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

1 Answers 1

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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.

  • 0
    Thank you for your answer; I will read the entire post soon, but what do you mean by "extending" $(3)$ to all of $\mathcal{D}(A^{\alpha}) \times \mathcal{D}(A^{-\alpha})$? $(3)$ already holds for all $x\in\mathcal D(A^\alpha)$ and $y\in\mathcal D(A^{-\alpha})$! (Note that $\mathcal D(A^\beta)\subseteq H$ for all $\beta\in\mathbb R$ by definition of $\mathcal D(A^\beta)$; but if $\beta\le0$ we've also got $\mathcal D(A^\beta)\subseteq H$, i.e. $\mathcal D(A^\beta)=H$.) – 2017-01-13
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    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. – 2017-01-13
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    There's still an isometric isomorphism $H \to \mathcal{D}(A^{\alpha})'$ because both are isometrically isomorphic to $\ell^2(\mathbb{N})$, but it's not a natural identification. If $(\lambda_n)$ is bounded, we have the boring situation that all $\mathcal{D}(A^{\beta})$ are $H$, and all the $A^{\beta}$-norms are equivalent. – 2017-01-13
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    So, the definition of $\mathcal D(A^\alpha)$ in the question is only useful for $\alpha\ge0$, isn't it? If I got you right, we should define $$\mathcal D(A^\alpha):=\overline H^{\langle\;\cdot\;,\;\cdot\;\rangle_{\alpha}}$$ for $\alpha\le0$ (where $\langle\;\cdot\;,\;\cdot\;\rangle_{\alpha}$ is defined as in the question and actually is an inner product on $H\times H$ by the previous argumentation). – 2017-01-13
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    Yes. Or to have a symmetrical definition we can use the identification with the weighted $\ell^2$-spaces. – 2017-01-13
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    I guess it's obvious, but how do we see that $\mathcal D(A^\alpha)$ (defined as in the question) equipped with $\langle\;\cdot\;,\;\cdot\;\rangle_{\alpha}$ is not complete, if $\alpha<0$? – 2017-01-13
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    If $\lambda_n \to +\infty$, the map $A^{-\alpha} \colon H \to H$ is a compact operator (with respect to the original topology on $H$). In particular, its range is not closed, and thus not complete. But when we view that as a map $(H,\langle\,\cdot\,,\,\cdot\,\rangle_{-\alpha}) \to (H,\langle\,\cdot\,,\,\cdot\,\rangle_H)$, it is an isometry, so $\mathcal{D}(A^{-\alpha}) \cap H$ is isometrically isomorphic to a non-closed subspace of $H$, hence not complete. – 2017-01-13
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    I will now read your post and write here again as soon as possible. But let me remark that your $\ell^2((\lambda_n^{\color{red}{2}\alpha}),\mathbb{N})$ actually should be $\ell^2((\lambda_n^{\alpha}),\mathbb{N})$, if we use the usual Definition of the $\ell^p$-spaces. – 2017-01-13
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    I used the weight sequence for the measure of each point, so $\ell^p((w_n),\mathbb{N}) = L^p(\zeta_w)$, where $$\zeta_w(A) = \sum_{n \in A} w_n.$$ Then we need the exponent $2\alpha$. If we use the sequence to multiply the $x_n$, then the exponent should be $\alpha$ indeed. – 2017-01-13
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    I've understood almost everything of what you've wrote. Let $\alpha\in\mathbb R$. It's clear that $(e_n)_{n\in\mathbb N}$ is an orthogonal system in $\mathcal D(A^\alpha)$. So, if $x\in\mathbb R^{\mathbb N}$, then $$\sum_{n\in\mathbb N}\left\|x_ne_n\right\|_\alpha^2=\sum_{n\in\mathbb N}\lambda_n^{2\alpha}\left|x_n\right|^2<\infty\tag4$$ if and only if $x\in\ell^2(\lambda^\alpha)$. Now, if $\mathcal D(A^\alpha)$ is defined as in the question, then $$\mathcal D(A^\alpha)=\left\{u\in H:(\langle u,e_n\rangle_H)_{n\in\mathbb N}\in\ell^2(\lambda^\alpha)\right\}\;.$$ – 2017-01-15
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    Now, with $(4)$ in mind, it's intuitively clear to me how you're planing to enrich this space, but I don't know how I need to understand the series occurring in your new definition of $\mathcal D(A^\alpha)$ (this might be an artificial problem only). In general, it is an abbreviating notation for a limit in a normed space. But it's not clear to me what this space is here. Most probably $\mathcal D(A^\alpha):=\overline H^{\langle\;\cdot\;,\;\cdot\;\rangle_{\alpha}}$. – 2017-01-15
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    Yes, it's the completion of $H$ with respect to the $\lVert\cdot\rVert_{\alpha}$-norm. – 2017-01-15
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    For some reason, I really got problems to conclude. Let $α\le0$ and $H_α:=\overline{H}^{\left\|\;\cdot\;\right\|_α}$. By definition, $H_α$ is some abstract thing. Let $(x_n)_{n∈ℕ}\subseteq H$ be $\left\|\;\cdot\;\right\|_α$-Cauchy. The usual approach to find out what $H_α$ really is would be to show that $(x_n)_{n∈ℕ}$ is Cauchy in another complete space $V_α$ and hence has a limit in this space; such that we're able to conclude $H_α=V_α$. Let's take a look what we've got here: Let $$μ^n:=\left(\langle x_n,e_k\rangle_H\right)_{k∈ℕ}\;\;\;\text{for }n∈ℕ\;.$$ – 2017-01-15
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    Then, $(μ^n)_{n∈ℕ}\subseteq\ell^2(λ^α)$ with $$\left\|μ^m-μ^n\right\|_{\ell^2(λ^α)}^2=\left\|x_m-x_n\right\|_α^2\xrightarrow{m,\:n→∞}0\;.\tag5$$ Since $\ell^2(λ^α)$ is complete, there is a $μ∈\ell^2(λ^α)$ with $$\left\|μ^n-μ\right\|_{\ell^2(λ^α)}\xrightarrow{n→∞}0\tag6$$ and $$\sum_{k∈ℕ}\left\|μ_ke_k\right\|_α^2=\left\|μ\right\|_{\ell^2(λ^α)}\tag7\;.$$ Now, if $U$ is a Hilbert space, $I$ is any index set and $(u_i)_{i∈I}$ is any orthogonal system in $U$, then $\sum_{i∈I}u_i$ exists as an element of $U$ if and only if $\sum_{i∈I}\left\|u_i\right\|_U^2<∞$. – 2017-01-15
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    The only thing which prevents me to conclude from $(7)$ is that I don't know what my $U$ is here. – 2017-01-15
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    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 – 2017-01-15
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    $\{ \lambda_n^{-\alpha}\cdot e_n : n \in \mathbb{N}\}$ is an ortho**normal** system in $U$, and it spans a dense subspace (since the linear span is $\lVert\cdot\rVert_{\alpha}$-dense in $H$), thus it is in fact an orthonormal basis of $U$. By general theory, we can write every element of $U$ uniquely as $\sum\limits_{n\in \mathbb{N}} \xi_n\cdot (\lambda_n^{-\alpha}e_n)$ where $\xi \in \ell^2(\mathbb{N})$. If you reassociate and write $\sum (\xi_n \lambda_n^{-\alpha})e_n$, you see that you can write every element of $U$ uniquely as $\sum \eta_n e_n$ with $\eta \in U_0$. – 2017-01-15
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    I guess something went wrong with the definition of your embedding (I didn't used the symbol $x$ in my description above). Maybe you mean $x^n\mapsto\mu^n$. In that case, why is the image dense in $U_0$? – 2017-01-16
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    Sorry. I thought you had defined $\mu^n = \langle x, e_n\rangle_H$ above. Misread it. The map I mean is $H \ni x \mapsto (\langle x, e_n\rangle_H)_{n\in\mathbb{N}}$. The image is dense because it contains a finitely supported sequences, and those form a dense subspace. – 2017-01-16
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    Yes, exactly that. – 2017-01-16
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    Maybe I'm just stupid, but there are still two things that I don't get: (a) Your reassociation should be impossible, unless the isometry between $H$ and $U$ is linear; or am I missing something? (b) You say that any element of $U$ can be written as $\sum_nx_ne_n$ with $x∈U_0$. Note that, by definition, $U_0=\ell^2(λ^α)$. Now, you say that $\ell^2(λ^α)$ is a completion of $H$; so we may choose $U=\ell^2(λ^α)$. But now your claim becomes that any element of $\ell^2(λ^α)$ can be written as $\sum x_ne_n$ with $x∈\ell^2(λ^α)$; which doesn't make sense (since $(e_n)$ is an orthonormal Basis of $H$). – 2017-01-16
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    (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. – 2017-01-16
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    (b) I used "a completion" in the more categorical sense (but with some abuse of language). If $X$ is a normed space, a completion of $X$ is by definition a pair $(Y,j)$, where $Y$ is a Banach space, and $j \colon X \to Y$ is an isometric embedding (linear of course) with dense image. [More categorically, $(Y,j)$ is a completion if $Y$ is a Banach space, $j \colon X \to Y$ is a linear isometry, and for every continuous linear $f\colon X \to Z$ where $Z$ is a Banach space, there is a *unique* continuous linear $\tilde{f}\colon Y \to Z$ such that $f = \tilde{f}\circ j$. – 2017-01-16
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    That's the universal property. By Hahn-Banach, the universal property is equivalent to $\overline{j(X)} = Y$.] The isometry $j$ is typically not explicitly mentioned (like one usually doesn't mention the topology in "let $X$ a topological space"). So a completion of $H$ doesn't necessarily contain $H$, and usually $H\not\subset U_0$. But any two completions are isometrically isomorphic, and one can always construct a completion for which the isometry $j$ is just the inclusion. If $U$ is a completion for which the isometric embedding is the inclusion, – 2017-01-16
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    we can look at the family $\{ \eta_n e_n : n \in \mathbb{N}\}$ for every $\eta \in \mathbb{R}^{\mathbb{N}}$ and ask when that family is summable. Since the family is orthogonal, it is summable in $U$ if and only if $\sum \lVert \eta_n e_n\rVert^2_{\alpha} < +\infty$, equivalently if and only if $\eta \in U_0$. Thus we obtain a map $T \colon U_0 \to U$, mapping $\eta$ to $\sum \eta_n e_n$. – 2017-01-16
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    One verifies that this map is a linear isometry and that $T\circ j_0 = \operatorname{id}_H$, where $j_0 \colon H \to U_0$ is the isometry $x \mapsto (\langle x,e_n\rangle_H)_{n\in\mathbb{N}}$. So $T(U_0)$ is a complete subspace of $U$ containing $H$, hence $T(U_0) = U$ and we have our isometric isomorphism between $U_0$ and $U$. – 2017-01-16
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    I'm sorry that this must seem to be as an endless discussion to you. The problem is, that (almost) everything you wrote is clear to me. It's clear to me that $$(H,\left\|\;\cdot\;\right\|_\alpha)\to\ell^2(\Lambda^\alpha)\;,\;\;\;x\mapsto(\langle x,e_n\rangle_H)_{n\in\mathbb N}$$ $(\alpha\le0$) is a dense isometric embedding. So, the functional analytic definition of a completion (which is unique up to isometric isomorphism) tells us, that $\ell^2(\Lambda^\alpha)$ is a completion of $(H,\left\|\;\cdot\;\right\|_\alpha)$. – 2017-01-16
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    My problem is what we're trying to make of that. The notation, $$\sum_{n\in\mathbb N}x_ne_n\tag8$$ for $x\in\mathbb R^{\mathbb N}$ has not to be understood with respect to $\left\|\;\cdot\;\right\|_H$, but somehow with respect to $\left\|;\cdot\;\right\|_\alpha$. The intuition is to enrich $\mathcal D(A^\alpha)$ to a space $H_\alpha$ such that "$A^\alpha u\in H$" for all $u\in H_\alpha$. However, the problem is $(8)$. – 2017-01-16
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    We should have $$H_\alpha=\left\{\sum_{n\in\mathbb N}x_ne_n:x\in\ell^2(\lambda^\alpha)\right\}\;,$$ where the series is with respect to $\left\|\;\cdot\;\right\|_\alpha$ in a space which can be equipped with that norm, but it's beyond my imagination what this space really is ... (And why can we always find a completion such that the isometry is the inclusion?) – 2017-01-16
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    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 $h$, – 2017-01-17
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    that is, define $z_1 + z_2 = h^{-1}(h(z_1) + h(z_2))$, $c\cdot z = h^{-1}(c\cdot h(z))$, and $\lVert z\rVert_Z = \lVert h(z)\rVert_Y$. These operations coincide with the originally given ones on $X$: $\lVert x\rVert_Z = \lVert h(x)\rVert_Y = \lVert j(x)\rVert_Y = \lVert x\rVert_X$, and $x_1 +_Z x_2 = h^{-1}(h(x_1) +_Y h(x_2)) = h^{-1}(j(x_1) +_Y j(x_2)) = h^{-1}(j(x_1 +_X x_2)) = h^{-1}(h(x_1 +_X x_2)) = x_1 +_X x_2$, and I'll leave the scalar multiplication to you. Since by definition $h$ is a bijective linear isometry, $Z$ is a Banach space, and $X$ a dense linear subspace of $Z$. – 2017-01-17
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    So we have a completion $Z$ of $X$ such that $X\subset Z$. Yes, it's an ugly construction, but that type of construction is useful and widespread for all sorts of extensions (extending a cancellative monoid to a group, extending an integral domain to its field of fractions, completion of metric or uniform spaces) whenever one wants the original thing to be a substructure of the extension and not only have an embedding (in whatever sense of embedding is pertinent). – 2017-01-17
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    Why does such a $T$ exist? And I'm not sure how the cardinality of an uncountable set is defined. – 2017-01-17
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    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 $\kappa \to S$. – 2017-01-17
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    If $ι_α$ (arbitrary $α\in\mathbb R$) is the dense isometric embedding from $\mathcal D(A^α)$ into $\ell^2(\lambda^α)$ (your $U_0$), $κ_α$ is the linear isometry between $\ell^2(\lambda^α)$ and $H_α:=U$ (your arbitrary completion) and $\tildeι_α$ is the dense isometric embedding from $\mathcal D(A^α)$ into $H_α$, then $$κ_α\circι_α=\tildeι_α\;.\tag9$$ You've chosen $H_α$ such that $ι_α$ is the inclusion; that's fine. But your conclusion (using your symbols) $T\circ j_0=\text{id}_H$ should be wrong; the right-hand side needs to be the inclusion and not the identity. Do you agree? – 2017-01-28
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    I'm still interested in an answer. Oh, and how does the duality pairing between $\mathcal D(A^\alpha)$ and $\mathcal D(A^{-\alpha})$ look like (without the identification)? – 2017-02-02
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    Sorry, I've been away and this kind of fell off the radar. I'm not sure whether I can untangle all the notation used in your previous comment, since we've used a lot of different (and inconsistent) notations in this comment thread. I'll try tomorrow. But as for now, consider the space $s = \mathbb{R}^{\mathbb{N}}$ of all real sequences. For $\alpha\in \mathbb{R}$, we have the linear map $M_{\alpha} \colon s \to s$ given by $M_{\alpha}\bigl((x_n)_{n\in\mathbb{N}}\bigr) = (\lambda_n^{\alpha} x_n)_{n\in \mathbb{N}}$. Since $\lambda_n > 0$ for all $n$, these are bijections and – 2017-02-02
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    $M_{\alpha}^{-1} = M_{-\alpha}$. Inside $s$, we have the subspace $\ell^2(\mathbb{N})$, and we can look at the spaces $E_{\alpha} = M_{\alpha}^{-1}\bigl(\ell^2(\mathbb{N})\bigr) = \{ x \in s : M_{\alpha}(x) \in \ell^2(\mathbb{N}\}$. Now for $H = \ell^2(\mathbb{N})$ and $\alpha \geqslant 0$, we have $E_{\alpha} = \mathcal{D}(A^{\alpha})$ and $A^{\alpha}= M_{\alpha}\lvert_{\mathcal{D}(A^{\alpha})}$. So it is not unreasonable to extend these identities to $\alpha < 0$. If the sequence $(\lambda_n)$ is bounded, we have $E_{\alpha}= \ell^2(\mathbb{N})$ for all $\alpha \in \mathbb{R}$ and everything – 2017-02-02
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    is boring. If $\lambda_n \to +\infty$, then we have $E_{\alpha} \subsetneq \ell^2(\mathbb{N}) \subsetneq E_{-\alpha}$ for $\alpha > 0$ and thus have a natural domain for $A^{-\alpha}$ that extends beyond $H = \ell^2(\mathbb{N})$. Furthermore, we have a natural dual pairing $E_{\alpha} \times E_{-\alpha} \to \mathbb{R}$ for all $\alpha \in \mathbb{R}$, since $E_{-\alpha} = \Bigl\{ y \in s : \sum x_n y_n \text{ converges for all } x \in E_{\alpha}\Bigr\}$, and if we endow $E_{\alpha}$ with the norm $\lVert x\rVert_{E_{\alpha}} = \sqrt{\sum \lambda_n^{2\alpha} x_n^2}$ – 2017-02-02
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    (which makes $E_{\alpha}$ a Hilbert space), this gives an isometric isomorphism between $E_{-\alpha}$ and $E_{\alpha}'$. So if we start with $H = \ell^2(\mathbb{N}) \subset s$, we have a concrete picture. And we can easily see that $\ell^2(\mathbb{N})$ is dense in $E_{\alpha}$ for $\alpha < 0$ with respect to $\lVert\cdot\rVert_{E_{\alpha}}$. We can always map things to this situation by identifying $x\in H$ with the sequence of its Fourier coefficients with respect to the ONB $(e_n)$. – 2017-02-02