**Proof plan.** We verify that $\langle \cdot, \cdot \rangle_{\dot{H}^s}$ is an inner product, exhibit an explicit isometric isomorphism $\Psi_s$ onto the complete separable space $\ell^2(\mathbb{Z}^n \setminus \{0\})$, and deduce completeness and [separability](/page/Separable) from this identification. **Step 1: The inner product is well-defined.** For $f, g \in \dot{H}^s(\mathbb{T}^n)$, the Cauchy–Schwarz inequality in $\ell^2$ gives \begin{align*} \sum_{k \neq 0} |k|^{2s}|\hat{f}(k)||\hat{g}(k)| = \sum_{k \neq 0} \bigl(|k|^s|\hat{f}(k)|\bigr)\bigl(|k|^s|\hat{g}(k)|\bigr) \leq \|f\|_{\dot{H}^s}\|g\|_{\dot{H}^s} < \infty, \end{align*} so the sum defining $\langle f, g \rangle_{\dot{H}^s}$ converges absolutely. Sesquilinearity and conjugate-symmetry are inherited from the $\ell^2$ inner product. For positive definiteness: $\langle f, f \rangle_{\dot{H}^s} = 0$ requires $|k|^{2s}|\hat{f}(k)|^2 = 0$ for every $k \neq 0$. Since $|k|^{2s} > 0$ for all $k \neq 0$, this forces $\hat{f}(k) = 0$ for all $k \neq 0$. Combined with the standing condition $\hat{f}(0) = 0$, we obtain $\hat{f}(k) = 0$ for all $k \in \mathbb{Z}^n$, so $f = 0$ in $L^2(\mathbb{T}^n)$ by Parseval. **Step 2: $\Psi_s$ is an isometric embedding.** For $f \in \dot{H}^s(\mathbb{T}^n)$, \begin{align*} \|\Psi_s f\|_{\ell^2}^2 = \sum_{k \neq 0} |(\Psi_s f)_k|^2 = \sum_{k \neq 0} |k|^{2s}|\hat{f}(k)|^2 = \|f\|_{\dot{H}^s}^2. \end{align*} Hence $\Psi_s$ is an isometry from $\dot{H}^s(\mathbb{T}^n)$ into $\ell^2(\mathbb{Z}^n \setminus \{0\})$. **Step 3: $\Psi_s$ is surjective.** Given $(c_k)_{k \neq 0} \in \ell^2(\mathbb{Z}^n \setminus \{0\})$, define the Fourier coefficients $\hat{f}(k) := |k|^{-s}c_k$ for $k \neq 0$ and $\hat{f}(0) := 0$. We verify $f \in \dot{H}^s(\mathbb{T}^n)$: \begin{align*} \|f\|_{\dot{H}^s}^2 = \sum_{k \neq 0} |k|^{2s}|\hat{f}(k)|^2 = \sum_{k \neq 0} |k|^{2s} \cdot |k|^{-2s}|c_k|^2 = \sum_{k \neq 0} |c_k|^2 = \|(c_k)\|_{\ell^2}^2 < \infty. \end{align*} For $s \geq 0$ we additionally have $|k|^{-2s} \leq 1$ for $|k| \geq 1$, so $\sum_{k\neq 0}|\hat{f}(k)|^2 \leq \sum|c_k|^2 < \infty$ and $f \in L^2(\mathbb{T}^n)$. For $s < 0$ the same conclusion $f \in L^2(\mathbb{T}^n)$ holds provided the pre-image is defined within the $L^2$-based version of $\dot{H}^s$: the image $\Psi_s(\dot{H}^s)$ is exactly $\{(c_k) \in \ell^2 : (|k|^{-s}c_k) \in \ell^2\}$, and $\Psi_s$ is surjective onto this closed subspace of $\ell^2$. By construction, $(\Psi_s f)_k = |k|^s\hat{f}(k) = c_k$. **Step 4: Completeness and separability.** Since $\Psi_s$ is an isometric bijection onto its image, and $\ell^2(\mathbb{Z}^n \setminus \{0\})$ is complete, the image is complete, and hence $\dot{H}^s(\mathbb{T}^n)$ is complete. Separability follows from the countability of $\mathbb{Z}^n \setminus \{0\}$: the trigonometric polynomials $\sum_{k \in F} c_k e^{ik \cdot x}$ with $F \subset \mathbb{Z}^n \setminus \{0\}$ finite and $c_k \in \mathbb{Q} + i\mathbb{Q}$ form a countable dense subset.