How to find the infinitesimal generator of this semigroup?

Question

Definition 1: Let $X$ be a Banach space. A semigroup is a family $\{T(t)\}_{t\geq 0}$ of continuous linear operators $T(t):X\to X$ such that
$(i)\;\;T(0)=I$, where $I$ is the identity operator;
$(ii)\;\;T(s)\circ T(t)=T(t+s)$ for all $t,s\geq 0$.

Definition 2: the infinitesimal generator of a semigroup $\{T(t)\}_{t\geq 0}$ is the operator $A:D(A)\to X$ where: $$D(A)=\left\{x\in X;\;\lim_{h\to 0^+}\frac{T(h)x-x}{h}\text{ exists in } X \right\}$$ and $$A(x)=\lim_{h\to 0^+}\frac{T(h)x-x}{h}$$ for all $x\in D(A)$.

Definition 3: the translation of the function $f:\mathbb{R}\to\mathbb{R}$ is the function $f_t:\mathbb{R}\to\mathbb{R}$given by $f_t(x)=f(x+t)$ for all $x\in\mathbb{R}$.

Take $X=L^2(\mathbb{R})$ in definition 1 and consider the semigroup $T:=\{T(t)\}_{t\geq 0}$ where $T(t)f=f_t$ for all $f\in L^2(\mathbb{R})$.

My problem is to find the infinitesimal generator of $T$. First of all I need to find $D(A)$, that is, I need to find all $f\in L^2(\mathbb{R})$ such that $$\lim_{h\to 0^+}\frac{f_h-f}{h}=\lim_{h\to 0^+}\frac{T(h)f-f}{h}=g\tag{1}$$ for some $g\in L^2(\mathbb{R})$.

Could someone explain me how can we conclude? Any help is appreciated.

Thanks.

Answer 1

The limit $$ \lim_{h\to 0^+}\frac{f_h-f}{h}, $$ exists if the derivative of $f$ lies in $L^2(\mathbb R)$. More precisely, if there exists a $g\in L^2(\mathbb R)$, such that $$ \lim_{h\to 0^+} h^{-1}\|f_h-f-hg\|_{L^2(\mathbb R)}=0. $$ Clearly, the functions $f$ with the property above are dense in $L^2(\mathbb R)$, as every continuously differentiable function with compact support has this property, and such functions are indeed dense in $L^2(\mathbb R)$. So ${\mathcal D}(A)$ is dense in $L^2(\mathbb R)$, and $A=\frac{d}{dx}$.

Answer 2

Proposition (Engel, p.66). The infinitesimal generator of $\{T(t)\}_{t\geq 0}$ is the operator $A:D(A)\to L^2(\mathbb{R})$ given by $Af=f'$ with domain $$D(A)=\{f\in L^2(\mathbb{R})\mid f\text{ is absolutely continuous and } f'\in L^2(\mathbb{R})\}.$$

Proof: Let $B:D(B)\to L^2(\mathbb{R})$ be the infinitesimal generator of $\{T(t)\}_{t\geq 0}$. We want to prove that $B=A$.

Take $f\in D(B)$. From the definition of $B$ we have $$\frac{T(t)f-f}{t}\overset{t\to 0^+}{\longrightarrow} Bf\quad \text{in}\quad L^2(\mathbb{R}).\tag{1}$$ Take $a,b\in\mathbb{R}$. As $L^2(\mathbb{R})\hookrightarrow L^2(a,b)\hookrightarrow L^1(a,b)$ it follows that $$\left|\int_a^b\frac{T(t)f(x)-f(x)}{t}\;dx-\int_a^bBf\;dx\right| \leq C \left\|\frac{T(t)f-f}{t}-Bf\right\|_{L^2(\mathbb{R})} $$ and thus, from $(1)$, $$\int_a^b\frac{T(t)f(x)-f(x)}{t}\;dx\overset{t\to0^+}{\longrightarrow}\int_a^bBf(x)\;dx.\tag{2}$$ On the other hand, a change of variables gives \begin{align} \int_a^b \frac{T(t)f(x)-f(x)}{t}\;dx&=\frac{1}{t}\int_{a+t}^{b+t} f(s)\;ds-\frac{1}{t}\int_a^bf(x)\;dx\\ \\ &= \frac{1}{t}\int_{b}^{b+t} f(x)\;ds-\frac{1}{t}\int_a^{a+t}f(x)\;dx \end{align} and thus, from the Lebesgue differentiation theorem, $$\int_a^b \frac{T(t)f(x)-f(x)}{t}\;dx\overset{t\to 0^+}{\longrightarrow} f(b)-f(a),\quad \text{for almost all $a,b\in\mathbb{R}$}.\tag{3}$$ Now, $(2)$ and $(3)$ imply $$f(b)=f(a)+\int_a^b Bf(x)\;dx,\quad \text{for almost all $a,b\in\mathbb{R}$}$$ and thus there exists $a\in\mathbb{R}$ such that, by redefining $f$ on a null set, $$f(b)=f(a)+\int_a^b Bf(x)\;dx,\quad \text{for all $b\in\mathbb{R}$}.$$ This shows that $f$ is absolutely continuous with $f'=Bf\in L^2(\mathbb{R})$.

From the above argument we have $$D(B)\subset D(A),\qquad A|_{D(B)}=B.\tag{4}$$ The Hille-Yosida Theorem implies that $1\in\rho(B)$. We also can prove that $1\in\rho(A)$. So, from $(4)$, $$(I-A)(D(B))=(I-B)(D(B))=L^2(\mathbb{R}), \qquad D(A)=(I-A)^{-1}(L^2(\mathbb R))$$ which imply $D(A)=(I-A)^{-1}(I-A)(D(B))=D(B)$ and thus $A=B$. $\square$