A noncomplete inner product space may have a nonempty closed convex subset which does not have a unique element of minimal norm

Question

It is well known that if $H$ is a Hilbert space and $E$ is a nonempty closed convex subset of $H$, then there is a unique element in $E$ of minimal norm, i.e., a unique element $x_0\in E$ such that $\|x_0\|=\min _{x\in E} \|x\|$. (cf. Rudin's Real and Complex Analysis, Theorem 4.10) Its proof crucially uses completeness of $H$. I'm wondering if this fails if $H$ is not complete, but equipped with an inner product.

A counterexample when $H$ is a Banach sapce, is given in Counterexamples to a theorem in Rudin's book on elements of smallest norm in a closed convex sets in a Hilbert space. But in this counterexample, $C[0,1]$ is not an inner product space.

Is there a counterexample for a noncomplete innerproduct space?

I guess it is similar https://math.stackexchange.com/questions/3452638/counter-example-to-minimal-norm-theorem-for-hilbert-spaces/3452884#3452884 — Maman, Apr 21 '22 at 22:45

score 1 · Answer 1 · answered Jun 14 '20 at 16:03

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Completeness needs to prove the existence part. I think that without completeness we still have the uniqueness part, see Brezis Theorem 5.2 Functional Analysis, Sobolev spaces and PDEs.

answered Jun 14 '20 at 16:03

Daniele Tiberio

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A noncomplete inner product space may have a nonempty closed convex subset which does not have a unique element of minimal norm

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