.
3. ${\displaystyle M\cap N=\{0\}}$ and ${\displaystyle M+N=X}$; in this case ${\displaystyle N}$ is called an algebraic complement or supplement to ${\displaystyle M}$ in ${\displaystyle X}$ and the two subspaces are said to be complementary or supplementary.[2]^[3]

When these conditions hold, the inverse ${\displaystyle S^{-1}:X\to M\times N}$ is well-defined and can be written in terms of coordinates as

The first coordinate ${\displaystyle P_{M}:X\to M}$ is called the canonical projection of ${\displaystyle X}$ onto ${\displaystyle M}$; likewise the second coordinate is the canonical projection onto ${\displaystyle N.}$^[4]

Equivalently, ${\displaystyle P_{M}(x)}$ and ${\displaystyle P_{N}(x)}$ are the unique vectors in ${\displaystyle M}$ and ${\displaystyle N,}$ respectively, that satisfy

As maps, where ${\displaystyle \operatorname {Id} _{X}}$ denotes the on ${\displaystyle X}$.^[2]

Motivation

Suppose that the vector space ${\displaystyle X}$ is the algebraic direct sum of ${\displaystyle M\oplus N}$. In the category of vector spaces, finite products and coproducts coincide: algebraically, ${\displaystyle M\oplus N}$ and ${\displaystyle M\times N}$ are indistinguishable. Given a problem involving elements of ${\displaystyle X}$, one can break the elements down into their components in ${\displaystyle M}$ and ${\displaystyle N}$, because the projection maps defined above act as inverses to the natural inclusion of ${\displaystyle M}$ and ${\displaystyle N}$ into ${\displaystyle X}$. Then one can solve the problem in the vector subspaces and recombine to form an element of ${\displaystyle X}$.

In the category of topological vector spaces, that algebraic decomposition becomes less useful. The definition of a topological vector space requires the addition map ${\displaystyle S}$ to be continuous; its inverse ${\displaystyle S^{-1}:X\to M\times N}$ may not be.^[1] The categorical definition of direct sum, however, requires ${\displaystyle P_{M}}$ and ${\displaystyle P_{N}}$ to be morphisms — that is, continuous linear maps.

The space ${\displaystyle X}$ is the topological direct sum of ${\displaystyle M}$ and ${\displaystyle N}$ if (and only if) any of the following equivalent conditions hold:

1. The addition map ${\displaystyle S:M\times N\to X}$ is a
   TVS-isomorphism (that is, a surjective linear homeomorphism).^[1]
2. ${\displaystyle X}$ is the algebraic direct sum of ${\displaystyle M}$ and ${\displaystyle N}$ and also any of the following equivalent conditions:
   1. The inverse of the addition map ${\displaystyle S^{-1}:X\to M\times N}$ is continuous.
   2. Both canonical projections ${\displaystyle P_{M}:X\to M}$ and ${\displaystyle P_{N}:X\to N}$ are continuous.
   3. At least one of the canonical projections ${\displaystyle P_{M}}$ and ${\displaystyle P_{N}}$ is continuous.
   4. The canonical
      quotient map
      ${\displaystyle p:N\to X/M;p(n)=n+M}$ is an isomorphism of topological vector spaces (i.e. a linear homeomorphism).[2]
3. ${\displaystyle X}$ is the direct sum of ${\displaystyle M}$ and ${\displaystyle N}$ in the category of topological vector spaces.
4. The map ${\displaystyle S}$ is
   open
   .
5. When considered as additive topological groups, ${\displaystyle X}$ is the topological direct sum of the subgroups ${\displaystyle M}$ and ${\displaystyle N.}$

The topological direct sum is also written ${\displaystyle X=M\oplus N}$; whether the sum is in the topological or algebraic sense is usually clarified through

.

Definition

Every topological direct sum is an algebraic direct sum ${\displaystyle X=M\oplus N}$; the converse is not guaranteed. Even if both ${\displaystyle M}$ and ${\displaystyle N}$ are closed in ${\displaystyle X}$, ${\displaystyle S^{-1}}$ may still fail to be continuous. ${\displaystyle N}$ is a (topological) complement or supplement to ${\displaystyle M}$ if it avoids that pathology — that is, if, topologically, ${\displaystyle X=M\oplus N}$. (Then ${\displaystyle M}$ is likewise complementary to ${\displaystyle N}$.)^[1] Condition 2(d) above implies that any topological complement of ${\displaystyle M}$ is isomorphic, as a topological vector space, to the

${\displaystyle X/M}$.

${\displaystyle M}$ is called complemented if it has a topological complement ${\displaystyle N}$ (and uncomplemented if not). The choice of ${\displaystyle N}$ can matter quite strongly: every complemented vector subspace ${\displaystyle M}$ has algebraic complements that do not complement ${\displaystyle M}$ topologically.

Because a

) spaces is the same as in topological vector spaces.

Equivalent characterizations

The vector subspace ${\displaystyle M}$ is complemented in ${\displaystyle X}$ if and only if any of the following holds:[1]

• There exists a continuous linear map ${\displaystyle P_{M}:X\to X}$ with image ${\displaystyle P_{M}(X)=M}$ such that ${\displaystyle P\circ P=P}$. That is, ${\displaystyle P_{M}}$ is a continuous linear projection onto ${\displaystyle M}$. (In that case, algebraically ${\displaystyle X=M\oplus \ker {P}}$, and it is the continuity of ${\displaystyle P_{M}}$ that implies that this is a complement.)
• For every TVS ${\displaystyle Y,}$ the restriction map ${\displaystyle R:L(X;Y)\to L(M;Y);R(u)=u|_{M}}$ is surjective.^[5]

If in addition ${\displaystyle X}$ is Banach, then an equivalent condition is

• ${\displaystyle M}$ is closed in ${\displaystyle X}$, there exists another closed subspace ${\displaystyle N\subseteq X}$, and ${\displaystyle S}$ is an
  abstract direct sum
  ${\displaystyle M\oplus N}$ to ${\displaystyle X}$.

Examples

• If ${\displaystyle Y}$ is a measure space and ${\displaystyle X\subseteq Y}$ has positive measure, then ${\displaystyle L^{p}(X)}$ is complemented in ${\displaystyle L^{p}(Y)}$.
• ${\displaystyle c_{0}}$, the space of sequences converging to ${\displaystyle 0}$, is complemented in ${\displaystyle c}$, the space of convergent sequences.
• By
  Lebesgue decomposition
  , ${\displaystyle L^{1}([0,1])}$ is complemented in ${\displaystyle \mathrm {rca} ([0,1])\cong C([0,1])^{*}}$.

Sufficient conditions

For any two topological vector spaces ${\displaystyle X}$ and ${\displaystyle Y}$, the subspaces ${\displaystyle X\times \{0\}}$ and ${\displaystyle \{0\}\times Y}$ are topological complements in ${\displaystyle X\times Y}$.

Every algebraic complement of ${\displaystyle {\overline {\{0\}}}}$, the closure of ${\displaystyle 0}$, is also a topological complement. This is because ${\displaystyle {\overline {\{0\}}}}$ has the

If ${\displaystyle X=M\oplus N}$ and ${\displaystyle A:X\to Y}$ is surjective, then ${\displaystyle Y=AM\oplus AN}$.[2]

Finite dimension

Suppose ${\displaystyle X}$ is Hausdorff and locally convex and ${\displaystyle Y}$ a

: for some set ${\displaystyle I}$, we have ${\displaystyle Y\cong \mathbb {K} ^{I}}$ (as a t.v.s.). Then ${\displaystyle Y}$ is a closed and complemented vector subspace of ${\displaystyle X}$.^{[proof 1]} In particular, any finite-dimensional subspace of ${\displaystyle X}$ is complemented.^[7]

In arbitrary topological vector spaces, a finite-dimensional vector subspace ${\displaystyle Y}$ is topologically complemented if and only if for every non-zero ${\displaystyle y\in Y}$, there exists a continuous linear functional on ${\displaystyle X}$ that separates ${\displaystyle y}$ from ${\displaystyle 0}$.^[1] For an example in which this fails, see § Fréchet spaces.

Finite codimension

Not all finite-codimensional vector subspaces of a TVS are closed, but those that are, do have complements.^[7][8]

Hilbert spaces

In a Hilbert space, the orthogonal complement ${\displaystyle M^{\bot }}$ of any closed vector subspace ${\displaystyle M}$ is always a topological complement of ${\displaystyle M}$. This property characterizes Hilbert spaces within the class of Banach spaces: every infinite dimensional, non-Hilbert Banach space contains a closed uncomplemented subspace, a deep theorem of Joram Lindenstrauss and Lior Tzafriri.^[9][3]

Fréchet spaces

Let ${\displaystyle X}$ be a Fréchet space over the field ${\displaystyle \mathbb {K} }$. Then the following are equivalent:^[10]

1. ${\displaystyle X}$ is not normable (that is, any continuous norm does not generate the topology)
2. ${\displaystyle X}$ contains a vector subspace TVS-isomorphic to ${\displaystyle \mathbb {K} ^{\mathbb {N} }.}$
3. ${\displaystyle X}$ contains a complemented vector subspace TVS-isomorphic to ${\displaystyle \mathbb {K} ^{\mathbb {N} }}$.

Properties; examples of uncomplemented subspaces

A complemented (vector) subspace of a Hausdorff space ${\displaystyle X}$ is necessarily a closed subset of ${\displaystyle X}$, as is its complement.^{[1][proof 2]}

From the existence of

Since any complemented subspace is closed, none of those subspaces is complemented.

Likewise, if ${\displaystyle X}$ is a complete TVS and ${\displaystyle X/M}$ is not complete, then ${\displaystyle M}$ has no topological complement in ${\displaystyle X.}$^[11]

Applications

If ${\displaystyle A:X\to Y}$ is a continuous linear

, then the following conditions are equivalent:

1. The kernel of ${\displaystyle A}$ has a topological complement.
2. There exists a "right inverse": a continuous linear map ${\displaystyle B:Y\to X}$ such that ${\displaystyle AB=\mathrm {Id} _{Y}}$, where ${\displaystyle \operatorname {Id} _{Y}:Y\to Y}$ is the identity map.^[5]

(Note: This claim is an erroneous exercise given by Trèves. Let ${\displaystyle X}$ and ${\displaystyle Y}$ both be ${\displaystyle \mathbb {R} }$ where ${\displaystyle X}$ is endowed with the usual topology, but ${\displaystyle Y}$ is endowed with the trivial topology. The identity map ${\displaystyle X\to Y}$ is then a continuous, linear bijection but its inverse is not continuous, since ${\displaystyle X}$ has a finer topology than ${\displaystyle Y}$. The kernel ${\displaystyle \{0\}}$ has ${\displaystyle X}$ as a topological complement, but we have just shown that no continuous right inverse can exist. If ${\displaystyle A:X\to Y}$ is also open (and thus a TVS homomorphism) then the claimed result holds.)

The Method of Decomposition

Topological vector spaces admit the following

:

Let ${\displaystyle X}$ and ${\displaystyle Y}$ be TVSs such that ${\displaystyle X=X\oplus X}$ and ${\displaystyle Y=Y\oplus Y.}$ Suppose that ${\displaystyle Y}$ contains a complemented copy of ${\displaystyle X}$ and ${\displaystyle X}$ contains a complemented copy of ${\displaystyle Y.}$ Then ${\displaystyle X}$ is TVS-isomorphic to ${\displaystyle Y.}$

The "self-splitting" assumptions that ${\displaystyle X=X\oplus X}$ and ${\displaystyle Y=Y\oplus Y}$ cannot be removed:

${\displaystyle X}$ and ${\displaystyle Y}$, each complemented in the other.[12]

In classical Banach spaces

Understanding the complemented subspaces of an arbitrary Banach space ${\displaystyle X}$ 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 ${\displaystyle L_{1}[0,1]}$.^[13]

For some Banach spaces the question is closed. Most famously, if ${\displaystyle 1\leq p\leq \infty }$ then the only complemented infinite-dimensional subspaces of ${\displaystyle \ell _{p}}$ are isomorphic to ${\displaystyle \ell _{p},}$ and the same goes for ${\displaystyle c_{0}.}$ 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.^[13]

The spaces ${\displaystyle L_{p}[0,1]}$ are not prime whenever ${\displaystyle p\in (1,2)\cup (2,\infty );}$ in fact, they admit uncountably many non-isomorphic complemented subspaces.^[13]

The spaces ${\displaystyle L_{2}[0,1]}$ and ${\displaystyle L_{\infty }[0,1]}$ are isomorphic to ${\displaystyle \ell _{2}}$ and ${\displaystyle \ell _{\infty },}$ respectively, so they are indeed prime.^[13]

The space ${\displaystyle L_{1}[0,1]}$ is not prime, because it contains a complemented copy of ${\displaystyle \ell _{1}}$. No other complemented subspaces of ${\displaystyle L_{1}[0,1]}$ are currently known.^[13]

Indecomposable Banach spaces

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 ${\displaystyle X}$ is always isomorphic to ${\displaystyle X,}$ 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.^[14]

See also

• Direct sum – Operation in abstract algebra composing objects into "more complicated" objects
• Direct sum of modules – Operation in abstract algebra
• Direct sum of topological groups

Proofs

References

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