Theorem 12.9.8 (Closed Complementation Theorem).label Let $E$ be a complete metric TVS, $\bracsn{0}\ne F \subsetneq E$ be a closed subspace, then the following are equivalent:
- (1)
$F$ admits a closed complementary subspace.
- (2)
There exists $P \in L(E; E)$ such that $P^{2} = P$ and $P(E) = F$.
Proof. (2) $\Rightarrow$ (1): Let $G = \ker(P)$, then the mapping
is a toplinaer isomorphism.
(1) $\Rightarrow$ (2): Let $G$ be a closed complementary subspace of $E$, so that the mapping
is a continuous, bijective linear map. As $F, G \subset E$ are both closed, they are complete by Proposition 6.7.3, and $E \times F$ is a complete metric TVS by Proposition 6.7.2. By the Open Mapping Theorem, the $T$ is a toplinear isomorphism. As such, $P$ is given by the following composition
$\square$
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