Let $X,Y$ be Banach spaces. We denote by $\mathcal{K}(X,Y)$ the space of all compact operators from $X$ into $Y$. For an operator $T:X\rightarrow Y$, we let $$\|T\|\_{e}:=\inf\{\|T-K\|:K\in \mathcal{K}(X,Y)\},$$ and $$\|T\|\_{m}:=\inf\{\|T|\_{M}\|:codim M<\infty\},$$ where $M$ represents the finite co-dimensional subspace of $X$. It is known that $T$ is compact if and only if $\|T\|\_{m}=0$.


Let $X$ be a Banach space. We define $J:X\rightarrow l\_{\infty}(B\_{X^{\*}})$ by $\langle Jx,x^{\*}\rangle=\langle x^{\*},x\rangle, x\in X, x^{\*}\in B\_{X^{\*}}.$ Let $T:Z\rightarrow X$ be an operator. My question is:


Question. Does $\|JT\|\_{e}=\|T\|\_{m}$ hold?



Two measures of noncompactness of operators
The answer is yes. Indeed, let us denote $Y\_h=\ell\_\infty(B\_{Y^\*})$. For every $L\in\mathcal{K}(X,Y\_h)$,


$\|T\|\_m=\|JT\|\_m=\|JT-L\|\_m\leq \|JT-L\|$. Hence $\|T\|\_m\leq \|JT\|\_e$.


Conversely, since $Y\_h$ is $1$-injective, for every finite codimensional closed subspace $M$ of $X$, then there exists an extension $\hat T:X\to Y\_h$ of $JT|\_M$ with $\|\hat T\|=\|T|\_M\|$. Since $JT-\hat T$ is compact, $\|JT\|\_e\leq \|T\|\_m$. 


