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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 whith $$\lim_{|\lambda|\to\infty}\|B-\lambda A\|=\infty$$ so by compactness argument, there exists $z_0\in \mathbb{C}$ s.t $$ \|B-z_0 A\|\leq \|(B-z_0 A)+\lambda A\|,\forall \lambda \in \mathbb{C}$$

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    $\begingroup$ In general, "by a compactness argument" means "we can show this with a proof what 'because a particular set is compact' is the core element". $\endgroup$
    – Arthur
    May 14 at 7:52

2 Answers 2

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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

$$\|B-z_0 A\|\leq \|(B-z_0 A)+\lambda A\|,\forall \lambda \in \mathbb{C}.$$

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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 $\|B-\lambda A\|\ge\|B-0\times A\|\ge\|B-z_0A\|$.
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  • $\begingroup$ why don't we take $K=\{\lambda:|\lambda|\leq M\}$ $\endgroup$
    – Mary Maths
    May 14 at 9:12
  • $\begingroup$ Isn't that what I did? $\endgroup$ May 14 at 9:21
  • $\begingroup$ no you take z\in C $\endgroup$
    – Mary Maths
    May 14 at 10:57
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    $\begingroup$ I see what what happened. I wrote $\{z\in\Bbb C:|\lambda|\le M\}$. But I meant to write $\{\lambda\in\Bbb C:|\lambda|\le M\}$. I've edited my answer. $\endgroup$ May 14 at 11:38

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