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In Buhler&Salamon's book "Functional Analysis" they state and prove the following conclusion by Phillips:

Theorem(Phillips) Assume $H$ is a Hilbert space, $A$ is a densely defined linear operator, then the following is equivalent:

  • The operator $A$ is the infinitesimal generator of a strongly continuous self-adjoint semigroup $S\colon [0,\infty )\to\mathcal{L} (H)$.
  • The operator $A$ is self-adjoint and $$ \sup_{x\in\mathrm{dom} (A)\setminus\{0\} }\frac{\langle x,Ax\rangle }{\Vert x\Vert^2 } <\infty $$ If these equivalent conditions are satisfied then $$ \frac{\log\Vert S(t)\Vert }{t} =\sup_{x\in\mathrm{dom} (A)\setminus\{0\} }\frac{\langle x,Ax\rangle}{\Vert x\Vert^2 } $$ for all $t>0$.

I think the second condition would obviously imply that $A$ must be bounded, but don't it be too strong for a strongly continuous semigroup? I'm confused.

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The important thing to notice is that there is no absolute value in the expression over which the supremum is taken. Hence the second does not imply that $A$ is bounded (one calls such operators "bounded above"). Another way to express this condition is that the spectrum of $A$ is contained in $(-\infty, \lambda]$ for some $\lambda\in\mathbb{R}$.

As a simple example, a multiplication operator on $L^2$ is bounded if the function is bounded, and bounded above if the function is bounded above.

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  • $\begingroup$ That's it, thanks. $\endgroup$ – TheWildCat Oct 4 at 23:03

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