# Unique minimizer of a quadratic functional $f(x)= \frac{1}{2}||x||^2-\phi(y)$

The question is the following:

Let $$\mathcal{H}$$ be a real Hilbert space, and $$\phi \in \mathcal{H}^*$$. Define the quadratic functional $$f: \mathcal{H} \to \mathbb{R}$$ by $$f(x)= \frac{1}{2}||x||^2-\phi(x)$$ Prove that there is a unique element $$\bar{x} \in \mathcal{H}$$ such that $$f(\bar{x}) = \inf_{x \in \mathcal{H}}f(x)$$

I know that

if a function $$f:C \to \mathbb{R}$$ is strongly lower semicontinuous, strictly convex on a strongly closed, convex, bounded subset $$C$$ of a Hilbert space. Then $$f$$ is bounded below and attains its infimum uniquely.

It also implies that

if a function $$f:\mathcal{H}\to \mathbb{R}$$ is coercive, strongly lower semicontinuous, convex function on a Hilbert space $$\mathcal{H}$$. Then $$f$$ is bounded below and attains its infimum.

It seems that the infimum is achieved from the second one because of the norm squared, but how can I get the uniqueness?

We can write $$\phi (x) =\langle x , x_0 \rangle$$ for some $$x_0$$. Let $$y$$ be any point where the minimum of $$f$$ is attained. Then $$\frac 1 2 \|y\|^{2} \leq \frac 1 2 \|x+y\|^{2}-\langle x, x_0 \rangle$$ for all $$x$$. This gives $$\|y\|^{2} \leq \|x\|^{2}+\|y\|^{2}+2 \langle x, y \rangle -2 \langle x, x_0 \rangle$$. So $$2\langle x , x_0-y \rangle \leq \|x\|^{2}$$. Put $$x =x_0-y$$ in this to get $$2\|x_0-y\|^{2} \leq \|y-x_0\|^{2}$$ This implies $$\|y-x_0\|=0$$ and $$y=x_0$$. So the only point where the infimum is attained is $$x_0$$.