Is $d^2(x,C)$ a convex function?

Question

Let $d(.,C):\mathbb{R}^n \rightarrow \mathbb R$ defined as $d(x,C)=\inf_{c \in C}||x-c||$ where $C$ is a nonempty closed convex subset of $\mathbb{R}^n$. We know that $d(.,C)$ is a convex function.

Is the function $d^2(.,C)$ convex?

I computed that function with $C$ is $R_+$ and $[0,1]$, then $d^2(.,C)$ is convex.

I also got the relation $d^2(x,C)=||x-P(x,C)||^2$ where $P(x,C)$ is projection mapping.

I guess it is convex. Do you have any idea?

Answer 1

It is easy to see that $d(x,C)$ is convex. Since $x \to x^{2}$ is an increasing convex function on $[0,\infty)$ the composition $d^{2}(x,C)$ is convex.

Answer 2

It's convex. For instance, its gradient is $x-P_Cx$, which is (firmly) nonexpansive and thus monotone. Or, you can argue that the function is an infimal convolution of the norm squared and the indicator function of $C$. This should be in any good book on Convex Analysis, starting with the trusted one by Rockafellar.