Adjoint of sum of two operators

Question

Let $A$ be self-adjoint and $B$ symmetric (which means densely defined for me as well) with $A$-bound less than $1$. Does this imply that $(A+iB)^*=A-iB$ ?

Answer 1

Under the stated conditions: If $L=A+iB$ on $\mathcal{D}(A)$, and if $M=A-iB$ on $\mathcal{D}(A)$, then $L$ and $M$ are closed, densely-defiined and satisfy $L^{\star}=M$, $M^{\star}=L$.

Start by observing that there are positive constants $a$, $b$ with $0 < b < 1$ such that $\|Bx\|\le a\|x\|+b\|Ax\|$, and $\mathcal{D}(A)\subseteq\mathcal{D}(B)$. Then, for any $\epsilon > 0$, and for $x\in\mathcal{D}(A)$, $$ \begin{align} \|Bx\|^{2} & \le (a\|x\|+b\|Ax\|)^{2}\\ & = a^{2}\|x\|^{2}+2ab\|x\|\|Ax\|+b^{2}\|Ax\|^{2} \\ & = a^{2}\|x\|^{2}+\frac{a^{2}}{\epsilon^{2}}\|x\|^{2}+\epsilon^{2}b^{2}\|Ax\|^{2}+b^{2}\|Ax\|^{2}\\ & = (1+\epsilon^{2})\left\|\pm i\frac{a}{\epsilon}x+bAx\right\|^{2} \end{align} $$ Therefore, if $\epsilon$ is chosen so that $b\sqrt{1+\epsilon^{2}} =\rho < 1$, then for $\mu=a(\epsilon b)^{-1}$, $$ \|Bx\| \le \rho\|(A\pm i\mu I)x\|,\;\; x \in\mathcal{D}(A), $$ or $$ \|B(A\pm i\mu I)^{-1}y\|\le \rho\|y\|,\;\; y \in X. $$ Therefore, $I+wB(A\pm i\mu I)^{-1}$ are invertible with bounded inverses $C_{\pm}(w)$ for $|w| \le 1$. Because $A$ is selfadjoint, then $A\pm i\mu I$ are surjective and injective, which carries over to the following operators on $\mathcal{D}(A)$: $$ C_{\pm}(w)(A\pm i\mu I)=(A\pm i\mu I)+wB=(A+wB)\pm i\mu I. $$ Let $L=A+iB$ on $\mathcal{D}(A)$ and let $M=A-iB$ on $\mathcal{D}(A)$. Then $L$ is densely-defined and closed because $L\pm i\mu I$ are injective and surjective with bounded inverses; the same is true of $M\pm i\mu I$. Suppose $y \in \mathcal{D}(L^{\star})$ so that $$ ((L-i\mu I)x,y) = (x,(L^{\star}+i\mu I)y),\;\;\; x \in \mathcal{D}(A). $$ Then there exists $z \in \mathcal{D}(A)$ such that $(M+i\mu I)z=(L^{\star}+i\mu I)y$, which gives $$ ((L-i\mu I)x,y) = (x,(M+i\mu I)z)=((L-i\mu I)x,z),\;\;\; x \in \mathcal{D}(A). $$ Because $L-i\mu I$ is surjective, then that forces $y=z \in \mathcal{D}(A)$. So $\mathcal{D}(L^{\star})\subseteq\mathcal{D}(A)$, which then forces $L^{\star}=M$, and establishes the desired result because the argument is symmetric in $L$ and $M$.