Infinitesimal generator of $(T(t)u)(s) $ is $(Au)(s)=\lambda\Big(u(s-\mu)-u(s)\Big)$

Question

Let $\mathscr{X}=C(-\infty,\infty)$, the space of bounded uniformly continous functions on $(-\infty,\infty)$. Define the linear operator $T(t)$ by $$(T(t)u)(s)=e^{-\lambda t} \sum_{n=0}^{\infty}\frac{(\lambda t)^n}{n!} u(s-n \mu), t>0 $$ and $t=0$, $(T(t)u)(s)=u(s)$, where $\lambda, \mu>0$.

I have proved the $\{T(t), t\geq 0\}$ is a strongly continuous contraction semi-group. But how to prove that the infinitesimal generator is the difference operator $A$: $(Au)(s)=\lambda \Big(u(s-\mu)-u(s)\Big)$

Answer 1

By definition, $$Au=\lim_{t\to 0^+}\frac{T(t)u-u}{t}\in \mathscr{X}$$ and thus you have to prove that $$\left\|\frac{T(t)u-u}{t}-\lambda(g_1-u)\right\|_\mathscr{X}\overset{t\to 0^+}\longrightarrow 0,$$ where $g_n$ ($n\in\mathbb N$) stands for the function $g_{n}(s)=u(s-n\mu)$ ($s\in\mathbb R$). To this end, note that \begin{align*} 0&\leq \left\|\frac{T(t)u-u}{t}-\lambda(g_1-u)\right\|_\mathscr{X}\\ &=\sup_{s\in\mathbb R}\left|\frac{e^{-\lambda t}\sum_{n=0}^{\infty}\frac{(\lambda t)^n}{n!} g_n(s)-u(s)}{t} -\lambda (g_1(s)-u(s))\right|\\ &=\sup_{s\in\mathbb R}\left|\frac{e^{-\lambda t}\left[u(s)+\lambda tg_1(s)+\sum_{n=2}^{\infty}\frac{(\lambda t)^n}{n!} g_n(s)\right]-u(s)}{t} -\lambda (g_1(s)-u(s))\right|\\ &=\sup_{s\in\mathbb R}\left|u(s)\left(\frac{e^{-\lambda t}-1}{t}+\lambda \right) +\lambda g_1(s)\left(e^{-\lambda t} - 1\right) +e^{-\lambda t}\sum_{n=2}^{\infty}\frac{\lambda^nt^{n-1}}{n!} g_n(s)\right|\\ &\leq C\left|\frac{e^{-\lambda t}-1}{t}+\lambda \right|+ C\left|e^{-\lambda t} - 1\right| +C\left|e^{-\lambda t}\right|\\ &\overset{t\to 0}\longrightarrow 0+0+0 \end{align*}

Answer 2

To clean up notation write $T_tu$. The infinitesimal generator is defined as:

$$A_tf :=\lim_{t\rightarrow 0}\frac{E[T_tu]-u(s)}{t}.$$

Take the derivative of $T(t)$ using the product rule, to get:

$$\frac{d}{dt}T_tu = -\lambda (T_tu)(s) + \lambda (T_tu)(s-\mu) = T_t\lambda (u(s-\mu)-\mu(s)).$$ Now take $t\rightarrow 0$, and since $u$ is bounded, $T_t$ reduces to the identity.