# 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!

• Are you asking because you are thinking about following up on a lemma in my paper with March Boedihardjo? Jun 13, 2020 at 18:58
• Yes. I want to improve Theorem 2.1 in your paper with March Boedihardjo and characterize the bounded compact approximation property. Jun 14, 2020 at 0:16

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.
• Thanks, Bill. I think that in you answer, $\|T^{*}x^{*}-T^{*}x^{*}_{\alpha}\|_{\alpha}<1/\textrm{dim} \alpha$ should be $\|x^{*}-x^{*}_{\alpha}\|_{\alpha}<1/\textrm{dim} \alpha$. Jun 15, 2020 at 1:54
• Moreover, I think that for each $\alpha$, we can pick $x^{*}_{\alpha}\in S$ such that $\|T^{*}x^{*}_{\alpha}-T^{*}x^{*}\|\geq \frac{\epsilon}{2}$. Can we do it? Jun 15, 2020 at 2:04
• But I am not sure that we can pick $x^{*}_{\alpha}\in S$ such that $\|T^{*}x^{*}_{\alpha}-T^{*}x^{*}\|\geq \epsilon/2$ for all $\alpha$. Jun 15, 2020 at 3:10
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.