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Let $X$ be a strictly convex Banach space, let $C\subseteq X$ be a nonempty closed convex set, and let $P_C$ be the set-valued metric projection $$P_C(x) = \{y\in C : \|x-y\| = d(x,C)\} . $$ We know that since $X$ is strictly convex, the metric projection onto a closed convex set is either empty or a singleton.

My question: Does anyone know of examples where $P_C(x)$ is empty? (assuming $X$ is a strictly convex Banach space and C is a nonempty closed convex set)

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    $\begingroup$ For example, consider the kernel of a linear functional which does not attain its norm. This is optimal because James's theorem says such a linear functional exists as long as $X$ is non-reflexive. $\endgroup$ – Narutaka OZAWA Apr 12 at 6:18
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To complete Narutaka OZAWA's answer in comment by a concrete example as asked in the OP, here is a bounded linear functional on $c_0$ not attaining its norm w.r.to (an equivalent) strictly convex norm.

On the space $c_0 $ consider a norm $\Vert x\Vert :=\Vert x\Vert _\infty +\Vert x\Vert _2$, which is obtained adding a (weighted) pre-Hilbert norm, hence strictly convex
$$\Vert x\Vert _2:=\sqrt{\sum_{k=1}^\infty 2^{-k}x_k^2}$$ to the standard norm $$\Vert x\Vert_\infty:=\max_{k\ge1}|x_k|.$$ Thus $ \Vert x\Vert _\infty\le \Vert x\Vert\le 2 \Vert x\Vert _\infty$, so $\|\cdot\|$ is equivalent to $\|\cdot\|_\infty$, and it is strictly convex.

Consider the linear form $f\in(c_0)^*=\ell_1$ defined by $f_k:=2^{-k}$. Then for any $x\in c_0\setminus\{0\}$ $$\langle f,x\rangle=\sum_{k=1}^\infty 2^{-k}x_k\le\sqrt{\sum_{k=1}^\infty 2^{-k}}\sqrt{\sum_{k=1}^\infty 2^{-k}x_k^2}=\|x\|_2 $$ and also $$\langle f,x\rangle<\Big( \sum_{k=1}^\infty 2^{-k}\Big)\|x\|_\infty=\|x\|_\infty =\|x\|-\|x\|_2.$$ Therefore $\displaystyle\langle f,x\rangle<\frac12\|x\|$ for all $x\neq 0$. On the other hand, for the sequence $x^n:=\sum_{k=1}^ne_k=({1,\dots,1},0,0\dots)\in c_0$ one has $$\frac{\langle f,x^n\rangle}{\|x^n\|}=\frac{ \sum_{k=1}^n 2^{-k}}{1+\sqrt{\sum_{k=1}^n 2^{-k}}}=\frac{ 1- 2^{-n}}{1+\sqrt{1- 2^{-n}}}=\frac12+o(1).$$ So $f$ has dual norm $\displaystyle \frac12$, but it is not attained on the closed unit ball.

And the affine closed hyperplane $C:=\{f=\frac12\}$ has no element of minimum norm (minumum distance from $0$)

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