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I am wondering about the following :

In $\mathbb{R}^n$, suppose you are given compact convex bodies $\left\{ C_k : k \geq 1 \right\}$ and $C$, such that for every $x \in \mathbb{R}^n$ $$ \mathbb{1}_{C_k}(x) \rightarrow \mathbb{1}_C(x).$$

EDIT suppose also that you have $$\lim \int |\mathbb{1}_C-\mathbb{1}_{C_k}| = 0.$$

Is it true that the convex bodies $C_k$ converges to $C$ in the Hausdorff metric ? i.e. do we have, for any $\epsilon > 0$, a rank $k_0 \gg 1$ such that for any $k \geq k_0$, $C_k$ is included in the $\epsilon$-neighborhood of $C$ and conversely ?

Any comments/suggestions is welcome ! Thanks !

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  • $\begingroup$ Did you try to use en.wikipedia.org/wiki/Blaschke_selection_theorem ? $\endgroup$
    – Jochen Wengenroth
    Commented Mar 22 at 8:25
  • $\begingroup$ @JochenWengenroth : Please see the answer below. $\endgroup$
    – Iosif Pinelis
    Commented Mar 22 at 14:26
  • $\begingroup$ @JochenWengenroth thx a lot for the reference ! I didn't know about it ! I guess that with the simple cvg hypothesis then we can obtain something.... $\endgroup$
    – Anthony
    Commented Mar 22 at 14:43

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A counterexample is given by $n=2$, $C=\{(0,0)\}$, and $C_k=\{(t,kt)\colon0\le t\le1\}$ (or one can instead take $C_k=\{(t,t/k)\colon0\le t\le1\}$).

Another counterexample, in the same spirit, is given by $n=1$, $C=\emptyset$, and $C_k=\{k\}$ (or one can instead take $C_k=\{1/k\}$) -- if you don't mind $C$ being empty.

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  • $\begingroup$ Thx for your answer ! Ok I see your point, you're making things escaping... then maybe I need to use a L1 convergence... $\endgroup$
    – Anthony
    Commented Mar 22 at 14:45
  • $\begingroup$ Also I was more thinking about convex bodies... but your example can be modified anyway. $\endgroup$
    – Anthony
    Commented Mar 22 at 14:56

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