**Proof plan.** The key mechanism is that $s > n/2$ makes the weight [sequence](/page/Sequence) $(|k|^{-s})_{k \neq 0}$ square-summable over $\mathbb{Z}^n \setminus \{0\}$, which via Cauchy–Schwarz forces the Fourier coefficients of any $f \in \dot{H}^s(\mathbb{T}^n)$ to be absolutely summable. Absolute summability of the [Fourier series](/page/Fourier%20Series) implies [uniform convergence](/page/Uniform%20Convergence), and uniform [limits](/page/Limit) of continuous [functions](/page/Function) are continuous. **Step 1: Absolute summability via Cauchy–Schwarz.** For $f \in \dot{H}^s(\mathbb{T}^n)$, insert the factor $|k|^{-s} \cdot |k|^s$ and apply the Cauchy–Schwarz inequality in $\ell^2(\mathbb{Z}^n \setminus \{0\})$: \begin{align*} \sum_{k \neq 0} |\hat{f}(k)| = \sum_{k \neq 0} |k|^{-s} \cdot |k|^s|\hat{f}(k)| \leq \left(\sum_{k \neq 0} |k|^{-2s}\right)^{1/2} \left(\sum_{k \neq 0} |k|^{2s}|\hat{f}(k)|^2\right)^{1/2}. \end{align*} The second factor is $\|f\|_{\dot{H}^s(\mathbb{T}^n)}$. The first factor is finite when $s > n/2$, as established in Step 2. **Step 2: Convergence of $\sum_{k \neq 0}|k|^{-2s}$.** Partition $\mathbb{Z}^n \setminus \{0\}$ into dyadic shells $A_j = \{k \in \mathbb{Z}^n : 2^{j-1} \leq |k| < 2^j\}$ for $j = 0, 1, 2, \ldots$ (with $A_0 = \{k : |k| < 1\} = \emptyset$, so the relevant shells start at $j = 1$). Each shell $A_j$ contains at most $C_n 2^{jn}$ lattice points (since the shell $\{x \in \mathbb{R}^n : 2^{j-1} \leq |x| < 2^j\}$ has volume $\sim 2^{jn}$, and each unit cube contains at most one lattice point up to a constant). For $k \in A_j$, $|k|^{-2s} \leq 2^{-2s(j-1)}$, so: \begin{align*} \sum_{k \neq 0} |k|^{-2s} \leq \sum_{j=1}^\infty \#A_j \cdot 2^{-2s(j-1)} \leq C_n \sum_{j=1}^\infty 2^{jn} \cdot 2^{-2s(j-1)} = C_n' \sum_{j=1}^\infty 2^{j(n-2s)}. \end{align*} The geometric series $\sum_{j=1}^\infty 2^{j(n-2s)}$ converges if and only if $n - 2s < 0$, i.e., $s > n/2$. Therefore $C_{s,n} := \left(\sum_{k \neq 0}|k|^{-2s}\right)^{1/2} < \infty$ whenever $s > n/2$. **Step 3: $L^\infty$ bound.** Combining Steps 1 and 2: \begin{align*} \sum_{k \neq 0} |\hat{f}(k)| \leq C_{s,n} \|f\|_{\dot{H}^s(\mathbb{T}^n)}. \end{align*} Since $\hat{f}(0) = 0$ for $f \in \dot{H}^s(\mathbb{T}^n)$, the Fourier series of $f$ is $\sum_{k \neq 0}\hat{f}(k)e^{ik\cdot x}$. For any $x \in \mathbb{T}^n$: \begin{align*} |f(x)| \leq \sum_{k \neq 0}|\hat{f}(k)||e^{ik \cdot x}| = \sum_{k \neq 0}|\hat{f}(k)| \leq C_{s,n}\|f\|_{\dot{H}^s(\mathbb{T}^n)}. \end{align*} Taking the supremum over $x$: $\|f\|_{L^\infty(\mathbb{T}^n)} \leq C_{s,n}\|f\|_{\dot{H}^s(\mathbb{T}^n)}$. **Step 4: [Continuity](/page/Continuity) of $f$.** The partial sums $S_N f(x) = \sum_{0 < |k| \leq N}\hat{f}(k)e^{ik\cdot x}$ are trigonometric polynomials and hence continuous. By Step 3, $\|S_N f - f\|_{L^\infty} \leq \sum_{|k| > N}|\hat{f}(k)| \to 0$ as $N \to \infty$ (the tail of an absolutely convergent series). A uniform limit of continuous functions is continuous, so $f \in C(\mathbb{T}^n)$.