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.


  • $\begingroup$ 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. $\endgroup$ – mathematician Dec 25 '13 at 0:22
  • $\begingroup$ Hint: $g$ wants to be $f'$.. $\endgroup$ – Berci Dec 25 '13 at 0:50
  • $\begingroup$ @Berci In your notation, is $f'$ the weak derivative of $f$? $\endgroup$ – Pedro Dec 26 '13 at 22:11
  • $\begingroup$ @mathematician Do you know some specific book that presents this example? $\endgroup$ – Pedro Dec 26 '13 at 22:52
  • 1
    $\begingroup$ @Pedro I know one parameter semigroups by nagel/engel has it. $\endgroup$ – 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 $A=\frac{d}{dx}$.

  • 1
    $\begingroup$ 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. $\endgroup$ – Post No Bulls Dec 25 '13 at 8:00
  • $\begingroup$ 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. $\endgroup$ – Pedro Dec 26 '13 at 22:27
  • $\begingroup$ 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. $\endgroup$ – Yiorgos S. Smyrlis Dec 26 '13 at 22:57
  • $\begingroup$ @YiorgosS.Smyrlis Could you give me an example of a function $f\in D(A)$ such that $f\notin H^1(\mathbb{R})$? $\endgroup$ – Pedro Dec 27 '13 at 12:52
  • $\begingroup$ 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. $\endgroup$ – 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 $A=B$. $\square$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.