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 $\|\widehat{j}\|\leq C\cdot \epsilon$?


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

2 Answers 2


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.

  • $\begingroup$ Thanks, Bill. I think that in you answer, $\|T^{*}x^{*}-T^{*}x^{*}_{\alpha}\|_{\alpha}<1/\textrm{dim} \alpha$ should be $\|x^{*}-x^{*}_{\alpha}\|_{\alpha}<1/\textrm{dim} \alpha$. $\endgroup$ Jun 15, 2020 at 1:54
  • $\begingroup$ 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? $\endgroup$ Jun 15, 2020 at 2:04
  • $\begingroup$ Yes; sorry for the typo. I'll fix it. $\endgroup$ Jun 15, 2020 at 2:53
  • $\begingroup$ 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 $\alpha$. $\endgroup$ Jun 15, 2020 at 3:10

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.


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