Question:
For a non-empty set $A\subseteq \mathbb{R}^n$ and $x\in \mathbb{R}^n$, define the distance from $x$ to $A$ by $d_A(x) = \inf_{y\in A}||x-y||$. Prove that if $A$ is convex, then $d_A(x)$ is a convex function.
Attempt:
We draw a straight line through points $x, x_1$ and $x_2$ so that $x = \lambda x_1 + (1-\lambda)x_2$. Then, we connect these dots to a circle, which represents set $A$, since our goal is to find the distance between $x$ and the circle. So, we will have:
$d_A(x) = d_A(\lambda x_1 + (1-\lambda)x_2) = \inf_{y\in A}||\lambda x_1 + (1-\lambda)x_2-y||$.
My challenge is, I can’t see where I will apply the condition that says “if $A$ is convex…”. If I should write $y$ in convex form (since $y\in A$), I suspect I will run into trouble.
I would really appreciate any help from you guys.
Let $\epsilon > 0$, and $y_1\in A$ and $y_2\in A$ such that,
\begin{align} \left\|x_1 - y_1\right\| &\le d_A(x_1) + \epsilon\\ \left\|x_2 - y_2\right\| &\le d_A(x_2) + \epsilon \end{align}
Then $\lambda y_1 + (1-\lambda) y_2\in A$ and,
\begin{align} d_A\left(\lambda x_1 + (1-\lambda) x_2\right) & \le \left\|\lambda x_1 + (1-\lambda)x_2 - (\lambda y_1 + (1-\lambda) y_2)\right\|\\ &\le \lambda \left\|x_1 - y_1\right\| + (1-\lambda) \left\|x_2 - y_2\right\|\\ &\le \lambda d_A\left(x_1\right) + (1-\lambda) d_A\left(x_2\right) + \epsilon \end{align}
Now $\epsilon \to 0$ and you will have what you are looking for