0
$\begingroup$

Let $X, Y$ be two different Banach spaces, and let $T: X \to Y$ be a compact linear operator. Suppose the identity $I : X \to Y$ is well-defined. (For example, we could have $X = L^2([0,1])$ and $Y = L^1([0,1])$, both equipped with the Lebesgue measure).

Do most/all elementary results in spectral theory hold in that setting? For example, is it true that the spectrum of $T$ is discrete and that the eigenvalues of $T$ only accumulate at $0$? Is there a simple way to see that without reproving the result? Practically all textbooks only deal with the case where $X = Y$...

$\endgroup$
  • $\begingroup$ As formulated the problem is ill posed. You're probably thinking of spectra of unbounded operators. $\endgroup$ – Liviu Nicolaescu Mar 8 '16 at 2:55
  • $\begingroup$ I may be just repeating @LiviuNicolaescu's statement, but what is the spectrum in this case? I guess that you mean that $T - \lambda I$ is not invertible—but not one-sided, or not two-sided invertible? If the latter, then it seems that most operators will have spectrum equal to most of $\mathbb C$. If the former, then which side? $\endgroup$ – LSpice Mar 8 '16 at 3:49
  • $\begingroup$ so we are considering $T = Id_{XY}T_{XX}$ where $T_{XX}$ is a compact operator $X\to X$, and $Id_{XY}$ is ... what do we know on $Id_{XY}$ ? $\endgroup$ – reuns Mar 8 '16 at 4:12
2
$\begingroup$

By "$I$ is well-defined", I presume you mean you have a continuous injection $\iota$ of $X$ into $Y$. If $X$ and $Y$ are not isomorphic it will not be a bijection, because there is no continuous linear bijection from $X$ to $Y$, and the same applies to $T - \lambda \iota$. Thus $T - \lambda \iota$ will never be invertible, and the "spectrum" of $T$ will be all of $\mathbb C$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I see... That answers my question. Thanks a lot! $\endgroup$ – Dom Mar 8 '16 at 13:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.