Intuition behind the generator of a semigroup

Question

This is how we defined the domain $D(A)$ and the operator $A$ of a semigroup:

Let $(P_t)_{t>0}$ be a semigroup of linear contractions on $L$. Let $$D(A) := \left\{ f\in L : \operatorname*{\mathit{s}-lim}_{h\downarrow 0} \frac 1 h [P_h f-f] \text{ exists in $L$}\right\}.$$ We define an operator $A$ on $D(A)$ by $$Af := \operatorname*{\mathit{s}-lim}_{h\downarrow 0} \frac 1 h [P_h f-f], \qquad\text{for $f\in D(A)$.}$$ Then $A$ is linear but not necessarily bounded with domain $D(A)\subset L$. The operator $A$ is called generator of $(P_t)_{t\ge 0}$.

Is there an intuitive reason/image why $A$ is the generator of $(P_t)$? And what is the reason?

Answer 1

A good reason to call $A$ the generator of the semigroup once you develop some of the theory is that it genuinely "generates" the semigroup in the sense that given the operator $A$ you can recover the entire semigroup.

This is the content of many generation for theorems for $C_0$-semigroups (which are slightly more general than contraction semigroups). For example, the most famous such result is the Hille-Yosida theorem, which I state for contraction semigroups below.

Theorem: (Hille-Yosida for contraction semigroups) Let $A$ be an (potentially unbounded) operator on a Banach space $E$. Then $A$ is the generator of a semigroup of contractions if and only if $A$ is closed and densely defined and for every $\lambda >0$, $\lambda - A$ is invertible and $$\|(\lambda - A)^{-1}\| \leq \lambda ^{-1}.$$ In this case, the semigroup $(P_t)$ with generator $A$ is given by either of the formulas $$ P_t x = \lim_{n \to \infty} \big( I - \frac{t}{n} A)^{-n} x$$ or $$P_t x = \lim_{n \to \infty} e^{-nt} \sum_{k=0}^\infty \frac{(n^2t)^k}{k!} (n-A)^{-k} x.$$ In particular, a contraction semigroup is uniquely determined by its generator.

These two defining formulas may look strange, but they are both inspired by the fact that we ought to have $P_t = e^{tA}$. When $A$ is bounded, $e^{tA}$ is defined by its power series and coincides with all of the other expressions. Unfortunately, in most interesting cases $A$ isn't bounded so we can't define the right hand side via its power series since that doesn't converge. This leads us to the expressions stated in Hille-Yosida.

Answer 2

A semigroup $P$ of linear operators has the exponential property on $[0,\infty)$: $$ P(s)P(t) = P(s+t),\;\; P(0)=I $$ So you expect there to be a sense in which $P(t)=e^{tA}$, where $A$ is a generator of $P$. And this is true if $P$ is strongly continuous from the right at $0$, meaning $$ \lim_{t\downarrow 0}P(t)x = x,\;\;\; x\in X. $$ Intuitively, the generator is the derivative of $P$ at $t=0$, i.e., $$ Ax=\lim_{t\downarrow 0} \frac{1}{t}(e^{tA}-I)x. $$ This turns out to be true, at least for a dense subspace $\mathcal{D}(A)$ of $x$. And $A: \mathcal{D}(A)\subset X\rightarrow X$ is a closed, densely-defined linear operator. It is a generator in that $P(t)$ can be constructed from $A$.