I'm studying the theory of semigroups from Pazy's book. I'm struggling to understand a specific inequality in the proof of a theorem stating that every bounded linear operator is an infinitesimal generator of some semigroup.
How does one get the second inequality, which implies that $A$ is indeed the generator of $T$?
Best Answer
If one is familiar with differentiating functions taking values in a Banach space, the second inequality can be obtained as follows.
Define $f:[0,\infty ) \to \mathcal{L}(X)$ by $$f(t) := T(t) - tA.$$ Note that as $T'(t) = AT(t)$ for each $t\in [0,\infty )$, $f$ is differentiable with $f'(t) = A(T(t) - I)$ for each such $t$.
By applying the mean value inequality from calculus in Banach spaces, for $t>0$,
$$\|f(t) - f(0)\| \leq t \sup_{s\in [0,t]}\|f'(s)\| = t \sup_{s\in [0,t]}\|A(T(s)-I)\| = t\|A\| \sup_{s\in [0,t]}\|T(s) - I\|.$$
Dividing both sides by $t$ and using $\frac{1}{t}(f(t) - f(0)) = \frac{1}{t}(T(t) - I) - A$ obtains the desired inequality.
If the mean value inequality is not a familiar result, it can be obtained in the Banach space case as follows:
$$\|f(t) - f(0)\| = \left\| \int_{0}^{t}f'(s)\,{\rm d}s \right\| \leq \int_{0}^{t}\|f'(s)\|\,{\rm d}s \leq t \sup_{s\in [0,t]}\|f'(s)\|.$$