0

Let $\mathcal{H}$ be a real Hilbert space. A function $f \colon \mathcal{H} \to \mathbb{R} \cup \left\lbrace + \infty \right\rbrace$ is inf-compact if for any $r > 0$ and $\kappa \in \mathbb{R}$, \begin{equation*} \textrm{Lev}_{\kappa}^{r} \left( f \right) := \left\lbrace x \in \mathcal{H} \colon \left\lVert x \right\rVert \leq r , f \left( x \right) \leq \kappa \right\rbrace \end{equation*} is relatively compact set in $\mathcal{H}$ (A relatively compact set is the set which its closure is compact).

In particular, I am interested to check if the following function is inf-compact or not

$$ f \left( x_{1} , x_{2} \right) := \dfrac{1}{2} \left\lVert x_{1} - x_{2} \right\rVert ^{2} , \forall \left( x_{1} , x_{2} \right) \in \mathcal{H}_{1} \times \mathcal{H}_{2} . $$

We can show that $\textrm{Lev}_{\kappa}^{r} \left( f \right)$ is always a closed and bounded set. Hence when $\mathcal{H}_{1} \times \mathcal{H}_{2}$ is a finite dimensional space, $\textrm{Lev}_{\kappa}^{r} \left( f \right)$ is a compact set and thus $f$ is inf-compact. However, for a infinite dimensional space we can not do the same. In fact, as I found we can only prove that ($\textrm{Lev}_{\kappa}^{r} \left( f \right)$ is weakly relative compact). How can I conclude from this?

ViktorStein
  • 5,024
JKay
  • 647

1 Answers1

1

I don't think $\operatorname{Lev}_\kappa^r(f)$ is relatively compact for your $f$ if one the Hilbert spaces (w.l.o.g. $\mathcal{H}^1$) is infinite dimensional, since the set $$\{(x_1,0)\in \mathcal{H}^1\times\mathcal{H}^2|\ \|x_1\|_{\mathcal{H}^1}\leq \sqrt{2\min{(\kappa,r)}}\}\subset \operatorname{Lev}_\kappa^r(f).$$ This set is basically the ball with radius $\sqrt{2\min{(\kappa,r)}}$ in $\mathcal{H}^1$, which is not compact (see e.g. Is any closed ball non-compact in an infinite dimensional space?).

humanStampedist
  • 3,632