Necessity of hypothesis in distance from a set in an inner product space

Question

In Kreyzig's Functional Analysis book, one theorem in inner product spaces is about the existence and uniqueness of a minimal point from a set.

3.3-1 Theorem (Minimizing vector). Let $X$ be an inner product space and $M\ne\varnothing$ a convex subset which is complete (in the metric induced by the inner product). Then for every given $x\in X$ there exists a unique $y\in M$ such that $\delta=\inf_{\widetilde{y}\in M}||x-\widetilde{y}||=||x-y||$.

First, the hypothesis of convexity is obviously necessary for the proof of uniqueness. (A counterexample would be a circle.) Is it also necessary for the proof of existence? I feel that completeness of $M$ should be enough but am unable to find a counterexample.

That is, I am looking for a space where $M$ is not convex but is complete and yet the $y$ mentioned in the theorem does not exist.

Second, is it necessary that the space be an inner product space and not merely a normed space? I confess I have not thought about the second question as much but any replies to either question will be much appreciated.

Answer 1

Let $H$ be a Hilbert space with complete orthonormal set $\{ e_{n}\}_{n=0}^{\infty}$. Define a curve $C(t)$ on $[1,\infty)$ in such a way that for $n=1,2,3,\ldots$, $$ C(t)=(1+2^{-t})\left[\cos(\pi (t-n)/2)e_{n}+\sin(\pi (t-n)/2)e_{n+1}\right], \;\;\; n \le t \le n+1. $$ Then $\|C(t)\|=(1+2^{-t})$ for all $t \ge 1$, and $$ \|e_{0}-C(t)\|^{2}=1+\|C(t)\|^{2} $$ satisfies $\inf_{t \ge 1}\|e_{0}-C(t)\|=2$, a value which is not achieved for any $t \in [1,\infty)$. So there is no closest point to $e_{0}$ on the curve $C$.

The image of the curve $C$ is a complete subset of $H$ because (a) $[1,\infty)$ is a complete subset of $\mathbb{R}$, and (b) $\{ C(t_{n})\}_{n=1}^{\infty}$ is a Cauchy sequence iff $\{ t_{n}\}_{n=1}^{\infty}$ is a Cauchy sequence.

Answer 2

The existence of such a point holds for closed subsets of complete metric spaces. If $(X,d)$ is a complete metric space, $F\subset X$ a closed subset and $y\in X$, there is a sequence $(x_n)$ in $F$ such that $d(x_n,y)\rightarrow d(y,F)$. You can use the triangle equality to show that $(x_n)$ is a Cauchy sequence, hence the limit $x$ exists and $x\in F$, since $F$ is closed. The continuity if the distance function guarantees $d(x,y)=d(y,F)$.

Edit: You'd need $F$ to be compact, since $(x_n)$ isn't necessarily Cauchy. In an inner product space this can be shown, but not for arbitrary metric spaces.