Weyl theorem, compact and relatively compact operators Weyl's Theorem says that if $A$ is a $T$-compact operator on a Banach space $X$, then $T$ and $T+A$ have the same essential spectrum.
To be $T$-compact, $A$ must satisfy two conditions
(1) $D(T)\subset D(A)$
(2) For any bounded sequence $\{u_n\}\subset D(T)$ such that $\{Tu_n\}$ is bounded then $\{Au_n\}$ contains a convergent sub-sequence.
I have a few questions:
(1) From what I understand, the definition of being compact is that for any bounded sequence $\{u_n\}\subset X$ then $\{Au_n\}$ contains a convergent sub-sequence. Thus it implies that the domain of a compact operator be all of $X$. Is that right?
(2) If $A$ is compact, it is automatically $T$-compact, thus satisfies the condition of the theorem. Is that right?
(3) If I have an operator $T+A$ with both $T$ and $A$ having each a "natural" domain such that the condition $D(T)\subset D(A)$ be not satisfied. However, $D(T+A)=D(T\cap A)$. Then one could reduce the domain $T$ and $A$ to $D(T+A)$. Then the condition (1) of relative compactness would automatically be satisfied and only condition (2) is needed to be checked. Then, in the end, one just need to check Condition (2) on $D(T+A)$ to apply Weyl's theorem. Is that right?
(4) Finally, I am in the following situation: I can check that for any bounded sequence $\{u_n\}\subset D(A)$, then $\{Au_n\}$ has a converging subsequence. So, "$A$ is compact on its domain". Does such a notion exist?
 A: For (1), one usually demands that a compact operator have domain all of $X$. Strictly speaking you don't need to do this, you could simply ask for "compact on its domain" as you write in (4), but this notion is rather pointless, as any densely defined operator that is "compact on its domain" admits a unique extension to a globally defined operator that is compact and continuous. For (2) any compact operator is $T$-compact for any $T$ as you remark.
For (3), what you write is correct, but note that if you change the domain of $T$ to $D(T)\cap D(A)$ then the resulting operator may not be densely defined. Further this operation of changing the domain can completely destroy the spectrum, so that the spectrum of your modified $T$ has nothing to do with the previous $T$. You will thus not gain insight into $T$ by modifiying its spectrum so that $A$ becomes $T$-compact.
For (4), as remarked above "compact on its domain" does not seem helpful, as any such operator admits a unique bounded extension with domain all of $X$ that is still compact. First note that if for every bounded sequence $u_n$ you have that $Au_n$ has a convergent sub-sequence, then $Au_n$ can never be unbounded for $u_n$ a bounded sequence. As such the operator $A$ is a bounded operator and you may uniquely extend its domain to be all of $X$ (assuming $A$ is densely defined). $A$ will still remain compact: If $u_n$ is an arbitrary bounded sequence define $\tilde u\in D(A)$ so that $\|\tilde u_n - u_n\|≤\frac1n$, then $A(\tilde u_n-u_n)\to0$ and if $A(\tilde u_n)$ admits a convergent subsequence then so does $A(u_n)$.
