[proofplan] We adopt the convention that $(H^s(\mathbb{T}^n))^*$ denotes the space of bounded $\mathbb{C}$-linear functionals $\Lambda : H^s(\mathbb{T}^n) \to \mathbb{C}$. For $g \in H^{-s}(\mathbb{T}^n)$ we define the functional $\Theta_g(f) := \langle f, g \rangle = \sum_k \hat{f}(k)\, \overline{\hat{g}(k)}$, which is linear in $f$ (since the Fourier coefficients $\hat{f}(k)$ are linear in $f$) and conjugate-linear in $g$. Throughout, the Fourier coefficients of an element of $H^t(\mathbb{T}^n)$ for any $t \in \mathbb{R}$ are interpreted via the isometric Fourier model (Theorem 3132): $H^t(\mathbb{T}^n)$ is identified with the weighted $\ell^2$-space $\{(c_k)_{k \in \mathbb{Z}^n} : \sum_k (1+|k|^2)^t |c_k|^2 < \infty\}$ via $f \leftrightarrow (\hat{f}(k))_k$. The proof has four steps: (1) verify the pairing is well-defined and bounded by $\|f\|_{H^s} \|g\|_{H^{-s}}$, so that $\Theta_g \in (H^s)^*$ with $\|\Theta_g\|_{(H^s)^*} \le \|g\|_{H^{-s}}$; (2) exhibit a maximiser to show this bound is sharp, so $g \mapsto \Theta_g$ is an isometry $H^{-s} \to (H^s)^*$; (3) for arbitrary $\Lambda \in (H^s)^*$, apply the Riesz Representation Theorem on the Hilbert space $H^s$ to produce $g \in H^{-s}$ with $\Theta_g = \Lambda$; (4) collect the conclusions, observing that $g \mapsto \Theta_g$ is a conjugate-linear isometric bijection (the theorem statement asks for an isometric identification, not for the identifying map to be linear). [/proofplan]

[step:Verify the pairing is well-defined and bounded]Fix $s \in \mathbb{R}$, $f \in H^s(\mathbb{T}^n)$, and $g \in H^{-s}(\mathbb{T}^n)$. Factor each summand of the pairing as \begin{align*} \hat{f}(k)\, \overline{\hat{g}(k)} = \big[ (1 + |k|^2)^{s/2}\, \hat{f}(k) \big] \cdot \big[ (1 + |k|^2)^{-s/2}\, \overline{\hat{g}(k)} \big]. \end{align*} The hypothesis $f \in H^s(\mathbb{T}^n)$ means, by the Fourier model of [Theorem 3132](/theorems/3132), that the sequence $a_k := (1+|k|^2)^{s/2}\, \hat{f}(k)$ lies in $\ell^2(\mathbb{Z}^n)$ with $\|a\|_{\ell^2}^2 = \|f\|_{H^s(\mathbb{T}^n)}^2$. Similarly, $g \in H^{-s}(\mathbb{T}^n)$ means the sequence $b_k := (1+|k|^2)^{-s/2}\, \overline{\hat{g}(k)}$ lies in $\ell^2(\mathbb{Z}^n)$ with $\|b\|_{\ell^2}^2 = \|g\|_{H^{-s}(\mathbb{T}^n)}^2$. Apply the Cauchy-Schwarz inequality on $\ell^2(\mathbb{Z}^n)$ to $(a_k)$ and $(b_k)$ — both are in $\ell^2$, so the hypothesis is met: \begin{align*} \sum_{k \in \mathbb{Z}^n} |\hat{f}(k)\, \overline{\hat{g}(k)}| = \sum_{k} |a_k|\, |b_k| \le \|a\|_{\ell^2} \|b\|_{\ell^2} = \|f\|_{H^s(\mathbb{T}^n)}\, \|g\|_{H^{-s}(\mathbb{T}^n)} < \infty. \end{align*} Absolute convergence implies the series defining $\langle f, g \rangle$ converges to a well-defined complex number. The triangle inequality applied to the absolutely convergent series gives \begin{align*} |\langle f, g \rangle| \le \sum_{k} |\hat{f}(k)|\, |\hat{g}(k)| \le \|f\|_{H^s(\mathbb{T}^n)}\, \|g\|_{H^{-s}(\mathbb{T}^n)}. \end{align*} For each fixed $g \in H^{-s}(\mathbb{T}^n)$, define the functional \begin{align*} \Theta_g : H^s(\mathbb{T}^n) &\to \mathbb{C}, \\ f &\mapsto \langle f, g \rangle = \sum_{k \in \mathbb{Z}^n} \hat{f}(k)\, \overline{\hat{g}(k)}. \end{align*} For $f_1, f_2 \in H^s(\mathbb{T}^n)$ and $\alpha, \beta \in \mathbb{C}$, the linearity of Fourier coefficients in the function gives $\widehat{\alpha f_1 + \beta f_2}(k) = \alpha \hat{f}_1(k) + \beta \hat{f}_2(k)$, hence \begin{align*} \Theta_g(\alpha f_1 + \beta f_2) = \sum_k (\alpha \hat{f}_1(k) + \beta \hat{f}_2(k))\, \overline{\hat{g}(k)} = \alpha\, \Theta_g(f_1) + \beta\, \Theta_g(f_2), \end{align*} so $\Theta_g$ is $\mathbb{C}$-linear. Combined with the bound above, $\Theta_g \in (H^s(\mathbb{T}^n))^*$ with operator norm \begin{align*} \|\Theta_g\|_{(H^s)^*} := \sup_{f \in H^s,\ f \ne 0} \frac{|\Theta_g(f)|}{\|f\|_{H^s(\mathbb{T}^n)}} \le \|g\|_{H^{-s}(\mathbb{T}^n)}. \end{align*} The assignment $g \mapsto \Theta_g$ is conjugate-linear: $\Theta_{\alpha g_1 + \beta g_2}(f) = \sum_k \hat{f}(k)\, \overline{\alpha \hat{g}_1(k) + \beta \hat{g}_2(k)} = \bar\alpha\, \Theta_{g_1}(f) + \bar\beta\, \Theta_{g_2}(f)$, so $\Theta_{\alpha g_1 + \beta g_2} = \bar\alpha\, \Theta_{g_1} + \bar\beta\, \Theta_{g_2}$.[/step]

[guided]We want to define a map from $H^{-s}(\mathbb{T}^n)$ to $(H^s(\mathbb{T}^n))^*$. The candidate is the rule that sends $g$ to the functional $\Theta_g(f) := \langle f, g \rangle$. Two things must be checked: that the pairing series converges (so $\Theta_g(f)$ is a number, not a divergent expression), and that $f \mapsto \Theta_g(f)$ is a bounded linear functional on $H^s$. **Convergence and bound via Cauchy-Schwarz.** The pairing is unweighted, but $f$ and $g$ live in weighted $\ell^2$-spaces with weights $(1+|k|^2)^s$ and $(1+|k|^2)^{-s}$ respectively. Since $s + (-s) = 0$, the weights are reciprocals — they are "$\ell^2$-Hölder conjugate." We exploit this by splitting each summand into two pieces of equal $\ell^2$-mass under the appropriate weights: \begin{align*} \hat{f}(k)\, \overline{\hat{g}(k)} = \big[ (1 + |k|^2)^{s/2}\, \hat{f}(k) \big] \cdot \big[ (1 + |k|^2)^{-s/2}\, \overline{\hat{g}(k)} \big]. \end{align*} Define $a_k := (1+|k|^2)^{s/2}\, \hat{f}(k)$ and $b_k := (1+|k|^2)^{-s/2}\, \overline{\hat{g}(k)}$. By the Fourier model of $H^t(\mathbb{T}^n)$ (the isometry of [Theorem 3132](/theorems/3132)), $a \in \ell^2(\mathbb{Z}^n)$ with $\|a\|_{\ell^2} = \|f\|_{H^s}$, and $b \in \ell^2(\mathbb{Z}^n)$ with $\|b\|_{\ell^2} = \|g\|_{H^{-s}}$ (using $|\overline{\hat{g}(k)}| = |\hat{g}(k)|$). The Cauchy-Schwarz inequality on $\ell^2(\mathbb{Z}^n)$ requires both sequences to lie in $\ell^2$ — verified — and yields \begin{align*} \sum_{k} |a_k|\, |b_k| \le \|a\|_{\ell^2}\, \|b\|_{\ell^2} = \|f\|_{H^s(\mathbb{T}^n)}\, \|g\|_{H^{-s}(\mathbb{T}^n)} < \infty, \end{align*} which is exactly the absolute summability of $\hat{f}(k)\, \overline{\hat{g}(k)}$. The pairing therefore defines a complex number, and the triangle inequality on the absolutely convergent series gives \begin{align*} |\langle f, g \rangle| \le \|f\|_{H^s(\mathbb{T}^n)}\, \|g\|_{H^{-s}(\mathbb{T}^n)}. \end{align*} **Linearity of $\Theta_g$ in $f$.** Fix $g \in H^{-s}$ and define $\Theta_g : H^s \to \mathbb{C}$ by $\Theta_g(f) := \langle f, g \rangle$. For $\alpha, \beta \in \mathbb{C}$ and $f_1, f_2 \in H^s$, the Fourier coefficient operation $f \mapsto \hat{f}(k)$ is itself linear (this is built into the Fourier model), so $\widehat{\alpha f_1 + \beta f_2}(k) = \alpha \hat{f}_1(k) + \beta \hat{f}_2(k)$. Substituting: \begin{align*} \Theta_g(\alpha f_1 + \beta f_2) = \sum_k (\alpha \hat{f}_1(k) + \beta \hat{f}_2(k))\, \overline{\hat{g}(k)} = \alpha\, \Theta_g(f_1) + \beta\, \Theta_g(f_2). \end{align*} Hence $\Theta_g$ is $\mathbb{C}$-linear. Combined with the bound, $\Theta_g \in (H^s)^*$ and $\|\Theta_g\|_{(H^s)^*} \le \|g\|_{H^{-s}}$. **Conjugate-linearity in $g$.** A similar substitution with $\overline{\alpha \hat{g}_1(k) + \beta \hat{g}_2(k)} = \bar\alpha\, \overline{\hat{g}_1(k)} + \bar\beta\, \overline{\hat{g}_2(k)}$ shows $\Theta_{\alpha g_1 + \beta g_2} = \bar\alpha\, \Theta_{g_1} + \bar\beta\, \Theta_{g_2}$. So $g \mapsto \Theta_g$ is conjugate-linear, not linear. This is fine: the theorem asks for an isometric identification of $H^{-s}$ with $(H^s)^*$, not for the identifying map to be linear. Conjugate-linearity in $g$ does not prevent each individual $\Theta_g$ from being a well-defined element of $(H^s)^*$.[/guided]

[step:Show that $g \mapsto \Theta_g$ is an isometry by exhibiting a maximiser]Fix $g \in H^{-s}(\mathbb{T}^n)$ with $g \ne 0$. We must produce $f_* \in H^s(\mathbb{T}^n) \setminus \{0\}$ achieving equality in \begin{align*} \|\Theta_g\|_{(H^s)^*} \ge \frac{|\Theta_g(f_*)|}{\|f_*\|_{H^s(\mathbb{T}^n)}}. \end{align*} The Cauchy-Schwarz inequality $|\sum_k a_k\, \overline{b_k}| \le \|a\|_{\ell^2} \|b\|_{\ell^2}$ for $a_k = (1+|k|^2)^{s/2} \hat{f}(k)$ and $b_k = (1+|k|^2)^{-s/2} \overline{\hat{g}(k)}$ attains equality iff there exists $c \in \mathbb{C}$ with $a_k = c\, b_k$ for every $k$. Pick $c = 1$; then $a_k = b_k$ amounts to $(1+|k|^2)^{s/2} \hat{f}(k) = (1+|k|^2)^{-s/2} \overline{\hat{g}(k)}$, i.e., \begin{align*} \hat{f}_*(k) := (1+|k|^2)^{-s}\, \overline{\hat{g}(k)}, \qquad k \in \mathbb{Z}^n. \end{align*} We verify $f_* \in H^s(\mathbb{T}^n)$. Compute \begin{align*} \sum_{k \in \mathbb{Z}^n} (1 + |k|^2)^s |\hat{f}_*(k)|^2 = \sum_{k} (1 + |k|^2)^s\, (1+|k|^2)^{-2s} |\hat{g}(k)|^2 = \sum_{k} (1 + |k|^2)^{-s} |\hat{g}(k)|^2 = \|g\|_{H^{-s}(\mathbb{T}^n)}^2. \end{align*} The Fourier model of $H^s(\mathbb{T}^n)$ ([Theorem 3132](/theorems/3132)) — which states that the map $f \leftrightarrow (\hat{f}(k))_k$ is an isometric bijection between $H^s(\mathbb{T}^n)$ and the weighted $\ell^2$-space with weight $(1+|k|^2)^s$ — has the hypothesis that the prescribed sequence $(\hat{f}_*(k))_k$ satisfies $\sum_k (1+|k|^2)^s |\hat{f}_*(k)|^2 < \infty$. The computation above verifies this hypothesis with the value $\|g\|_{H^{-s}}^2 < \infty$. Hence there exists a unique $f_* \in H^s(\mathbb{T}^n)$ with these Fourier coefficients, and \begin{align*} \|f_*\|_{H^s(\mathbb{T}^n)}^2 = \|g\|_{H^{-s}(\mathbb{T}^n)}^2. \end{align*} Compute the pairing using $\overline{\hat{f}_*(k)} = (1+|k|^2)^{-s} \hat{g}(k)$ (since $(1+|k|^2)^{-s}$ is real): \begin{align*} \Theta_g(f_*) = \sum_{k} \hat{f}_*(k)\, \overline{\hat{g}(k)} = \sum_{k} (1+|k|^2)^{-s}\, \overline{\hat{g}(k)}\, \overline{\hat{g}(k)} = \sum_{k} (1+|k|^2)^{-s}\, \overline{\hat{g}(k)}^{\,2}. \end{align*} Since the sum involves $\overline{\hat g(k)}^{\,2}$ rather than $|\hat g(k)|^2$, take instead $c = 1$ for the absolute value of Cauchy-Schwarz: take $|\hat f_*(k)| = (1+|k|^2)^{-s} |\hat g(k)|$ with phase chosen so that $\hat f_*(k) \overline{\hat g(k)} \ge 0$ for every $k$. The clean choice is \begin{align*} \hat{f}_*(k) := (1+|k|^2)^{-s}\, \hat{g}(k), \end{align*} which satisfies $|a_k| = |b_k|$ pointwise and gives \begin{align*} \Theta_g(f_*) = \sum_{k} \hat{f}_*(k)\, \overline{\hat{g}(k)} = \sum_{k} (1+|k|^2)^{-s}\, \hat{g}(k)\, \overline{\hat{g}(k)} = \sum_{k} (1+|k|^2)^{-s}\, |\hat{g}(k)|^2 = \|g\|_{H^{-s}(\mathbb{T}^n)}^2. \end{align*} The same computation as before shows $\sum_k (1+|k|^2)^s |\hat{f}_*(k)|^2 = \|g\|_{H^{-s}}^2$, so $f_* \in H^s$ exists by Theorem 3132 with $\|f_*\|_{H^s} = \|g\|_{H^{-s}}$. Therefore \begin{align*} \frac{|\Theta_g(f_*)|}{\|f_*\|_{H^s(\mathbb{T}^n)}} = \frac{\|g\|_{H^{-s}(\mathbb{T}^n)}^2}{\|g\|_{H^{-s}(\mathbb{T}^n)}} = \|g\|_{H^{-s}(\mathbb{T}^n)}. \end{align*} Combined with the upper bound from Step 1, \begin{align*} \|\Theta_g\|_{(H^s(\mathbb{T}^n))^*} = \|g\|_{H^{-s}(\mathbb{T}^n)}. \end{align*} This identity also holds when $g = 0$ (both sides are zero). Hence $g \mapsto \Theta_g$ is an isometry from $H^{-s}(\mathbb{T}^n)$ into $(H^s(\mathbb{T}^n))^*$.[/step]

[guided]The Cauchy-Schwarz bound from Step 1 has a known sharpness condition, and exploiting it produces the required maximiser explicitly. Consider the inequality $|\sum_k a_k\, \overline{b_k}| \le \|a\|_{\ell^2}\, \|b\|_{\ell^2}$ applied to the sequences $a_k := (1+|k|^2)^{s/2}\, \hat{f}(k)$ and $b_k := (1+|k|^2)^{-s/2}\, \overline{\hat{g}(k)}$. Equality in Cauchy-Schwarz on $\ell^2$ holds if and only if there is a single complex constant $c \in \mathbb{C}$ (not depending on $k$) such that $a_k = c\, b_k$ for all $k$. **Solving the equality condition.** We choose $c = 1$ for normalisation. Then $a_k = b_k$ becomes \begin{align*} (1+|k|^2)^{s/2}\, \hat{f}(k) = (1+|k|^2)^{-s/2}\, \overline{\hat{g}(k)}, \end{align*} which solves to $\hat{f}(k) = (1+|k|^2)^{-s}\, \overline{\hat{g}(k)}$. However, our pairing is $\sum_k \hat{f}(k)\, \overline{\hat{g}(k)}$, and we want the summand to be a non-negative real number so that the absolute value of the sum equals the sum of the moduli. Substituting this candidate gives $\hat f(k) \overline{\hat g(k)} = (1+|k|^2)^{-s} \overline{\hat g(k)}^{\,2}$, which is generally complex. To force the summand to be $(1+|k|^2)^{-s}\, |\hat g(k)|^2 \ge 0$, the correct normalisation is to instead take \begin{align*} \hat{f}_*(k) := (1+|k|^2)^{-s}\, \hat{g}(k). \end{align*} Then $|\hat f_*(k)| = (1+|k|^2)^{-s}\, |\hat g(k)|$, which still saturates Cauchy-Schwarz pointwise (since $|a_k| = |b_k|$ everywhere), and the phase is aligned so that $\hat f_*(k)\, \overline{\hat g(k)} = (1+|k|^2)^{-s}\, |\hat g(k)|^2 \in \mathbb{R}_{\ge 0}$. **Verification that $f_* \in H^s$.** We invoke the Fourier model of $H^s(\mathbb{T}^n)$ from [Theorem 3132](/theorems/3132): the assignment $f \leftrightarrow (\hat f(k))_k$ is an isometric isomorphism between $H^s(\mathbb{T}^n)$ and the weighted $\ell^2$-space $\{(c_k) : \sum_k (1+|k|^2)^s |c_k|^2 < \infty\}$. Theorem 3132 thus tells us that to construct $f_* \in H^s$ from a prescribed Fourier sequence, we must verify the sequence lies in this weighted $\ell^2$-space. We compute \begin{align*} \sum_k (1+|k|^2)^s\, |\hat f_*(k)|^2 = \sum_k (1+|k|^2)^s\, (1+|k|^2)^{-2s}\, |\hat g(k)|^2 = \sum_k (1+|k|^2)^{-s}\, |\hat g(k)|^2 = \|g\|_{H^{-s}}^2, \end{align*} which is finite by hypothesis. The hypothesis of Theorem 3132 is met, and $f_* \in H^s(\mathbb{T}^n)$ exists with $\|f_*\|_{H^s} = \|g\|_{H^{-s}}$. **Computing the pairing at $f_*$.** Substituting: \begin{align*} \Theta_g(f_*) = \sum_k \hat f_*(k)\, \overline{\hat g(k)} = \sum_k (1+|k|^2)^{-s}\, \hat g(k)\, \overline{\hat g(k)} = \sum_k (1+|k|^2)^{-s}\, |\hat g(k)|^2 = \|g\|_{H^{-s}}^2. \end{align*} The ratio is $|\Theta_g(f_*)| / \|f_*\|_{H^s} = \|g\|_{H^{-s}}^2 / \|g\|_{H^{-s}} = \|g\|_{H^{-s}}$. Combined with $\|\Theta_g\|_{(H^s)^*} \le \|g\|_{H^{-s}}$ from Step 1, we conclude $\|\Theta_g\|_{(H^s)^*} = \|g\|_{H^{-s}}$. The case $g = 0$ is trivial (both norms vanish). The map $g \mapsto \Theta_g$ is thus an isometry — norm-preserving, hence injective. **Why this argument works.** The maximiser $f_*$ is essentially unique up to a positive scalar: the equality condition in Cauchy-Schwarz on $\ell^2$ pins down the proportionality $a_k = c\, b_k$, and the requirement that $\Theta_g(f_*)$ be real and positive (so that the absolute value of the sum equals the sum of moduli, saturating the triangle inequality as well as Cauchy-Schwarz) fixes the phase. The construction $\hat f_*(k) = (1+|k|^2)^{-s}\, \hat g(k)$ realises both: it makes the weighted sequences $(a_k), (b_k)$ have the same moduli pointwise, and it aligns the phase of each summand of the pairing so that $\hat f_*(k)\, \overline{\hat g(k)} = (1+|k|^2)^{-s}\, |\hat g(k)|^2$ is real and non-negative for every $k$. The supremum defining the operator norm is therefore an attained maximum, and the upper bound from Step 1 cannot be loose.[/guided]

[step:Show $g \mapsto \Theta_g$ is surjective via Riesz representation]Let $\Lambda \in (H^s(\mathbb{T}^n))^*$ be an arbitrary bounded $\mathbb{C}$-linear functional. We produce $g \in H^{-s}(\mathbb{T}^n)$ with $\Theta_g = \Lambda$. By [Theorem 3132](/theorems/3132), $H^s(\mathbb{T}^n)$ is a Hilbert space with inner product \begin{align*} (f, h)_{H^s(\mathbb{T}^n)} := \sum_{k \in \mathbb{Z}^n} (1 + |k|^2)^s\, \hat{f}(k)\, \overline{\hat{h}(k)}, \end{align*} which is linear in the first argument $f$ and conjugate-linear in the second argument $h$. The Riesz Representation Theorem applies to a complex Hilbert space $H$ and a bounded $\mathbb{C}$-linear functional $\Lambda : H \to \mathbb{C}$; its hypotheses are (i) $H$ is a Hilbert space — verified by Theorem 3132 with $H := H^s(\mathbb{T}^n)$ — and (ii) $\Lambda \in H^*$ — given. The conclusion is that there exists a unique $h \in H^s(\mathbb{T}^n)$ with \begin{align*} \Lambda(f) = (f, h)_{H^s(\mathbb{T}^n)} = \sum_{k \in \mathbb{Z}^n} (1 + |k|^2)^s\, \hat{f}(k)\, \overline{\hat{h}(k)} \qquad \text{for all } f \in H^s(\mathbb{T}^n), \end{align*} and moreover $\|h\|_{H^s(\mathbb{T}^n)} = \|\Lambda\|_{(H^s)^*}$. Define a candidate sequence \begin{align*} \hat{g}(k) := (1 + |k|^2)^s\, \hat{h}(k), \qquad k \in \mathbb{Z}^n. \end{align*} We verify the prescribed sequence lies in the Fourier model of $H^{-s}(\mathbb{T}^n)$, i.e., $\sum_k (1+|k|^2)^{-s} |\hat g(k)|^2 < \infty$: \begin{align*} \sum_{k \in \mathbb{Z}^n} (1 + |k|^2)^{-s} |\hat{g}(k)|^2 = \sum_{k} (1 + |k|^2)^{-s} (1 + |k|^2)^{2s} |\hat{h}(k)|^2 = \sum_{k} (1 + |k|^2)^{s} |\hat{h}(k)|^2 = \|h\|_{H^s(\mathbb{T}^n)}^2 < \infty. \end{align*} By Theorem 3132 applied with index $-s$, there exists a unique $g \in H^{-s}(\mathbb{T}^n)$ with these Fourier coefficients, and \begin{align*} \|g\|_{H^{-s}(\mathbb{T}^n)}^2 = \|h\|_{H^s(\mathbb{T}^n)}^2 = \|\Lambda\|_{(H^s)^*}^2. \end{align*} It remains to verify $\Theta_g = \Lambda$. From $\hat g(k) = (1+|k|^2)^s\, \hat h(k)$, taking complex conjugates and using that $(1+|k|^2)^s$ is real: \begin{align*} \overline{\hat g(k)} = (1+|k|^2)^s\, \overline{\hat h(k)}. \end{align*} Substituting into the pairing, for every $f \in H^s(\mathbb{T}^n)$: \begin{align*} \Theta_g(f) = \sum_{k \in \mathbb{Z}^n} \hat{f}(k)\, \overline{\hat{g}(k)} = \sum_{k} \hat{f}(k)\, (1 + |k|^2)^s\, \overline{\hat{h}(k)} = (f, h)_{H^s(\mathbb{T}^n)} = \Lambda(f). \end{align*} Hence $\Theta_g = \Lambda$. Since $\Lambda \in (H^s)^*$ was arbitrary, the map $g \mapsto \Theta_g$ is surjective.[/step]

[guided]The strategy is to convert the abstract bounded linear functional $\Lambda$ into a concrete $H^s$-element via Riesz, then translate that element into an $H^{-s}$-element by absorbing the Hilbert-space weight into the coefficients. **Step a: Riesz representation on $H^s(\mathbb{T}^n)$.** The Riesz Representation Theorem in the complex setting requires (i) a Hilbert space $H$ and (ii) a bounded $\mathbb{C}$-linear functional $\Lambda : H \to \mathbb{C}$; it concludes that there is a unique $h \in H$ representing $\Lambda$ via the inner product, with $\|h\|_H = \|\Lambda\|_{H^*}$. We apply it with $H = H^s(\mathbb{T}^n)$. Hypothesis (i) is the content of [Theorem 3132](/theorems/3132): $H^s(\mathbb{T}^n)$ is a Hilbert space with explicit inner product \begin{align*} (f, h)_{H^s(\mathbb{T}^n)} := \sum_k (1+|k|^2)^s\, \hat{f}(k)\, \overline{\hat{h}(k)}, \end{align*} linear in $f$ and conjugate-linear in $h$ (the standard convention on Androma; see the notation guide §2). Hypothesis (ii) is the assumption on $\Lambda$. The conclusion gives a unique representer $h \in H^s$ with $\Lambda(f) = (f, h)_{H^s}$ for all $f \in H^s$ and $\|h\|_{H^s} = \|\Lambda\|_{(H^s)^*}$. **Step b: Why we cannot just take $g = h$.** The Riesz formula has a weight $(1+|k|^2)^s$ inside the sum, while the natural pairing $\Theta_g(f) = \sum_k \hat f(k)\, \overline{\hat g(k)}$ has no weight. The two formulas match if and only if $\overline{\hat g(k)} = (1+|k|^2)^s\, \overline{\hat h(k)}$, equivalently $\hat g(k) = (1+|k|^2)^s\, \hat h(k)$ (using that the weight is real). So $g$ is obtained from $h$ by multiplying its Fourier coefficients pointwise by the weight $(1+|k|^2)^s$. **Step c: Verify $g \in H^{-s}$ via the Fourier model.** [Theorem 3132](/theorems/3132) applies with index $-s$: it states that the Fourier model identifies $H^{-s}(\mathbb{T}^n)$ with the weighted $\ell^2$-space $\{(c_k) : \sum_k (1+|k|^2)^{-s}|c_k|^2 < \infty\}$. To construct $g \in H^{-s}$ from prescribed coefficients $\hat g(k) := (1+|k|^2)^s\, \hat h(k)$, we must verify the prescribed sequence is in this weighted $\ell^2$-space. Compute: \begin{align*} \sum_k (1+|k|^2)^{-s}\, |\hat g(k)|^2 = \sum_k (1+|k|^2)^{-s}\, (1+|k|^2)^{2s}\, |\hat h(k)|^2 = \sum_k (1+|k|^2)^s\, |\hat h(k)|^2 = \|h\|_{H^s}^2, \end{align*} which is finite (and equal to $\|\Lambda\|_{(H^s)^*}^2$ by Riesz). The hypothesis of Theorem 3132 is verified, and $g \in H^{-s}$ exists with \begin{align*} \|g\|_{H^{-s}} = \|h\|_{H^s} = \|\Lambda\|_{(H^s)^*}. \end{align*} The bookkeeping confirms the norm identity that Step 4 will collect. **Step d: $\Theta_g = \Lambda$.** Substituting the definition of $\hat g$ into the pairing, and using $\overline{(1+|k|^2)^s\, \hat h(k)} = (1+|k|^2)^s\, \overline{\hat h(k)}$: \begin{align*} \Theta_g(f) = \sum_k \hat f(k)\, \overline{\hat g(k)} = \sum_k \hat f(k)\, (1+|k|^2)^s\, \overline{\hat h(k)} = (f, h)_{H^s(\mathbb{T}^n)} = \Lambda(f), \end{align*} identically in $f \in H^s$. Hence $\Theta_g = \Lambda$ as elements of $(H^s)^*$. The redistribution of weights between $g$ and $h$ is the heart of the duality: the $H^s$-norm is the $\ell^2$-norm with weight $(1+|k|^2)^s$, and the $H^{-s}$-norm is the $\ell^2$-norm with the reciprocal weight $(1+|k|^2)^{-s}$. These are precisely the conjugate weights for the unweighted $\ell^2$-pairing, which is why the Riesz representer in $H^s$ corresponds bijectively to a representer in $H^{-s}$ for the unweighted pairing.[/guided]

[step:Conclude that $g \mapsto \Theta_g$ is a conjugate-linear isometric bijection] Step 1 establishes that for each $g \in H^{-s}(\mathbb{T}^n)$, the functional $\Theta_g(f) := \langle f, g \rangle$ defines an element of $(H^s(\mathbb{T}^n))^*$ with $\|\Theta_g\|_{(H^s)^*} \le \|g\|_{H^{-s}(\mathbb{T}^n)}$, and that the map $g \mapsto \Theta_g$ is conjugate-linear. Step 2 strengthens the bound to an equality $\|\Theta_g\|_{(H^s)^*} = \|g\|_{H^{-s}(\mathbb{T}^n)}$, so $g \mapsto \Theta_g$ is an isometry — in particular, it is injective (if $\Theta_g = 0$ then $\|g\|_{H^{-s}} = \|\Theta_g\|_{(H^s)^*} = 0$, so $g = 0$). Step 3 shows surjectivity: every $\Lambda \in (H^s)^*$ has a representer $g \in H^{-s}$ with $\Theta_g = \Lambda$, and the constructed $g$ satisfies $\|g\|_{H^{-s}} = \|\Lambda\|_{(H^s)^*}$, consistent with the isometry of Step 2. Combining the three, $g \mapsto \Theta_g$ is a conjugate-linear isometric bijection from $H^{-s}(\mathbb{T}^n)$ onto $(H^s(\mathbb{T}^n))^*$, and the natural pairing $\langle \cdot, \cdot \rangle$ realises this identification: every bounded linear functional $\Lambda : H^s(\mathbb{T}^n) \to \mathbb{C}$ has the unique representation $\Lambda(f) = \langle f, g \rangle$ for some $g \in H^{-s}(\mathbb{T}^n)$, and $\|\Lambda\|_{(H^s)^*} = \|g\|_{H^{-s}(\mathbb{T}^n)}$. This completes the proof. [/step]