Consider two positive unbounded operators $A$ and $B$ densely defined on a Hilbert space $H$ self-adjoint on a domain $\mathcal{D}(A) = \mathcal{D}(B) = H_1$. By the spectral theorem, we can define the fractional powers of $A$ and $B$ as self-adjoint linear operators on $H$. My question is, is $\mathcal{D}(A^{\alpha}) = \mathcal{D}(B^\alpha)$, where $\alpha \in (0, 1)$ necessarily? Is this part of the spectral theorem? If yes, where can I find such a statement?
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$\begingroup$ Hmm seems not expectable but interesting... $\endgroup$– C-star-W-starJun 17, 2015 at 3:12
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If $A$ and $B$ are positive definite densely-defined selfadjoint linear operators on a complex Hilbert space $H$, then $AB^{-1}$ and $BA^{-1}$ are bounded if $\mathcal{D}(A)=\mathcal{D}(B)$. Then $A^{s}B^{-s}$ and $B^{s}A^{-s}$ are bounded for $0 < s \le 1$ with $$ \|A^{s}B^{-s}\| \le \|AB^{-1}\|^{s} \\ \|B^{s}A^{-s}\| \le \|BA^{-1}\|^{s}. $$ This implies that $\mathcal{D}(A^{s})=\mathcal{D}(B^{s})$ for $0 < s \le 1$.
Reference (Inequality a3): https://www.encyclopediaofmath.org/index.php/Heinz-Kato_inequality
answered Jun 17, 2015 at 5:14
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$\begingroup$ How do you deduce boundedness from positive definite? I mean one has $\mathcal{W}(H)>0$ but only $\sigma(H)=\overline{\mathcal{W}(H)}$ for $H=H^*$. $\endgroup$ Jun 22, 2015 at 15:41
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$\begingroup$ Oh, I'm sorry, I missed the precise definition of positive definite. $\endgroup$ Jun 22, 2015 at 15:45
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1$\begingroup$ @Freeze_S : positive definite means $A \ge \delta I$ with $\delta > 0$, which gives boundedness of $A^{-1}$. $\endgroup$ Jun 22, 2015 at 16:03