Suppose $A$ is a self-adjoint operator on the Hilbertspace $\mathcal{H}$. Furthermore let $p,q \in \mathbb{R}$ with $0 \leq q < p$. I want to show that $A^q$ is relatively bounded w.r.t $A^p$ with $A^p$-bound zero. So basically I have to show the following:
$$\exists a,b\ge 0: \|A^qx\|^2 \leq a \|x\|^2 + b \|A^px\|^2$$
With the infimum of all $b´s$ satisfying this inequality being equal to zero. My thoughts so far: From functional calculus, I know that I can define $A^q$ as follows:
$A^q = \int t^q dE(t)$, where $E$ is the spectal measure associated with $A$. Furthermore I know that I have:
$$\|A^q x\|^2 = \int t^{2q} \langle dE(t)x,x\rangle \leq \int t^{p+q} \langle dE(t)x,x\rangle $$
At this point I don´t see how to proceed. Can anyone help me out with this please? Is there any (maybe Hölder-like??) inqulity that I can use for this problem?
Using functional calculus as indicated in the question, the problem boils down to showing that for every $b>0$ there exists $a>0$ such that $t^{2q}\leq a+bt^{2p}$ for all $t\geq 0$.
Since $\lim_{t\to\infty}(t^{2q}-bt^{2p})=-\infty$, there exists $T>0$ such that $t^{2q}-bt^{2p}\leq 0$ for $t\geq T$. On the compact interval $[0,T]$, the continuous function $t\mapsto t^{2q}-bt^{2p}$ is bounded above by a constant $a>0$. Altogether, $t^{2q}-bt^{2p}\leq a$ for all $t\geq 0$.