Fractional powers of a diagonal operator Let $H_1$ and $H_2$ be Hilbert spaces and $(A_1,D(A_1))$ and $(A_2, D(A_2))$ two definite positive self-adjoint operators on $H_1$ and $H_2$ respectively. Consider the diagonal operator:
$$A=\begin{pmatrix}
A_1 & 0\\ 
0 & A_2
\end{pmatrix}, \qquad D(A)=D(A_1)\times D(A_2).$$
Is it true that $D(A^\theta)=D(A_1^\theta)\times D(A_2^\theta)$ and
$$A^\theta=\begin{pmatrix}
A_1^\theta & 0\\ 
0 & A_2^\theta
\end{pmatrix},$$
where $B^\theta$ is the fractional power of $B$ for $\theta\in (0,1)$.
I tried the definition using the spectral decomposition but I didn't get the final result. Any hint or reference would be helpful.
 A: In order to set notation:
For a self-adjoint $A: D(A)\to H$ you may define for $F:\sigma(A)\to \Bbb R$ measurable the operator $F(A)$ by:
$$F(A)v= \int_{\Bbb R} F(\lambda)\, dP(\lambda)v $$
where $dP(\lambda)$ is the spectral measure of $A$ and $v\in D(F(A))$, which is given by all $v\in H$ for which the above integral exists in the Bochner sense.
Now for your example what you need to see is:
$$dP(\lambda) = \begin{pmatrix} dP_1(\lambda) &0 \\ 0& dP_2(\lambda)\end{pmatrix}$$
From this everything else follows, for example:
$$F(A)\begin{pmatrix}v_1\\ v_2\end{pmatrix} = \int F(A) \begin{pmatrix} dP_1(\lambda) v_1 \\dP_2(\lambda)v_2\end{pmatrix} = \int F(\lambda) \begin{pmatrix} dP_1(\lambda) v_1 \\0\end{pmatrix}+\int F(\lambda) \begin{pmatrix} 0 \\dP_2(\lambda)v_2\end{pmatrix}$$
which gives both that the question of existence reduces to the question on both components (giving $D(F(A)) = D(F(A_1))\times D(F(A_2))$)  and that the evaluation is the same as the sum of the evaluations of the components (giving $F(A) = F(A_1)\oplus F(A_2)$).
Now the conditions that the spectral measure must fulfil are:


*

*$P(\lambda)$ is projection valued.

*$P(\lambda)$ is monotone in $\lambda$.

*$P(\lambda)$ is right continuous. (Or was it left continuous?)

*$\int \lambda dP(\lambda) = A$.


It is elementary to check that the suggested form of $dP(\lambda)$ satisfies these conditions in this case.
