I have a question about compact operators on Banach spaces.


Let $B$ be a real Banach space and $L$ a closed linear operator on $B$.
We assume that $L$ generates a contraction semigroup $\{T\_t\}\_{t>0}$ on $B$ .


**If $B$ is a Hilbert space** and $L$ is self-adjoint, the following assertions are equivalent:


(1) The spectrum of $L$ is discrete (the essential spectrum $\sigma\_{ess}(L)=\emptyset$).


(2) $T\_t$ is compact for any $t>0$.


(3) $T\_t$ is compact for some $t>0$.


(4) $R\_{\lambda}:=(\lambda-L)^{-1}$ is compact for any $\lambda \in \rho(L)$.


(5) $R\_{\lambda}$ is compact for some $\lambda \in \rho(L)$.


Here, $\rho(L)$ is the resolvent set of $L$.


Even if $B$ is not a Hilbert space, (2)$\Rightarrow$(4), (4)$\Leftrightarrow$(5), (5)$\Rightarrow$(1).


**My question**


In what follows, we further assume that $\{T\_t\}\_{t>0}$ is strongly continuous and $B$ is a $L^1$ space on a measure space.


Does (1)$\Rightarrow$(5) hold? or


**Under what conditions, does (1)$\Rightarrow$(5) hold?**


By the way, I am particularly interested in situations where $\{T\_t\}\_{t>0}$ is generated by a **symmetric** Markov process on a locally compact metric measure space $(X,\mu)$. In this case, for each $1\le p <\infty$, $\{T\_t\}\_{t>0}$ is extended to a strongly continuous contraction semigroup $\{T\_t^p\}\_{t>0}$ on $L^{p}(X,\mu)$ and it holds that $T\_t^p f=T\_tf$ for any $t>0$ and $f \in L^{1}(X,\mu) \cap L^{p}(X,\mu)$.



Compact operators on Banach spaces and their spectra
The essential spectrum (and even the spectrum) of the generator of a contractive $C\_0$-semigroup on an $L^1$-space can be empty even if the generator does not have compact resolvent.


**Example.** Endow $[0,1]^2$ with the Lebesgue measure and define a $C\_0$-semigroup $(T\_t)\_{t \ge 0}$ on $L^1([0,1]^2)$ by
\begin{align\*}
 (T\_tf)(x,y) = 
 \begin{cases}
 f(x+t,y) \quad & \text{if } x+t\le 1, \\
 0 \quad & \text{if } x+t>1
 \end{cases}
\end{align\*}
for all $f \in L^1([0,1]^2)$ and all times $t \ge 0$. Then $(T\_t)\_{t \ge 0}$ is nilpotent (since $T\_t = 0$ for $t \ge 1$), so the generator of the semigroup has empty spectrum; in particular, the generator has empty essential spectrum.


However, the resolvent of the generator is not compact (this follows from the fact that the semigroup action is trivial along the $y$-axis).


**Remark.** I'm not sure whether there is a natural set of additional assumptions which make the implication (1) $\Rightarrow$ (5) true on $L^1$.


