H, If $K$ is a non-empty convex and closed subset of a uniformly convex Banach space $X$ (Hilbert for example) and $v \notin K$, we know that there exists a unique $k_0\in K$ such that $|v-k_0|=d(v,K):=\inf_{k\in K}|v-k|$.
1) My first question is: does this property hold for non uniformly convex Banach spaces, maybe by some additional assumptions on the convex $K$ ?
2) My second question is: Is there some results in the case $K$ is not convex, but convex on the "area" near $v$ as we can see it on $\mathbb{R}^2$ in this picture:
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1) If $X$ is not uniformly convex, there exist $x,y\in S_X$ ($S_X$ being the unit sphere of $X$) with $\|x-y\|\ge\epsilon$ for which $\|x+y\|/2\ge 1$. Take $v=x+y$. Then $d(v,S_X)=1$ as for every $u\in S_X$ we have $d(v,u)=\|v-u\|\ge |\|x+y\|-\|u\||\ge 2-1=1$. But we also have $d(v,x)=\|y\|=1=\|x\|=d(v,y)$. So the inf of the distance is reached twice.
As an example you can take $\mathbb{R}^2$ with the $\|\cdot\|_\infty$, $x=(1,0), y=(1,1)$.
2)If there is r>0 such that $B_X(r)=\{x\in X: \|x\|<r\}$ and $K\cap \overline{B_X(r)}$ is convex you can still apply the result as every point $p\in K\setminus \overline{B_X(r)}$ would have $d(v,p)\ge r$.