Does the approximation property pass to subspaces?

Question

A Banach space $X$ has the approximation property (AP) if for any compact subset $K$ of $X$ and any $\epsilon>0$, there exists a bounded finite rank operator $R$ such that $\| x - R(x) \| < \epsilon$ for every $x \in K$.

Question: If $M \subset X$ is a subspace which does not have AP, then $X$ does not have AP too.

I have the feeling that it is true. But I don't know how to show it. Any idea?

@soup: so you think that the statement holds? – Idonknow Jun 17 '16 at 02:11 — Idonknow, Jun 17 '16 at 02:11

score 2 · Answer 1 · answered Jun 16 '16 at 12:39

2

Every $\ell_p$, $p\neq 2$, and $c_0$ have closed subspaces without the ap. So the answer to your question is no.

answered Jun 16 '16 at 12:39

Ben W

5,236

score 2 · Accepted Answer · answered Jun 19 '16 at 22:20

Ben has answered your question, but here is an even easier route to a counter-example (modulo the hard part of the existence of a space without the AP).

Take your favourite separable space without the AP. You may embed it into $C[0,1]$, which actually has a basis (the Schauder system, a prototypical example of a Schauder basis), so in particular it has the AP.

Does the approximation property pass to subspaces?

2 Answers2