# Isomorphic generators

Given the two generators $$(A,D(A))$$ and $$(B,D(B))$$ of two $$C_0$$-semigroups on $$X$$ and $$Y$$ ( Banach spaces), respectively. We assume that there exists an isomorphism $$V:D(A)\longrightarrow D(B)$$ such that $$A= V^{-1} BV.$$ Are necessarily the $$C_0$$-semigroups isomorphic?

• Assume that $A,B$ are invertible (otherwise we shift both), then $S=BVA^{-1}$ is an isomorphism of $X$ with inverse $S^{-1}=AV^{-1}B^{-1}$ and $A=S^{-1}BS$. Jul 11, 2021 at 8:05

If $$V$$ is merely an isomorphism from $$D(A)$$ to $$D(B)$$, the operator $$V^{-1}BV$$ is not well-defined (since $$BV$$ maps $$D(A)$$ to $$Y$$ rather than to $$D(B)$$).
The "right" notion is a follows: Let's call the operators $$A$$ and $$B$$ similar if there exists a Banach space isomorphism $$V: X \to Y$$ that satisfies $$VD(A) = D(B)$$ and $$A = V^{-1}BV$$ (one might say that "$$V$$ intertwines $$A$$ and $$B$$").
If $$A$$ and $$B$$ are similar, then so are the semigroups generated by $$A$$ and $$B$$ (with the same isomorphism $$V$$). That's very easy to prove. Here are two different arguments:
• Use that $$V$$ intertwines the resolvents of $$A$$ and $$B$$ and then use Euler's formula for the semigroups.
• Use the isomorphism $$V$$ to "transport" the semigroup generated by $$B$$ to the space $$X$$ and show that the semigroup obtained this way has generator $$V^{-1}BV: X \supseteq V^{-1}D(B) \to X$$, which is precisely the operator $$A$$. Then use that a semigroup is uniquely determined by its generator.