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?
Best Answer
Suppose $C$ is also closed, there exists $z_1,z_2\in C$ such that $g(x)=||x-z_1||$, $g(y)=||y-z_2||$. Then $$\lambda g(x)+(1-\lambda)g(y)\ge ||\lambda x + (1-\lambda)y - \lambda z_1-(1-\lambda)z_2||.$$ Using the convexity of $C$, $z=\lambda z_1+ (1-\lambda)z_2\in C$. Hence, $$\lambda g(x)+(1-\lambda)g(y)\ge ||\lambda x + (1-\lambda)y - z||\ge g(\lambda x + (1-\lambda)y).$$ If $C$ is not closed, probably some limit argument may plus the above reasoning should work.