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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?

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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 $x_0$.

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