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, $$\frac{d}{dt}\|T(t)x\|^2=2\left\langle \frac{d}{dt}T(t)x, T(t)x\right\rangle =2\langle AT(t)x, T(t)x\rangle.$$

  • $\begingroup$ In $(4)$, shouldn't I exclude the point $t=0$? Otherwise, this would imply that $\mathcal D(A)=H$. $\endgroup$ – 0xbadf00d Mar 22 at 10:31
  • $\begingroup$ 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. $\endgroup$ – S. Maths Mar 22 at 13:44
  • 1
    $\begingroup$ Yes, sorry. I'd actually meant $(5)$, not $(4)$. $\endgroup$ – 0xbadf00d Mar 22 at 15:50

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