[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*}
[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]
[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*}
[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]
[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 $X\ge 1$.
[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]
[/step]
