finding a specific operator

**Zeke** · January 13th 2012, 07:21 PM

Hi! I am kinda going mad trying to find the following operator. Here goes the problem: Let $(z_n)$ be a complex seq. such that $\lim z_n=0$ and let $(x_n)$ be a seq. of pos. numbers such that $\lim x_n=\infty$ . Show that there exists (i.e. find) a compact operator $P$ on $l^2$ so that the following two are satisfied:
(a) $\lim_n \frac{\|(z_n I-P)^{-1}\|}{ x_n}=\infty$
and
(b) All the $z_n$ lie in the resolvent of $P$ .

Thank You!

**girdav** · January 14th 2012, 04:45 AM

I've not done the computations, but did you try with $P$ diagonal?

**Zeke** · January 14th 2012, 08:14 AM

yes I did try... I'm not sure how that helps.

**girdav** · January 14th 2012, 09:44 AM

It would reduce the problem to find a sequence instead of an operator.

**Zeke** · January 15th 2012, 12:58 AM

I'm sorry but I just can't get it to work. How am I supposed to find it?

**girdav** · January 15th 2012, 04:44 AM

There is something I don't understand: if $z_n=0$ for all $n$ , then $P$ has to be both compact and invertible in order to make the expression have a sense. It's impossible since $\ell^2$ is infinite dimensional.

**Zeke** · January 15th 2012, 05:28 AM

The assumptio is not $z_n=0$ $\forall n$ , it is that $\lim_{n\rightarrow \infty} z_n=0$

**girdav** · January 15th 2012, 05:48 AM

Yes, but $z_n=0$ for all n is a particular case of $\lim_{n\to \infty}z_n=0$ . So we have to assume that $z_n\neq 0$ for all $n$ otherwise $P$ would be invertible.

**Zeke** · January 15th 2012, 06:21 AM

I see what you mean. Ok, how would I proceed after removing that special case.

**Opalg** · January 15th 2012, 10:32 AM

Originally Posted by Zeke

Hi! I am kinda going mad trying to find the following operator. Here goes the problem: Let $(z_n)$ be a complex seq. such that $\lim z_n=0$ and let $(x_n)$ be a seq. of pos. numbers such that $\lim x_n=\infty$ . Show that there exists (i.e. find) a compact operator $P$ on $l^2$ so that the following two are satisfied:
(a) $\lim_n \frac{\|(z_n I-P)^{-1}\|}{ x_n}=\infty$
and
(b) All the $z_n$ lie in the resolvent of $P$ .

Thank You!

As girdav pointed out, none of the $z_n$ can be 0. So make that assumption.

Following girdav again, let P be the operator on $\l^2(\mathbb{N})$ whose matrix (with respect to the standard basis) is diagonal, with diagonal elements $\alpha_n.$ Let $Z = \{z_n\}_{n\in\mathbb{N}}\cup\{0\}.$ For each n, choose $\alpha_n$ so that $\alpha_n\notin Z$ and $|\alpha_n-z_n| < (nx_n)^{-1}.$

Then P is compact (because $\alpha_n\to0$ ). The spectrum of P is $A = \{\alpha_n\}_{n\in\mathbb{N}}\cup\{0\}.$ Since, for each n, $z_n\notin A,$ it follows that (b) holds. Finally, $\|(z_nI-P)^{-1}\| = \max_{k\in\mathbb{N}}|z_n-\alpha_k|^{-1}\geqslant|z_n-\alpha_n|^{-1}>nx_n.$

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