Problem: Let $A$ be self adjoint and $B$ symmetric. Suppose $B$ is $A$ bounded with bound $\alpha$. Prove that $$ \lim _{n \rightarrow \infty}\left\|B(A+i n)^{-1}\right\| \leq \alpha $$
Definition: Let $\mathscr{H}$ be a Hilbert space, $A: D(A) \subset \mathscr{H} \rightarrow \mathscr{H}$ a self-adjoint operator and $B: D(B) \subset \mathscr{H} \rightarrow \mathscr{H}$ a symmetric operator with $D(A) \subset D(B)$. We say that $B$ is $A$-bounded if there are positive constants $\alpha, \beta$ such that $$ \|B x\| \leq \alpha\|A x\|+\beta\|x\| $$ for all $x \in D(A)$, and we say that $\alpha$ is an $A$-bound for $B$.
Definition: A densely defined linear operator $A: \mathscr{D}(A) \subset \mathscr{H} \rightarrow \mathscr{H}$ on a Hilbert space $\mathscr{H}$ is a $\underline{\text { Hermitian }}$ or symmetric operator if $(A x, y)=(x, A y)$ for all $x, y \in \mathscr{D}(A)$. This means that the adjoint $A^*$ of $A$ is defined at least on $\mathscr{D}(A)$ and that its restriction to that set coincides with $A$. This fact is often denoted by $A \subset A^*$.
Theorem. (Kato-Rellich) If $B$ is $A$-bounded with $A$-bound smaller than 1 , then $A+B$ is self-adjoint on $D(A)$, and essentially self-adjoint on any core of $A$. Moreover, if $A$ is bounded below, then so is $A+B$.
Definition: The operator A is self-adjoint if it coincides with its adjoint, i.e. if $A=A^*$. If A is closable and its closure coincides with its adjoint,i.e. $\bar{A}=A^*$, then A is said to be essentially self-adjoint.
Not sure how to manipulate around the definition, appreciate any help.
