[proofplan] The proof converts a large pointwise value of the Weyl sum into a positive-measure set of large values, then compares this with the assumed moment bound. If $A=|S_k(\theta;X)|$ is at least $1$, local stability gives a box in the $k$-torus on which the sum remains at least $A/2$, with volume bounded below by a power of $A$ and $X$. Markov's inequality bounds the measure of the same large-value set using the $2s$-th moment. Comparing the two estimates gives an inequality for $A^{2s+k}$, and taking roots yields the required pointwise estimate. [/proofplan]

[step:Work on the torus and discard the small-value case] Let $\mathbb T^k:=(\mathbb R/\mathbb Z)^k$, and let $\mu_k$ denote its Haar probability measure, identified with $\mathcal L^k$ on the fundamental domain $[0,1)^k$. Define the continuous map $S_{k,X}:\mathbb T^k\to\mathbb C$ by \begin{align*} S_{k,X}(\theta):=\sum_{1\le n\le N_X} e(\theta_1n+\theta_2n^2+\cdots+\theta_kn^k). \end{align*} This agrees with $S_k(\theta;X)$ on $[0,1)^k$. Fix $\theta\in\mathbb T^k$ and set \begin{align*} A:=|S_{k,X}(\theta)|. \end{align*} If $A<1$, then since $X\ge 1$ and the exponent \begin{align*} \frac{s+\Delta+k(k+3)/2}{2s+k} \end{align*} is non-negative, the desired estimate follows after increasing the implicit constant. Hence assume $A\ge 1$. [/step]

[step:Use local stability to create a positive-measure large-value box]Apply [[Local Stability Of Polynomial Weyl Sums](/theorems/9063)][citetheorem:9063] with fixed parameter $c=1/2$ to the point $\theta$ and the value $A=|S_{k,X}(\theta)|$. Its hypotheses are satisfied because $A\ge 1$ and $S_{k,X}$ is precisely the polynomial Weyl sum in $k$ variables with summation range $1\le n\le N_X$. In the notation of that result, the conclusion gives a rectangular box $B\subset\mathbb T^k$, centred at $\theta$ modulo $1$, on which the sum has size at least $A/2$, and whose side lengths are bounded below by constants times $AX^{-j-1}$ in the coordinate directions. Thus \begin{align*} |S_{k,X}(\theta')|\ge A/2 \end{align*} for every $\theta'\in B$, and the side length in the $\theta_j$ direction is bounded below by \begin{align*} c_{j,k}AX^{-j-1} \end{align*} for constants $c_{j,k}>0$ depending only on $j$ and $k$. These side lengths are legitimate torus side lengths after reducing the constants if necessary: since $A=|S_{k,X}(\theta)|\le \lfloor X\rfloor\le X$, one has $AX^{-j-1}\le X^{-j}\le 1$ for every $1\le j\le k$. Define \begin{align*} c_k:=\prod_{j=1}^k c_{j,k}>0. \end{align*} Since Haar measure on $\mathbb T^k$ agrees locally with product [Lebesgue measure](/page/Lebesgue%20Measure) and the box is interpreted modulo $1$, the measure of $B$ satisfies \begin{align*} \mu_k(B)\ge c_k A^k X^{-\sum_{j=1}^k(j+1)}. \end{align*} The finite sum is \begin{align*} \sum_{j=1}^k(j+1)=\frac{k(k+1)}{2}+k=\frac{k(k+3)}{2}. \end{align*} Therefore \begin{align*} \mu_k(B)\ge c_k A^k X^{-k(k+3)/2}. \end{align*}[/step]

[guided]The role of local stability is to prevent a large value at one point from being isolated. We apply [Local Stability Of Polynomial Weyl Sums][citetheorem:9063] to the polynomial Weyl sum \begin{align*} S_{k,X}(\theta)=\sum_{1\le n\le N_X} e(\theta_1n+\theta_2n^2+\cdots+\theta_kn^k). \end{align*} The theorem requires a point where the Weyl sum has size at least $A$, and we have chosen $A=|S_{k,X}(\theta)|$. Since we are in the non-trivial case $A\ge 1$, the hypotheses are satisfied. With the fixed parameter $c=1/2$, local stability gives a rectangular box $B\subset\mathbb T^k$ centred at $\theta$ modulo $1$ such that \begin{align*} |S_{k,X}(\theta')|\ge A/2 \end{align*} for every $\theta'\in B$. It also gives side lengths comparable to $AX^{-j-1}$ in the $\theta_j$ direction. For the lower bound we only need the lower comparability constants, so we choose constants $c_{j,k}>0$, depending only on $j$ and $k$, such that the side length in the $\theta_j$ direction is at least \begin{align*} c_{j,k}AX^{-j-1}. \end{align*} These are legitimate side lengths on the torus after reducing the constants if needed. Indeed, the triangle inequality gives $A=|S_{k,X}(\theta)|\le \lfloor X\rfloor\le X$, and therefore $AX^{-j-1}\le X^{-j}\le 1$ for every $1\le j\le k$. Because the box is taken on the torus, no boundary difficulty arises when $\theta$ is close to the edge of the fundamental domain $[0,1)^k$. Define \begin{align*} c_k:=\prod_{j=1}^k c_{j,k}. \end{align*} The measure of a rectangular box is the product of its side lengths, so Haar measure gives \begin{align*} \mu_k(B)\ge c_k\prod_{j=1}^k AX^{-j-1}. \end{align*} Multiplying the powers of $A$ gives $A^k$, and multiplying the powers of $X$ gives \begin{align*} X^{-\sum_{j=1}^k(j+1)}. \end{align*} Finally, \begin{align*} \sum_{j=1}^k(j+1)=\frac{k(k+1)}{2}+k=\frac{k(k+3)}{2}. \end{align*} Hence \begin{align*} \mu_k(B)\ge c_k A^k X^{-k(k+3)/2}. \end{align*}[/guided]

[step:Bound the same large-value set from above by the moment estimate]Define the measurable large-value set \begin{align*} E_A:=\{\theta'\in\mathbb T^k: |S_{k,X}(\theta')|\ge A/2\}. \end{align*} The box constructed above satisfies $B\subset E_A$, so \begin{align*} \mu_k(B)\le \mu_k(E_A). \end{align*} By Markov's inequality applied to the non-negative [measurable function](/page/Measurable%20Function) $\theta'\mapsto |S_{k,X}(\theta')|^{2s}$ with threshold $(A/2)^{2s}$, \begin{align*} \mu_k(E_A)\le \left(\frac{2}{A}\right)^{2s}\int_{\mathbb T^k}|S_{k,X}(\theta')|^{2s}\,d\mu_k(\theta'). \end{align*} By the definition of $J_{s,k}(X)$, equivalently the moment identity in [[Vinogradov Mean Value Identity](/theorems/9061)][citetheorem:9061], \begin{align*} \int_{\mathbb T^k}|S_{k,X}(\theta')|^{2s}\,d\mu_k(\theta')=J_{s,k}(X). \end{align*} Using the assumed moment bound, for every $\varepsilon>0$ there is a constant $C_{s,k,\varepsilon}>0$ such that \begin{align*} \mu_k(E_A)\le 2^{2s}C_{s,k,\varepsilon}A^{-2s}X^{s+\Delta+\varepsilon}. \end{align*}[/step]

[guided]The large-value box gives a lower bound for the measure of a set on which the Weyl sum is large. To compare it with the moment hypothesis, define \begin{align*} E_A:=\{\theta'\in\mathbb T^k: |S_{k,X}(\theta')|\ge A/2\}. \end{align*} Because every point of $B$ satisfies $|S_{k,X}(\theta')|\ge A/2$, we have $B\subset E_A$, and hence \begin{align*} \mu_k(B)\le \mu_k(E_A). \end{align*} We now apply Markov's inequality to the non-negative measurable function $\theta'\mapsto |S_{k,X}(\theta')|^{2s}$ on the probability space $(\mathbb T^k,\mu_k)$ with threshold $(A/2)^{2s}$. This gives \begin{align*} \mu_k(E_A)\le \left(\frac{2}{A}\right)^{2s}\int_{\mathbb T^k}|S_{k,X}(\theta')|^{2s}\,d\mu_k(\theta'). \end{align*} The measure $\mu_k$ agrees with $\mathcal L^k$ on the fundamental domain $[0,1)^k$, so the integral is exactly the Vinogradov mean value quantity defined in the statement, equivalently the identity supplied by [Vinogradov Mean Value Identity][citetheorem:9061]: \begin{align*} \int_{\mathbb T^k}|S_{k,X}(\theta')|^{2s}\,d\mu_k(\theta')=J_{s,k}(X). \end{align*} Finally, the assumed moment estimate says that for every $\varepsilon>0$ there is a constant $C_{s,k,\varepsilon}>0$, independent of $X$, such that \begin{align*} J_{s,k}(X)\le C_{s,k,\varepsilon}X^{s+\Delta+\varepsilon}. \end{align*} Substituting this into the Markov bound yields \begin{align*} \mu_k(E_A)\le 2^{2s}C_{s,k,\varepsilon}A^{-2s}X^{s+\Delta+\varepsilon}. \end{align*}[/guided]

[step:Compare the lower and upper bounds and take roots]Combining $B\subset E_A$ with the lower and upper estimates gives \begin{align*} c_k A^k X^{-k(k+3)/2}\le 2^{2s}C_{s,k,\varepsilon}A^{-2s}X^{s+\Delta+\varepsilon}. \end{align*} Multiplying by $A^{2s}X^{k(k+3)/2}$ yields \begin{align*} A^{2s+k}\le c_k^{-1}2^{2s}C_{s,k,\varepsilon}X^{s+\Delta+k(k+3)/2+\varepsilon}. \end{align*} Taking the $(2s+k)$-th root, we obtain \begin{align*} A\le C'_{s,k,\varepsilon}X^{\frac{s+\Delta+k(k+3)/2+\varepsilon}{2s+k}}, \end{align*} where \begin{align*} C'_{s,k,\varepsilon}:=(c_k^{-1}2^{2s}C_{s,k,\varepsilon})^{1/(2s+k)}. \end{align*} Given a target exponent loss $\varepsilon>0$, apply the preceding estimate with $(2s+k)\varepsilon$ in place of $\varepsilon$ in the moment hypothesis. This gives \begin{align*} |S_k(\theta;X)|=A\lesssim_{s,k,\varepsilon}X^{\frac{s+\Delta+k(k+3)/2}{2s+k}+\varepsilon}. \end{align*} The constants are independent of $\theta$ and $X$, so the bound is uniform in $\theta\in[0,1)^k$ and $X\ge 1$.[/step]

[guided]We combine the two estimates for the same set. The local-stability box gave \begin{align*} \mu_k(B)\ge c_k A^k X^{-k(k+3)/2}, \end{align*} while the Markov and moment estimate gave \begin{align*} \mu_k(E_A)\le 2^{2s}C_{s,k,\varepsilon}A^{-2s}X^{s+\Delta+\varepsilon}. \end{align*} Since $B\subset E_A$, the lower bound for $\mu_k(B)$ is bounded above by the upper bound for $\mu_k(E_A)$: \begin{align*} c_k A^k X^{-k(k+3)/2}\le 2^{2s}C_{s,k,\varepsilon}A^{-2s}X^{s+\Delta+\varepsilon}. \end{align*} Because $A\ge 1$, all powers of $A$ are finite and positive. Multiplying both sides by $A^{2s}X^{k(k+3)/2}$ gives \begin{align*} A^{2s+k}\le c_k^{-1}2^{2s}C_{s,k,\varepsilon}X^{s+\Delta+k(k+3)/2+\varepsilon}. \end{align*} Taking the $(2s+k)$-th root gives \begin{align*} A\le C'_{s,k,\varepsilon}X^{\frac{s+\Delta+k(k+3)/2+\varepsilon}{2s+k}}, \end{align*} where \begin{align*} C'_{s,k,\varepsilon}:=(c_k^{-1}2^{2s}C_{s,k,\varepsilon})^{1/(2s+k)}. \end{align*} The exponent still contains the auxiliary loss divided by $2s+k$. To obtain a prescribed final loss $\varepsilon>0$, apply the previous bound with $(2s+k)\varepsilon$ in the moment hypothesis. Then \begin{align*} |S_k(\theta;X)|=A\lesssim_{s,k,\varepsilon}X^{\frac{s+\Delta+k(k+3)/2}{2s+k}+\varepsilon}. \end{align*} All constants depend only on $s$, $k$, and the chosen $\varepsilon$, not on $\theta$ or $X$. This proves the required uniform pointwise bound.[/guided]