[proofplan] We prove that the Weyl sum is locally Lipschitz on the torus, with coordinate Lipschitz constants of size $X^{j+1}$ in the $\theta_j$ direction. The derivative bound follows by differentiating the finite exponential sum term by term and bounding the resulting power sum. If $\theta'$ is represented as $\theta+h$ with $|h_j|\le \rho_j$, the [mean value inequality](/theorems/328) along the line segment from $\theta$ to $\theta+h$ shows that the total change in $S_k$ is at most $cA/2\le A/2$. The lower bound then follows from the [reverse triangle inequality](/theorems/2300). [/proofplan] [step:Differentiate the finite Weyl sum in each coordinate] Let $N:=\lfloor X\rfloor$. The sum defining $S_k(\theta;X)$ is finite and equals \begin{align*} S_k(\theta;X)=\sum_{n=1}^{N} e(\theta_1n+\cdots+\theta_kn^k). \end{align*} For each $1\le j\le k$, the [partial derivative](/page/Partial%20Derivative) with respect to $\theta_j$ exists on $\mathbb R^k$ and termwise differentiation gives \begin{align*} \partial_{\theta_j}S_k(\theta;X)=2\pi i\sum_{n=1}^{N} n^j e(\theta_1n+\cdots+\theta_kn^k). \end{align*} Since $|e(t)|=1$ for every $t\in\mathbb R$, the triangle inequality gives \begin{align*} |\partial_{\theta_j}S_k(\theta;X)|\le 2\pi\sum_{n=1}^{N} n^j. \end{align*} Because $1\le n\le N\le X$, one has $n^j\le X^j$, and hence \begin{align*} \sum_{n=1}^{N}n^j\le NX^j\le X^{j+1}. \end{align*} Therefore, uniformly for every $\theta\in\mathbb R^k$, \begin{align*} |\partial_{\theta_j}S_k(\theta;X)|\le 2\pi X^{j+1}. \end{align*} [/step] [step:Control the change along a short coordinate box] Let $\mathcal L^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb R$. Let $\theta'\in B$. By the definition of $B$, there exists $h=(h_1,\dots,h_k)\in\mathbb R^k$ such that $|h_j|\le \rho_j$ for every $1\le j\le k$ and such that $\theta'$ is the class of $\theta+h$ modulo $\mathbb Z^k$. Since each summand $e(\theta_1n+\cdots+\theta_kn^k)$ is $1$-periodic in every coordinate, it is enough to estimate $S_k(\theta+h;X)-S_k(\theta;X)$ in $\mathbb R^k$. Define the path $\gamma:[0,1]\to\mathbb R^k$ by $\gamma(t):=\theta+th$. The one-variable chain rule applied to the [differentiable map](/page/Differentiable%20Map) $t\mapsto S_k(\gamma(t);X)$ gives \begin{align*} S_k(\theta+h;X)-S_k(\theta;X)=\int_0^1 \sum_{j=1}^{k} h_j\partial_{\theta_j}S_k(\theta+th;X)\,d\mathcal L^1(t). \end{align*} Taking absolute values, applying the triangle inequality, and using the derivative bound from the previous step, we obtain \begin{align*} |S_k(\theta+h;X)-S_k(\theta;X)|\le \int_0^1\sum_{j=1}^{k}|h_j|2\pi X^{j+1}\,d\mathcal L^1(t). \end{align*} Since $|h_j|\le \rho_j$, this becomes \begin{align*} |S_k(\theta+h;X)-S_k(\theta;X)|\le \sum_{j=1}^{k}2\pi X^{j+1}\frac{cA}{4\pi k}X^{-j-1}. \end{align*} The right-hand side simplifies to \begin{align*} \frac{cA}{2}. \end{align*} Because $0<c<1$, we have \begin{align*} |S_k(\theta+h;X)-S_k(\theta;X)|\le \frac{A}{2}. \end{align*} [guided] We now turn the derivative estimates into a uniform stability statement on a box. Choose $\theta'\in B$. The definition of $B$ means that $\theta'$ can be represented by a point $\theta+h$ in $\mathbb R^k$, where $h=(h_1,\dots,h_k)$ satisfies \begin{align*} |h_j|\le \rho_j \end{align*} for every coordinate $1\le j\le k$. If the box crosses a face of the fundamental cube $[0,1)^k$, this representative statement is exactly what the reduction modulo $1$ means. The Weyl sum is periodic in each coordinate: replacing $\theta_j$ by $\theta_j+m_j$ with $m_j\in\mathbb Z$ multiplies the $n$th summand by $e(m_jn^j)=1$. Thus the value of $S_k$ depends only on the class of $\theta$ in $(\mathbb R/\mathbb Z)^k$, and it is enough to estimate the change from $\theta$ to $\theta+h$ in $\mathbb R^k$. Define the line segment $\gamma:[0,1]\to\mathbb R^k$ by $\gamma(t):=\theta+th$. The map $t\mapsto S_k(\gamma(t);X)$ is continuously differentiable because $S_k(\cdot;X)$ is a finite sum of smooth exponential functions. By the one-variable [fundamental theorem of calculus](/theorems/632) applied to this composite function, \begin{align*} S_k(\theta+h;X)-S_k(\theta;X)=\int_0^1 \frac{d}{dt}S_k(\theta+th;X)\,d\mathcal L^1(t). \end{align*} The chain rule identifies the derivative as \begin{align*} \frac{d}{dt}S_k(\theta+th;X)=\sum_{j=1}^{k}h_j\partial_{\theta_j}S_k(\theta+th;X). \end{align*} Substituting this into the integral gives \begin{align*} S_k(\theta+h;X)-S_k(\theta;X)=\int_0^1 \sum_{j=1}^{k}h_j\partial_{\theta_j}S_k(\theta+th;X)\,d\mathcal L^1(t). \end{align*} Now we estimate the absolute value. The derivative bound proved in the previous step says that, for every $1\le j\le k$ and every $\vartheta\in\mathbb R^k$, \begin{align*} |\partial_{\theta_j}S_k(\vartheta;X)|\le 2\pi X^{j+1}. \end{align*} Applying the triangle inequality for the integral and then this derivative bound with $\vartheta=\theta+th$ gives \begin{align*} |S_k(\theta+h;X)-S_k(\theta;X)|\le \int_0^1\sum_{j=1}^{k}|h_j|2\pi X^{j+1}\,d\mathcal L^1(t). \end{align*} The point of choosing $\rho_j$ proportional to $AX^{-j-1}$ is that it cancels the derivative scale $X^{j+1}$. Indeed, since $|h_j|\le \rho_j$, \begin{align*} |h_j|2\pi X^{j+1}\le \frac{cA}{4\pi k}X^{-j-1}2\pi X^{j+1}. \end{align*} Thus each coordinate contributes at most $cA/(2k)$, and summing over the $k$ coordinates gives \begin{align*} |S_k(\theta+h;X)-S_k(\theta;X)|\le \frac{cA}{2}. \end{align*} Finally $0<c<1$, so this is bounded by $A/2$. This is the desired uniform Lipschitz control on the whole box. [/guided] [/step] [step:Apply the reverse triangle inequality to preserve half the size] Using the periodicity of $S_k$ again, $S_k(\theta';X)=S_k(\theta+h;X)$. The reverse triangle inequality gives \begin{align*} |S_k(\theta';X)|=|S_k(\theta+h;X)|\ge |S_k(\theta;X)|-|S_k(\theta+h;X)-S_k(\theta;X)|. \end{align*} By hypothesis $|S_k(\theta;X)|\ge A$, and by the previous step the second term is at most $A/2$. Therefore \begin{align*} |S_k(\theta';X)|\ge A-\frac{A}{2}=\frac{A}{2}. \end{align*} Since $\theta'\in B$ was arbitrary, the estimate holds for every point of $B$. It remains to justify that the displayed coordinate interval gives an actual side length on the torus rather than wrapping around. From the first step and the triangle inequality, \begin{align*} A\le |S_k(\theta;X)|\le \sum_{n=1}^{N}|e(\theta_1n+\cdots+\theta_kn^k)|=N\le X. \end{align*} Therefore, for every $1\le j\le k$, \begin{align*} 2\rho_j=\frac{cA}{2\pi k}X^{-j-1}\le \frac{c}{2\pi k}X^{-j}<1. \end{align*} Thus the coordinate map $h_j\in[-\rho_j,\rho_j]\mapsto \theta_j+h_j\bmod 1$ is injective, and the side length of $B$ in the $\theta_j$ direction is exactly $2\rho_j=\frac{cA}{2\pi k}X^{-j-1}$. The claimed comparability follows with constants depending only on $k$. [/step]