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Let $1\leq p<\infty$. Suppose that an operator $T:X\rightarrow Y$ has a factorization $T=RS$, where $S:X\rightarrow l_{p}$, $R:l_{p}\rightarrow Y$ are compact operators.

Question: Let $\epsilon>0$. Are there operators $B:X\rightarrow l_{p}$, $A:l_{p}\rightarrow Y$ and $\lambda=(\lambda_{j})_{j}\in c_{0}$ such that $T=AM_{\lambda}B$, where $M_{\lambda}:l_{p}\rightarrow l_{p}, (t_{j})_{j}\rightarrow (\lambda_{j}t_{j})_{j}$ is the diagonal operator, and $\|A\|\cdot \sup_{j}|\lambda_{j}|\cdot \|B\|\leq (1+\epsilon)\|R\|\cdot \|S\|$?

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Yes. If $T$ has finite rank this follows from the fact that every finite dimensional subspace of $\ell_p$ is contained in a finite dimensional superspace of $\ell_p$ that is $1+\epsilon$-isomorphic to the $L_p$ space of its dimension.

For the general case of compact $T$, write $T= \sum_{n=0}^\infty T_n$ with each $T_n$ finite rank and $ \|T_n\| < 4^{-n}\epsilon$ for $n=1,2,\dots$. Apply the finite rank case to each $T_n$ and take an $\ell_p$ sum.

Notice that every compact $T$ on $\ell_p$ satisfies the hypothesis you put on $T$.

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  • $\begingroup$ Why the compact operator $T$ can be written as $\sum_{n}T_{n}$ with each $T_{n}$ finite rank and the norm of $T_{n}$ is less than $4^{-n}\epsilon$? Is this true for every compact operator? $\endgroup$
    – Dongyang Chen
    Commented Jul 25, 2017 at 11:17
  • $\begingroup$ Sure, if the target space has the approximation property. It is elementary. Notice that the estimate applies to the \|T_n\| with $n>0$, not to $T_0$, which must have norm close to $\|T\|$. $\endgroup$
    – Bill Johnson
    Commented Jul 25, 2017 at 16:29

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