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I quote here the proof of a result given in Haim Brezis Functional Analysis, Sobolev Spaces and partial differential Equations:

I haven't been able to conclude where exactly is this hypothesis used:

A is closed and convex.

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As far as I understand, one concludes the existence of minimum in the convex, closed and bounded set $\tilde{A}$, but this is done without using the fact that $A$ is closed and convex,

The fact that $\tilde{A}$ is closed and convex simply follows from the lower semicontinuity and convexity of the function $\varphi$ and we obtain the existence of minimum in the set $\tilde{A}$ Then, as $\tilde{A} \subset A$ we obtain the minimum in the whole $A$.

D1X
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  • the minimum in a small part can't prove that it will be the minimum in the whole set for example $\varphi=x$ on $\hat{A}=[1,2]$ and $A=[0,2]$ – Hamza Jul 12 '16 at 00:05
  • But $\tilde{A}$ cannot be that set. It would be necessarily $[0,2]$ in that case due to the definition of it. – D1X Jul 12 '16 at 00:08
  • I give this as remark to your last line but here everything is ok – Hamza Jul 12 '16 at 00:09

1 Answers1

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No the assumption for $A$ to be closed and convex are important, because if we put $$ B=\{x\in A \; , \; \varphi(x)\leq \varphi(a) \} $$ so using the fact that $\varphi$ is lsc prove that $B$ is closed in the trace topology of $A$, since $A$ is closed $B$ is closed in $E$, and the fact that $A$ is convex and $\varphi$ is convex prove that $B$ is convex, and finally $(5)$ prove that $B$ is bounded so in particular $B$ is a compact in $\sigma(E,E^*)$.

If $A$ is not closed the theorem will fall in fact : $E=\mathbb{R}$ and $A=]0,+\infty[$ and $\varphi(t)=t$ it's clear that $\inf_{t\in A}\varphi(t)=0$ who is not achieved in any point of $A$.

If $A$ is closed but not convex the question is more complicated because every sci in a compact set achieve the minimum, so we need an infinite dimensional example,but the theorem fall also in this case in fact : $E=l^2(\mathbb{N})$ and $A=\{(x_n)_n\; , \; \sum_{k\geq0}|x_k|^2=1\}$ and $\varphi((x_n)_n)=x_1$ it's clear that $\inf_{x\in A}\varphi(x)=-\infty$ who is not achieved in any point of $A$.

Hamza
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  • Why do you consider the set $B$? $\varphi$ is defined on $A$, so then $\tilde{A}$ is closed and convex by lower semicontinuity and convexity of $\varphi$, and then compact in $\sigma(E, E^*)$. Then $\varphi$ attains its minimum on $\tilde{A}$, and as for every $x_0 \in C - \tilde{C}$ $\varphi(x) < \varphi(x_0)$, so it attains minimum at $C$. What is wrong with this argument? because I don't get it. – D1X Jul 12 '16 at 10:43
  • Can't we conclude without using the hypotesis I question from lsc and convexity that the set $\tilde{A}={x\in C ; , ; \varphi(x)\leq \varphi(a) }$ is closed and convex? $\varphi$ is defined on $C$. – D1X Jul 12 '16 at 13:10
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    if your set $C$ is not convex $\hat A$ will be not necessary convex think about $\varphi=x$ and $C=[0,1[\cup ]1,2]$ – Hamza Jul 12 '16 at 15:08
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    and the assumption that $C$ is closed is also important because using lsc we can proof that $\hat{A}$ is closed in the topology of $C$ (trace topology) only!!!, if you assume that $C$ closed then every closed set in the topology of $C$ will be closed in the topology of $X$ – Hamza Jul 12 '16 at 15:12
  • i will edit the answer now – Hamza Jul 12 '16 at 15:16
  • Thank you, that solved the question. – D1X Jul 12 '16 at 15:53