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 $B=\{0,1\}\times[0,1]$.



Condition for Banach space that the distance function can always reach its minimum at a unique point
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 $f(x\_0)=\inf\_n f(x\_n)=m$.


