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Let

  • $\Lambda:=(0,1)$
  • $\mathcal D(A):=H_0^1(\Lambda)\cap H^2(\Lambda)$ and $$Au:=-\Delta u\;\;\;\text{for }u\in\mathcal D(A)$$

Note that $$e_n(x):=\sqrt2\sin\frac{n\pi x}a\;\;\;\text{for }n\in\mathbb N\text{ and }x\in\Lambda$$ is an orthonormal basis of $H:=L^2(\Lambda)$ with $$Ae_n=\underbrace{\left(\frac{n\pi}a\right)^2}_{=:\:\lambda_n}e_n\;\;\;\text{for all }n\in\mathbb N\;.\tag 1$$

How can we show that $$\mathcal D(A^{r/2}):=\left\{u\in H:\sum_{n\in\mathbb N}\lambda_n^r\left|\langle u,e_n\rangle_H\right|^2<\infty\right\}=H_0^r(\Lambda)$$ for all $r\in\mathbb N$?

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    Depends on how $A^{r/2}$ is defined. If it's defined via the Fourier transform, this becomes a question about the domain of a multiplication operator on $\ell^2$. – 2017-01-05
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    @zaq If $\alpha\in\mathbb R$, then $$A^\alpha u:=\sum_{n\in\mathbb N}\lambda_n^\alpha\langle u,e_n\rangle_He_n\;\;\;\text{for }u\in\mathcal D(A^\alpha)\;.$$ – 2017-01-07
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    Can we write that $(-\Delta)^{\frac{1}{2}}=i \frac{d}{dx}$? – 2017-09-16

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