# 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!

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.

• This is a nice answer but it probably would have been better to edit your other answer (which was really a comment anyway) to include the content of this answer rather than posting a new one. Jul 3, 2019 at 13:57
• Thank you! The corresponding eigenvalues of $K$ should be non-decreasing, right?
– user525192
Jul 3, 2019 at 14:16
• Glad to help! That depends on how you order them, but if you mean the only limit point can be infinity, then yes, indeed.
– Teun
Jul 3, 2019 at 14:29
• Can you explain, Why we need the continuity of the function?
– user525192
Jul 3, 2019 at 15:18
• Proving that " $\sigma(K)$ not discrete $\Rightarrow$ $\sigma(K_t)\setminus\{0\}$ not discrete " (using whatever characterization of `discrete') uses continuity. E.g., using sequences, we can find a point $x\in\sigma(K)$ and a sequence $(x_n)$ in $\sigma(K)\setminus\{x\}$ converging to $x$. By continuity of $x\mapsto e^{-tx}$, we find a sequence $(y_n)$ in $\sigma(K_t)\setminus\{0\}$ converging to a point $y\in\sigma(K_t)\setminus\{0\}$, and by injectivity $y_n\neq y$.
– Teun
Jul 4, 2019 at 12:07

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.

• This identity should be true if $(K-id)$ is invertible...
– user525192
Jul 2, 2019 at 18:58
• Can you explain how it follows from this identity, that the resolvent is compact?
– user525192
Jul 2, 2019 at 19:05
• I don't know, sorry. My idea was that since the $K_t$'s are compact and commute, they can be expanded w.r.t. a common orthonormal set of eigenvectors. If you could explicitly calculate the eigenvalues, you'd find the eigenvalues of the resolvent. But I don't know if this path works.
– Teun
Jul 3, 2019 at 10:50
• Now that I think about it, you can't explicitly calculate the eigenvalues because they're not continuous in t. It winds down to the problem that your semigroup is not a $C_0$-semigroup.
– Teun
Jul 3, 2019 at 11:31
• I changed the assumption!
– user525192
Jul 3, 2019 at 12:44

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.