Semigroup of operators and spectrum Let $K_{t}=e^{-tK}, t>0$ be a self adjoint semigroup of operators that extends as $C_{0}$-semigroup with $||K_{t}||\leq 1$ where $K$ is a self-adjoint and not necessarily bounded operator. Is the following implication true:
$$K_{t}~\text{is compact for every}~t\Longrightarrow K~\text{ has only discrete spectrum}?$$ Thanks in advance!
 A: I can't comment unfortunately. Have you tried to relate $K_t$ to $(K-i)^{-1}$, maybe by checking whether
$$(K-i)^{-1}=\int_0^\infty e^{it}K_t dt$$
is true? Because if $(K-i)^{-1}$ is compact you are done.
A: Yes. According to Theorem IV.3.6 in Engel and Nagel, we have (for all $t$) that
$$e^{-t\sigma(K)}\subseteq\sigma(K_t)\,.$$
Note that the left-hand side doesn't contain 0, and that $x\mapsto e^{-tx}$ is injective and continuous. Hence, by general topology, if $\sigma(K)$ is not discrete, then $\sigma(K_t)\setminus\{0\}$ is not discrete. But this isn't possible if $K_t$ is compact, by the spectral theorem for compact operators.
Engel, Klaus-Jochen; Nagel, Rainer, One-parameter semigroups for linear evolution equations, Graduate Texts in Mathematics. 194. Berlin: Springer. xxi, 586 p. (2000). ZBL0952.47036.
A: Hint: Let $e^{tA}$ be a $C_0$-semigroup. Then, are equivalent:
1) $e^{tA}$ is immediately compact.
2) $e^{tA}$ is immediately norm continuous, and its generator $A$ has compact resolvent.
See the book of Engel & Nagel for more details.
