$D((A+B)^*)= D(A^*)$ if $B$ is $A$-bounded with $A$-bound $0$

Let $$A:D(A) \subseteq H \to H$$ and $$B:D(B) \subseteq H \to H$$ be linear operators on a Hilbert space $$H$$ such that $$A$$ is a closed densely defined operator and $$B$$ is relatively bounded with respect to $$A$$ with relative bound $$0$$. We have to show that $$D((A+B)^*)= D(A^*)$$.

Since $$B$$ is $$A$$-bounded with $$A$$-bound $$0$$ and $$A$$ is closed, we have that $$A+B$$ is a densely defined closed operator with $$D(A+B)=D(A)$$. We know that $$D(A^*) \subseteq D((A+B)^*)$$ because $$A^*+B^* \subseteq (A+B)^*$$, but how can we show that $$D((A+B)^*) \subseteq D(A^*)$$.

Can you give me any hint or a reference for the adjoint of a relatively bounded perturbation, please?

• According to the definition I found in Kato, $B$ is $A$ bounded with relative bound $\xi$ if $D(B)\subset D(A)$ and there is a $\beta$ so that $\|Bu\|≤\beta \|u\| + \xi\|Au\|$. If the relative bound is $0$ then $B$ must be bounded by this definition. What definition are you working with? – s.harp Jul 30 at 20:53
• @s.harp According to the definition of Kato, if $B$ is $A$- bounded with relative bound $\xi$, we can not guarantee that there exists $\beta>0$ such that $\|Bu\| \leq \beta \|u\| + \xi \| Au\|$ for $u \in D(A)$. – Mainkit Jul 30 at 21:44
• Then what is the definition you are using? – s.harp Jul 30 at 21:47
• I'm using definition from Kato: B is $A$-bounded if $D(A) \subseteq D(B)$ and there exists $a,b \geq 0$ such that $\|Bu \|\leq a \| u\|+ b\|Au \| ->(1)$ for all $u \in D(A)$. In that case we say that the relative bound is the infimum of $b\geq0$ such that (1) holds for some $a\geq 0$. – Mainkit Jul 30 at 21:52