How to show that $ B $ is bounded with respect to $ A $.

Question

Assume that $ A $ is a densely defined and closed operator and $ B $ is a closable operator. Assume that the definite fields of $ A,B $ denoted by $ D(A) $ and $ D(B) $ satisfies that $ D(A)\subset D(B) $. Show that $ B $ is bounded with respect to $ A $.

I want to prove by contradiction. If $ B $ is not bounded with respect to $ A $. Then there exists a sequence $ \{x_n\} $ such that $ \|Bx_n\|=1 $ and $ 1\geq n(\|x_n\|+\|Ax_n\|) $. However, I cannot know how to go on. Can you give me some hints or references?

Answer 1

WLOG we may assume that $B$ is closed. Consider the linear operator $T:\Gamma_A\to \Gamma_B$ defined by $T(x,Ax)=(x,Bx).$ The graph of $T$ is closed. Indeed assume that $(x_n,Ax_n)\to (x,y)$ and $(x_n,Bx_n)=T(x_n,Ax_n)\to (x,z).$ As $\Gamma_A$ and $\Gamma_B$ are closed we get $y=Ax$ and $z=Bx.$ Therefore $T(x_n,Ax_n)\to (x,Bx),$ i.e. $T(x,Ax)=(x,Bx).$ By the closed graph theorem ($\Gamma_A$ and $\Gamma_B$ are Banach spaces) the operator $T$ is bounded. Therefore $$\|x\|+\|Bx\|\le C(\|x\|+\|Ax\|),\quad x\in D(A)$$