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.


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.

  • $\begingroup$ That's it, thanks. $\endgroup$
    – TheWildCat
    Oct 4 '19 at 23:03

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