Let $(X,\|\cdot\|)$ be a Banach space and $B\subset X$ be a bounded subset. Now define a function $f_B:X\to [0,\infty)$ by $f_B(x)=\sup_{b\in B}\|b-x\|$.
My question is in which kind of Banach space, $f_B$ can always reach its minimum at a unique point of $X$ for any given bounded subset $B$.
I find that the norm does matter: if $X=\mathbb{R}^2$ equipped with $l^2$ norm, it is true. However, if $X=\mathbb{R}^2$ equipped with $l^\infty$ norm, it is not true when $B=\{0,1\}\times[0,1]$.
I think the condition you need is the uniformly convexity: For any fixed $M>0$, $\forall \epsilon>0$, $\exists \delta>0$ such that $$\|x\|\le M,\|y\|\le M,\|x-y\|\ge \epsilon\Rightarrow\left\|\frac{x+y}{2}\right\|\le M-\delta.$$
Since $B$ is bounded, suppose $B\subset B_M=\{x\in X:\|x\|\le M\}$ for some $M>0$. It is clear that we only need to focus on the value of $f(x)$ in $B_{3M}$, because for $\|x\|>3M$, $\|x-b\|\ge\|x\|-\|b\|> 2M$. However, since $B\subset B_M$, the minimum value of $f$ should be $\le 2M$.
Let $m=\inf_{x\in X}f(x)=\inf_{x\in B_{3M}}f(x)$. Then for any fixed $\epsilon>0$, there exists $\delta>0$ such that $$\|x\|\le m,\|y\|\le m,\|x-y\|\ge \epsilon\Rightarrow\left\|\frac{x+y}{2}\right\|\le m-\delta.$$
Denote $T=\max\{\|x\|,\|y\|\}$ hence $\|x\|,\|y\|\le T$. If $m\le T\le 4M$. Denote $x'=\frac{mx}{T}, y'=\frac{my}{T}$, so $\|x'\|,\|y'\|\le m$. If $\|x-y\|\ge \frac{4M}{m}\epsilon$, i.e., $\|x'-y'\|\ge \frac{4M}{T}\epsilon\ge \epsilon$, then \begin{align*} \left\|\frac{x+y}{2}\right\|= \frac{T}{m}\left\|\frac{x'+y'}{2}\right\| \le\frac{T}{m}(m-\delta) = T-\frac{T}{m}\delta \le T-\delta =\max\{\|x\|,\|y\|\}-\delta \end{align*}
To sum up, for any $\epsilon>0$, there exists $\delta>0$ for any $x,y\in B_M$ with $m\le\max\{\|x\|,\|y\|\}\le 4M$ and $\|x-y\|\ge \epsilon$, such that $$\left\|\frac{x+y}{2}\right\|\le \max\{\|x\|,\|y\|\}-\delta.$$
Given $\epsilon>0$, for fixed $x,y\in B_{3M}$, $\|x-y\|\ge\epsilon$, let $S=\{b\in B: \|x-b\|\ge m\text{ or }\|y-b\|\ge m\}$. Then for all $b\not\in S$, $$\left\|\frac{x+y}{2}-b\right\|=\left\|\frac{x-b}{2}+\frac{y-b}{2}\right\|\le \frac{\|x-b\|+\|y-b\|}{2}<m,$$ so we do not need to focus on the value on $B\setminus S$.
For $b\in S$, since $\|x-b\|\le \|x\|+\|b\|\le 4M$, $\|y-b\|\le 4M$, hence $m\le\max\{\|x-b\|,\|y-b\|\}\le 4M$ and $\|(x-b)-(y-b)\|=\|x-y\|\ge\epsilon$, by above inequality, we have \begin{align*} \left\|\frac{x+y}{2}-b\right\|&\le \max\{\|x-b\|,\|y-b\|\}-\delta\\ &\le \max\{\sup_{b\in S}\|x-b\|,\sup_{b\in S}\|y-b\|\}-\delta\\ & = \max\{\sup_{b\in B}\|x-b\|,\sup_{b\in B}\|y-b\|\}-\delta\\ & = \max\{f(x),f(y)\}-\delta. \end{align*} Thus \begin{align*} f\left(\frac{x+y}{2}\right)=\sup_{b\in B}\left\|\frac{x+y}{2}-b\right\|&=\sup_{b\in S}\left\|\frac{x+y}{2}-b\right\|\le \max\{f(x),f(y)\}-\delta. \end{align*}
Uniqueness: If $x,y\in X$ such that $f(x)=f(y)=m$, then $\exists \delta>0$ such that
$$f\left(\frac{x+y}{2}\right)\le \max\{f(x),f(y)\}-\delta=m-\delta<m,$$
which is a contradiction.
Existence: Suppose $(x_n)$ is a sequence in $B_M$ such that $f(x_n)\to m$. We shall prove that $(x_n)$ is a Cauchy sequence. If not, there exists $\epsilon>0$ and a subsequence $(x_{n_k})$ such that $\|x_{n_{k+1}}-x_{n_k}\|\ge\epsilon$. Thus $$f\left(\frac{x_{n_{k}}+x_{n_{k+1}}}{2}\right)\le \max\{f(x_{n_{k}}),f(x_{n_{k+1}})\}-\delta.$$ Since $f(x_n)\to m$, there exists $N>0$ such that $\forall n>N$, $f(x_n)<m+\frac{\delta}{2}$, which follows that for $n_k>N$ $$m\le f\left(\frac{x_{n_{k}}+x_{n_{k+1}}}{2}\right)\le m+\frac{\delta}{2}-\delta<m,$$ which is a contradiction.
Thus $(x_n)$ is a Cauchy sequence. Since $X$ is a Banach space hence complete, $(x_n)$ converges to a unique element $x_0\in X$ and $f(x_0)=\inf_n f(x_n)=m$.
