0
$\begingroup$

(Cross-post from Math Stackexchange, where some work has been done in the comments)

Let $T,K$ be unbounded operators on a Hilbert space $H$. I've seen the following definition of a relatively compact operator:

(i) The operator $K$ is called relatively compact with respect to $T$, if for some $z$ in the resolvent set of $T$, $KR_T(z)$ is compact, where $R_T(z):=(T-z)^{-1}.$

I've also seen:

(ii) The operator $K$ is called relatively compact with respect to $T$, if for every sequence $(x_n)_{n \in \mathbb{N}}\subseteq H$ such that $(Tx_n)_{n \in \mathbb{N}}$ is bounded, $(Kx_n)_{n \in \mathbb{N}}$ contains a convergent subsequence.

All of this is in the context of spectral theory and $T$ can be assumed to be self-adjoint. Do definitions (i) and (ii) have something to do with each other, or are they distinct? What is the intuition behind these definitions? Definition (ii) looks like a generalisation of a compact operator, but definition (i) is just weird.

$\endgroup$
  • 3
    $\begingroup$ What about the trivial case $T=0$? (i) holds if and only if $K$ is compact but (ii) only holds for $K=0$. $\endgroup$ – Jochen Wengenroth Jul 17 '19 at 9:04
4
$\begingroup$

Both definitions aim to make the same statement -- the operator $KT^{-1}$ is compact. But we cannot really to do that, because $T^{-1}$ does not necessarily exist, so in (i) we replace the inverse by the resolvent and in (ii) we write the more usual definition of compactness; if $(Tx_n)$ is bounded then $KT^{-1}(Tx_n)= Kx_n$ should contain a convergent subsequence.

Actually, (i) is equivalent to a version of (ii), where we only allow bounded sequences $(x_n)$. Indeed, to see that (ii) implies (i), pick a bounded sequence $(y_n)$. Then the sequence $x_n:= (T-z)^{-1}y_n$ is bounded as well, and $Tx_n= (T-z)x_n + zx_n = y_n + zx_n$ is bounded. By (ii), $(Kx_n)$ contains a convergent subsequence, but $Kx_n = K(T-z)^{-1}y_n$, so we showed compactness of $K(T-z)^{-1}$.

For the other direction, note that if the sequences $(x_n)$ and $(Tx_n)$ are bounded, then $((T-z)x_n)$ is bounded as well, so by (i) $K(T-z)^{-1} (T-z)x_n = Kx_n$ contains a convergent subsequence.

As stated in your question, the definitions are not equivalent (but (ii) still implies (i)). Note that (i) does not imply (ii), because any compact operator satisfies (i), but not necessarily (ii) (look at $T=0$).

$\endgroup$
  • $\begingroup$ @Hannes: yes, indeed, thanks a lot. :) $\endgroup$ – Mateusz Wasilewski Jul 17 '19 at 9:29
  • $\begingroup$ Very nice, thank you! Maybe, since I've originally posted this question on Stack Exchange, you could post this answer there as well? (I don't know about the standard procedure regarding this) $\endgroup$ – Jannik Pitt Jul 17 '19 at 10:35
  • $\begingroup$ Ok, I posted the answer on Stack Exchange as well. $\endgroup$ – Mateusz Wasilewski Jul 17 '19 at 11:48

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.