The definitions I am working with are as follows:
- A linear operator $T:X\rightarrow X$ is compact if for every bounded sequence $\\\{x_n\\\}_n$, $\\\{Tx_n\\\}_n$ has a convergent subsequence.
- A bounded linear operator $T:X\rightarrow X$ is quasi-compact if there is a direct sum decomposition $X=F\oplus H$ and $0<\rho<\rho(T)$ where: $F$ and $H$ are closed and $T$-invariant ($T(F)\subset F$, $T(H)\subset H$), $\text{dim}(F)<\infty$ and all eigenvalues of $T|_F:F\rightarrow F$ have modulus larger than $\rho$, and the spectral radius of $L|_H$ is smaller than $\rho$.
I assumed that compactness would imply quasi-compactness, however I came across a theorem by Hennion (generalizing one of Doeblin & Fortet) which states the following:
Suppose $(X,||\cdot||)$ is a Banach space and $T:X\rightarrow X$ is a bounded linear operator with spectral radius $\rho(T)$. Assume that there exists a semi-norm $||\cdot||'$ with the following properties
- $||\cdot||'$ is continuous on $X$
- For any sequence of $f_n\in X$, if $\sup ||f_n||<\infty$ then there exists a subsequence $n_k$ and $g\in X$ such that $||T(f_{n_k})-g||'\rightarrow 0$ as $k\rightarrow \infty$
- There is an $M>0$ such that $||T(f)||'\leq M||f||'$ for all $f\in X$
- There are $k\geq 1$, $0<r<\rho(T)$, and $R>0$ such that $$||T^kf||\leq r^k||f||+R||f||'$$ Then $T$ is quasi-compact.
But if $T$ is compact, then the first three conditions seem to be met just by having $||\cdot||'=||\cdot||$. Condition 1 is given by the reverse triangle inequality. Condition 2 I believe is given by compactness. Condition 3 is given by boundedness (which in turn is given by compactness). So if compactness implies quasi-compactness, why is the fourth condition needed? I'm not really seeing the relation between compact and quasi-compact operators.