If $S$ is a strongly continuous semigroup on a Hilbert space $H$ and $u\in C^1((0,T),H)$, show that $\lim_{h\to0+}S(h)\frac{u(s)-u(s-h)}h=u'(s)$

Question

Let

• $T>0$
• $H$ be a $\mathbb R$-Hilbert space
• $S:[0,\infty)\to H$ be a strongly continuous semigroup on $H$
• $A$ be the infinitesimal generator of $A$
• $u\in C^1((0,T),H)$ and $$u(t)\in\mathcal D(A)\;\;\;\text{for all }t\in(0,T)$$

Let $s\in(0,T)$ and $h\in(0,\min(s,t-s))$. I want to show that $$\left\|S(h)\frac{u(s)-u(s-h)}h-u'(s)\right\|_H\xrightarrow{h\to0+}0\tag1\;.$$

If $S$ would be uniformly continuous, we could simply write

\begin{equation} \begin{split} \left\|S(h)\frac{u(s)-u(s-h)}h-u'(s)\right\|_H&=\left\|S(h)\left(\frac{u(s)-u(s-h)}h-u'(s)\right)+\left(S(h)-S(0)\right)u'(s)\right\|_H\\ &\le\left\|S(h)\right\|_{\mathfrak L(H)}\left\|\frac{u(s)-u(s-h)}h-u'(s)\right\|_H+\left\|S(h)-S(0)\right\|_{\mathfrak L(H)}\left\|u'(s)\right\|_H\xrightarrow{h\to0+}0\;. \end{split}\tag2 \end{equation}

But what can we do in general?

Answer 1

For $\delta >0$ there exists $C$ such that $\|S(h)\|\le C$ whenever $0 \le h \le \delta$. This follows from uniform boundedness, because $\sup_{0 \le h \le \delta}\|S(h)x\| < \infty$ for every fixed $x$. Therefore, for $0 < h \le \delta$ and fixed $s$, \begin{align} &\left\|S(h)\frac{1}{h}\{u(s)-u(s-h)\}-u'(s)\right\| \\ & = \left\|S(h)\left[\frac{1}{h}\{u(s)-u(s-h)\}-u'(s)\right]+\{S(h)-I\}u'(s)\right\| \\ & \le C\left\|\frac{1}{h}\{u(s)-u(s-h)\}-u'(s)\right\|+\|\{S(h)-I\}u'(s)\|. \end{align} Both terms on the far right side tends to $0$ as $h\downarrow 0$.