Compact operators on Banach spaces and their spectra

Question

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 $f \in L^{1}(X,\mu) \cap L^{p}(X,\mu)$.

Answer 1

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 $L^1$.