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I recently started to study about Hilbert space and the following is a Lemma of approximation:

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I understand the whole proof except for one part. Why is the infimum $d$ of $A$ defined? We know $A$ is a closed convex subset of $H$, does this imply it is also bounded?

I have the feeling , this is quite elementary but I've been stuck here. I would appreciate any help. Thanks in advance!!!

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No, $A$ may not bounded. And it is not "the infimum of $A$", but the infimum of the set $\{\lVert x-a\rVert\,:\,a\in A\}$. The map $\lVert \bullet -x\rVert:A\to \Bbb R$ takes values in $[0,\infty)$. So $\inf_{a\in A}\lVert a-x\rVert$ exists and it is a non-negative real number.

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  • $\begingroup$ It was that simple...! Thank you very much for your quick response. $\endgroup$ Commented Mar 8, 2017 at 13:41
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No. If $A$ is a closed and convex subset of $H$, then $A$ need not to be bounded. Example: $A=H$.

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  • $\begingroup$ If $ A \subset H\;$ then could be bounded? If it's not bounded , why infimum of $A$ is defined? $\endgroup$ Commented Mar 8, 2017 at 13:36

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