Then every approximately compact set in this space is proximal.

Question

Let $(X, \langle \cdot, \cdot \rangle)$ be a vector space over $\mathbb{R}$ with an inner product. Then every approximately compact set in this space is proximal. Now, I want to find an example where the converse of the above statement does not necessarily hold. My attempt: consider the set $K = \{y \in l_2 \mid \|y\| = 1\}$ and the sequence $(e_n)$ defined by $e_n(i)=$ $$ \delta_{n,i} = \begin{cases} 0, & \text{if } n \neq i, \\ 1, & \text{if } n = i \end{cases} \quad \forall n, i \in \mathbb{N}. $$ Observe the following:

• $e_n \in K$ and $d(0, K) = 1 = \|e_n\|$ for all $n \in \mathbb{N}$, hence the sequence $(e_n)$ in $K$ is minimizing for the point $0$.
• $\|e_n - e_m\| = \sqrt{2}$ for all $n, m \in \mathbb{N}, n \neq m$, so no subsequence of $(e_n)$ is convergent, and thus the set $K$ is not approximately compact.

Now I want to prove:

1. The set $K$ is not convex.
2. $P_K(x) = \frac{x}{\|x\|}$ for all $x \neq 0$ and $P_K(0) = K$.
3. The set $K$ is proximal but not Chebyshev.

For 1. point it seems obvious that $K$ in not convex, but how would I prove it formally. Point 2. I just don't know how to even approach. Any for point 3. why does point 1. and 2. imply point 3.? Why do we even need that $K$ is not Chebyshev here? Thanks for all your help in advance.

Note: Let $K$ be a nonempty subset of the inner product space $X$ and let $x \in X$. An element $y_0 \in K$ is called a best approximation, or nearest point, to $x$ from $K$ if $$ \|x - y_0\| = d(x, K), $$ where $d(x, K) := \inf_{y \in K} \|x - y\|$. The number $d(x, K)$ is called the distance from $x$ to $K$, or the error in approximating $x$ by $K$.

The (possibly empty) set of all best approximations from $x$ to $K$ is denoted by $P_K(x)$. Thus $$ P_K(x) := \{ y \in K \mid \|x - y\| = d(x, K) \}. $$

This defines a mapping $P_K$ from $X$ into the subsets of $K$ called the metric projection onto $K$.

If each $x \in X$ has at least (respectively exactly) one best approximation in $K$, then $K$ is called a proximal (respectively Chebyshev) set. Thus $K$ is proximal (respectively Chebyshev) if and only if $P_K(x) \neq \emptyset$ (respectively $P_K(x)$ is a singleton) for each $x \in X$.

Answer 1

The first point follows from $\pm e_1 \in K$ and $\frac{1}{2}e_1+\frac{1}{2}(-e_1)=0\notin K$. The third point follows trivially from the second point. We are left to prove the second point. It's clear that $P_K(0)=K$ as all elements in $K$ have distance $1$ to the origin. Thus, we need to show that $P_K(x)=x/\Vert x \Vert$ for $x\neq 0$. We first consider the special case $x=c e_1$ for some $c>0$ and then deduce the general case from it.

Special case: $P_K(ce_1)=\{e_1\}$ for $c>0$: We need to find $x=(x_j)_j \in K$ which minimizes $\Vert x-ce_1 \Vert^2$. Using that $x\in K$ we can write $$ \Vert x-ce_1\Vert^2 = \vert x_1-c \vert^2 + \sum_{j\geq 2} \vert x_j\vert^2 = \vert x_1-c \vert^2 + 1-\vert x_1\vert^2. $$ Write $x_1=a+ib$. Then we need to minimize $$ f(a,b) = (a-c)^2+b^2+1-a^2-b^2=1-a^2+(a-c)^2 $$ under the constraint $a^2+b^2\leq 1$ (remember that $x\in K$ and hence $a^2+b^2=\vert x_1\vert^2 \leq 1$). The function $f$ does not depend on $b$ at all, so we really only need to minimize $$ g(a) = 1-a^2+(a-c)^2$$ under the constraint $a^2\leq 1$. Basic calculus gives us that the minimum is at $a=1$ (as $c>0$) and hence the overall minimizer is $x_1=1$ (as $\vert x_1\vert \leq 1$ which forces $b=0$), respectively $x=e_1$.

General case: We note that $P_K(x)$ is the points $y\in K$ which have minimal distance to $x$. Fix $x_0\neq 0$ and pick a unitary map $U: \ell^2 \rightarrow \ell^2$ such that $U(x_0)=\Vert x_0\Vert e_1$. How do we get this unitary map? Well, you can start with $f_1=x_0/\Vert x_0\Vert$ and extend it to a Hilbert basis $(f_n)_{n\in \mathbb{N}}$, see here How to extend an orthonormal set to a basis on a Hilbert space?. Now define $U(f_n)=e_n$ and extend this linearly. It's easy to check that this is unitary on the span of $(f_n)_{n\in \mathbb{N}}$ (as it maps an orthonormal basis to another orthonormal basis) and thus extend to a unitary map on all of $\ell^2$.

Note that $U(K)=K$ (as $U$ is unitary we have $\Vert U(x)\Vert = \Vert x\Vert$).

Now, if there exists a single point $z\in K$ which minimizes the distance to $U(x_0)$, then $U^{-1}(z)$ is the unique minimizer for $x_0$ (here we use again that $U$ is unitary and hence $\Vert z-U(x_0) \Vert= \Vert U^{-1}(z)-x_0\Vert$). In formulas, we get $$ P_K(x_0) = U^{-1}P_{U(K)}(U(x_0))=U^{-1}P_K(\Vert x_0\Vert e_1) = \{U^{-1}(e_1) \} = \{ x_0/\Vert x_0\Vert \}, $$ where we have used the previously established special case $P_K(ce_1)=\{ e_1\}$ for $c>0$.