If $T(t)$ is an immediately differentiable semigroup on $H$ with generator $A$, does $\frac{d}{dt}\|T(t)x\|_H^2=2⟨AT(t)x,T(t)x⟩_H$ hold for all $x∈H$?

Let $$(T(t))_{t\ge0}$$ be a semigroup on a $$\mathbb R$$-Hilbert space $$H$$ with $$\sup_{s\in[0,\:t)}\left\|T(s)\right\|_{\mathfrak L(H)}<\infty\tag1$$ for some (and hence all) $$t>0$$ and $$(\mathcal D(A),A)$$ denote the generator of $$(T(t))_{t\ge0}$$. By $$(1)$$, $$[0,\infty)\ni t\mapsto T(t)\tag2$$ is (locally uniformly) differentiable with derivative $$AT(t)x=T(t)Ax\tag3$$ for all $$t\ge0$$ and $$x\in\mathcal D(A)$$. In particular, $$\frac\partial{\partial t}\left\|T(t)x\right\|_H^2=2\langle AT(t)x,T(t)x\rangle_H\tag4$$ by the chain rule for all $$t>0$$ and $$x\in\mathcal D(A)$$.

If $$(T(t))_{t\ge0}$$ is immediately differentiable, i.e. $$T(t)H\subseteq\mathcal D(A)\;\;\;\text{for all }t\ge0\tag5,$$ does $$(4)$$ hold for all $$t>0$$ and $$x\in H$$?

The fact that $$T(t)$$ is immediately differentiable is equivalent to: for all $$x\in H$$, $$t\longmapsto T(t)x$$ is differentiable for $$t>0$$.
Then for all $$x\in H$$ we have, $$\frac{d}{dt}\|T(t)x\|^2=2\left\langle \frac{d}{dt}T(t)x, T(t)x\right\rangle =2\langle AT(t)x, T(t)x\rangle.$$
• In $(4)$, shouldn't I exclude the point $t=0$? Otherwise, this would imply that $\mathcal D(A)=H$. – 0xbadf00d Mar 22 at 10:31
• In $(4)$, if $t=0$ you obtain $0=\langle Ax,x\rangle$, for all $x\in D(A)$, and this is not true in general. In $(5)$ if you include $t=0$, you obtain what you said. But since the operator $A$ is closed, closed graph theorem implies that $A$ is bounded, and this case is not of much interest. – S. Maths Mar 22 at 13:44
• Yes, sorry. I'd actually meant $(5)$, not $(4)$. – 0xbadf00d Mar 22 at 15:50