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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!

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3 Answers 3

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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.

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  • $\begingroup$ 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. $\endgroup$ Jul 3, 2019 at 13:57
  • $\begingroup$ Thank you! The corresponding eigenvalues of $K$ should be non-decreasing, right? $\endgroup$
    – user525192
    Jul 3, 2019 at 14:16
  • $\begingroup$ Glad to help! That depends on how you order them, but if you mean the only limit point can be infinity, then yes, indeed. $\endgroup$
    – Teun
    Jul 3, 2019 at 14:29
  • $\begingroup$ Can you explain, Why we need the continuity of the function? $\endgroup$
    – user525192
    Jul 3, 2019 at 15:18
  • $\begingroup$ 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 $y_n\neq y$. $\endgroup$
    – Teun
    Jul 4, 2019 at 12:07
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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.

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  • $\begingroup$ This identity should be true if $(K-id)$ is invertible... $\endgroup$
    – user525192
    Jul 2, 2019 at 18:58
  • $\begingroup$ Can you explain how it follows from this identity, that the resolvent is compact? $\endgroup$
    – user525192
    Jul 2, 2019 at 19:05
  • $\begingroup$ 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. $\endgroup$
    – Teun
    Jul 3, 2019 at 10:50
  • $\begingroup$ 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. $\endgroup$
    – Teun
    Jul 3, 2019 at 11:31
  • $\begingroup$ I changed the assumption! $\endgroup$
    – user525192
    Jul 3, 2019 at 12:44
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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.

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