Equality of two operators using (inverse) resolvent

Question

I am trying to understand the proof of the Hille Yosida Theorem on $C_0$ semigroups and their generators but I get stuck on one detail.

Suppose we have shown that an operator $A'$ extends $A$ and that $$ (\alpha I - A') D(A) = (\alpha I-A)D(A'). $$ The author now states that we may conclude $D(A)=D(A')$, but I don't understand why exactly. Can we perhaps somehow use some closure property of the resolvent to get this?

Any advice much appreicated.

Answer 1

The equality $(\alpha I - A') D(A) = (\alpha I-A)D(A')$ makes no sense if you do not have the inclusion $D(A')\subset D(A)$, which is exactly what you want to prove.

The right equality is $(\alpha I - A') D(A) = (\alpha I-A)D(A)$ because $$A'\text{ extends }A\tag{1}.$$

In the context of the Hille-Yosida Theorem, we also have the hypothesis $\alpha\in\rho(A)$. Thus $$\alpha I-A:D(A)\to X\text{ is surjective}.\tag{2}$$ As a result, $$(\alpha I-A')D(A)\overset{(1)}=(\alpha I-A)D(A)\overset{(2)}=X.\tag{3}$$

On the other hand, $\alpha\in \rho(A')$ (because $A'$ is the infinitesimal generator). Thus $$\alpha I-A':D(A')\to X\text{ is injective}.\tag{4}$$ As a result, $$D(A)\overset{(4)}=(\alpha I-A')^{-1}(\alpha I-A')D(A)\overset{(3)}=(\alpha I-A')^{-1}X=D(A').\tag{5}$$

So, in general we have the following

Proposition. If $A'$ extends $A$ and $\rho(A)\cap\rho(A')\neq \varnothing$ then $A'=A$.

Proof: Take $\alpha \in\rho(A)\cap\rho(A')$. Then we have $(5)$. $\square$