[proofplan] We transfer Hilbert structure from the standard sequence space $\ell^2(\mathbb{Z}^n)$ to $H^s(\mathbb{T}^n)$ through the weighted Fourier coefficient map $\Psi_s$ defined by $\Psi_s(f)_k := (1 + |k|^2)^{s/2}\,\hat{f}(k)$. The proof has four parts: (1) verify that $\Psi_s$ is an isometry from $H^s(\mathbb{T}^n)$ into $\ell^2(\mathbb{Z}^n)$ by direct computation of the norm; (2) prove surjectivity by exhibiting, for each $\ell^2$-sequence, a tempered distribution on $\mathbb{T}^n$ whose Fourier coefficients realise it; (3) deduce the Hilbert space structure of $H^s(\mathbb{T}^n)$ by transfer through the isometric isomorphism; (4) establish separability by exhibiting a countable dense subset built from trigonometric polynomials with Gaussian-rational coefficients. [/proofplan]

[step:Set up the weighted Fourier coefficient map] Recall that, by definition, $H^s(\mathbb{T}^n)$ is the space of tempered distributions $f \in \mathcal{D}'(\mathbb{T}^n)$ with Fourier coefficients $\hat{f}: \mathbb{Z}^n \to \mathbb{C}$ satisfying \begin{align*} \|f\|_{H^s(\mathbb{T}^n)}^2 := \sum_{k \in \mathbb{Z}^n} (1 + |k|^2)^s |\hat{f}(k)|^2 < \infty, \end{align*} endowed with the inner product \begin{align*} (f, g)_{H^s(\mathbb{T}^n)} := \sum_{k \in \mathbb{Z}^n} (1 + |k|^2)^s\, \hat{f}(k)\, \overline{\hat{g}(k)}. \end{align*} We adopt the following Fourier conventions throughout the proof. For a test function $\varphi \in \mathcal{D}(\mathbb{T}^n) = C^\infty(\mathbb{T}^n)$ we set \begin{align*} \hat{\varphi}(k) := \frac{1}{(2\pi)^n}\int_{\mathbb{T}^n} \varphi(x)\, e^{-i k \cdot x}\, d\mathcal{L}^n(x), \qquad k \in \mathbb{Z}^n, \end{align*} and for a distribution $T \in \mathcal{D}'(\mathbb{T}^n)$ we set \begin{align*} \hat{T}(k) := T(e_{-k}), \qquad \text{where } e_{-k}(x) := e^{-i k \cdot x}. \end{align*} This convention is consistent with the regular case: if $T = T_f$ is the distribution induced by an $L^1$ function $f$ via $T_f(\varphi) := \frac{1}{(2\pi)^n}\int_{\mathbb{T}^n} f\, \varphi\, d\mathcal{L}^n$, then $\hat{T_f}(k) = T_f(e_{-k}) = \hat{f}(k)$, so the two notions of Fourier coefficient agree. In particular, the test function $e_{-k}$ has Fourier coefficients $\hat{e_{-k}}(j) = \frac{1}{(2\pi)^n}\int_{\mathbb{T}^n} e^{-i k \cdot x}\, e^{-i j \cdot x}\, d\mathcal{L}^n(x) = \delta_{j, -k}$, where $\delta$ is the Kronecker delta. Fix $s \in \mathbb{R}$ and define the weighted Fourier coefficient map \begin{align*} \Psi_s: H^s(\mathbb{T}^n) &\to \mathbb{C}^{\mathbb{Z}^n} \\ f &\mapsto \big( (1 + |k|^2)^{s/2}\, \hat{f}(k) \big)_{k \in \mathbb{Z}^n}. \end{align*} The map $\Psi_s$ is $\mathbb{C}$-linear because Fourier coefficients are linear in $f$ and the weight $(1 + |k|^2)^{s/2}$ is independent of $f$. [/step]

[step:Verify that $\Psi_s$ takes values in $\ell^2(\mathbb{Z}^n)$ and is an isometry]Let $f \in H^s(\mathbb{T}^n)$. Then \begin{align*} \|\Psi_s(f)\|_{\ell^2(\mathbb{Z}^n)}^2 = \sum_{k \in \mathbb{Z}^n} \big| (1 + |k|^2)^{s/2}\, \hat{f}(k) \big|^2 = \sum_{k \in \mathbb{Z}^n} (1 + |k|^2)^s\, |\hat{f}(k)|^2 = \|f\|_{H^s(\mathbb{T}^n)}^2 < \infty. \end{align*} Hence $\Psi_s(f) \in \ell^2(\mathbb{Z}^n)$ and $\|\Psi_s(f)\|_{\ell^2(\mathbb{Z}^n)} = \|f\|_{H^s(\mathbb{T}^n)}$. Polarising the identity (both inner products are sesquilinear and continuous in each argument) yields \begin{align*} (\Psi_s(f), \Psi_s(g))_{\ell^2(\mathbb{Z}^n)} = (f, g)_{H^s(\mathbb{T}^n)}, \end{align*} so $\Psi_s$ is a linear isometry $H^s(\mathbb{T}^n) \to \ell^2(\mathbb{Z}^n)$.[/step]

[guided]The verification is essentially a redefinition: the very norm placed on $H^s(\mathbb{T}^n)$ was designed to make $\Psi_s$ an isometry. To see this concretely, fix $f \in H^s(\mathbb{T}^n)$. By construction \begin{align*} \|\Psi_s(f)\|_{\ell^2(\mathbb{Z}^n)}^2 = \sum_{k \in \mathbb{Z}^n} | \Psi_s(f)_k |^2 = \sum_{k \in \mathbb{Z}^n} (1 + |k|^2)^{s} | \hat{f}(k) |^2 = \|f\|_{H^s(\mathbb{T}^n)}^2. \end{align*} Since $f \in H^s(\mathbb{T}^n)$, the right-hand side is finite, so $\Psi_s(f) \in \ell^2(\mathbb{Z}^n)$. Why does this norm identity also imply preservation of inner products? Both $(\cdot,\cdot)_{H^s(\mathbb{T}^n)}$ and $(\cdot,\cdot)_{\ell^2(\mathbb{Z}^n)}$ are conjugate-linear in the second argument and linear in the first, and the polarisation identity \begin{align*} 4(u, v) = \|u + v\|^2 - \|u - v\|^2 + i\|u + iv\|^2 - i\|u - iv\|^2 \end{align*} recovers the inner product from the norm in any complex inner product space. Applying polarisation on both sides and using the norm identity for the four arguments $f \pm g$, $f \pm ig$ gives \begin{align*} (\Psi_s(f), \Psi_s(g))_{\ell^2(\mathbb{Z}^n)} = (f, g)_{H^s(\mathbb{T}^n)}. \end{align*}[/guided]

[step:Show that $\Psi_s$ is surjective onto $\ell^2(\mathbb{Z}^n)$]Let $(c_k)_{k \in \mathbb{Z}^n} \in \ell^2(\mathbb{Z}^n)$. We exhibit $f \in H^s(\mathbb{T}^n)$ with $\Psi_s(f) = (c_k)_{k}$. Define the candidate Fourier coefficients \begin{align*} a_k := (1 + |k|^2)^{-s/2}\, c_k, \qquad k \in \mathbb{Z}^n. \end{align*} Define the linear functional \begin{align*} T: \mathcal{D}(\mathbb{T}^n) &\to \mathbb{C} \\ \varphi &\mapsto \sum_{k \in \mathbb{Z}^n} a_k\, \hat{\varphi}(-k). \end{align*} Note the index $-k$: under the convention from Step 1 the test function $e_{-k}$ has $\hat{e_{-k}}(j) = \delta_{j,-k}$, so the index reflection is what produces the desired identity $\hat{T}(k) = a_k$ below. We first verify that the defining sum converges absolutely for every $\varphi \in \mathcal{D}(\mathbb{T}^n)$. By the Cauchy-Schwarz inequality applied to the pair of $\ell^2$-sequences $\big( (1+|k|^2)^{s/2} a_k \big)_k$ and $\big( (1+|k|^2)^{-s/2} \hat{\varphi}(-k) \big)_k$, \begin{align*} \sum_{k \in \mathbb{Z}^n} |a_k|\, |\hat{\varphi}(-k)| \le \left( \sum_{k} (1+|k|^2)^{s} |a_k|^2 \right)^{1/2} \left( \sum_{k} (1+|k|^2)^{-s} |\hat{\varphi}(-k)|^2 \right)^{1/2}. \end{align*} The first factor equals $\big( \sum_k |c_k|^2 \big)^{1/2} = \|(c_k)\|_{\ell^2}$. For the second factor, since $\varphi \in C^\infty(\mathbb{T}^n)$, integration by parts on $\mathbb{T}^n$ gives, for every multi-index $\alpha$, \begin{align*} (ik)^\alpha\, \hat{\varphi}(k) = \widehat{\partial^\alpha \varphi}(k), \qquad |\hat{\varphi}(k)| \le \|\partial^\alpha \varphi\|_{L^\infty(\mathbb{T}^n)}\, |k|^{-|\alpha|} \quad (k \ne 0). \end{align*} Hence the Fourier coefficients of $\varphi$ decay faster than any polynomial in $|k|$, which (combined with the index symmetry $|k| = |-k|$) makes $\sum_{k} (1+|k|^2)^{-s} |\hat{\varphi}(-k)|^2$ finite for every $s \in \mathbb{R}$, and $T(\varphi)$ is well-defined. We now exhibit an explicit Schwartz-seminorm bound for $T$. Pick an integer $N$ with $2N > n - 2s$, so that \begin{align*} \sum_{k \in \mathbb{Z}^n} (1 + |k|^2)^{-s} (1 + |k|^2)^{-N} = \sum_{k \in \mathbb{Z}^n} (1 + |k|^2)^{-N - s} =: C_{N,s} < \infty. \end{align*} For each $k \in \mathbb{Z}^n$ and any multi-index $\alpha$ with $|\alpha| \le 2N$, the Fourier-coefficient bound $|k^\alpha\, \hat{\varphi}(k)| \le \|\partial^\alpha \varphi\|_{L^\infty}$ summed over $|\alpha| \le 2N$ yields \begin{align*} (1 + |k|^2)^N |\hat{\varphi}(k)|^2 \le C_n \sum_{|\alpha| \le 2N} \|\partial^\alpha \varphi\|_{L^\infty(\mathbb{T}^n)}^2, \end{align*} where $C_n$ depends only on $n$ and accounts for the multinomial expansion of $(1+|k|^2)^N$. Multiplying through by $(1+|k|^2)^{-s-N}$ and summing over $k$, \begin{align*} \sum_{k} (1+|k|^2)^{-s} |\hat{\varphi}(k)|^2 \le C_n\, C_{N,s} \sum_{|\alpha| \le 2N} \|\partial^\alpha \varphi\|_{L^\infty(\mathbb{T}^n)}^2. \end{align*} Combining with Cauchy-Schwarz gives the explicit estimate \begin{align*} |T(\varphi)| \le C\, \|(c_k)\|_{\ell^2}\, \sum_{|\alpha| \le 2N} \|\partial^\alpha \varphi\|_{L^\infty(\mathbb{T}^n)}, \end{align*} with $C = C(n, s) > 0$ depending only on $n$ and $s$. This is precisely the seminorm estimate required for $T$ to belong to $\mathcal{D}'(\mathbb{T}^n)$, so $T \in \mathcal{D}'(\mathbb{T}^n)$. We now identify the Fourier coefficients of $T$. Using the convention $\hat{T}(k) = T(e_{-k})$ from Step 1 and the fact that $\hat{e_{-k}}(j) = \delta_{j,-k}$, \begin{align*} \hat{T}(k) = T(e_{-k}) = \sum_{j \in \mathbb{Z}^n} a_j\, \hat{e_{-k}}(-j) = \sum_{j} a_j\, \delta_{-j,-k} = \sum_{j} a_j\, \delta_{j,k} = a_k. \end{align*} So $T$ has Fourier coefficients $(a_k)_k$, and \begin{align*} \sum_{k} (1 + |k|^2)^s |\hat{T}(k)|^2 = \sum_{k} (1 + |k|^2)^s\, (1+|k|^2)^{-s}\, |c_k|^2 = \sum_{k} |c_k|^2 < \infty. \end{align*} Hence $T \in H^s(\mathbb{T}^n)$. Setting $f := T$ gives $\Psi_s(f)_k = (1 + |k|^2)^{s/2}\, a_k = c_k$, so $\Psi_s(f) = (c_k)$. This shows $\Psi_s$ is surjective.[/step]

[guided]The strategy is to invert the procedure. Given a target sequence $(c_k) \in \ell^2(\mathbb{Z}^n)$, we must produce a distribution on $\mathbb{T}^n$ whose Fourier coefficients are precisely \begin{align*} a_k := (1 + |k|^2)^{-s/2}\, c_k, \qquad k \in \mathbb{Z}^n. \end{align*} Why divide by the weight $(1+|k|^2)^{s/2}$? The defining relation $\Psi_s(f)_k = (1 + |k|^2)^{s/2}\, \hat{f}(k) = c_k$ forces \begin{align*} \hat{f}(k) = (1 + |k|^2)^{-s/2}\, c_k = a_k. \end{align*} The challenge for general $s \in \mathbb{R}$ is that $a_k$ may fail to be the Fourier coefficients of an $L^2$ function. When $s < 0$ the weight $(1+|k|^2)^{-s/2}$ grows polynomially, so $|a_k|$ can grow polynomially in $|k|$ without obstructing membership of $(c_k)$ in $\ell^2$. The candidate target $f$ must therefore be constructed as a distribution rather than a function. A natural choice is to declare the action of $f$ on a test function $\varphi$ by \begin{align*} T: \mathcal{D}(\mathbb{T}^n) &\to \mathbb{C} \\ \varphi &\mapsto \sum_{k \in \mathbb{Z}^n} a_k\, \hat{\varphi}(-k). \end{align*} The index $-k$ in $\hat{\varphi}(-k)$ is dictated by the convention $\hat{e_{-k}}(j) = \delta_{j,-k}$ from Step 1: it is exactly what makes $\hat{T}(k) = a_k$ rather than $a_{-k}$. We must verify three things. **(i) Convergence of the defining sum.** Apply Cauchy-Schwarz to the pair of $\ell^2$-sequences $\big((1+|k|^2)^{s/2} a_k\big)_k$ and $\big((1+|k|^2)^{-s/2} \hat{\varphi}(-k)\big)_k$: \begin{align*} \sum_k |a_k|\, |\hat{\varphi}(-k)| \le \left( \sum_k (1+|k|^2)^s |a_k|^2 \right)^{1/2} \left( \sum_k (1+|k|^2)^{-s} |\hat{\varphi}(-k)|^2 \right)^{1/2}. \end{align*} The first factor evaluates to \begin{align*} \Big( \sum_k (1+|k|^2)^s\, (1+|k|^2)^{-s}\, |c_k|^2 \Big)^{1/2} = \|(c_k)\|_{\ell^2(\mathbb{Z}^n)} < \infty \end{align*} by hypothesis on $(c_k)$. For the second factor, integration by parts on $\mathbb{T}^n$ against $e^{-i k \cdot x}$ gives, for every multi-index $\alpha$ and $k \ne 0$, \begin{align*} (ik)^\alpha\, \hat{\varphi}(k) = \widehat{\partial^\alpha \varphi}(k), \qquad |\hat{\varphi}(k)| \le \|\partial^\alpha \varphi\|_{L^\infty(\mathbb{T}^n)}\, |k|^{-|\alpha|}. \end{align*} Hence $\hat{\varphi}(k)$ decays faster than any polynomial in $|k|$, and using $|-k| = |k|$ the inverse-weighted sum $\sum_k (1+|k|^2)^{-s} |\hat{\varphi}(-k)|^2$ is finite for every $s \in \mathbb{R}$. **(ii) Continuity in the test-function topology.** We must produce an explicit Schwartz-seminorm bound for $T(\varphi)$. Choose an integer $N$ with $2N > n - 2s$, so that \begin{align*} C_{N,s} := \sum_{k \in \mathbb{Z}^n} (1+|k|^2)^{-N-s} < \infty. \end{align*} The threshold $2N > n - 2s$ matches the lattice analogue of $\int_{\mathbb{R}^n} (1+|x|^2)^{-N-s}\, d\mathcal{L}^n(x) < \infty$, which holds precisely when $2(N+s) > n$. The multinomial expansion of $(1+|k|^2)^N$ together with the integration-by-parts bound from (i) gives \begin{align*} (1+|k|^2)^N\, |\hat{\varphi}(k)|^2 \le C_n \sum_{|\alpha| \le 2N} \|\partial^\alpha \varphi\|_{L^\infty(\mathbb{T}^n)}^2, \end{align*} where $C_n > 0$ depends only on $n$. Multiplying by $(1+|k|^2)^{-N-s}$ and summing, \begin{align*} \sum_k (1+|k|^2)^{-s} |\hat{\varphi}(k)|^2 \le C_n\, C_{N,s} \sum_{|\alpha| \le 2N} \|\partial^\alpha \varphi\|_{L^\infty(\mathbb{T}^n)}^2. \end{align*} Combining with the Cauchy-Schwarz bound from (i), \begin{align*} |T(\varphi)| \le C\, \|(c_k)\|_{\ell^2}\, \sum_{|\alpha| \le 2N} \|\partial^\alpha \varphi\|_{L^\infty(\mathbb{T}^n)}, \end{align*} with $C = C(n, s) > 0$. This is exactly the form of bound required by the definition of $\mathcal{D}'(\mathbb{T}^n)$, so $T \in \mathcal{D}'(\mathbb{T}^n)$. The exponent $N$ grows linearly with $|s|$: large negative $s$ amplifies $|a_k|$, forcing $\hat{\varphi}(k)$ to absorb that growth through more derivatives. This is the precise way in which the distributional order of $T$ scales with the Sobolev index. **(iii) The Fourier coefficients of $T$ equal $a_k$.** By the convention from Step 1, $\hat{T}(k) := T(e_{-k})$, and we computed there that the test function $e_{-k}(x) = e^{-i k \cdot x}$ has Fourier coefficients $\hat{e_{-k}}(j) = \delta_{j,-k}$. Substituting $\varphi = e_{-k}$ in the definition of $T$, \begin{align*} \hat{T}(k) = T(e_{-k}) = \sum_{j \in \mathbb{Z}^n} a_j\, \hat{e_{-k}}(-j) = \sum_{j} a_j\, \delta_{-j,-k} = \sum_j a_j\, \delta_{j,k} = a_k. \end{align*} This is why we used $\hat{\varphi}(-k)$ rather than $\hat{\varphi}(k)$ in the definition of $T$: the index reflection compensates for the reflection in $\hat{e_{-k}}(j) = \delta_{j,-k}$, yielding $\hat{T}(k) = a_k$ on the nose. Once (i), (ii), (iii) are verified, the norm identity \begin{align*} \sum_k (1+|k|^2)^s |\hat{T}(k)|^2 = \sum_k (1+|k|^2)^s\, (1+|k|^2)^{-s}\, |c_k|^2 = \sum_k |c_k|^2 < \infty \end{align*} shows $T \in H^s(\mathbb{T}^n)$, and $\Psi_s(T)_k = (1+|k|^2)^{s/2} a_k = c_k$, completing the surjectivity proof.[/guided]

[step:Conclude that $H^s(\mathbb{T}^n)$ is a Hilbert space] Combining the previous two steps, $\Psi_s: H^s(\mathbb{T}^n) \to \ell^2(\mathbb{Z}^n)$ is a bijective $\mathbb{C}$-linear isometry between inner-product spaces. The space $\ell^2(\mathbb{Z}^n)$ is a Hilbert space — it is complete, by the Riesz–Fischer theorem applied to the counting measure on $\mathbb{Z}^n$, with the standard inner product $((a_k), (b_k))_{\ell^2} = \sum_k a_k \overline{b_k}$. Completeness is preserved under isometric isomorphism: if $(f_n) \subset H^s(\mathbb{T}^n)$ is Cauchy, then $\Psi_s(f_n) \subset \ell^2(\mathbb{Z}^n)$ is Cauchy because $\Psi_s$ is an isometry, hence converges to some $(c_k) \in \ell^2(\mathbb{Z}^n)$ by completeness of $\ell^2$. By surjectivity, $(c_k) = \Psi_s(f)$ for some $f \in H^s(\mathbb{T}^n)$, and then $\|f_n - f\|_{H^s} = \|\Psi_s(f_n) - (c_k)\|_{\ell^2} \to 0$. So $H^s(\mathbb{T}^n)$ is complete. Together with the inner product structure, this makes $H^s(\mathbb{T}^n)$ a Hilbert space. [/step]

[step:Establish separability via trigonometric polynomials with Gaussian-rational coefficients]Define the countable set \begin{align*} \mathcal{Q} := \left\{ \sum_{|k|_\infty \le N} q_k\, e^{i k \cdot x} : N \in \mathbb{N},\ q_k \in \mathbb{Q} + i\mathbb{Q} \right\} \subset H^s(\mathbb{T}^n), \end{align*} where $|k|_\infty := \max_i |k_i|$. Each element of $\mathcal{Q}$ is a trigonometric polynomial, hence smooth, hence in $H^s(\mathbb{T}^n)$ for every $s$. The set $\mathcal{Q}$ is countable: its elements are parameterised by a finite truncation level $N \in \mathbb{N}$ and a finite tuple of Gaussian-rational coefficients $q_k \in \mathbb{Q} + i\mathbb{Q}$; the union over $N$ of finite-dimensional Cartesian powers of the countable set $\mathbb{Q} + i\mathbb{Q}$ is a countable union of countable sets, hence countable. We show $\mathcal{Q}$ is dense in $H^s(\mathbb{T}^n)$. Fix $f \in H^s(\mathbb{T}^n)$ and $\varepsilon > 0$. Since $\sum_{k \in \mathbb{Z}^n} (1 + |k|^2)^s |\hat{f}(k)|^2 < \infty$, choose $N \in \mathbb{N}$ large enough that the tail satisfies \begin{align*} \sum_{|k|_\infty > N} (1 + |k|^2)^s |\hat{f}(k)|^2 < \frac{\varepsilon^2}{4}. \end{align*} The set of multi-indices $\{k \in \mathbb{Z}^n : |k|_\infty \le N\}$ is finite, with cardinality $(2N+1)^n$. The set $\mathbb{Q} + i\mathbb{Q}$ is dense in $\mathbb{C}$, so we may choose $q_k \in \mathbb{Q} + i\mathbb{Q}$ for each $|k|_\infty \le N$ satisfying \begin{align*} |q_k - \hat{f}(k)|^2 < \frac{\varepsilon^2}{4 (2N+1)^n (1 + |k|^2)^s}. \end{align*} Set $g := \sum_{|k|_\infty \le N} q_k\, e^{i k \cdot x} \in \mathcal{Q}$. The Fourier coefficients of $g$ are $\hat{g}(k) = q_k$ for $|k|_\infty \le N$ and $\hat{g}(k) = 0$ for $|k|_\infty > N$. Therefore \begin{align*} \|f - g\|_{H^s}^2 &= \sum_{|k|_\infty \le N} (1 + |k|^2)^s |\hat{f}(k) - q_k|^2 + \sum_{|k|_\infty > N} (1 + |k|^2)^s |\hat{f}(k)|^2 \\ &< \sum_{|k|_\infty \le N} (1 + |k|^2)^s \cdot \frac{\varepsilon^2}{4 (2N+1)^n (1 + |k|^2)^s} + \frac{\varepsilon^2}{4} \\ &= \frac{\varepsilon^2}{4 (2N+1)^n} \cdot (2N+1)^n + \frac{\varepsilon^2}{4} = \frac{\varepsilon^2}{2} < \varepsilon^2. \end{align*} Thus $\|f - g\|_{H^s} < \varepsilon$. Since $\varepsilon > 0$ and $f \in H^s(\mathbb{T}^n)$ were arbitrary, $\mathcal{Q}$ is dense, so $H^s(\mathbb{T}^n)$ is separable.[/step]

[guided]We construct a countable dense subset $\mathcal{Q} \subset H^s(\mathbb{T}^n)$. The natural candidate is the set of trigonometric polynomials \begin{align*} \mathcal{Q} := \left\{ \sum_{|k|_\infty \le N} q_k\, e^{i k \cdot x} : N \in \mathbb{N},\ q_k \in \mathbb{Q} + i\mathbb{Q} \right\}, \end{align*} where $|k|_\infty := \max_i |k_i|$. Why this choice? Trigonometric polynomials are smooth (hence in $H^s$ for every $s$), and they have only finitely many non-zero Fourier coefficients — so they are exactly the finite-dimensional approximations of an arbitrary $H^s$ element in the Fourier basis. Restricting the coefficients to $\mathbb{Q} + i\mathbb{Q}$ ensures countability: $\mathcal{Q}$ is a countable union (over $N \in \mathbb{N}$) of finite Cartesian powers of $\mathbb{Q} + i\mathbb{Q}$, hence countable. To prove density, fix $f \in H^s(\mathbb{T}^n)$ and $\varepsilon > 0$. We split the approximation error into two contributions. **(a) Tail error.** Even matching the low-frequency Fourier coefficients of $f$ exactly leaves the high-frequency tail $|k|_\infty > N$ entirely undescribed. Since $f \in H^s(\mathbb{T}^n)$, the weighted sum $\sum_k (1+|k|^2)^s |\hat{f}(k)|^2$ converges, so the tail beyond level $N$ shrinks to $0$ as $N \to \infty$. We choose $N$ large enough that \begin{align*} \sum_{|k|_\infty > N} (1 + |k|^2)^s |\hat{f}(k)|^2 < \frac{\varepsilon^2}{4}. \end{align*} **(b) Coefficient quantisation error.** Even within $|k|_\infty \le N$, the true coefficient $\hat{f}(k) \in \mathbb{C}$ is a continuous parameter, but our countable set $\mathcal{Q}$ only allows Gaussian-rational coefficients $q_k \in \mathbb{Q} + i\mathbb{Q}$. The number of relevant indices is $(2N+1)^n$, finite. Using density of $\mathbb{Q} + i\mathbb{Q}$ in $\mathbb{C}$, we approximate each individual coefficient to within precision \begin{align*} |q_k - \hat{f}(k)|^2 < \frac{\varepsilon^2}{4 (2N+1)^n (1 + |k|^2)^s}. \end{align*} Why this precise specification? When $s > 0$ the coefficient at high $k$ is amplified by a large weight in the $H^s$ norm, requiring tighter precision; when $s < 0$ the weight is small and the precision can be looser. The factor $(1+|k|^2)^s$ in the precision exactly cancels the same factor in the norm computation, leaving a uniform contribution $\frac{\varepsilon^2}{4(2N+1)^n}$ per index. Summing over the $(2N+1)^n$ indices contributes $\varepsilon^2/4$ to the squared error. **Combining (a) and (b).** Let $g := \sum_{|k|_\infty \le N} q_k\, e^{i k \cdot x} \in \mathcal{Q}$, which has $\hat{g}(k) = q_k$ for $|k|_\infty \le N$ and $\hat{g}(k) = 0$ otherwise. Then \begin{align*} \|f - g\|_{H^s}^2 = \underbrace{\sum_{|k|_\infty \le N} (1 + |k|^2)^s |\hat{f}(k) - q_k|^2}_{\text{quantisation, } < \varepsilon^2/4} + \underbrace{\sum_{|k|_\infty > N} (1 + |k|^2)^s |\hat{f}(k)|^2}_{\text{tail, } < \varepsilon^2/4} < \frac{\varepsilon^2}{2} < \varepsilon^2. \end{align*} Thus $\|f - g\|_{H^s} < \varepsilon$. Since $f$ and $\varepsilon$ were arbitrary, $\mathcal{Q}$ is a countable dense subset of $H^s(\mathbb{T}^n)$, so $H^s(\mathbb{T}^n)$ is separable.[/guided]

[step:Combine the four results to complete the proof] By Steps 2 and 3, $\Psi_s: H^s(\mathbb{T}^n) \to \ell^2(\mathbb{Z}^n)$ is a bijective linear isometry. By Step 4, $H^s(\mathbb{T}^n)$ inherits completeness from $\ell^2(\mathbb{Z}^n)$ via $\Psi_s$, and combined with the inner product structure declared in Step 1 this makes $H^s(\mathbb{T}^n)$ a Hilbert space. By Step 5, the countable set $\mathcal{Q}$ is dense, so $H^s(\mathbb{T}^n)$ is separable. The map $f \mapsto (\hat{f}(k))_{k \in \mathbb{Z}^n}$ is then an isometric isomorphism from $H^s(\mathbb{T}^n)$ to $\ell^2_s(\mathbb{Z}^n)$ — defined as the weighted sequence space $\{(a_k) : \sum (1+|k|^2)^s |a_k|^2 < \infty\}$ with the inner product $\sum (1+|k|^2)^s a_k \bar{b_k}$ — because $\Psi_s$ factors as the composition of $f \mapsto (\hat{f}(k))_k$ followed by multiplication by the weights. This completes the proof. [/step]