If $f$ is bounded linear functional, then $\|x\|^2-f(x)$ is bounded from below Can someone give me a hint of how to do the following:
Let $H$ be a real, Hilbert space and let $f: H \to \mathbb R$ be a bounded, linear functional.
Prove that the function $g: H \to \mathbb R$ defined with $g(x)=\|x\|^2-f(x)$ is bounded from below. Moreover, for any nonempty, convex and closed set, $g$ achieves its infimum precisely in one point of that set.
 A: Since $f$ is continuous $f(x)\le\|f\|\|x\|$, then $g(x)\ge \|x\|^2-\|f\|\|x\|$ that is a parabola bounded from below in the variable $\|x\|$, so $g$ is bounded from below.
Now take $C$ nonempty, covex and closed, then it is also weakly closed. Consider $x_0\in C$ and let $\alpha=g(x_0)$. The set $S:=\{x\in H|g(x)\le\alpha \}$ is bounded, in fact if $x\in S$ then $\|x\|^2-f(x)\le \alpha$ and then $\|x\|^2\le\alpha+\|f\|\|x\|$, so there is $R$ such that $\|x\|\le R$, that is $S\subset B_R$ (the ball is taken closed). Finally $C\cap B_R$ is nonempty and weakly compact (since $B_R$ is weakly compact by the fact that $H$ is reflexive) and also $\inf_C g(x)=\inf_{C\cap B_R}g(x)$. Since $g$ is weakly lower semicontinuous (because $\|\cdot\|$ is weakly lower semicontinuous) it admits a minimum on the weakly compact set $C\cap B_R$, that is also a minimum on $C$.
The minimizer is unique. In fact if by contradiction $y,z\in C$ are different minimizers (so $g(y)=g(z)=\beta)$, by convexity of $C$ we have $y/2+z/2\in C$ and by strict convexity of $(\cdot)^2$:
$g(y/2+z/2)\le (\frac{1}{2}\|y\|+\frac{1}{2}\|z\|)^2-\frac{1}{2}f(y)-\frac{1}{2}f(z)<\frac{1}{2}\|y\|^2+\frac{1}{2}\|z\|^2 -\frac{1}{2}f(y)-\frac{1}{2}f(z)=\beta=\inf_C g(x) $
that is impossible.
