Addison Trujillo

Answered

2022-07-13

Let ${\varphi}_{1},...,{\varphi}_{k}$ be linear functionals on ${\mathbb{R}}^{n}$, ${c}_{1},...{c}_{k}\in \mathbb{R}$ and

$H=\{x\in {\mathbb{R}}^{n}|\phantom{\rule{thickmathspace}{0ex}}|{\varphi}_{j}(x)|\le {c}_{j},\phantom{\rule{thickmathspace}{0ex}}j=1,..,n\}$

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:

$|{\varphi}_{1}(x)|\le {c}_{1}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|{\alpha}_{11}{x}_{1}+...+{\alpha}_{1n}{x}_{n}|\le {c}_{1}$

$|{\varphi}_{2}(x)|\le {c}_{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|{\alpha}_{21}{x}_{1}+...+{\alpha}_{2n}{x}_{n}|\le {c}_{2}$

...

$|{\varphi}_{k}(x)|\le {c}_{k}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|{\alpha}_{k1}{x}_{1}+...+{\alpha}_{kn}{x}_{n}|\le {c}_{k}$

$H=\{x\in {\mathbb{R}}^{n}|\phantom{\rule{thickmathspace}{0ex}}|{\varphi}_{j}(x)|\le {c}_{j},\phantom{\rule{thickmathspace}{0ex}}j=1,..,n\}$

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:

$|{\varphi}_{1}(x)|\le {c}_{1}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|{\alpha}_{11}{x}_{1}+...+{\alpha}_{1n}{x}_{n}|\le {c}_{1}$

$|{\varphi}_{2}(x)|\le {c}_{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|{\alpha}_{21}{x}_{1}+...+{\alpha}_{2n}{x}_{n}|\le {c}_{2}$

...

$|{\varphi}_{k}(x)|\le {c}_{k}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|{\alpha}_{k1}{x}_{1}+...+{\alpha}_{kn}{x}_{n}|\le {c}_{k}$

Answer & Explanation

Jamarcus Shields

Expert

2022-07-14Added 17 answers

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

2022-07-15Added 5 answers

Take the elements of the dual of the standard basis, so ${\varphi}_{i}(x)={x}_{i}$ for $x=({x}_{1},\dots ,{x}_{n})$. 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.

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