Consider a nonnegative unbounded self-adjoint operator $A$ on a Hilbert space $\mathcal{H}$, with domain $\mathcal{D}(A)$; by the spectral theorem, the operator $A^r$ is well-defined for all $r\geq0$. Let us focus on the cases $r\in[1/2,1]$. By construction, the domain of $A^r$ is larger than the domain of $A$; consequently, $A^r$ is relatively bounded with respect to $A$, that is, there exist $a_r,b_r\geq0$ such that, for all $\psi\in\mathcal{D}(A)$, \begin{equation} \|A^r\psi\|\leq a_r\|A\psi\|+b_r\|\psi\|, \end{equation} with the infimum of all $a_r$ as above being the $A$-bound of $A^r$.
I am interested in estimating the $A$-bound of $A^r$. I know what happens in the two cases $r=1$ and $r=1/2$: clearly, in the first case the $A$-bound is $1$, while in the second case it can be easily proven, in full generality, that the $A$-bound of $A^{1/2}$ is zero; in other words, $A^{1/2}$ is infinitesimally bounded with respect to $A$. This can be proven by using the inequality ($\psi\in\mathcal{D}(A)$) \begin{equation} \|A^{1/2}\psi\|^2=\left\langle\psi,A\psi\right\rangle\leq\|\psi\|\|A\psi\| \end{equation} and noticing that, for all $\epsilon>0$, \begin{equation} 0\leq\epsilon\left(\|A\psi\|^{1/2}-\frac{\|\psi\|^{1/2}}{2\epsilon}\right)^2=\epsilon\|A\psi\|-\|A\psi\|^{1/2}\|\psi\|^{1/2}+\frac{1}{4\epsilon}\|\psi\|. \end{equation}
This leads me to the following (naive) guess: the $A$-bound $a_r$ of $A^r$ for $r\in[1/2,1]$ must be some function of $r$ interpolating (monotonically?) between the values $a_{1/2}=0$ and $a_1=1$.
In order to check whether that is true, I was thinking to apply the following property: the $A$-bound of $A^r$ must be equal to \begin{equation} \lim_{\lambda\to\infty}\left\|A^r(A\pm i\lambda)^{-1}\right\|, \end{equation} where, in the latter equation, $\|\cdot\|$ is to be interpreted as the operator norm. However, right now I have not managed to evaluate $a_r$ by means of the equation above.
The $A$-bound $a_r$ is zero for all $r<1$: Let $\mu$ be the spectral measure of $\psi$. By the spectral theorem, $$ \lVert A^r\psi\rVert^2=\int\lambda^{2r}\,d\mu. $$ By Hölder's inequality with $p=1/r$ and $q=1/(1-r)$ we have $$ \lVert A^r\psi\rVert^2\leq \left(\int\lambda^2\,d\mu\right)^r\left(\int\,d\mu\right)^{1-r}=\lVert A\psi\rVert^{2r}\lVert \psi\rVert^{2(1-r)}. $$ Thus $\lVert A^r\psi\rVert\leq \lVert A\psi\rVert^r\lVert\psi\rVert^{1-r}$. By Young's inequality with $p$ and $q$ as above, $$ \lVert A\psi\rVert^r\lVert\psi\rVert^{1-r}\leq \epsilon r \lVert A\psi\rVert+(1-r)\epsilon^{-r/(1-r)}\lVert\psi\rVert $$ for all $\epsilon>0$.