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Let $V$ be a Hilbert space, $Q \subset V$ be convex and compact and $Q_n \subset V$ be convex and compact for $n\in \mathbb{N}$ such that $Q_n \rightarrow Q$ for $n\rightarrow \infty$ in Hausdorff distance.

Let $L$ be a linear subspace of $V$ with $Q_n \cap L \neq \emptyset$ for all $n\in \mathbb{N}$ and $Q \cap L \neq \emptyset$.

Does it then also hold that $Q_n \cap L \rightarrow Q \cap L$ for $n\rightarrow \infty$ in Hausdorff distance?

I would also be thankful for any pointers to relevant literature.

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  • $\begingroup$ This already fails in $\mathbb{R}^2$. $\endgroup$
    – Nik Weaver
    Commented Jul 26, 2019 at 14:17
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    $\begingroup$ Let $Q$ be the line segment joining $(0,0)$ and $(1,0)$ and let $Q_n$ be the line segment joining $(0,1/n)$ and $(1,0)$. $\endgroup$
    – Nik Weaver
    Commented Jul 26, 2019 at 14:50
  • $\begingroup$ @NikWeaver That was simpler than expected, thanks! $\endgroup$
    – Steve
    Commented Jul 26, 2019 at 15:28
  • $\begingroup$ No problem, you're welcome. $\endgroup$
    – Nik Weaver
    Commented Jul 26, 2019 at 16:34
  • $\begingroup$ For compact convex sets, I think Nik Weaver's example is typical, that is for any such sequence $Q_n$ in an infinite dimensional $V$, there exists a subspace $L$ for which the intersection is not continuous. If the $Q_n$ are non-empty, closed, "equi-uniformly convex" sets (hence they are not compact, unless $V$ is finite dimensional), then I guess the continuity of intersection with $L$ is true; maybe also with any closed convex $L$. $\endgroup$
    – Pietro Majer
    Commented Jul 27, 2019 at 13:07

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