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Let $H$ be a $\mathbb R$-Hilbert space and $-A$ be the generator of a strongly continuous semigroup $(T(t))_{t\ge0}$ on $H$.

I've read that it would be a consequence of a suitable version of the spectral theorem that if $A$ is dissipative and self-adjoint, then $(T(t))_{t\ge0}$ is immediately differentiable$^1$. How can we prove that?


$^1$ i.e. $T(t)x\in\mathcal D(A)$ for all $x\in H$ and $t>0$.

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  • $\begingroup$ My guess is that if you use the spectral theorem in its multiplication operator on $L^2$ spaces form, the fact that $A$ is dissipative will give you good behaviour of the multiplying function and you will left be checking something along the lines of "if $f \in L^2$ and $m$ is a non-positive function then $m e^{tm} f \in L^2$". $\endgroup$ Jun 4 '21 at 9:45
  • $\begingroup$ @RhysSteele Thank you for your comment. I'll need to reconsider this version of the spectral theorem and try to figure something out, but maybe this actually isn't the best (or most direct) approach to solve the problem in the particular setting (of heat-type semigroups) which I've got in mind. I've asked a separate question for this particular setting: math.stackexchange.com/q/4163362/47771. Would be great if you could take a look. $\endgroup$
    – 0xbadf00d
    Jun 4 '21 at 19:37
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You have even a stronger result. A self-adjoint dissipative operator generates an analytic semigroup, which is, in particular, immediately differentiable.

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