# Thread: finding a specific operator

1. ## finding a specific operator

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

Thank You!

2. ## Re: finding a specific operator

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

3. ## Re: finding a specific operator

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

4. ## Re: finding a specific operator

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

5. ## Re: finding a specific operator

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

6. ## Re: finding a specific operator

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

7. ## Re: finding a specific operator

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

8. ## Re: finding a specific operator

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

9. ## Re: finding a specific operator

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

10. ## Re: finding a specific operator

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

Thank You!
As girdav pointed out, none of the $\displaystyle z_n$ can be 0. So make that assumption.

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

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