2022-07-13

Let ${\varphi }_{1},...,{\varphi }_{k}$ be linear functionals on ${\mathbb{R}}^{n}$, ${c}_{1},...{c}_{k}\in \mathbb{R}$ and
$H=\left\{x\in {\mathbb{R}}^{n}|\phantom{\rule{thickmathspace}{0ex}}|{\varphi }_{j}\left(x\right)|\le {c}_{j},\phantom{\rule{thickmathspace}{0ex}}j=1,..,n\right\}$
Under what conditions is H compact? So my intuition tells me we must have $k\ge n$, and ${\varphi }_{1},...,{\varphi }_{k}$ and at least n of them must be linearly independent in the dual space, not sure how to prove it though, since we're working with a system of inequalities rather then equalities. That is, my idea was to write:
$|{\varphi }_{1}\left(x\right)|\le {c}_{1}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}|{\alpha }_{11}{x}_{1}+...+{\alpha }_{1n}{x}_{n}|\le {c}_{1}$
$|{\varphi }_{2}\left(x\right)|\le {c}_{2}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}|{\alpha }_{21}{x}_{1}+...+{\alpha }_{2n}{x}_{n}|\le {c}_{2}$
...
$|{\varphi }_{k}\left(x\right)|\le {c}_{k}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}|{\alpha }_{k1}{x}_{1}+...+{\alpha }_{kn}{x}_{n}|\le {c}_{k}$

Jamarcus Shields

Expert

Hint: The set $H$ is a closed subset of ${\mathbb{R}}^{n}$ by construction, so by the Heine Borel theorem, it is compact if and only if it is bounded. This is the case if and only if the functionals ${\varphi }_{1},\dots ,{\varphi }_{k}$ span the dual space. If they don't span the dual, then they have a non-zero common kernel and that common kernel is contained in $H$ which thus is unbounded. On the other hand, if they span the dual, then one easily concludes that the image of $H$ under each linear functional is bounded, which easily implies that $H$ is bounded.

ScommaMaruj

Expert

Take the elements of the dual of the standard basis, so ${\varphi }_{i}\left(x\right)={x}_{i}$ for $x=\left({x}_{1},\dots ,{x}_{n}\right)$. So you get information like that $x\in H$ implies $|{x}_{i}|\le {a}_{i}$ for some constant ai and all $i=1,\dots ,n$ and this implies that $H$ is bounded.