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Let $H$ 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 $$\lambda_{n+1}\ge\lambda_n\;\;\;\text{for all }n\in\mathbb N\;.\tag 2$$

Now, let $\alpha\in\mathbb R$ and $$A^\alpha x:=\sum_{n\in\mathbb N}\lambda_n^\alpha\langle x,e_n\rangle_He_n\;\;\;\text{for }x\in\mathcal D(A^\alpha)\;.\tag 3$$

$A^\alpha$ is called a fractional power of $A$. How is $\mathcal D(A^\alpha)$ usually defined?

We need to make sure that the series in $(3)$ exists for all $x\in\mathcal D(A^\alpha)$. Since it doesn't make sense to restrict oneself unnecessarily, we should choose $\mathcal D(A^\alpha)$ to be "maximal" with respect to this property. Since we know that a series $\sum_{n\in\mathbb N}x_n$ in $H$ exists if and only if $\sum_{n\in\mathbb N}\left\|x_n\right\|_H<\infty$, I would define $$\mathcal D(A^\alpha):=\left\{x\in H:\sum_{n\in\mathbb N}\lambda_n^\alpha\left|\langle x,e_n\rangle_H\right|<\infty\right\}\;.\tag 4$$

However, I've seen that many authors write $$\mathcal D(A)=\left\{x\in H:\sum_{n\in\mathbb N}\lambda_n^{\color{red}{2}}\left|\langle x,e_n\rangle_H\right|^{\color{red}{2}}<\infty\right\}\tag 5$$ (While this doesn't need to be the case, let's assume that $\mathcal D(A)$ is "maximal" too).

Maybe there is a relation of $(5)$ and Parseval's identity. In any case, I don't understand why $\mathcal D(A)$ in $(5)$ is not defined as my suggested definition of $\mathcal D(A^1)$ in $(4)$.

So, what am I missing and is there anything wrong with my definition of $\mathcal D(A^\alpha)$?

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We need to make sure that the series in $(3)$ exists for all $x\in\mathcal D(A^\alpha)$. Since it doesn't make sense to restrict oneself unnecessarily, we should choose $\mathcal D(A^\alpha)$ to be "maximal" with respect to this property.

Right, one usually chooses $\mathcal{D}(A^{\alpha})$ as the largest subspace on which the definition makes sense, i.e.

$$\mathcal{D}(A^{\alpha}) = \left\{ x \in H : \sum_{n\in\mathbb{N}} \lambda_n^{\alpha}\langle x, e_n\rangle_H e_n\text{ is convergent}\right\}.\tag{$\ast$}$$

Since we know that a series $\sum_{n\in\mathbb N}x_n$ in $H$ exists if and only if $\sum_{n\in\mathbb N}\left\|x_n\right\|_H<\infty$

That is incorrect. In a nontrivial normed space, there are plenty of convergent series - here we're working with $\mathbb{N}$-indexed families of elements of $H$, so a natural order of summation is given and it makes sense to define convergence of a series as convergence of the sequence of partial sums - that aren't absolutely convergent. Even if we want more than just (conditional) convergence of the series and require unconditional convergence or summability (in Fréchet spaces, in particular in Banach spaces, unconditional convergence and summability are the same for countable families), in infinite-dimensional (Banach) spaces we usually don't have the equivalence of absolute convergence and summability. Absolute convergence still implies summability in Banach spaces, but the converse generally doesn't hold.

Our situation is particularly nice, we are looking at an orthogonal family of elements of a Hilbert space, and for this there is an easy characterisation of summability:

Let $\mathscr{H}$ be a Hilbert space, and $\mathscr{F} = \{ x_{\alpha} : \alpha \in A\}$ an orthogonal family in $\mathscr{H}$ (i.e., $\langle x_{\alpha}, x_{\beta}\rangle_{\mathscr{H}} = 0$ for $\alpha\neq \beta$). Then $\mathscr{F}$ is a summable family if and only if $$\sum_{\alpha \in A} \lVert x_{\alpha}\rVert^2 < +\infty.$$

The proof is easy, using

$$\Biggl\lVert\sum_{\alpha \in B} x_{\alpha} \Biggr\rVert^2 = \sum_{\alpha\in B} \lVert x_{\alpha}\rVert^2$$

for finite $B\subset A$.

So here the family $\{ \lambda_n^{\alpha} \langle x, e_n\rangle_H e_n : n \in \mathbb{N}\}$ is summable if and only if

$$\sum_{n \in \mathbb{N}} \lVert\lambda_n^{\alpha} \langle x, e_n\rangle_H e_n\rVert^2 = \sum_{n \in \mathbb{N}} \lambda_n^{2\alpha} \lvert \langle x, e_n\rangle_H\rvert^2 < +\infty.$$

This is also equivalent to the convergence of the sequence of partial sums

$$s_N = \sum_{n = 0}^N \lambda_n^{\alpha} \langle x, e_n\rangle_H e_n$$

due to the orthogonality, hence we wouldn't get a larger $\mathcal{D}(A^{\alpha})$ by considering convergence of the sequence of partial sums instead of summability. Thus we can rewrite $(\ast)$:

$$\mathcal{D}(A^{\alpha}) = \left\{ x \in H : \sum_{n\in \mathbb{N}} \lambda_n^{2\alpha} \lvert\langle x, e_n\rangle_H\rvert^2 < +\infty\right\}. \tag{$\ast\ast$}$$

For $\alpha = 1$, this is precisely $(5)$, and generally this properly contains the space you defined in $(4)$.

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  • $\begingroup$ I'm not sure if I understood what you mean by this is also equivalent to the convergence of the sequence of partial sums [..] due to the orthogonality. What I have understood is that $$\left\|\sum_{n=M}^N\lambda_n^\alpha\langle x,e_n\rangle e_n\right\|^2=\sum_{n=M}^N\lambda_n^{2\alpha}\left|\langle x,e_n\rangle\right|^2\xrightarrow{M,\:N\to\infty}0$$ by Pythagoras' theorem. And since $H$ is complete, this implies the convergence of $(s_N)_{N\in\mathbb N}$. $\endgroup$ – 0xbadf00d Dec 29 '16 at 20:11
  • $\begingroup$ By that, I mean that if we have an orthogonal family $\mathscr{F} = \{ x_n : n \in \mathbb{N}\}$, then the sequence of partial sums $s_N = \sum_{n = 0}^N x_n$ converges if and only if $\mathscr{F}$ is a summable family (if and only if the series is unconditionally convergent, that is, for every bijection $\pi \colon \mathbb{N}\to \mathbb{N}$ the sequence $s^{(\pi)}_N = \sum_{n = 0}^N x_{\pi(n)}$ converges to the same limit). Without the orthogonality, that doesn't follow, of course, a series can be (truly) conditionally convergent [already in $\mathbb{R}$], for arbitrary families $\endgroup$ – Daniel Fischer Dec 29 '16 at 20:14
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    $\begingroup$ $\{x_n : n \in \mathbb{N}\}$, summability is a more restrictive condition than the convergence of the (ordered) series. But for orthogonal families, the weaker condition implies the stronger. $\endgroup$ – Daniel Fischer Dec 29 '16 at 20:16

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