# How to find the infinitesimal generator of this semigroup?

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.

• This is a standard example that you can probably find with a google search of "infinitesimal generator of translation semigroup on l2" or something similar. – mathematician Dec 25 '13 at 0:22
• Hint: $g$ wants to be $f'$.. – Berci Dec 25 '13 at 0:50
• @Berci In your notation, is $f'$ the weak derivative of $f$? – Pedro Dec 26 '13 at 22:11
• @mathematician Do you know some specific book that presents this example? – Pedro Dec 26 '13 at 22:52
• @Pedro I know one parameter semigroups by nagel/engel has it. – mathematician Dec 27 '13 at 0:09

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 $A=\frac{d}{dx}$.

• Clarification for the readers: one has to be careful with the part "the derivative of $f$ lies in $L^2$". (It's appropriately clarified in the following sentence.) E.g., the Cantor staircase function has derivative a.e., which is equal to 0, and thus defines an element of $L^2$. But it does not have a derivative in the $L^2$ sense described by the display formula here. – Post No Bulls Dec 25 '13 at 8:00
• Yiorgos S. Smyrlis, your answer is $$D(A)=\left\{f\in L^2(\mathbb{R}); \text{ exists } g\in L^2(\mathbb R)\text{, such that }\lim_{h\to 0^+} h^{-1}\|f_h-f-hg\|_{L^2(\mathbb R)}=0\right\}$$ Is it possible to prove that this space is the sobolev space $H^1(\mathbb{R})$? If so, how can we do it? Probably it's a elementary question, but I need help. – Pedro Dec 26 '13 at 22:27
• Well, what is true is that $H^1(\mathbb R)\subset {\mathcal D}(A)$. The elements of $H^1(\mathbb R)$ are continuous functions and if $f\in H^1(\mathbb R)$, then $\frac{f(x+h)-f(x)-hf'(x)}{h}=\frac{1}{h}\int_0^h (f'(x+t)-f'(x))dt$ and hence $h^{-2}\|f_h-f-hf'\|^2\le h^{-2}\int_{\mathbb R}(\int_0^h (f'(x+t)-f(x))dt)^2dx\le \frac{1}{h}\int_0^h \|f_t'-f'\|^2dt\to 0$, as $h\to 0$. It requires some work to become totally rigorous. – Yiorgos S. Smyrlis Dec 26 '13 at 22:57
• @YiorgosS.Smyrlis Could you give me an example of a function $f\in D(A)$ such that $f\notin H^1(\mathbb{R})$? – Pedro Dec 27 '13 at 12:52
• No I can't think of one. But if there is a $f\in L^2$ with $f'_+\in L^2$, but $f'\not\in L^2$, then that is the example. – Yiorgos S. Smyrlis Dec 27 '13 at 13:01

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 $A=B$. $\square$