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A closed subspace is complemented exactly when it is the range of a bounded projection

Statement

Let $X$ be a normed space and let $M \leq X$ . Then $M$ is complemented if and only if there is a bounded linear operator $P : X \to X$ such that

$P^{2} = P, ran (P) = M .$

Facts & Assumptions

Given: A normed space $X$ , a linear subspace $M \leq X$ , and a bounded linear operator $P : X \to X$ .

[L1]

If $M$ is complemented, then $X = M \oplus N$ for some closed subspace $N$ and the coordinate maps $P_{M}$ and $P_{N}$ are bounded linear operators (A complemented closed subspace of a normed space).

[L2]

A bounded linear operator is linear and satisfies norm estimates (A bounded linear operator between normed spaces).

Proof

technique · direct

1.1L1

Assume $M$ is complemented, and write $x = m + n$ with $m \in M$ , $n \in N$ as in [L1]. Let $P : = P_{M}$ . Then $P (x) = m$ , so $P^{2} (x) = P (m) = m = P (x)$ and $ran (P) = M$ . Thus a complemented subspace is the range of a bounded projection.

1.2

Conversely, assume $P^{2} = P$ and $ran (P) = M$ . For every $x \in X$ ,

$x = P x + (x - P x) .$

Here $P x \in M$ , and $P (x - P x) = P x - P^{2} x = 0$ , so $x - P x \in \ker P$ . [L2, algebra]

1.3L2choose

The kernel $\ker P$ is a closed linear subspace of $X$ . It is linear because $P$ is linear by [L2]. If $x_{n} \in \ker P$ and $x_{n} \to x$ , let $C$ be a bound for $P$ from [L2]. Then $∥ P x ∥ = ∥ P (x - x_{n}) ∥ \leq C ∥ x - x_{n} ∥ \to 0$ , so $P x = 0$ and $x \in \ker P$ .

2.1step 1.2L2algebra

The sum in step 1.2 is direct: if $z \in M \cap \ker P$ , then $z = P w$ for some $w$ and also $P z = 0$ , so $z = P z = P^{2} w = 0$ . Therefore $X = M \oplus \ker P$ .

3.1step 2.1step 1.3L1L2algebra

The coordinate projections for the direct sum $X = M \oplus \ker P$ are $P$ and $I - P$ . The first is bounded by hypothesis, and the second is bounded because $∥ (I - P) x ∥ \leq ∥ x ∥ + ∥ P x ∥$ for every $x$ . Together with steps 2.1 and 1.3, this is exactly the complemented-subspace condition of [L1].

A closed subspace is complemented exactly when it is the range of a bounded projection

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources