Operator semigroup: Notation Ambiguity

Question

When talking about operator semigroups, some books use the notation $(e^{tA})_{t \geq 0}.$ Doesn't this create an ambiguity? Why are results proved for this semigroup instead of taking a general semigroup, say $(T(t))_{t \geq 0}$?

Answer 1

When talking about operator semigroups, some books use the notation $(e^{tA})_{t \geq 0}.$ Doesn't this create an ambiguity?

No, because for these authors the notation $(e^{tA})_{t \geq 0}$ means the semigroup $(T(t))_{t\geq 0}$ whose generator is $A$.

• This notation makes sense because each semigroup has exactly one generator and no operator generates more than one semigroup.

• This notation is natural because for the particular case in which $A$ is bounded, the semigroup generated by $A$ is exactly $\sum_{n=0}^\infty\frac{(tA)^n}{n!}$. In other words, "the semigroup generated by $A$" is the generalization of "exponential of $A$" and thus it is natural to use the same notation for both.

Why are results proved for this semigroup instead of taking a general semigroup, say $(T(t))_{t \geq 0}$?

The results are proved for a general semigroup. But, in order to be concise, this general semigroup is called $(e^{tA})_{t \geq 0}$. Then we mention the semigroup and the generator in the same notation and can write things like

As $\mathbf i\mathbb R\subset \rho(A)$, we conclude that $(e^{tA})_{t \geq 0}$ is strongly stable.

Consider two semigroups $(e^{tA})_{t \geq 0}$ and $(e^{tB})_{t \geq 0}$.

$e^{t\sigma(A)}\subset \sigma(e^{tA})$.

instead of

As $\mathbf i\mathbb R\subset \rho(A)$, we conclude that the semigroup $(T(t))_{t\geq 0}$ generated by $A$ is strongly stable.

Consider two semigroups $(T(t))_{t \geq 0}$ and $(S(t))_{t \geq 0}$ generated by $A$ and $B$, respectively.

$e^{t\sigma(A)}\subset \sigma(T(t))$ where $(T(t))_{t\geq 0}$ is the semigroup generated by $A$.

In fact, many authors do not use this notation. But some authors (of books and papers) do.

Answer 2

The two are equivalent under weak continuity assumptions. In particular if $(T(t))$ is a semigroup and $||T(t)-I||\to0$ as $t\to0$ then there exists a bounded $A$ such that $T(t)=e^{tA}$.

Edit: Or so I'd read. Too stubborn to look at the proof - I think this works:

If $h>0$ is small enough then $$||I-\frac1h\int_0^h T(t)dt||<1,$$so $\frac1h\int_0^hT(t)dt$ is invertible.

Since $T$ is a semigroup it seems like the standard method of evaluating a finite geometric series should say something about that integral. Say $J=\int_0^h T(t)dt$, just because "$I$" is taken. For small $\epsilon>0$ we have $$ (T(\epsilon)-I)J=\int_h^{h+\epsilon}T-\int_0^\epsilon T$$. So $$\lim_{\epsilon\to0^+}\frac{T(\epsilon)-I}{\epsilon}=(T(h)-I)J^{-1}:=A.$$

Hence the semigroup property shows that $$D_RT(t)=AT(t),$$where $D_R$ denotes the right-hand derivative. It's easy to see that this implies that the two-sided derivative exists, but never mind that. The product rule shows that $$D_R(T(t)e^{-tA})=0,$$hence $T(t)e^{-tA}$ is constant, hence $$T(t)=e^{tA}.$$

Heh: That has very little to do with operator theory; in fact the same argument works if $B$ is a Banach algebra with identity and $T:(0,\infty)\to B$ is a continuous semigroup.