Property of uniformly continuous semigroup

Question

let $A$ be a bounded operator on a Hilbert space $H$ and let $t \mapsto e^{tA}$ denote the associated uniformly continuous semigroup, we may also assume that $\|e^{tA}\| \leq 1$ for all $t \in \mathbb{R}$. Does the equality ($x \in H$) $$\|e^{tA}x\|^2 = (x, e^{2tA}x)$$ hold even if $A$ is not necessarily self-adjoint? My textbook seems to claim this is true, but I fail to see why it should be.

Answer 1

If this did hold, then $$ (e^{tA}x,e^{tA}x)=(x,e^{2tA}x) \\ \frac{d}{dt}(e^{tA}x,e^{tA}x)=\frac{d}{dt}(x,e^{2tA}x) \\ (e^{tA}Ax,e^{tA}x)+(e^{tA}x,e^{tA}Ax)=(x,e^{2tA}2Ax) $$ Setting $t=0$ would then give $$ (Ax,x)+(x,Ax)=(x,2Ax), \\ (Ax,x)=(x,Ax),\;\;\; x \in H. $$

Answer 2

It is a good idea to have a few easy examples of semigroups to check such things out. For example, on $H=\mathbb{R}^2$ the semigroup $e^{tA}=\pmatrix{\cos(t)&\sin(t)\\-\sin(t)&\cos(t)}$ satisfies $\|e^{tA}\|\leq 1$. But at time $t=\pi/2$ we have $(x,e^{2tA}x)=-\|x\|$ which is strictly negative if $\|x\|>0.$