prove that if there exists a $v \neq 0$ with $Av \preceq 0$ then domain of $f_0$ is unbounded

Question

I want to prove that if there exists a $v \neq 0$ with $Av \preceq 0$ then domain of $f_0$ is unbounded. This is a problem in Boyd's convex optimization book.

The answer uses a sequence $x_k$ such that $||x_k||_2 \to \infty$. It then defines $v_k = x_k / ||x_k||_2$. The sequence has a convergent subsequence becasue $||v_k||_2 = 1$ for all $k$. Let $v$ be its limit, then we have $||v||_2 = 1$ and, since $a_i^T v_k < b_i / ||x_k||_2$ for all $k$, and $a_i^T v \le 0$. Therefore $Av \preceq 0$ and $v \neq 0$.

What I don't get is how do we know that there exists a convergent subsequence? And how do we get $a_i^T v \le 0$ from $a_i^T v_k < b_i / ||x_k||_2$ ? I'm guessing they use the limit and got $\lim a_i^T v_k < \lim b_i / ||x_k||_2 \implies a_i^T v < 0$ but this doesn't include $0$.

Answer 1

What I don't get is how do we know that there exists a convergent subsequence?

I am assuming that this is in a finite dimensional space.

The sequence is contained in the closed unit ball, which is a compact set. In a compact set in a metric space, each sequence has a convergent subsequence. See also here and here.

And how do we get $a_i^T v \le 0$ from $a_i^T v_k < b_i / ||x_k||_2$ ? I'm guessing they use the limit and got $\lim a_i^T v_k < \lim b_i / ||x_k||_2 \implies a_i^T v < 0$ but this doesn't include $0$.

We have $\|x_k\|_2\to\infty$. Then $b_i / \|x_k\|_2\to0$ follows for $k\to\infty$. Note that when taking the limit in an "$<$" inequality, the limit only satisfies "$\leq$" (for example, $0<1/n$ holds, but in the limit we have $0\leq 0$).