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Problem

Given a Banach space $E$. Denote compact sets by $\mathcal{C}$, compact operators by $\mathcal{C}(X,Y)$, and finite rank operators by $\mathcal{F}(X,Y)$.

Suppose it has the approximation property: $$\forall C\in\mathcal{C}\,\exists T_N\in\mathcal{F}(E):\quad\|T_N-1\|_C:=\sup_{x\in C}\|T_Nx-x\|\stackrel{N\to\infty}{\to}0$$

Then every compact operator is of almost finite rank: $$\overline{\mathcal{F}(X,E)}=\mathcal{C}(X,E)\subseteq\mathcal{B}(X,E)$$

How do I prove this actual equivalence?

Attempt

As the image of the unit ball is precompact one has: $$C\in\mathcal{C}(X,E):\quad\|T_NC-C\|=\|T_N-1\|_{C(B)}\to0\quad(T_NC\in\mathcal{F}(X,E))$$ For the converse one might try to smuggle in a compact operator: $$C\subseteq rB:\quad\|T_N-1\|_C\leq r\|T_N-C\|_B+\|C-1\|_C<r\delta_T+\delta_C\quad(C\in\mathcal{C}(E))$$ But how to construct one that approximates the identity?

Reference

This thread originated from MSE: Approximation Property: Characterization

As reference the german wiki: Approximationseigenschaft

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  • $\begingroup$ Can you define some of the notation here? What are $X$, $E$, $\mathcal{C}$, $\mathcal{F}(E)$, $\mathcal{C}(X,E)$? Also, it would help if you make the quantifiers explicit (is your first display asserting the existence of such a sequence $T_n$, or saying it holds for all $T_n \in \mathcal{F}(E)$, or what?) $\endgroup$
    – Nate Eldredge
    Commented Feb 9, 2015 at 18:25
  • $\begingroup$ @NateEldredge: Sure. One moment please. $\endgroup$
    – C-star-W-star
    Commented Feb 9, 2015 at 18:26
  • $\begingroup$ @NateEldredge: Hope it is clearer now. $\endgroup$
    – C-star-W-star
    Commented Feb 9, 2015 at 18:38
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    $\begingroup$ The converse is a non trivial result of Grothendieck's. It is in his AMS Memoirs. $\endgroup$
    – Bill Johnson
    Commented Feb 9, 2015 at 18:57
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    $\begingroup$ @BillJohnson: Great great great!!! =D $\endgroup$
    – C-star-W-star
    Commented Feb 9, 2015 at 19:11

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