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Show that

$$ g(x)=\inf_{z \in C}\|x-z\| $$ where $g:\mathbb{R}^n \rightarrow \mathbb{R}$, $C$ is a convex set in $\mathbb{R}^n$ (nor close neither bounded), and $\|\cdot\|$ is a norm on $\mathbb{R}^n$. Let $x,y$ be in $\mathbb{R}^n$. We need to show that

$$ g(\lambda x +(1-\lambda)y) \leq \lambda g(x)+ (1-\lambda)g(y) \tag{1} $$

I tried the following:

$$ \|\lambda x +(1-\lambda)y-z\| \leq \lambda\| x -z\| + (1-\lambda)\| y-z\| \,\, \forall {z \in C} $$ Since

$$ g(\lambda x +(1-\lambda)y)=\inf_{z \in C}\|\lambda x +(1-\lambda)y-z\| \leq \|\lambda x +(1-\lambda)y-z\| \,\, \forall {z \in C} $$

So

$$ g(\lambda x +(1-\lambda)y)=\inf_{z \in C}\|\lambda x +(1-\lambda)y-z\| \leq \lambda\| x -z\| + (1-\lambda)\| y-z\| \,\, \forall {z \in C} $$

I do not know how to handle the right hand side and apply infimum in a right way because the following is not correct in general

$$ \inf_{z \in C}\|\lambda x +(1-\lambda)y-z\| \nleq \lambda \inf_{z \in C} \| x -z\| + (1-\lambda) \inf_{z \in C} \| y-z\| $$

Or maybe my initial way to prove the convexity is wrong. Can you complete my proof or show the claim using another way?

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  • $\begingroup$ Hint: you can prove this for any jointly convex function $f(x,z)$, not just for $f(x,z) = ||x-z||$. $\endgroup$ – LinAlg Dec 2 '18 at 23:44
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Hint: try starting from the opposite direction, consider the subadditivity of the infimum and see if you can show it that way

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  • $\begingroup$ Your statement is not an answer, please write it as a comment and delete your answer. $\endgroup$ – Saeed Dec 3 '18 at 4:17

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