The Calkin representation for Banach spaces 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!
 A: 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.
A: The answer to question 2 is yes. 
The proof you already know since you proved that $T$ compact implies $\hat{T}$ is zero.  (For someone who has not thought about this, it is immediate from the elementary fact that a bounded net in $X^*$ that converges to zero weak$^*$ must converge uniformly to zero on compact subsets of $X$.) So if 
$(x^*_\alpha)_\alpha$ is in $\widehat{X}$ 
with $\sup \|x_\alpha^\alpha \| \le 1$ and 
$x^*_\alpha \to 0$ weak$^*$, then 
$x^*_\alpha \to 0$
uniformly on $K$. Since the unit ball of $Y$ is contained in 
$K+\epsilon B_{X}$, 
it follows that 
$\|\hat{j}\| \le \epsilon$,
so $C$ can be one.
