Let $X$ be an infinite dimensional Banach space. Let $\Lambda\_{0}$ be the set of all finite dimensional subspaces of $X$ directed by the inclusion $\subseteq$. For each $\alpha\in \Lambda\_{0}$, let $I\_{\alpha}:=\{\beta\in\Lambda\_{0}:\alpha\subseteq \beta\}$. Then $\{I\_{\alpha}:\alpha\in \Lambda\_{0}\}$ is a filter basis and hence is contained in some ultrafilter $\mathcal{U}$. 


For an infinite dimensional Banach space $Y$, let $(Y^{\*})\_{\mathcal{U}}$ be the ultrapower of $Y^{\*}$ with respect to $\mathcal{U}$. Let $\widehat{Y}$ be the subspace of $(Y^{\*})\_{\mathcal{U}}$ defined by $$\widehat{Y}:=\{(y^{\*}\_{\alpha})\_{\mathcal{U}}\in (Y^{\*})\_{\mathcal{U}}:w^{\*}-\lim\_{\mathcal{U}}y^{\*}\_{\alpha}=0\}.$$ For an operator $T:Y\rightarrow X$, we define $\widehat{T}:\widehat{X}\rightarrow \widehat{Y}$ by $\widehat{T}((x^{\*}\_{\alpha})\_{\mathcal{U}})=(T^{\*}x^{\*}\_{\alpha})\_{\mathcal{U}}.$ It is easy to see that $\widehat{T}=0$ if $T$ is compact.


Question 1. Is $T$ compact if $\widehat{T}=0$?


Question 2. Let $K$ be a compact, convex and balanced subset of $B\_{X}$ and let $\epsilon>0$. We set $A:=K+\epsilon B\_{X}$ and define the gauge of $A$ by $$\|x\|\_{A}:=\inf\{t>0:x\in tA\}, \quad x\in X.$$
It is easy to see that $$\epsilon\|x\|\_{A}\leq \|x\|\leq (1+\epsilon)\|x\|\_{A}, \quad x\in X.$$ We set $Y:=(X,\|\cdot\|\_{A})$ and let $j:Y\rightarrow X$ be the formal identity. Is there a constant $C$ such that $\|\widehat{j}\|\leq C\cdot \epsilon$?


Thanks!



The Calkin representation for Banach spaces
The answer to question 1 is yes.


Suppose that $T^\*B\_{X^\*}$ is not compact. Since it is norm closed, there is 
$\epsilon >0$ 
and an infinite subset $S$ of $B\_{X^\*}$ so that 
$\|T^{\*}x\_1^\*-T^{\*}x\_2^\*\| > \epsilon$ 
for all 
$x\_1^\*\not= x\_2^\*$ 
in 
$S$. Let $x^\*$ be any weak$^\*$ limit point of $S$. For $\alpha$ in $\Lambda\_0$ pick $x\_\alpha^\*$ in $S$ with 
$T^\*x^\*\_\alpha \not= T^\*x^\*$ 
so that 
$\|x^\* - x\_\alpha^\*\|\_\alpha < 1/\dim \alpha$, where 
$\|z^\*\|:= \|z^\*\_{|\alpha}\|$. Then by the choice of $\Lambda\_0$,
$x^\*\_\alpha \to x^\*$ weak$^\*$
and hence 
$\widehat{T}(x^{\*}\_\alpha -x^{\*})\_\alpha =0$, which means
$\|T^\*x\_\alpha^\* - T^\*x^\*\| \to 0$.
Since $S$ is $\epsilon$-separated, this forces $T^\*x\_\alpha^\* = T^\*x^\*$ eventually, which is a contradiction.


