# Condition for Banach space that the distance function can always reach its minimum at a unique point

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 $$b\not\in S$$, $$\left\|\frac{x+y}{2}-b\right\|=\left\|\frac{x-b}{2}+\frac{y-b}{2}\right\|\le \frac{\|x-b\|+\|y-b\|}{2} so we do not need to focus on the value on $$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 $$f\left(\frac{x+y}{2}\right)\le \max\{f(x),f(y)\}-\delta=m-\delta
Existence: Suppose $$(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), which follows that for $$n_k>N$$ $$m\le f\left(\frac{x_{n_{k}}+x_{n_{k+1}}}{2}\right)\le m+\frac{\delta}{2}-\delta which is a contradiction.
Thus $$(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$$.