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Let $A$ be a normed vector space, and $C$ a convex subset of $A$.
Let $M=\{c\in C:||c||=\inf_{a\in C}||a||\}$ be the set of all minimum norm elements of $C$ and suppose this has at least two elements.

  1. What is an example of such an $C$?
  2. How do I prove that $M$ has infinitely many elements?

I would greatly appreciate it if somebody would give me an example and let me check whether it suffices, for the proof however I have no idea and would really appreciate all hints.

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In order to get an example, you can consider $\mathbb{R}^2$ with e.g. the $\infty$-norm (see e.g. here). Draw a picture of the "unit sphere" w.r.t this norm and take $C$ to be an adjacent square.

Let $x,y \in M$ then for any $\lambda \in (0,1)$ we have $\left\| z \right\| = \left\| \lambda x + (1-\lambda)y \right\| \leq \lambda \left\| x \right\| + (1-\lambda)\left\| y \right\|$. Since $C$ is convex, $z \in C$. Since $x$ and $y$ minimize the norm in $C$, $\left\| z \right\| = \left\| x \right\|=\left\| y \right\|$ and $z \in M$. There are infinitely many such $z$.

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  • $\begingroup$ I dont see why $||z||=||x||=||y||$. I understand that $x$ and $y$ minimize the norm, but where does the equality come from? $\endgroup$ – user368886 Oct 3 '16 at 12:26
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    $\begingroup$ Since $x$ and $y$ minimize the norm, $\left\| x \right\| = \left\| y \right\|$. Triangle inequality gives you $\left\| z \right\| \leq \left\|x \right\| = \left\| y\right\|$. Strict inequality would contradict the fact that $x$ and $y$ minimize the norm, hence we have equality. $\endgroup$ – Matias Heikkilä Oct 3 '16 at 12:27
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    $\begingroup$ Yes, thank you. I'm not fully awake yet ;) $\endgroup$ – user368886 Oct 3 '16 at 12:30
  • $\begingroup$ I still fixed some typos, it should be correct now. $\endgroup$ – Matias Heikkilä Oct 3 '16 at 12:32
  • $\begingroup$ What do you mean by adjacent square? $\endgroup$ – user368886 Oct 3 '16 at 12:38

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