How to prove that $f(x)+\|z-x\|_H^2$ posesses a strict minimum? Let $H$ be a Hilbert space, suppose $f:H \to (-\infty,\infty]$ is lower semi-continuous and convex, for arbitrary $z\in H$,let $g(x)=f(x)+\|z-x\|_H^2$, how to prove that $g$ possesses a strict minimum?
 A: hint: Check out the properties of strictly convex functions, and what happens when you add them to another (not necessarily strict) convex function.
A: Hint:

*

*Prove that $g(x) := f(x) + |z-x|^2$ is coercive, i.e. $\lim_{|x|\to +\infty} g(x) = +\infty$. (Remember that, being $f$ convex, it is bounded from below by a hyperplane.)


*Take any point $x_0\in H$ where $f(x_0) < +\infty$, and consider the set
$$
C := \{ x \in H:\ g(x) \leq g(x_0)\}.
$$
This set is non-empty (since $x_0\in C$), convex (since $g$ is convex), closed (since $g$ is l.s.c.) and bounded (since $g$ is coercive).
Then it is compact in the weak topology. Moreover, $g$ is l.s.c. also with respect to the weak topology (being convex and l.s.c. w.r.t the strong topology).


*By Weierstrass's theorem, $g$ has a minimum point in $C$ (and then in $H$). By strict convexity, this minimum point is unique.

A: I am assuming that $\operatorname{dom} f $ is non empty.
It is straightforward to show using the parallelogram identity that the map $x \mapsto \|z-x\|^2$ is strictly convex and we have $\operatorname{dom} f = \mathbb{H}$.
It is straightforward to show that the sum of a convex function and a strictly convex function is strictly convex on the intersection of their (convex) domains.
It is straightforward to show by contradiction that a strictly convex function has a unique minimiser.
