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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 $X = Y$...

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  • $\begingroup$ As formulated the problem is ill posed. You're probably thinking of spectra of unbounded operators. $\endgroup$
    – Liviu Nicolaescu
    Commented Mar 8, 2016 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
    Commented Mar 8, 2016 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
    Commented Mar 8, 2016 at 4:12

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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 $\mathbb C$.

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  • $\begingroup$ I see... That answers my question. Thanks a lot! $\endgroup$
    – Dom
    Commented Mar 8, 2016 at 13:50

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