I've not done the computations, but did you try with diagonal?
Hi! I am kinda going mad trying to find the following operator. Here goes the problem: Let be a complex seq. such that and let be a seq. of pos. numbers such that . Show that there exists (i.e. find) a compact operator on so that the following two are satisfied:
(b) All the lie in the resolvent of .
There is something I don't understand: if for all , then has to be both compact and invertible in order to make the expression have a sense. It's impossible since is infinite dimensional.
Following girdav again, let P be the operator on whose matrix (with respect to the standard basis) is diagonal, with diagonal elements Let For each n, choose so that and
Then P is compact (because ). The spectrum of P is Since, for each n, it follows that (b) holds. Finally,