A question about Semigroups and Operator theory

Question

I was going through semigroups theory, where they always give the Domain of the considered operator (often differential operator),

I would like to know how can we identify the Domain of such an operator, let's take for example the operator $A=\frac{d^2}{dx^2}$ on the Hilbert space $H=L^2(0,1)$, I can see that the related equation is $$u_t(t,x)= u_{xx}(t,x)=Au(t,x)$$ Where the semigroup associated is $$S(t)=e^{tA}= \sum_{n=0}^\infty \frac{t^n}{n!}A^n$$ Now,let's identify the domaine by it's definition, which consist of all elements $u\in H$ such that the following limits exists $$\lim_{h \rightarrow 0} \frac{S(h)-I}{h}u$$

My question is: In this limits there terms involving high derivative of $u$, have $u$ to be infinite differentiable? How can we calculate the Domain of an operator in general?

NB. I would like to get a constructive questions and hints, instead of full answer.

Thanks in advance?

Answer 1

This question results from some misunderstandings about unbounded operators, but it is not a bad question, because many of these misunderstandings are shared by other beginners in the field.

1. The semigroup $e^{tA}$ is defined in terms of $A$, including its domain. Different domains yield different different semigroups. So if you want to define the domain of $A$ in terms of $e^{tA}$ and $e^{tA}$ in terms of $A$, you're in a vicious circle. You either need the domain of $A$ first or an independent definition of $e^{tA}$.
2. The semigroup $S(t)$ generated by an unbounded operator is not given by $$ S(t)=\sum_{n=0}^\infty \frac{t^n}{n!}A^n. $$ This formula only holds for bounded $A$.
3. Whenever $V$ is a subspace of $L^2$ such that you can make sense of $Af$ for all $f\in V$, the operator with domain $V$ that acts as $f\mapsto Af$ is a (possibly unbounded) operator in $L^2$. In this sense, you really get to choose the domain of $A$ as an operator on $L^2$. Of course, some choices of $V$ bring more desirable properties with them than others (for example, the operator will not always generate a semigroup), but on bounded domains, there is typically not one preferred choice. Instead, you need to choose boundary conditions. In your example, both $$ V_1=\{f\in H^2(0,1)\mid f(0)=f(1)=0\} $$ and $$ V_2=\{f\in H^2(0,1)\mid f'(0)=f'(1)=0\} $$ are good choices as domains in the sense that the resulting operator is self-adjoint. And there are many more such domains. This corresponds to the fact that the heat equation on $L^2(0,1)$ without boundary conditions is not well-posed.
4. Related to the three points above: The choice of the domain of an operator is not just a formality to make the functional analysis work, but really matters. This can also be seen in your example: The semigroup $e^{tA_1}$ (corresponding to the domain $V_1$) will dissipate mass in the (very strong) sense that $e^{tA_1}f\to 0$ as $t\to \infty$ for all $f\in L^2$. In contrast, $e^{tA_2}$ preserves mass in the sense that $$ \int_0^1 e^{tA_2}f(x)\,dx=\int_0^1 f(x)\,dx $$ for all $t\geq 0$, $f\in L^2$.