Assume $(T(t))_{t\ge 0}$ is a $\mathcal C_0-$semigroup on the Banach space $X$ and let $A$ be its generator. Then
$\forall x\in X$ we have: $\quad\lim\limits_{h\to 0} \frac{1}{h} \int_t^{t+h} T(s)x\;ds=T(t)x \quad(*)$
Now, I am studying a proof regarding the existence and uniqueness of strong solution for the inhomogeneous linear hyperbolic Cauchy problem and I've stuck to the following part:
Let $f\in C^1(I;X)$ then: $\quad \lim\limits_{h\to 0} \frac{1}{h} \int_t^{t+h} T(t+h-s)f(s)\;ds=f(t) $ and this follows immediately from $(*)$
QUESTION: Why does this follow from $(*)$? Ho do we use $(*)$ since in the last expression $h$ appears also in $T(\cdot)$ in contrast to $(*)$?
Any help is much appreciated!
Thanks in advance!
Write $\frac 1 h \int_t^{t+h} T(t+h-s)f(s)\,ds$ as $$ \frac 1 h \int_t^{t+h}T(t+h-s)(f(s)-f(t))\,ds+\frac 1 h \int_t^{t+h}T(t+h-s)f(t)\,ds. $$ The first summand goes to zero since $\|T(\cdot)\|$ is locally bounded and $f$ is continuous. In the second summand substitute $r=t+h-s$ to get $$ \frac 1 h \int_0^h T(r)f(t)\,dr, $$ which converges to $f(t)$ by $(\ast)$.