Is a semigroup always an exponential?

Question

Let $H$ be a Banach space, $\mathscr{B}(H)=\{T:H\to H: \text{where $T$ is a bounded linear operator}\}$, and $S:[0,\infty)\to \mathscr{B}(H)$, a map with the following properties: $$ S(0)=I, \quad S(s+t)=S(s)S(t), \quad \text{for all $s,t\in[0,\infty)$}, $$ and $$ \lim_{t\to s^+}S(t)x=S(s)x, \quad \text{for all $x\in H$.} $$ Is there a general theorem which allows us to express $S(t)=\exp(tA)$, for a suitable unbounded operator $A$? If yes, what kind of operator is $A$? In particular, what can we tell about its spectrum?

Answer 1

Yes. For every $C_0$-semigroup, $S(t)=\exp(At)$ in the sense that $$S(t)x=\lim_{n\to\infty} (I-{At\over n})^{-n}x.$$ The relationship between the spectrum of $A$ and the spectrum of $S(t)$ is not quite so straightforward. Pazy's book (Semigroups of Linear Operators and Applications to Partial Differential Equations) is a good place to start reading about the subject.

Answer 2

To expand the answer of Michael Renardy: your question concerns the important question when can we "insert" an operator into a function and what properties can we deduce from the properties of the function only. This is the main idea behind various "functional calculus concepts", and I recommend the excellent monograph and papers (for example, see this or this) by Markus Haase.

The most widely used is the so-called Hille-Phillips calculus, which is based on Laplace transform techniques. This and the Post-Widder inversion formula makes the connection Michael mentions.