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?

## 1 Answer

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.

More information can be found, as usual, in the book of Engel and Nagel (2000).