Definition of perturbation of operator $B$ is $A$-bounded

Question

The operator $A$ and $B$ are closable on Hilbert space, and the norm on $D(A)$ is $\Vert x \Vert_{A}= \Vert x \Vert+\Vert Ax \Vert$, if

(1)$D(A)\subset D(B)$

(2)If there exists $a,b>0$, s.t. $\Vert Bx \Vert \leq a\Vert Ax \Vert +b\Vert x \Vert,$

we say $B$ is $A$-bounded. And $\inf a$ is called the bound of $B$ about $A$.

In textbook by Kato, condition (2) is equivalent to (2)': $$\Vert Bx \Vert^2 \leq c^2\Vert Ax \Vert^2 +d^2\Vert x \Vert^2.$$

Note that if (2)' is true then (2) obviously true, since let $a=c, b=d$. But how to prove the inverse? I do not know that let $$ c= \sqrt{1+\delta} a, d=\sqrt{1+\delta^{-1}} b,$$ then $$\Vert Bx \Vert^2 \leq (1+\delta)a^2\Vert Ax \Vert^2 +(1+\delta^{-1})b^2\Vert x \Vert^2$$ But how to do next? We also need to make $\inf c$ still the bound of $B$.

Thanks for any advanced details.

Answer 1

Okay, I get a solution:

$$ \Vert Bx \Vert ^2 \leq (a\Vert Ax \Vert + b\Vert x \Vert )^2 \leq a^2\Vert Ax \Vert^2+b^2\Vert x \Vert ^2+2ab\Vert Ax \Vert \Vert x \Vert,$$

and we consider the $2ab\Vert Ax \Vert \Vert x \Vert=2a\sqrt{\delta} \sqrt{\delta}^{-1}b\Vert Ax \Vert \Vert x \Vert \leq a^2\delta\Vert Ax \Vert+b^2\delta^{-1}\Vert x \Vert.$ Hence, $$ \Vert Bx \Vert ^2 \leq a^2(1+\delta)\Vert Ax \Vert+b^2(1+\delta^{-1})\Vert x \Vert,$$ let $c=a\sqrt{1+\delta}, d=b\sqrt{1+\delta^{-1}}.$

Answer 2

For nonnegative $u,v$, then $(u+v)^{2}\leq(2\max\{u,v\})^{2}=4\max\{u^{2},v^{2}\}\leq 4(u^{2}+v^{2})$, so

$\|Bx\|^{2}\leq 4(a^{2}\|Ax\|^{2}+b^{2}\|x\|^{2})=c^{2}\|Ax\|^{2}+d^{2}\|x\|^{2}$ by letting $c=2a$ and $d=2b$.

If $\|Bx\|^{2}\leq c^{2}\|Ax\|^{2}+d^{2}\|x\|^{2}$, then $\|Bx\|\leq\sqrt{c^{2}\|Ax\|^{2}+d^{2}\|x\|^{2}}\leq c\|Ax\|+d\|x\|$ because of the inequality that $\sqrt{s^{2}+t^{2}}\leq s+t$ for nonnegative $s,t$.