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!