# Relative $A$-Boundedness of Operator

Let $A, B$ operators on a Banach space $X$ with $\operatorname{dom}(A) = \operatorname{dom}(B)$. Futher let $A$ and $A + B$ closed. Then there exist $\alpha, \beta \geq 0$ with $$\Vert B x\Vert \leq \alpha \Vert A x\Vert + \beta \Vert x \Vert$$ for all $x \in \operatorname{dom}(A)$. I don't see how to achieve this inequality. Since $A$ and $A + B$ are closed I know that the graph norms are complete but I dont see how to use that. I also tried to use the sequence characterization of closedness but i struggle to facilitate that. I would appreciate some hints on the topic :)

Let $X \oplus X$ be the Banach space of pairs $(z_1, z_2) \in X\times X$ with the norm $$||(z_1, z_2)|| = ||z_1|| + ||z_2||.$$ Consider the operator $(A, A+B) : D \to X \oplus X$ where $D = dom(A) = dom(B) \subset X$. We claim it is a closed operator. To see that, suppose $$x_n \to x \in X;$$ $$(A, A+B)(x_n) \to (y, z) \in X \oplus X.$$ Considering how we defined the norm on $X \oplus X$, the second equation gives us $$Ax_n \to y$$ and $$(A+B)x_n \to z.$$ Now it follows from the closeness of $A$ and $A+B$ that $y=Ax$ and $z = (A+B)x$.
Next observe that the projection $(x, Ax, (A+B)x) \to (x, Ax)$ from $\mathrm{Graph}\, (A, A+B)$ to $\mathrm{Graph}\, A$ is obviously continuous so the open mapping theorem gives us an estimate $$||Bx|| \le ||x|| + ||Ax|| + ||(A+B)x|| \le C(||x|| + ||Ax||).$$
• I don't see how you get the estimate to be honest. I see that you get by continuity an estimate of the form: $\Vert Ax \Vert + \Vert x \Vert \leq C(\Vert(A + B)x\Vert + \Vert Ax \Vert + \Vert x\Vert)$. But how does this help you? – Yaddle Oct 19 '17 at 13:57