Given a closed linear subspace, is there always a projection that maps onto it?

Question

Given a closed linear subspace, is there always a projection that maps onto it?

Here, a projection $P$ should be a linear and continuous mapping and satisfies $P^2 = P$.

Answer 1

No. In fact, there is a theorem by Lindenstrauss and Tzafriri, stating that an infinite dimensional Banach space $X$ is isomorphic to a Hilbert space if and only if there is a continuous linear projection on every closed subspace of $X$. It follows that whenever $X$ is a Banach space not isomorphic to Hilbert space, there will be some (closed) subspace of $X$ on which there will be no continuous linear projection from $X$. Such a subspace must necessarily be infinite dimensional, too.

There are also concrete examples, of course.

There does not exist a continuous linear projection from $\ell_{\infty}$ onto $c_0$

For a proof, see here.

Answer 2

Some remarks:

On incomplete spaces even if the projection exists, it doesn't have to be continuous. Consider the space of all finitely-supported sequences $(c_{00}, \|\cdot\|_\infty)$ and $P : c_{00} \to c_{00}$ with defined as $$P(a_n)_n = (a_1+a_2, 0, a_3+2a_4, 0, a_5+3a_6, 0, \ldots)$$

$P$ is a projection on the closed subspace $$\{(a_n)_n \in c_{00} : a_{2n} = 0, \forall n\in\mathbb{N}\}$$ but it is clearly unbounded.

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On the other hand, assume $X$ is a Banach space and $X = M \dotplus N$ where $M$ and $N$ are closed subspaces. If $P : X \to X$ is projection onto $M$ (a linear map such that $\operatorname{Im} P = M$ and $P^2 = P$), then $P$ is continuous.

Indeed, assume $x_n \to x \in X$ and $Px_n \to a' \in X$. In fact $a' \in M$ since $M$ is closed. We can find unique $a_n, a \in M$ and $b_n, b \in N$ such that $x_n = a_n + b_n$, $x = a+b$. Then $Px_n = a_n$ so $a_n \to a$. Hence $N \ni b_n = x_n - a_n \to x-a'$, and this vector is in $N$ since $N$ is closed.

We have $$x=\underbrace{a'}_{\in M} + \underbrace{x-a'}_{\in N}$$ so $a' = a$ and $x-a' = b$. The Closed graph theorem implies that $P$ is bounded.