Existence of $q\in M$ that minimizes distance

Question

Problem: Let $g: \mathbb{R}^{n+1}\rightarrow\mathbb R$ differentiable with regular point $0$. Given a regular surface $M=g^{-1}(0)$ and $p \notin M$ ,

a) show $\exists\; q \in M$ such that

$$ \ \rvert \rvert{p-q} \rvert \rvert=\min\{ \rvert \rvert{p-q'} \rvert \rvert\;;\; q' \in M\}$$

b) show that $p-q\perp M$.

Context: This problem comes from the section of Lagrange multipliers. The last part $(b)$ was okay, but I'm a bit lost in the first one $(a)$. That not seems to be that hard (actually its quite intuitive that it should happen, since $M$ is regular).

But I'm cannot see how to write any good proof. Can someone give me a hint to start ?

Answer 1

A continuous function on a non-empty compact set attains its minimum. The only potential issue is that $M$ may not be compact.

Let $r$ be any point on $M$ and let $B$ be the closed ball centered at $p$ of radius $\|p-r\|.$

Argue that (1) $B \cap M$ is compact and non-empty; (2) the function $d(q) = \|p-q\|$ is continuous; (3) a minimizer $q$ on $B\cap M$ is also the minimizer on all of $M$.