Looking for a Banach space in which projection on a closed convex set is empty?

Question

It is well-known that projection on closed convex sets is nonempty in Hilbert spaces. I am interested in seeing a Banach space where the projection is empty on a closed convex set.

Definition of Projection of $x $ on set $C$:

$$ P_{C} (x) := \{ y \in C ~| ~ d(x , C ) = \|x -y\| \} $$

Answer 1

This MO answer solves your question. First, a small technicality: if $C$ is empty, it is a closed convex set such that all projections on $C$ are empty. This is technically a solution, but we want to exclude it.

Let $X$ be a non-reflexive Banach space (e.g. $\ell^1$). Let $f : X \to \mathbb{R}$ be a continuous linear functional of norm $1$ on $X$ which does not attain a maximum on the closed unit ball of $X$ (e.g. $(x_n) \mapsto \sum_{n \geq 1} \frac{n-1}{n} x_n : \ell^1 \to \mathbb{R}$). Let $C = \{x \in X : f(x) = 1, \lVert x \rVert \leq 2\}$.

$C$ is closed and convex, since it is the intersection of the closed convex sets $f^{-1}(1)$ and $\{x \in X : \lVert x \rVert \leq 2\}$.

Since $\lVert f \rVert = \sup_{\lVert x \rVert = 1} \lvert f(x) \rvert = 1$, there is some $a \in X$ such that $f(a) = 1/2$ and $\lVert a \rVert \leq 1$. Then $2a \in C$, so $C$ is nonempty.

Now suppose for contradiction that $P_C(0)$ is nonempty; let $y \in P_C(0)$ -- in other words, $y \in C$ has minimal norm.

Since $f(y) = 1$, we have $y \neq 0$. Thus, let $z = \frac{1}{\lVert y \rVert} y$. Since $f$ does not attain a maximum on the unit ball, there exists some $z'$ such that $\lVert z' \rVert = 1$ and $f(z') > f(z)$. Now let $y' = \frac{1}{f(z')}z'$, so that $f(y') = 1$. Moreover, $$\lVert y \rVert \leq 2 \implies f(z) = \frac{1}{\lVert y \rVert} f(y) = \frac{1}{\lVert y \rVert} \geq \frac{1}{2},$$ so $$\lVert y' \rVert = \frac{1}{\lvert f(z') \rvert} \lVert z' \rVert = \frac{1}{f(z')} < \frac{1}{f(z)} \leq 2.$$ Thus, $y' \in C$. Finally, $$\lVert y' \rVert = \frac{1}{\lvert f(z') \rvert} \lVert z' \rVert = \frac{1}{f(z')} < \frac{1}{f(z)} = \frac{1}{\frac{1}{\lVert y \rVert} f(y)} = \lVert y \rVert.$$ This contradicts the prior conclusion that $y \in C$ has minimal norm.