Show that the fractional power of a linear operator is closed 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
  
  
*
  
*$x\in\mathcal D(A^\alpha)$
  
*$y=A^\alpha x$
  
  
  How can we do that?


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 $H\times H$.
