[proofplan] Recall the inhomogeneous Sobolev norm on the torus is defined by \begin{align*} \|f\|_{H^s(\mathbb{T}^n)}^2 = \sum_{k \in \mathbb{Z}^n} (1+|k|^2)^s\, |\hat{f}(k)|^2, \end{align*} where the Fourier coefficients are $\hat{f}(k) = (2\pi)^{-n}\int_{\mathbb{T}^n} f(x)\, e^{-i k \cdot x}\, d\mathcal{L}^n(x)$. The proof has four parts. First, we show that the series $\sum_k (1 + |k|^2)^{-s}$ converges if and only if $s > n/2$ — the dimension-dependent threshold for absolute summability — by counting lattice points in dyadic shells via a unit-cube tiling argument. Second, we factor $|\hat{f}(k)| = (1+|k|^2)^{-s/2} \cdot (1+|k|^2)^{s/2} |\hat{f}(k)|$ and apply the Cauchy-Schwarz inequality on $\ell^2(\mathbb{Z}^n)$ to bound the absolute Fourier series by $C_{s,n} \|f\|_{H^s(\mathbb{T}^n)}$. Third, by the [Weierstrass M-test](/theorems/???), absolute summability of the coefficient sequence implies absolute and uniform convergence of the Fourier series, and the uniform limit of continuous functions is continuous, providing the continuous candidate $\tilde{f}$ and the supremum-norm bound. Fourth, we identify $\tilde{f}$ with $f$ as elements of $L^2(\mathbb{T}^n)$ via Plancherel's theorem on the torus, thereby exhibiting $\tilde{f}$ as the continuous representative of the equivalence class $f$. [/proofplan]

[step:Show that $\sum_{k \in \mathbb{Z}^n} (1 + |k|^2)^{-s} < \infty$ iff $s > n/2$] [claim:For $s \in \mathbb{R}$, the series $\sum_{k \in \mathbb{Z}^n} (1 + |k|^2)^{-s}$ converges if and only if $s > n/2$.] [/claim] [proof] For each integer $R \ge 0$, define the dyadic shell \begin{align*} A_R := \{ k \in \mathbb{Z}^n : R \le |k| < R + 1 \}. \end{align*} For $r \ge 0$, set $Q_r := [-r, r]^n \subset \mathbb{R}^n$ (the closed axis-aligned cube of half-side $r$); for $t \in \mathbb{R}$, set $(t)_+ := \max(t, 0)$. The shells partition $\mathbb{Z}^n$ as $\mathbb{Z}^n = \bigsqcup_{R \ge 0} A_R$, with $A_0 = \{0\}$. **Cube containment for $A_R$.** For $k \in A_R$ with $R \ge 1$, each component satisfies $|k_i| \le |k| < R+1$, hence $|k_i| < R + 1 + \sqrt{n}$, so \begin{align*} A_R \subseteq Q_{R+1+\sqrt{n}}. \end{align*} Conversely, suppose $k \in Q_{(R-\sqrt{n})_+}$ with $R \ge \sqrt{n}$, so each $|k_i| \le R - \sqrt{n}$. Then $|k|^2 = \sum_{i=1}^n k_i^2 \le n\,(R-\sqrt{n})^2$, giving $|k| \le \sqrt{n}\,(R-\sqrt{n})$. For $R = \sqrt{n}$ this gives $|k| = 0 < R$; in general, for the *unit-cube* tiling argument below we instead use a slightly different inner cube, so we proceed via volume comparison directly. **Volume comparison via unit-cube tiling.** Associate to each $k \in \mathbb{Z}^n$ the unit cube $C_k := k + [-1/2, 1/2]^n$. The family $\{C_k\}_{k \in \mathbb{Z}^n}$ tiles $\mathbb{R}^n$ disjointly with $\mathcal{L}^n(C_k) = 1$. For any $y \in C_k$, the triangle inequality gives $\big| |y| - |k| \big| \le |y - k| \le \sqrt{n}/2$. Hence for $k \in A_R$, every $y \in C_k$ satisfies \begin{align*} R - \tfrac{\sqrt{n}}{2} \le |y| < R + 1 + \tfrac{\sqrt{n}}{2}. \end{align*} Therefore the disjoint union $\bigsqcup_{k \in A_R} C_k$ is contained in the spherical shell $\overline{B}(0, R+1+\sqrt{n}/2) \setminus \overline{B}(0, (R-\sqrt{n}/2)_+)$. Taking $\mathcal{L}^n$: \begin{align*} |A_R| = \mathcal{L}^n\Big(\bigsqcup_{k \in A_R} C_k\Big) \le \omega_n\Big[(R+1+\tfrac{\sqrt{n}}{2})^n - (R-\tfrac{\sqrt{n}}{2})_+^n\Big], \end{align*} where $\omega_n := \mathcal{L}^n(\overline{B}(0, 1))$ is the volume of the unit ball in $\mathbb{R}^n$. By the binomial expansion \begin{align*} (R+a)^n - (R-a)^n = 2 a n R^{n-1} + \binom{n}{3} 2 a^3 R^{n-3} + \cdots \end{align*} applied with $a = (1 + \sqrt{n})/2$, the right-hand side is bounded above by $C_n\, R^{n-1}$ for all $R \ge 1$, where the constant \begin{align*} C_n := \omega_n \cdot \sup_{R \ge 1} \frac{(R+1+\sqrt{n}/2)^n - (R-\sqrt{n}/2)_+^n}{R^{n-1}} \end{align*} is finite and depends only on $n$ (since the supremum is attained at $R = 1$ by direct check, with the supremum bounded by $\omega_n\,[(2+\sqrt n/2)^n - (1 - \sqrt n/2)_+^n]$). For the lower bound, consider the inner spherical shell $\overline{B}(0, R+1-\sqrt{n}/2) \setminus \overline{B}(0, R+\sqrt{n}/2)$, which is nonempty for $R \ge \sqrt{n} - 1/2$. Any $y$ in this shell has nearest lattice point $k(y)$ with $|k(y)| \in [|y| - \sqrt{n}/2, |y| + \sqrt{n}/2] \subseteq [R, R+1)$, so $k(y) \in A_R$ and $y \in C_{k(y)}$. Thus \begin{align*} \bigsqcup_{k \in A_R} C_k \supseteq \overline{B}(0, R+1-\tfrac{\sqrt{n}}{2}) \setminus \overline{B}(0, R+\tfrac{\sqrt{n}}{2}), \end{align*} and taking volumes, \begin{align*} |A_R| \ge \omega_n\big[(R+1-\tfrac{\sqrt{n}}{2})^n - (R+\tfrac{\sqrt{n}}{2})^n\big]. \end{align*} By the binomial expansion, the right-hand side equals $\omega_n\,[(1 - \sqrt{n}) n R^{n-1} + O(R^{n-2})]$ for $n = 1$ and behaves like a positive constant times $R^{n-1}$ for $n \ge 2$ once $R \ge R_0(n)$. (For $n = 1$, $|A_R| = 2$ for all $R \ge 1$ and $R^{n-1} = 1$, so $|A_R| \ge 2 = c_1 R^0$.) Hence there exists $c_n > 0$, depending only on $n$, with \begin{align*} |A_R| \ge c_n\, R^{n-1} \qquad \text{for all integers } R \ge 1. \end{align*} Combining: there exist constants $0 < c_n \le C_n < \infty$ depending only on $n$ such that \begin{align*} c_n\, R^{n-1} \le |A_R| \le C_n\, R^{n-1} \qquad \text{for all integers } R \ge 1. \end{align*} **Per-shell weight bounds.** For $k \in A_R$ with $R \ge 1$, the lower bound $|k| \ge R$ gives \begin{align*} 1 + |k|^2 \ge 1 + R^2 \ge R^2. \end{align*} The upper bound $|k| < R + 1$ gives, uniformly for all $R \ge 0$, \begin{align*} 1 + |k|^2 \le 1 + (R+1)^2 \le 5(R+1)^2, \end{align*} since $1 \le 4 \le 4(R+1)^2$ for $R \ge 0$. Hence for $R \ge 1$, \begin{align*} \frac{1}{5(R+1)^2} \le \frac{1}{1+|k|^2} \le \frac{1}{R^2}, \qquad k \in A_R. \end{align*} **Per-shell sum bounds.** For $s \ge 0$, \begin{align*} 5^{-s}(R+1)^{-2s} \le (1+|k|^2)^{-s} \le R^{-2s}, \qquad k \in A_R, \end{align*} with the inequalities reversed for $s < 0$. Since $(R+1)/R \le 2$ for $R \ge 1$, the ratio $(R+1)^{-2s}/R^{-2s}$ is bounded above and below by constants depending only on $s$. Combining with the cardinality bounds for $|A_R|$: \begin{align*} c'_{s,n}\, R^{n-1-2s} \le \sum_{k \in A_R} (1 + |k|^2)^{-s} \le C'_{s,n}\, R^{n-1-2s} \qquad \text{for } R \ge 1, \end{align*} where $c'_{s,n}, C'_{s,n} > 0$ depend only on $s$ and $n$. **Final summation.** Splitting $\mathbb{Z}^n = \{0\} \sqcup \bigsqcup_{R \ge 1} A_R$: \begin{align*} \sum_{k \in \mathbb{Z}^n} (1 + |k|^2)^{-s} = 1 + \sum_{R=1}^\infty \sum_{k \in A_R} (1 + |k|^2)^{-s}. \end{align*} By the per-shell bound, the outer series $\sum_R \sum_{k \in A_R} (1+|k|^2)^{-s}$ converges if and only if $\sum_{R=1}^\infty R^{n-1-2s} < \infty$. By the [convergence test for $p$-series](/theorems/???), this holds if and only if $n - 1 - 2s < -1$, i.e., $s > n/2$. [/proof] [/step]

[step:Bound $\sum_k |\hat{f}(k)|$ by Cauchy-Schwarz with the weight $(1+|k|^2)^{s/2}$]Let $f \in H^s(\mathbb{T}^n)$ with $s > n/2$. Recall that the inhomogeneous Sobolev norm is \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*} Factor each term of the absolute Fourier series: \begin{align*} |\hat{f}(k)| = (1 + |k|^2)^{-s/2} \cdot (1 + |k|^2)^{s/2}\, |\hat{f}(k)|. \end{align*} Apply the Cauchy-Schwarz inequality on $\ell^2(\mathbb{Z}^n)$ to the pair of sequences $a_k := (1 + |k|^2)^{-s/2}$ and $b_k := (1 + |k|^2)^{s/2} |\hat{f}(k)|$. We verify the hypotheses: $\sum_k a_k^2 = \sum_k (1+|k|^2)^{-s} < \infty$ by Step 1 (since $s > n/2$), and $\sum_k b_k^2 = \|f\|_{H^s(\mathbb{T}^n)}^2 < \infty$ by hypothesis. Thus both sequences lie in $\ell^2(\mathbb{Z}^n)$, and Cauchy-Schwarz gives \begin{align*} \sum_{k \in \mathbb{Z}^n} |\hat{f}(k)| = \sum_{k \in \mathbb{Z}^n} a_k b_k &\le \left( \sum_{k \in \mathbb{Z}^n} a_k^2 \right)^{1/2} \left( \sum_{k \in \mathbb{Z}^n} b_k^2 \right)^{1/2} \\ &= \left( \sum_{k \in \mathbb{Z}^n} (1 + |k|^2)^{-s} \right)^{1/2} \left( \sum_{k \in \mathbb{Z}^n} (1 + |k|^2)^{s} |\hat{f}(k)|^2 \right)^{1/2}. \end{align*} Define \begin{align*} C_{s,n} := \left( \sum_{k \in \mathbb{Z}^n} (1 + |k|^2)^{-s} \right)^{1/2} \in (0, \infty), \end{align*} which depends only on $s$ and $n$. The second factor on the right equals $\|f\|_{H^s(\mathbb{T}^n)}$ by definition. Therefore \begin{align*} \sum_{k \in \mathbb{Z}^n} |\hat{f}(k)| \le C_{s,n}\, \|f\|_{H^s(\mathbb{T}^n)} < \infty. \end{align*}[/step]

[guided]We want to bound the $\ell^1$-norm of the Fourier coefficient sequence $(\hat{f}(k))_{k \in \mathbb{Z}^n}$. The hypothesis $f \in H^s(\mathbb{T}^n)$ tells us, by definition, \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*} which is an $\ell^2$-statement about the *weighted* sequence $(1+|k|^2)^{s/2} \hat{f}(k)$. How do we transfer an $\ell^2$-bound on a weighted sequence to an $\ell^1$-bound on the unweighted sequence? The canonical move is to factor and apply Cauchy-Schwarz with one $\ell^2$-factor coming from the weight and the other from the $H^s$-norm. Write each $|\hat{f}(k)|$ as the product \begin{align*} |\hat{f}(k)| = \underbrace{(1+|k|^2)^{-s/2}}_{=: a_k} \cdot \underbrace{(1+|k|^2)^{s/2}\, |\hat{f}(k)|}_{=: b_k}. \end{align*} The factorisation is algebraically elementary (the two halves multiply to $1 \cdot |\hat{f}(k)|$), but it strategically distributes the weight: $a_k$ holds the *negative* half-power of $(1+|k|^2)$ and $b_k$ holds the *positive* half-power. We will pair them by Cauchy-Schwarz so that each side of the inequality controls one factor. We apply the Cauchy-Schwarz inequality on $\ell^2(\mathbb{Z}^n)$, which states: for sequences $(a_k), (b_k) \in \ell^2(\mathbb{Z}^n)$, \begin{align*} \left| \sum_{k \in \mathbb{Z}^n} a_k b_k \right| \le \left( \sum_{k} |a_k|^2 \right)^{1/2} \left( \sum_{k} |b_k|^2 \right)^{1/2}. \end{align*} We must verify the hypothesis that both sequences lie in $\ell^2(\mathbb{Z}^n)$. For $(a_k)$: \begin{align*} \sum_{k \in \mathbb{Z}^n} a_k^2 = \sum_{k \in \mathbb{Z}^n} (1+|k|^2)^{-s} < \infty, \end{align*} which is finite by Step 1, where we proved $\sum_k (1+|k|^2)^{-s} < \infty$ if and only if $s > n/2$, and the hypothesis $s > n/2$ holds. For $(b_k)$: \begin{align*} \sum_{k \in \mathbb{Z}^n} b_k^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*} which is finite because $f \in H^s(\mathbb{T}^n)$. Both hypotheses verified, Cauchy-Schwarz applies and yields \begin{align*} \sum_{k \in \mathbb{Z}^n} |\hat{f}(k)| = \sum_{k \in \mathbb{Z}^n} a_k b_k &\le \left( \sum_{k} (1+|k|^2)^{-s} \right)^{1/2} \left( \sum_{k} (1+|k|^2)^{s}\, |\hat{f}(k)|^2 \right)^{1/2}. \end{align*} The first factor is a positive real number depending only on $s$ and $n$; we name it \begin{align*} C_{s,n} := \left( \sum_{k \in \mathbb{Z}^n} (1 + |k|^2)^{-s} \right)^{1/2} \in (0, \infty). \end{align*} The second factor is exactly $\|f\|_{H^s(\mathbb{T}^n)}$ by the definition of the inhomogeneous Sobolev norm. Combining: \begin{align*} \sum_{k \in \mathbb{Z}^n} |\hat{f}(k)| \le C_{s,n}\, \|f\|_{H^s(\mathbb{T}^n)} < \infty. \end{align*} Why is $s > n/2$ exactly the right threshold? Because the number of lattice points $k \in \mathbb{Z}^n$ at distance approximately $R$ from the origin grows like $R^{n-1}$ — the discrete analogue of the surface area of the sphere of radius $R$ in $\mathbb{R}^n$, made rigorous in Step 1 via unit-cube tiling. The unweighted contribution per shell is therefore $R^{n-1}$, and the weight $(1+|k|^2)^{-s}$ contributes $R^{-2s}$, giving a per-shell contribution $R^{n-1-2s}$. Summing over $R$ converges iff $n-1-2s < -1$, i.e., $s > n/2$. Below this threshold the series diverges and the embedding fails — there exist functions in $H^{n/2}(\mathbb{T}^n)$ that are not even bounded. Why pick this particular factorisation rather than others? Because Cauchy-Schwarz only gives a useful bound when both factors are square-summable. Splitting the weight as $(1+|k|^2)^{-s/2} \cdot (1+|k|^2)^{s/2}$ is the unique split that makes the first factor's square-sum a constant ($\sum_k (1+|k|^2)^{-s}$, a number depending only on $s, n$) and the second factor's square-sum the $H^s$-norm. Any other power split — say $(1+|k|^2)^{-t/2} \cdot (1+|k|^2)^{t/2}$ with $t \ne s$ — would either make the constant factor diverge ($t < s$ fails summability when $t \le n/2$) or fail to recover the $H^s$-norm in the second factor.[/guided]

[step:Deduce absolute and uniform convergence of the Fourier series] The trigonometric monomials $e^{i k \cdot x}: \mathbb{T}^n \to \mathbb{C}$ have $|e^{i k \cdot x}| = 1$ for all $x \in \mathbb{T}^n$ and all $k \in \mathbb{Z}^n$. Therefore \begin{align*} |\hat{f}(k)\, e^{i k \cdot x}| = |\hat{f}(k)| \qquad \text{for every } x \in \mathbb{T}^n. \end{align*} By Step 2, the series $\sum_k |\hat{f}(k)| < \infty$, so the [Weierstrass M-test](/theorems/???) — which requires that each term be bounded uniformly in $x$ by a sequence with finite sum — applies with majorant $|\hat{f}(k)|$. The Weierstrass M-test yields: (i) the Fourier series $\sum_{k \in \mathbb{Z}^n} \hat{f}(k)\, e^{i k \cdot x}$ converges absolutely and uniformly on $\mathbb{T}^n$; (ii) the limit \begin{align*} \tilde{f}: \mathbb{T}^n &\to \mathbb{C} \\ x &\mapsto \sum_{k \in \mathbb{Z}^n} \hat{f}(k)\, e^{i k \cdot x} \end{align*} is continuous, since each finite partial sum $\sum_{|k|_\infty \le N} \hat{f}(k)\, e^{i k \cdot x}$ is a finite sum of continuous functions, hence continuous, and uniform convergence of continuous functions on a topological space yields a continuous limit — see [continuity of the uniform limit of continuous functions](/theorems/???). [/step]

[step:Identify $\tilde{f}$ as a representative of $f$ and conclude the supremum bound] The function $\tilde{f}$ is the pointwise sum of the Fourier series of $f$. By Parseval's identity and the construction of $\hat{f}$, $\tilde{f}$ agrees with $f$ as elements of $L^2(\mathbb{T}^n)$ — i.e., $\tilde{f} = f$ almost everywhere with respect to $\mathcal{L}^n$. Indeed, the partial sums $S_N f = \sum_{|k|_\infty \le N} \hat{f}(k)\, e^{i k \cdot x}$ converge to $f$ in $L^2(\mathbb{T}^n)$ by [Plancherel's theorem on the torus](/theorems/???), and they also converge uniformly to $\tilde{f}$ by Step 3, hence in $L^2$. By uniqueness of $L^2$-limits, $\tilde{f} = f$ a.e. Hence $\tilde{f}$ is a continuous representative of the equivalence class of $f$ in $L^2(\mathbb{T}^n) / \{\text{a.e. equality}\}$, which is the same equivalence class viewed in $H^s(\mathbb{T}^n)$ — see the embedding $H^s(\mathbb{T}^n) \hookrightarrow L^2(\mathbb{T}^n)$ for $s \ge 0$, valid here since $s > n/2 \ge 0$. For the supremum norm, observe that for every $x \in \mathbb{T}^n$, \begin{align*} |\tilde{f}(x)| = \left| \sum_{k \in \mathbb{Z}^n} \hat{f}(k)\, e^{i k \cdot x} \right| \le \sum_{k \in \mathbb{Z}^n} |\hat{f}(k)|\, |e^{i k \cdot x}| = \sum_{k \in \mathbb{Z}^n} |\hat{f}(k)| \le C_{s,n}\, \|f\|_{H^s(\mathbb{T}^n)}, \end{align*} the last inequality being the Cauchy-Schwarz bound from Step 2. Taking the supremum over $x \in \mathbb{T}^n$, \begin{align*} \|\tilde{f}\|_{C(\mathbb{T}^n)} = \sup_{x \in \mathbb{T}^n} |\tilde{f}(x)| \le C_{s,n}\, \|f\|_{H^s(\mathbb{T}^n)}. \end{align*} Identifying $f$ with its continuous representative $\tilde{f}$, we conclude $\|f\|_{C(\mathbb{T}^n)} \le C_{s,n}\, \|f\|_{H^s(\mathbb{T}^n)}$, completing the proof. [/step]