Every bounded linear operator is an infinitesimal generator

Question

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 $T$?

Answer 1

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:

$$\|f(t) - f(0)\| = \left\| \int_{0}^{t}f'(s)\,{\rm d}s \right\| \leq \int_{0}^{t}\|f'(s)\|\,{\rm d}s \leq t \sup_{s\in [0,t]}\|f'(s)\|.$$

Answer 2

Disclaimer: After a previously wrong solution, I wasn't smart enough to obtain the claimed estimate, but something else which also proves to claim that $A$ is the generator. Hopefully, it's now correct.

So let's calculate the left hand side of your desired inequality. We have

\begin{align*} \frac{\sum_{k=0}^\infty \frac{(tA)^k}{k!}-I}{t}-A&=\frac 1t \left(\sum_{k=1}^\infty \frac{(tA)^k}{k!}+I-I\right)-A\\&= \sum_{k=1}^\infty \frac{t^{k-1}A^k}{k!}-A\\ &=\sum_{k=2}^\infty\frac{t^{k-1}A^k}{k!}+A-A\\ \end{align*}

This follows from writing down terms explicitly for $k=0,1$.

Now, we take norms, use the boundedness of $A$ to take it outside together with the triangle inequality to obtain

\begin{align*} \left\|\sum_{k=2}^\infty \frac{t^{k-1}A^k}{k!}\right\|&\leq\|A\|\sum_{k=2}^\infty \frac{t^{k-1}\|A\|^{k-1}}{k!}\leq \|A\| \sum_{k=1}^\infty \frac{t^k\|A\|^k}{k!}\\&\leq t\|A\|^2 \sum_{k=1}^\infty \frac{t^{k-1}\|A\|^{k-1}}{k!}\leq t\|A\|^2 \mathrm{e}^{t\|A\|} \end{align*}

which for $t\to 0$ tends to 0, what we wanted to show.