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Let $H$ be a $\mathbb R$-Hilbert space and $(\mathcal D(A),A)$ be a linear operator.

Assume $(e_n)_{n\in\mathbb N}\subseteq\mathcal D(A)$ is an orthonormal basis of $H$ with $$Ae_n=\lambda_ne_n\;\;\;\text{for all }n\in\mathbb N\tag 1$$ for some $(\lambda_n)_{n\in\mathbb N}\subseteq(0,\infty)$ with $$\lambda_{n+1}\ge\lambda_n\;\;\;\text{for all }n\in\mathbb N\;.\tag 2$$

Let $\alpha\in\mathbb R$, $$\mathcal D(A^\alpha):=\left\{x\in H:\sum_{n\in\mathbb N}\lambda_n^{2\alpha}\left|\langle x,e_n\rangle_H\right|^2<\infty\right\}$$ and $$A^\alpha x:=\sum_{n\in\mathbb N}\lambda_n^\alpha\langle x,e_n\rangle_He_n\;\;\;\text{for }x\in\mathcal D(A^\alpha)\;.$$

Let $(x_n)_{n\in\mathbb N}\subseteq\mathcal D(A^\alpha)$ and $x,y\in H$ with $$\left\|x_n-x\right\|_H\xrightarrow{n\to\infty}0\tag 3$$ and $$\left\|A^\alpha x_n-y\right\|_H\xrightarrow{n\to\infty}0\;.\tag 4$$ I want to show that

  1. $x\in\mathcal D(A^\alpha)$
  2. $y=A^\alpha x$

How can we do that?

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The projection of $x_n -x$ onto the $e_k$ component must converge to zero. Multiply this projection by $\lambda_k^\alpha$ to get that $$\lambda_k^\alpha \langle x_n,e_k\rangle\to\lambda_k^\alpha \langle x,e_k\rangle .\tag{1}$$

The left-hand side of $(1)$ is the same as $\langle A^\alpha x_n,e_k\rangle$ which must converge to the $e_k$ component of $y$. This means $\langle y,e_k\rangle = \lambda_k^\alpha\langle x,e_k\rangle$. Since $\sum_k \langle\cdot,e_k\rangle e_k$ is the identity (convergence of sum in SOT) this gives:

$$y=\sum_k \langle y,e_k\rangle e_k=\sum_k \lambda_k^\alpha\langle x,e_k\rangle e_k\tag{2}$$ most importantly: the right-hand side exists. This means $x\in \mathcal D(A^\alpha)$ and also $y=A^\alpha x$ as can be read off of $(2)$.

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If you let $\{ e_n \}$ be an orthonormal basis of Hilbert space $H$, and if you let $\{ \mu_n \}$ be a sequence of non-negative real numbers, then $Ax = \sum_{n}\mu_n\langle x,e_n\rangle e_n$ is defined on its natural domain $$ \mathcal{D}(A)=\left\{ x\in H : \sum_{n}\mu_n^2|\langle x,e_n\rangle|^2 < \infty \right\}. $$ If $\lambda < 0$, then the following defines a bounded linear operator on $H$: $$ R(\lambda)x = \sum_{n}\frac{1}{\mu_n-\lambda}\langle x,e_n\rangle e_n. $$ You can check that the range of $R(\lambda)$ is contained in $\mathcal{D}(A)$, with $(A-\lambda I)R(\lambda)=I$. And $R(\lambda)(A-\lambda I)=I$ on $\mathcal{D}(A)$. $R(\lambda)$ is closed because it is defined everywhere and is bounded on $H$. So $A-\lambda I = R(\lambda)^{-1}$ is closed because the graph of $A-\lambda I$ is the transpose of the graph of $R(\lambda)$ in $H\times H$.

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