Let $H$ be a Hilbert space and $D:H\rightarrow H$ be a densely-defined, unbounded self-adjoint operator, such that $D$ is a bounded operator when viewed as an operator
$$D:\text{dom}(D)\rightarrow H,$$
where $\text{dom}(D)$ is given the graph norm. Then one knows that
$$(D^2+1)^{-1}:H\rightarrow \text{dom}(D^2)$$
is a positive bounded operator with a square root, which we can denote by $(D^2+1)^{-1/2}$. I believe it follows from functional analysis that the range of $(D^2+1)^{-1/2} = \text{dom}(D)$.
Question: Is $(D^2+1)^{-1/2}$ a bounded operator $H\rightarrow \text{dom}(D)$, where $\text{dom}(D)$ is given the graph norm as above?
$(D^2+I)^{-1}$ is a bounded positive operator, and $((D^2+I)^{-1})^{1/2}$ is the unique positive square root of that bounded operator, which makes it positive. To see that $(D^2+I)^{-1}$ is bounded, note that, for all $f\in\mathcal{D}(D^2)$, $$ \langle f,f \rangle \le \langle (D^2+I)f,f\rangle \le \|(D^2+I)f\|\|f\| \\ \|f\| \le \|(D^2+I)f\|. $$ Setting $f = (D^2+I)^{-1}g$ for an arbitrary $g$ leads to the conclusion that $(D^2+I)$ is bounded by $1$ in operator norm: $$ \|(D^{2}+I)^{-1}g\| \le \|g\|. $$