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Let $(H,\langle\;\cdot\;,\;\cdot\;\rangle)$ be a $\mathbb R$-Hilbert space. We say that $(\mathcal D(A),A)$ is a linear operator, if $\mathcal D(A)$ is a subspace of $H$ and $A:\mathcal D(A)\to H$ is linear.

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 $$A^\alpha x:=\sum_{n\in\mathbb N}\lambda_n^\alpha\langle x,e_n\rangle_He_n\;\;\;\text{for }x\in\mathcal D(A^\alpha)\;.$$

How can we show that $\mathcal D(A^1)\subseteq\mathcal D(A^{1/2})$?$^1$

I would prove the claim in the following way: Let $x\in\mathcal D(A^1)$ and $$\lambda_{\text{sup}}:=\sup_{n\in\mathbb N}\lambda_n\;.$$

  • Case 1: $\lambda_{\text{sup}}<\infty$ and hence $$\sum_{n=1}^N\lambda_n\left|\langle x,e_n\rangle\right|^2\le\lambda_{\text{sup}}\sum_{n=1}^N\left|\langle x,e_n\rangle\right|^2\xrightarrow{N\to\infty}\lambda_{\text{sup}}\left\|x\right\|^2\tag 3$$ by Parseval's identity
  • Case 2: $\lambda_{\text{sup}}=\infty$, hence $$\lambda_n\ge 1\;\;\;\text{for all }n\ge n_1\tag 4$$ for some $n_1\in\mathbb N$ and thereby $$\sum_{n=1}^N\lambda_n\left|\langle x,e_n\rangle\right|^2\le\sum_{n=1}^{n_1-1}\left|\langle x,e_n\rangle\right|^2+\sum_{n=n_1}^n\lambda_n^2\left|\langle x,e_n\rangle\right|^2\tag 5$$ where the second sum on the right-hand side of $(4)$ is convergent for $N\to\infty$ by definition of $\mathcal D(A^1)$

In both cases, we obtain $x\in\mathcal D(A^{1/2})$. However, for some reason I think that my argumentation is too complicated. Is there a simpler one?


$^1$ Note that I've explicitly written $\mathcal D(A^1)$; not $\mathcal D(A)$. However, it should be clear that $A$ can be extended to $\mathcal D(A^1)$ if $\mathcal D(A)$ is a proper subset of $\mathcal D(A^1)$.

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You want to show that $$ \sum_{n}\lambda_n^2|\langle x,e_n\rangle|^2 < \infty \implies \sum_n \lambda_n|\langle x,e_n\rangle|^2 < \infty. $$ You have assumed $0 < \lambda_{n} \le \lambda_{n+1}$ for all $n\in\mathbb{N}$. There are two cases to consider. If $\lambda_n$ is uniformly bounded, then both sums are finite for all $x$. The other case is where $\lambda_n\rightarrow\infty$ as $n\rightarrow\infty$. In that case there exists $N$ such that $1 \le \lambda_n$ for all $n\ge N$, which forces $\lambda_n \le \lambda_n^2$ for $n \ge N$. Hence, $$ \sum_{n\ge N} \lambda_n |\langle x,e_n\rangle|^2 \le \sum_{n\ge N}\lambda_n^2|\langle x,e_n\rangle|^2 < \infty \;\;\; \forall x\in\mathcal{D}(A)\\ \implies \mathcal{D}(A)\subseteq\mathcal{D}(A^{1/2}). $$

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  • $\begingroup$ Your argumentation is the same as mine in the question, isn't it? Note that $\displaystyle\lambda_{\text{max}}=\lim_{n\to\infty}\lambda_n$. $\endgroup$ – 0xbadf00d Dec 30 '16 at 11:45
  • $\begingroup$ Yes, I'm confirming what you're saying. $\endgroup$ – DisintegratingByParts Dec 30 '16 at 11:59
  • $\begingroup$ I've realized that we can prove a way more general result and provided an own answer. Maybe you can read it and tell me, if I made any mistake. $\endgroup$ – 0xbadf00d Jan 1 '17 at 20:45
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We can prove a way more general result: Let $$\lambda_{\text{sup}}:=\sup_{n\in\mathbb N}\lambda_n\in(0,\infty]$$ as in the question.

  1. If $\lambda_{\text{sup}}<\infty$, then $$\sum_{n\in\mathbb N}\lambda_n^{2\alpha}\left|\langle x,e_n\rangle_H\right|^2\le\lambda_{\text{sup}}^{2\alpha}\sum_{n\in\mathbb N}\left|\langle x,e_n\rangle_H\right|^2=\lambda_{\text{sup}}^{2\alpha}\left\|x\right\|^2_H<\infty\tag 6$$ by Parseval's identity for all $\alpha\ge0$ and $x\in H$, i.e. $$H\subseteq\mathcal D(A^\alpha)\;\;\;\text{for all }\alpha\ge0\tag 7$$
  2. Otherwise, $$\lambda_n\ge1\;\;\;\text{for all }n>n_1\tag 8$$ for some $n_1\in\mathbb N_0$ and hence $$\sum_{n\in\mathbb N}\lambda_n^{2\alpha}\left|\langle x,e_n\rangle_H\right|^2\le\sum_{n=1}^{n_1}\left|\langle x,e_n\rangle_H\right|^2+\sum_{n>n_1}\lambda_n^{2\beta}\left|\langle x,e_n\rangle_H\right|^2<\infty\tag 9$$ for all $\alpha,\beta\in\mathbb R$ with $\alpha\le\beta$ and $x\in\mathcal D(A^\beta)$, i.e. $$\mathcal D(A^\beta)\subseteq\mathcal D(A^\alpha)\;\;\;\text{for all }\alpha,\beta\in\mathbb R\text{ with }\alpha\le\beta\tag{10}$$
  3. In either case, $$\lambda_n^\alpha\le\lambda_1^\alpha\;\;\;\text{for all }\alpha<0\text{ and }n\in\mathbb N\tag{11}$$ and hence $$\sum_{n\in\mathbb N}\lambda_n^{2\alpha}\left|\langle x,e_n\rangle_H\right|^2\le\lambda_1^{2\alpha}\left\|x\right\|_H\tag{12}$$ by Parseval's identity for all $\alpha<0$ and $x\in H$, i.e. $$H\subseteq\mathcal D(A^\alpha)\;\;\;\text{for all }\alpha<0\tag{13}$$
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