The following looks like a strengthening of the approximation property, but I don't know, maybe this is equivalent. I would be grateful if somebody could explain this.

Let $X$ be a Banach space with the approximation property, and $K$ a compact set in $X$.

Is it true that for each $\varepsilon>0$ there exists an (bounded) operator of finite rank $T:X\to X$ such that

1) $T$ approximates the identity operator on $K$: $$ \forall x\in K\qquad ||Tx-x||<\varepsilon, $$ and

2) the range of $T$ belongs to the linear span of $K$ $$ T(X)\subseteq\text{span}\ K $$ ?

EDIT. Yes, I see that this is not equivalent, because otherwise we obtain that the (usual) approximation property is inherited by closed subspaces. OK, I change a bit my question:

Is anything known about this variant of the approximation property?

  • $\begingroup$ This sounds to me as this is mainly or possibly about retraction(?). $\endgroup$ Nov 12 '16 at 5:20

Since every separable closed subspace of $X$ is the closed linear span of a compact set, your property is equivalent to the HAP (hereditary approximation property).

Approximate $K$ by a finite rank bounded linear operator $T$ on the closed linear span of $K$. A small perturbation argument shows that WLOG $T$ ranges in the linear span of $K$. Since $T$ has finite rank, $T$ can be extended to bounded linear operator on $X$ that has the same range.

If $X$ is non separable, use the fact that if every separable subspace of $X$ has the AP (or is even contained in a subspace that has the AP), then $X$ has the AP.

Spaces with the HAP are rare but there are some that are not isomorphic to Hilbert spaces. I constructed the first one in the 1970s and Pisier's weak Hilbert spaces are a nice class of such spaces. More recent work on them is contained in my papers with Figiel and with Szankowski.


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