# Counter Example to Minimal Norm Theorem for Hilbert Spaces

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.

• A subspace of $X$ contains $0$ which has minimal norm. Did you mean something different? – Umberto P. Nov 27 '19 at 2:33
• My apologies, I meant a subset $A$. Ill edit it. – The math god Nov 27 '19 at 2:34
• I'm confused, isn't any subset of $\mathbb{Q}$ not convex? – The math god Nov 27 '19 at 2:43
• What about $X=\mathbb{R}^2$ and $A=\{(x,y))\,\mid\, y\geq \exp(x)\}$? – Jens Schwaiger Nov 27 '19 at 3:11
• 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? – The math god Nov 27 '19 at 4:34

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 $$L^2[0, 1]$$.)
• Why is $A$ closed in $L^2$? – Stefan Lafon Nov 30 '19 at 5:58
• @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. – Christopher Gadzinski Nov 30 '19 at 13:09
• 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 $[0, \frac 1 2]$. – Stefan Lafon Dec 1 '19 at 14:23
• 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. – Christopher Gadzinski Dec 1 '19 at 14:36
• 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. – Stefan Lafon Dec 1 '19 at 16:51