Different definitions of a relatively compact operator 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.

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.
This question has also been posted on Math Overflow.
 A: 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 $T=0$).
Answer cross-posted from Math Overflow.
