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Let $C \subset \mathbb{R^{n}}$ a convex set. Fixed $p \in \mathbb{R}^{n}$, let $\varphi: C \longrightarrow \mathbb{R}$ the function defined by $\varphi(x) = |x-p|=\sqrt{\langle x-p,x-p\rangle}$. Theres exists as most one point $a \in C$ such that $\varphi(a) = \inf \lbrace \varphi(x) | x \in C\rbrace$.

I don't know how to relate convexity to other hypothesis. I thought suppose $a \neq b$ such that $\varphi(a) = \varphi(b) = \inf\lbrace \varphi(x) | x \in C\rbrace$ and use $\varphi(a) \leq \varphi(b)$, $\varphi(b) \leq \varphi(a)$ so $\varphi(a) = \varphi(b)$. But I can't do anything with it anymore. I appreciate any hint.

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    $\begingroup$ Does the convexity help? The line segment between $a$ and $b$ must be contained in $C$. And yet the minimum is achieved on both sides of that line segment. $\endgroup$ – Matthew Leingang Mar 19 '18 at 0:17
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You need C to be closed btw. Notice that $\phi(x)$ is strictly convex. So if there were two minimizers then there would be a point $z$ between them with strictly lower value $\phi(z)$. Also, there must be a minimum on C if C is closed.

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As mentioned, we need $C$ to be closed for this. Assume $a, b \in C$ are both distance minimizers to $p$. Let $\varphi(a) = \varphi(b) = d$ be the minimum value.

Using the paralellogram identity, we obtain:

\begin{align} 4d^2 &= 2\|a - p\|^2 + 2\|b-p\|^2 \\ &= \|(a-p)+(b-p)\|^2 + \|(a-p)-(b-p)\|^2 \\ &= 4\underbrace{\left\|\frac{a+b}2 - p\right\|^2}_{\ge d^2} + \|a - b\|^2\\ &\ge 4d^2 + \|a - b\|^2 \end{align}

Because $C$ is convex, we also have $\frac{a+b}2 \in C$ so $\left\|\frac{a+b}2 - p\right\| = \varphi\left(\frac{a+b}2\right) \ge d$.

We conlude $\|a-b\| = 0$ which implies $a = b$.

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