Converse of existence of minimizers.

Question

Given $(V, \|\cdot\|)$ a real normed linear space, suppose it has the property that given any closed convex set $K$, there exists a $u_0 \in K$ such that $\|u_0\| \leq \|u\|$ for any $u \in K$. Does it imply the space is complete? What if we have a real inner product space instead? Does it imply our space is actually a Hilbert Space? The proof in establishing the existence of the minimizers in Hilbert space uses completeness at a crucial step but what I would like to know if that is actually necessary in general.

Edit. I think the following argument can show that if $(V,\langle \cdot , \cdot \rangle)$ is an inner product space, then existence of these minimizers implies that the space is complete: Given any closed subspace $M \subset V$, We can still define the orthogonal projection operator $P: V \rightarrow M$ such that $P(x)$ is the closest point to $x$ in that subspace. This will tell us that the decomposition of $V$ into $M \bigoplus M^{\perp}$ still goes through, and thus Riesz Representation Theorem still holds for any such space. Thus as $V^* \cong V$, the space is complete.

Does there exist anything for general Banach Spaces? (Perhaps it is necessary to add the existence of a unique minimizer)

Answer 1

We have the affirmative answer in the following partial case.

Let $(V',\|\cdot\|)$ be a completion of the space $(V,\|\cdot\|)$, that is $(V',\|\cdot\|)$ is a Banach space and $(V,\|\cdot\|)$ is a dense normed subspace of $(V',\|\cdot\|)$. Suppose that the space $V'$ has the following property: for each points $x,y\in V'$ there exists a unique point $z\in V$ such that $$\|z-x\|=\|z-y\|=\|x-y\|/2.$$

In particular, $V'$ has the property if $(V',\langle \cdot , \cdot \rangle)$ is a Hilbert space. Indeed, if $x,y,z\in V'$ are any points of $V'$ such that $z=(x+y)/2$ then $$\|z-x\|=\|z-y\|=\|x-y\|/2.$$ On the other hand, if $z'\in V'$ is any point such that $$\|z'-x\|=\|z'-y\|=\|x-y\|/2$$ then by the parallelogram equality we have $$\|x-y\|^2/2= \|z'-x\|^2+\|z'-y\|^2=$$ $$\|z'-z+(y-x)/2\|^2+\|z'-z-(y-x)/2\|^2=$$ $$2\|z'-z\|^2+2\|(y-x)/2\|^2,$$ so $\|z'-z\|=0$ and thus $z'=z$.

Now let $x$ be any nonzero point of $V'$ and $$B=\{y\in V':\|y-2x\|\le\|x\|\}.$$ Clearly, $x\in B$. On the other hand, by the triangle inequality for any $y\in B$ we have $$\|y\|\ge \|2x\|-\|y-2x\|\ge \|2x\|-\|x\|=\|x\|.$$ On the other hand, since the interior of $B$ is dense in $B$, and $V$ is dense in $V'$, $B\cap V$ is dense in $B$. So $$\|x\|=\inf_{y\in B} \|y\|=\inf_{y\in B\cap V} \|y\|.$$ Since the set $B$ is closed and convex in $V'$, the set $B\cap V$ is closed and convex in $V$. By the existence of minimizers in $V$, there exists a point $x'\in B\cap V$ such that $$\|x'\|=\inf_{y\in B\cap V} \|y\|=\|x\|.$$ Since $x'\in B$, $\|2x-x'\|\le \|x\|$. On the other hand, by the triangle inequality, $\|2x-x'\|\ge \|2x\|-\|x'\|=\|x\|$, so $\|2x-x'\|=\|x\|$. By the property of the space $V'$, $x'=x\in V$, so $V'=V$ and the space $V$ is complete.