# Topological Complements

#### – Introduction –

The first steps outside the comforts of the category of Hilbert spaces, the safe space for of functional analysis, into the unruly world of topological vector spaces, can be a troubling experience for any student, myself included. To easy the passage, here are  a few tips and results regarding the existence of complementary subspaces in the general setting of topological vector spaces. For Hilbert spaces it is known that every closed subspace has a preferred (topologically) complementary subspace, namely the orthogonal complement, but any two (algebraically) complementary closed subspaces are automatically (topologically) complementary (by Theorem 1).

First, lets get the definitions out of the way. Two subspaces $U, V \subset X$ of a vector spaces $X$ are said to be algebraic complements (denote $X = U \oplus_A V$) if any $x \in X$ can be written uniquely as a sum $x = u + v$, where $u \in U$ and $v \in V$. This is equivalent to the condition $U \cap V = \{0\}$ and $X = U + V$ (as a set).

Here two subspaces $U, V \subset X$ of a toplogical vector space $X$ are (topologically) complementary, written $X = U \oplus V$, if either of the following equivalent conditions hold:

• $X = U \oplus_A V$ and there exists a continuous projection $P$ onto $U$ along $V$ (or vice versa of course) . That is a continuous projection onto $U$ with $ker(P) = V$.
• There exists a homeomorphism $X \simeq V \times U$, where the latter space is endowed with the usual product topology.
Exercise The above equivalence can be proved…
Proof
If $P:X\to X$ is a projection onto $U$ along $V$, then $x \mapsto (P(x), (I-P))(x))$ is a homeomorphism onto $U\times V$, with inverse $(u, v) \mapsto u+v$, since both are clearly continuous. Conversely if $f: X\to U \times V$ is a homeomorphism, then compose $f$ with the projection map $\pi_1: U\times V \to U$ gives the requires continuous projection onto $U$ along $V$.

By the codimension of a subspace $V \subset X$, we mean the dimension of the quotient space $X/V$. This is equivalent to the, slightly more intuitive, definition of the dimension of the algebraic complement of a subspace since these have the same dimension, which can be verified using the first isomorphism theorem for vector spaces. All subspaces have algebraic complements,  just augment the basis of the subspace to a basis of the whole space.  I don’t intend to make a habit of pointing out what result relies on the axiom of choice here,  since one really cannot have much fun without it, but yes… it does rely on the axiom of choice.

#### Some Results

Theorem 1  (Closed algebraic complements)
If $X = U \oplus_A V$ and $U, V$ are closed, then the $X = U \oplus V$ if the closed graph theorem holds. That is, if $X$ is an $F$-space.
Proof
Let $X$ be an F-space, and let $P$ be the projection onto $V$ along $U$, where $X = U \oplus_A V$.  The graph $\{ (x, P(x)) ~| ~ x\in X\}$ is closed in $X\times X$, so $P$ is continuous by the closed graph theorem.

To the best of my knowledge the next theorem requires the Hahn-Banach separation theorem, so we are restricted to locally convex spaces.

Theorem 2  (Finite subspace)
If $U \subset X$ is a finite dimensional subspace of a locally convex topological vector space $X$, then there exists a closed subspace $V\subset X$ such that
$$X = U \oplus V$$
Proof
We here employ a corollary to the Hahn-Banach separation theorem (for locally convex spaces), stated here without proof.

Lemma 1
Let $X$ be a locally convex space (over $\mathbb{C}$ or $\mathbb{R}$) and $V \subset X$ a closed subspace, and $x \not\in V$. Then there exists a non-zero $\phi \in V^\perp$ (the anihilator of $V$) such that
$$\phi(x) = 1$$

Let $\{ e_i\}_{i=1}^n$ be a basis for the finite dimentional subspace $U \subset X$, with the notation
$$\{ e_1, …\hat{e_k}, … e_n \} = \{e_i\}_{i=1}^n\backslash \{ e_k \}$$ we pick
$\phi_k \in span(\{ e_1, …\hat{e_k}, … e_n \})^\perp$ with $\phi_{k}(e_k) = 1$. It is now easy to check that the linear operator $P_U$ defined by by the equation
$$P_U(x) = \sum_{i=1}^n \phi_i(x)e_i$$ for all $x\in X$ is a bounded projection onto $U$. Setting $$V = Ker(P)$$ we have our complementing subspaces.

Note the generality of the following statement.

Theorem 3 (Finite codimension)
Let $U \subset X$ be a cosed subspace of finite codimension, and $X$ an arbitrary topological vector space. Then there exits a subspace $V$ with $dim(V) = codim(U)$ such that
$$X = U \oplus V$$
Proof
Let $\pi: X \to X/U$ be the projection map. Let $\{ b_i \}_{i=1}^n$ be a basis for $X/U$, and let $\ \{e_i\}_{i=1}^n \subset X$ be such that $e_k \in \pi^{-1}(\{b_k \})$. Define $\tilde{P}: X/U \to X$ as the map
$\tilde{P}(b_i) = e_i$ and extend it by linearity. The composition $P = \tilde{P} \circ \pi$ can be seen to be continuous since $\pi$ is a quotient map and $\tilde{P}$ is a linear map of finite vector spaces. It is also a projection onto $span\{e_i\}_{i=1}^n$, with kernel $Ker(P) = Im(I – P) = U$. Setting $V = span\{e_i\}_{i=1}^n$ we thus have:
$$X = U \otimes V$$
Corollary 3.1
If $T: X \to Y$ is a continuous linear map and $Y$ is finite dimensional, then $Ker(T)$ has a complement $V$, that is
$$X = Ker(T) \oplus V$$
with $dim(V) = dim(Y)$
Proof
using the isomorphism theorem from linear algebra we know that $$X/Ker T \simeq Im(T) \subset Y$$ so it is of finite codimention. The result then follows from theorem 3.

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