# Example of empty projection in strictly convex Banach space

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)

• 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. – Narutaka OZAWA Apr 12 at 6:18

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 $$0$$)