What is a "compactness argument"? I'm working on this paper and I don't know what is meant by compactness argument in the proof of corrollary 4 page 226 which said that:

the function $\lambda \to \|B-\lambda A\|$ (where A and B are in L(H) ) is continuous
with  $$\lim_{|\lambda|\to\infty}\|B-\lambda A\|=\infty$$ so by  compactness argument, there exists $z_0\in \mathbb{C}$ s.t $$ \|B-z_0 A\|\leq \|(B-z_0 A)+\lambda A\|,\forall \lambda \in \mathbb{C}$$

 A: Since $\lim_{|\lambda|\to\infty}\|B-\lambda A\|=\infty$, there is some $M\in(0,\infty)$ such that$$
|\lambda|>M\implies\|B-\lambda A\|\ge\|B\|=\|B-0\times A\|.
$$The set $K=\{\lambda\in\mathbb{C}:|\lambda|\le M\}$ is compact, and therefore the function
$$
\begin{array}{ccc}K&\longrightarrow&[0,\infty)\\\lambda&\mapsto&\|B-\lambda A\|\end{array}
$$
has a minimum, at some point $z_0\in K$. But then, for each $\lambda\in\Bbb C$, $\|B-z_0A\|\le\|B-\lambda A\|$, since:

*

*If $|\lambda|\le M$, $\lambda\in K$, and therefore $\|B-z_0A\|\le\|B-\lambda A\|$.

*If $\lambda>M$, then $\|B-\lambda A\|\ge\|B-0\times A\|\ge\|B-z_0A\|$.

A: Hint
This follows from the following general fact, that is not difficult and instructive to prove using compactness. A continuous map $f : E \to \mathbb R$ where $E$ is a finite dimensional normed space such that $$\lim\limits_{\lVert x \rVert \to \infty} \lvert f(x) \rvert =\infty$$ attains its minimum.
Remember that a closed bounded subset of a finite dimensional normed space is compact.
Applied to your question, the result implies that $\lambda \to \lVert B-\lambda A \rVert$ attains its minimum at a point that we can name $z_0 \in \mathbb C$ and therefore
$$\|B-z_0 A\|\leq \|(B-z_0 A)+\lambda A\|,\forall \lambda \in \mathbb{C}.$$
