Relation between resolvent set and closure of an operator

Question

Consider an operator $A$. Is this statement true?

"If the resolvent of $A$ is bounded, then $A$ is closed."

I am not an expert in this field and I don't if it is a standard result or not, hence apologize me for the question. Thank you!

Answer 1

Let $\lambda \in \mathbb{C}$ be such that $A+\lambda I:D(A)\to H$ is invertible with bounded inverse $(A+\lambda I)^{-1}:H\to D(A)$.

Let $\left\{x_n\right\}\subset D(A)$ with $x_n\to x$, $A x_n\to y$ for some $x,y\in H$. Then $(A+\lambda I)x_n=Ax_n+\lambda x_n\to \lambda x+y$. Thus $$x_n=(A+\lambda I)^{-1}(A+\lambda I)x_n\to (A+\lambda I)^{-1}(\lambda x+y) $$ Since $x_n\to x$, therefore $x=(A+\lambda I)^{-1}(\lambda x+y)\in D(A)$, and so $$Ax+\lambda x=(A+\lambda I)x=\lambda x+ y\implies Ax=y. $$ Hence $A$ is closed.