# If $(\mathcal D(A),A)$ is a linear operator on a Hilbert space and $\alpha\ge0$, then $\mathcal D(A^{-\alpha})=\mathcal D(A^{\alpha})'$

Let

• $H$ be a $\mathbb R$-Hilbert space
• $(\mathcal D(A),A)$ be a densely-defined, linear, positive definite and symmetric operator on $H$
• $(e_n)_{n\in\mathbb N}\subseteq\mathcal D(A)$ be an orthonormal basis of $H$ with $$Ae_n=\lambda_ne_n\tag1$$ for some $(\lambda_n)_{n\in\mathbb N}\subseteq(0,\infty)$ with $$\lambda_n\le\lambda_{n+1}\;\;\;\text{for all }n\in\mathbb N\tag2$$

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

Now, let $\alpha\ge0$. I've read that $\mathcal D(A^{-\alpha})$ can be "identified" with the topological dual space $\mathcal D(A^\alpha)'$ of $\mathcal D(A^\alpha)$. How is this identification been done?

Note that $\mathcal D(A^\alpha)$ equipped with $$\langle x,y\rangle_\alpha:=\langle A^\alpha x,A^\alpha y\rangle_H\;\;\;\text{for }x,y\in\mathcal D(A^\alpha)$$ is a $\mathbb R$-Hilbert space and hence $\mathcal D(A^\alpha)'\cong\mathcal D(A^\alpha)$ in the sense of Riesz' representation theorem.

Moreover, $$\langle x,y\rangle_H=\langle A^\alpha x,A^{-\alpha}y\rangle_H\;\;\;\text{for all }x\in\mathcal D(A^\alpha)\text{ and }y\in H\subseteq\mathcal D(A^{-\alpha})\tag3\;.$$

You basically have everything. We just need to extend $(3)$ to all of $\mathcal{D}(A^{\alpha}) \times \mathcal{D}(A^{-\alpha})$.

We endow $\mathcal{D}(A^{\alpha})$ with the inner product $\langle\,\cdot\,,\,\cdot\,\rangle_{\alpha}$ and obtain a Hilbert space that is in an obvious way isometrically isomorphic to the weighted sequence space

$$\ell^2((\lambda_n^{2\alpha}),\mathbb{N}) = \Biggl\{ x \in \mathbb{R}^{\mathbb{N}} : \sum_{n\in\mathbb{N}} \lambda_n^{2\alpha}\lvert x_n\rvert^2 < +\infty\Biggr\}.$$

We can also consider the weighted sequence space

$$\ell^2((\lambda_n^{-2\alpha}),\mathbb{N}) = \Biggl\{ y \in \mathbb{R}^{\mathbb{N}} : \sum_{n\in\mathbb{N}} \lambda_n^{-2\alpha}\lvert y_n\rvert^2 < +\infty\Biggr\}.$$

We have a natural bilinear pairing

$$\langle\,\cdot\,,\,\cdot\,\rangle_{\alpha,-\alpha} \colon \ell^2((\lambda_n^{2\alpha}),\mathbb{N}) \times \ell^2((\lambda_n^{-2\alpha}),\mathbb{N}) \to \mathbb{R};\quad \langle x,y\rangle_{\alpha,-\alpha} = \sum_{n\in \mathbb{N}} x_n y_n.$$

One verifies easily that this bilinear pairing gives isometric isomorphisms $$\ell^2((\lambda_n^{-2\alpha}),\mathbb{N}) \to \ell^2((\lambda_n^{2\alpha}),\mathbb{N})'$$ and $$\ell^2((\lambda_n^{2\alpha}),\mathbb{N}) \to \ell^2((\lambda_n^{-2\alpha}),\mathbb{N})'.$$

On the other hand, we can identify $\mathcal{D}(A^{-\alpha})$ with $\ell^2((\lambda_n^{-2\alpha}),\mathbb{N})$, since we can write

$$\mathcal{D}(A^{-\alpha}) = \Biggl\{ \sum_{n\in\mathbb{N}} y_n e_n : y \in \ell^2((\lambda_n^{-2\alpha}),\mathbb{N})\Biggr\}.$$

Note: the sum is formed in $\mathcal{D}(A^{-\alpha})$, where $\lVert e_n\rVert = \lambda_n^{-\alpha}$, not in $H$, where $\lVert e_n\rVert_H = 1$.

The situation is analogous to the case of Sobolev spaces. We have $H^{-s} = (H^s)'$ by definition for $s > 0$, and the chain of inclusions (where each inclusion is not only strict as an inclusion of sets, also the topology on each space is strictly finer than the subspace topology induced by a larger space in the chain)

$$H^t \subsetneq H^s \subsetneq H^0 = L^2 \subsetneq H^{-s} \subsetneq H^{-t}$$

for $0 < s < t$, but as Hilbert spaces, all the $H^r$ are of course also isometrically isomorphic to their duals. Here, we have for $0 < \alpha < \beta$ the chain of inclusions

$$\mathcal{D}(A^{\beta}) \subset \mathcal{D}(A^{\alpha}) \subset \mathcal{D}(A^0) = H \subset \mathcal{D}(A^{-\alpha}) \subset \mathcal{D}(A^{-\beta}),$$

and if $\lambda_n \to +\infty$, all inclusions are strict and the topology on each space is strictly finer than the subspace topology induced by a larger space in the chain. And like for the Sobolev spaces, we have $\mathcal{D}(A^{-\alpha}) \cong \mathcal{D}(A^{\alpha})'$ by one natural bilinear pairing, and $\mathcal{D}(A^{\alpha}) \cong \mathcal{D}(A^{\alpha})'$ by the Riesz map.

• By the principle of "make the domain as large as possible", if $\lambda_n \to +\infty$, we must go beyond $H$ for the domain of $A^{-\alpha}$ when $\alpha > 0$. If we don't, then the assertion is wrong and we can't (naturally) identify $\mathcal{D}(A^{-\alpha})$ with $\mathcal{D}(A^{\alpha})'$, since $H$ isn't complete with respect to the $A^{-\alpha}$-norm. – Daniel Fischer Jan 13 '17 at 14:04
• Consider the sequence space $$U_0 = \Biggl\{ x\in \mathbb{R}^{\mathbb{N}} : \sum_{n\in \mathbb{N}} \lambda_n^{2\alpha} \lvert x_n\rvert^2 < +\infty\Biggr\},$$ endowed with the inner product $$\langle x, y\rangle_{U_0} = \sum_{n\in\mathbb{N}} \lambda_n^{2\alpha} x_n y_n.$$ This is a Hilbert space, and $x \mapsto (\mu^n)_{n\in\mathbb{N}}$ is an isometric embedding $(H,\langle\cdot,\cdot\rangle_{\alpha}) \to U_0$. The image of that embedding is dense, so $U_0$ is a completion. Now if $U \supset H$ is a completion of $(H,\langle\cdot,\cdot\rangle_{\alpha})$, then – Daniel Fischer Jan 15 '17 at 21:29
• (a) The isometry is linear, but that's not important for the reassociation. At that point we're working with elements of $U$, which is assumed to be a completion of $H$ (for the $\alpha$ norm) that contains $H$, so $e_n \in U$ for all $n$. And since $\xi_n$ and $\lambda_n^{-\alpha}$ are real numbers, we have $\xi_n(\lambda_n^{-\alpha} e_n) = (\xi_n\lambda_n^{-\alpha})e_n$ just from the vector space axioms. – Daniel Fischer Jan 16 '17 at 21:56
• Sorry, I'll have to think about how to explain that, will take a while. But as for why we can always have a completion such that the isometry is the inclusion: Let $j \colon X \hookrightarrow Y$ be an isometric embedding with dense image, where $Y$ is a Banach space. Let $T$ be a set with $T\cap X = \varnothing$ that has the same cardinality as $Y\setminus j(X)$, and $t\colon T \to Y\setminus j(X)$ a bijection. Let $Z = X \cup T$, and $h\colon Z \to Y$ the bijection such that $h(z)= j(z)$ if $z\in X$ and $h(z) = t(z)$ if $z\in T$. Transport the Banach space structure from $Y$ to $Z$ using $h$, – Daniel Fischer Jan 17 '17 at 20:05
• The existence of such a $T$ is guaranteed by the axioms of set theory. At least if one uses a sensible set theory. The standard is $\mathsf{ZFC}$. The cardinality of a set $S$ (in $\mathsf{ZFC}$; things get weird without choice) is the smallest ordinal $\kappa$ such that there is a bijection $\kappa \to S$. – Daniel Fischer Jan 17 '17 at 21:32