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 $\|A\|\cdot \sup\_{j}|\lambda\_{j}|\cdot \|B\|\leq (1+\epsilon)\|R\|\cdot \|S\|$?



A question on p-factorable operators
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 $T$.


