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 $x \in B_1$?
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 $\partial B_1$.