On compact set defined by functionals

Question

Let $\phi_1,...,\phi_k$ be linear functionals on $\mathbb{R}^n$, $c_1,...c_k\in\mathbb{R}$ and $$H = \{x\in \mathbb{R}^n |\;|\phi_j(x)|\leq c_j,\;j=1,..,n\}$$ Under what conditions is $H$ compact?
So my intuition tells me we must have $k\geq n$, and $\phi_1,...,\phi_k$ 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: $$|\phi_1(x)|\leq c_1\implies |\alpha_{11}x_1+...+\alpha_{1n}x_n| \leq c_1$$ $$|\phi_2(x)|\leq c_2\implies |\alpha_{21}x_1+...+\alpha_{2n}x_n| \leq c_2$$ ... $$|\phi_k(x)|\leq c_k\implies |\alpha_{k1}x_1+...+\alpha_{kn}x_n| \leq c_k$$ Though not sure where to go from there...

Answer 1

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 $\phi_1,\dots,\phi_k$ 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.