# Compact operators on Banach spaces and their spectra

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

• Why on earth would (1) and (5) be equivalent even in a Hilbert space? The identity operator has discrete spectrum... – András Bátkai Mar 1 at 9:54
• (2) and (3) also not equivalent, there are eventually compact semigroups (e.g. nilpotent shift) – András Bátkai Mar 1 at 9:56
• Sorry. I forgot to assume $L$ is self-adjoint. – sharpe Mar 1 at 9:59
• @JohnDoe I understood. – sharpe Mar 1 at 15:26
• @sharpe I don't think you did ... – 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 $$L^1$$.
• 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. – sharpe Mar 1 at 16:59