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This thread originated from MSE: Compact Approximation

This is meant as lemma for: Approximation Property

Given a Banach space $E$.

Denote compact domains by $\mathcal{C}$.

Denote compact operators by $\mathcal{C}(E)$.

Then there is a compact approximate identity: $$\forall C\in\mathcal{C}\,\exists C_N\in\mathcal{C}(E,E):\quad\|C_N-1\|_C:=\sup_{x\in C}\|C_Nx-x\|\stackrel{N\to\infty}\to0$$

How to construct such compact operators?

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  • $\begingroup$ To clarify: your display is asserting that there exist compact operators $C_N$ such that this holds? $\endgroup$
    – Nate Eldredge
    Commented Feb 9, 2015 at 18:26
  • $\begingroup$ @NateEldredge: Yes exactly. Wait let me add this. $\endgroup$
    – C-star-W-star
    Commented Feb 9, 2015 at 18:47
  • $\begingroup$ @NateEldredge: I added it to the question. $\endgroup$
    – C-star-W-star
    Commented Feb 9, 2015 at 18:50
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    $\begingroup$ There is not always a compact approximate identity. If there is, we say that $E$ has the compact approximation property. Enflo actually showed that not every Banach space has the compact approximation approximation property. Davie's examples are given in volume one of the book of Lindenstrauss and Tzafriri. However, they only prove the failure of the approximation property. You have to go back to Davie's paper to read a proof of the failure of the compact approximation property. $\endgroup$
    – Bill Johnson
    Commented Feb 9, 2015 at 19:08
  • $\begingroup$ A.M. Davie, The Banach approximation problem, J. Approx. Theory 13, 392-394, 1975. $\endgroup$
    – Bill Johnson
    Commented Feb 9, 2015 at 19:09

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