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

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    $\begingroup$ Why on earth would (1) and (5) be equivalent even in a Hilbert space? The identity operator has discrete spectrum... $\endgroup$ – András Bátkai Mar 1 at 9:54
  • $\begingroup$ (2) and (3) also not equivalent, there are eventually compact semigroups (e.g. nilpotent shift) $\endgroup$ – András Bátkai Mar 1 at 9:56
  • $\begingroup$ Sorry. I forgot to assume $L$ is self-adjoint. $\endgroup$ – sharpe Mar 1 at 9:59
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    $\begingroup$ @JohnDoe I understood. $\endgroup$ – sharpe Mar 1 at 15:26
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    $\begingroup$ @sharpe I don't think you did ... $\endgroup$ – Nik Weaver Mar 1 at 15:32

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 $L^1$.

  • $\begingroup$ Thank you for your reply. By the way, under what conditions, the implication (1) $\Rightarrow$(5) holds? I would appreciate it if you let me know. $\endgroup$ – sharpe Mar 1 at 16:59
  • $\begingroup$ @sharpe: My apologies, the wording in my final remark was not well chosen. What I actually wanted to say is that there might be some sufficient conditions, but (a) if so, then I do not know about them and (b) it is questionable whether such conditions would be fulfilled in the applications that you have in mind (which seem to be related to Markov processes). I edited my answer accordingly. $\endgroup$ – Jochen Glueck Mar 2 at 22:27
  • $\begingroup$ Thank you for your kind reply. $\endgroup$ – sharpe Mar 3 at 10:54

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