Let $(V, \lVert \cdot \rVert)$ be a separable Banach space and let $B_{V^*}$ denote the closed ball in the dual $V^*$. Suppose we have a family $C \subseteq B_{V^*}$ such that $\bigcap_{\Lambda \in C} \mathrm{ker}(\Lambda) = \{ 0 \}$. Is it always possible to find a subfamily $\widetilde{C} \subseteq C$ such that $\bigcap_{\Lambda \in \widetilde{C}} \mathrm{ker}(\Lambda) = \{ 0 \}$, but removing any element of $\widetilde{C}$ makes this equality false?
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$\begingroup$ Isn't this wrong for the euclidean unit ball in $\mathbb R^2$? $\endgroup$– Jochen WengenrothCommented Dec 4, 2021 at 14:14
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$\begingroup$ Right, it is definitely wrong... I need to rework the whole question... $\endgroup$– Kacper KurowskiCommented Dec 4, 2021 at 14:21
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$\begingroup$ (The above comments were about the previous version of a question) $\endgroup$– Kacper KurowskiCommented Dec 4, 2021 at 14:26
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2$\begingroup$ It is false, though. We can take $V=C[0,1]$ and $\Lambda$ as the collection of the Dirac measures $\delta_x$, $0\le x\le 1$. Then a subcollection will have zero intersection of the kernels if and only if the corresponding points form a dense set, but then you can of course always remove one more point. $\endgroup$– Christian RemlingCommented Dec 4, 2021 at 16:42
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1$\begingroup$ @KacperKurowski: It's not quite the unit ball since such measures (are positive and) have total mass one, but you can indeed approximate any such measure $\mu$ by convex combinations $\sum c_j \delta_{j/n}$, say, with $c_j=\mu([(j-1)/n, j/n))$. $\endgroup$– Christian RemlingCommented Dec 4, 2021 at 21:14
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