TITLE: Distance to the boundary function in a disk
QUESTION [5 upvotes]: Let $B_1 = B(0, 1)$ denote the open unit disk in $\mathbb{R}^d$. For $x \in B_1$, the "distance to the boundary $\partial B_1$" function is defined as
\begin{equation}
d(x) = \inf_{y\in \partial B_1} \|x-y\|.
\end{equation}
This definition is of course more general and not specific to $B_1$. 
My question is the following:

In the special case of the unit disk $B_1$, is it true (and if yes, why) that 
  \begin{equation}
d(x) = 1-\|x\|
\end{equation}
  for any $x \in B_1$?

REPLY [3 votes]: It's true. For all $x\in B_1$ you have
$$ \|x-y\|\ge \big|\|x\|-\|y\|\big|= \big|\|x\|-1\big|=1-\|x\| \quad \text{for all $y\in \partial B_1$,}$$
and hence
$$\inf_{y\in \partial B_1} \|x-y\|\ge 1-\|x\|.$$
On the other hand, for $x\neq 0$ we have $x/\|x\|\in\partial B_1$ which implies
$$\inf_{y\in \partial B_1} \|x-y\|\le \|x-x/\|x\|\|=\big|1-1/\|x\|\big|\cdot\|x\|=1-\|x\|.$$
For $x=0$ the infimum is attained at every point in $\partial B_1$.