Showing two conditions are equivalent for a non-empty convex set $K$ of a Hilbert Space $X$.

Question

My question is the following:

Let $K$ be a non-empty convex set of a Hilbert space $X$, and let $y\in K$. Prove that the following are equivalent:

(1) $||x-y||\le||x-z||$ for all $x\in X, z\in K$;

(2) $\langle x-y,z-y\rangle \le 0$ for all $x\in X, z\in K$.

I have been given the clue of considering vectors $(1-t)y+tz$ for $t\in (0,1)$, which will be in $K$ as this is a convex set. I assume that the expression ought to be substituted for $z$ somewhere, but I can't work out how, or which value of $t$ to take.

Moreover, the question might be related to the Closest Point Theorem, which guarantees that, if $K$ is closed, there is a unique point satisfying (1).

Answer 1

You will find the proof (with even a figure !) in Bourbaki Topological Vector Spaces Ch V § 5 Theorem 1.

Note : You don't need $K$ to be closed for the equivalence, only if $K$ is closed it necessarily exists that closest point $y$.