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Is there an example of a closed convex non-empty subset $A$ of an inner product space $X$ that does not contain an element of minimal norm?

I have absolutely no clue how to come up with an example for which the theorem does not hold when $X$ is not a Hilbert space. This is a practice test question, so I do apologize for the lack of work.

Any hints would be much appreciated.

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  • $\begingroup$ A subspace of $X$ contains $0$ which has minimal norm. Did you mean something different? $\endgroup$ – Umberto P. Nov 27 '19 at 2:33
  • $\begingroup$ My apologies, I meant a subset $A$. Ill edit it. $\endgroup$ – The math god Nov 27 '19 at 2:34
  • $\begingroup$ I'm confused, isn't any subset of $\mathbb{Q}$ not convex? $\endgroup$ – The math god Nov 27 '19 at 2:43
  • $\begingroup$ What about $X=\mathbb{R}^2$ and $A=\{(x,y))\,\mid\, y\geq \exp(x)\}$? $\endgroup$ – Jens Schwaiger Nov 27 '19 at 3:11
  • $\begingroup$ But if $A$ is a non-empty closed convex subset of a Hilbert space ($\mathbb{R}^2$ is Hilbert with the usual norm), then there is a theorem that says that there exists a unique element that has minimal norm. So doesn't $A$ have minimal norm? $\endgroup$ – The math god Nov 27 '19 at 4:34
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Consider the space of continuous functions from $[0, 1]$ to $\mathbb{R}$ with the usual $L^2$ inner product. Let $A$ be the set of functions $f$ in this space with $f(x) = 1$ when $0 \le x \le 1/2$. It is simple to verify that $A$ is non-empty and convex. (In fact, it is an affine subspace.) Furthermore, $A$ is closed; whenever a function $f$ is not identically $1$ on $[0, 1/2]$, there must be an open subset of $[0, 1/2]$ on which the values of $f$ are bounded away from $1$, which guarantees that $f$ is bounded away from $A$ with the $L^2$ norm.

However, while no element of $A$ has $L^2$-norm $1/2$, there are elements of $A$ with norms arbitrarily close to $1/2$; $A$ has no element of minimal norm.

(Indeed, $C[0,1]$ is not complete; it lives as a dense but proper subspace of the Hilbert space $L^2[0, 1]$.)

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    $\begingroup$ Why is $A$ closed in $L^2$? $\endgroup$ – Stefan Lafon Nov 30 '19 at 5:58
  • $\begingroup$ @StefanLafon Good question. Actually, it was not---oops. I even argued (correctly) that elements of $A$ could have arbitrarily small norm, which meant that $0$, an element outside of $A$, was adherent to $A$. I've fixed the example now and added a brief argument for closedness. $\endgroup$ – Christopher Gadzinski Nov 30 '19 at 13:09
  • $\begingroup$ I don't think that $A$ is closed, even with that new definition. For instance, you can find a sequence of elements of that set that converges to the indicator function of $[0, \frac 1 2]$. $\endgroup$ – Stefan Lafon Dec 1 '19 at 14:23
  • $\begingroup$ Convergence where? Such sequences prove that $A$ is not closed in $L^2[0, 1]$, but I claim that $A$ is closed in $C[0, 1]$. I've re-checked the argument I gave in the answer and it does hold water. $\endgroup$ – Christopher Gadzinski Dec 1 '19 at 14:36
  • $\begingroup$ It wasn't clear you meant closed in $C([0, 1])$. You might want to edit your answer to clarify closed in which space with which norm. $\endgroup$ – Stefan Lafon Dec 1 '19 at 16:51

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