[proofplan] We apply the [van der Corput inequality](/theorems/9057) to the unimodular sequence $z_n=e(f(n))$ on the integer interval $M<n\le M+N$. The only work is to identify the shifted correlation $z_{n+h}\overline{z_n}$ with the exponential of the finite difference $f(n+h)-f(n)$. This gives exactly the displayed A-process estimate, with the absolute constant inherited from van der Corput's inequality. [/proofplan] [step:Apply van der Corput to the exponential sequence] Define the finite index set $I:=\{n\in\mathbb Z:M<n\le M+N\}$, and define the sequence $z:I\to\mathbb C$ by \begin{align*} z(n):=e(f(n)). \end{align*} For readability, write $z_n:=z(n)$ for $n\in I$. For every integer $n\in I$, the value $f(n)$ is real, so \begin{align*} |z_n|=|e(f(n))|=1. \end{align*} We use the Van Der Corput Inequality [citetheorem:9057], which applies to complex numbers $(z_n)_{M<n\le M+N}$ satisfying $|z_n|\le 1$ and to an integer $H$ satisfying $1\le H\le N$. The bound $|z_n|\le 1$ follows from $|z_n|=1$, and the condition $1\le H\le N$ is part of the theorem statement. Therefore \begin{align*} \left|\sum_{M<n\le M+N} z_n\right|^2 \lesssim \frac{N^2}{H}+\frac{N}{H}\sum_{1\le h<H}\left|\sum_{M<n\le M+N-h}z_{n+h}\overline{z_n}\right|. \end{align*} Substituting $z_n=e(f(n))$ into the left-hand side gives \begin{align*} \left|\sum_{M<n\le M+N} z_n\right|^2=\left|\sum_{M<n\le M+N} e(f(n))\right|^2. \end{align*} [guided] The A-process estimate is a direct application of the Van Der Corput Inequality [citetheorem:9057], so we first verify that its hypotheses match the present situation exactly. Define the finite index set $I:=\{n\in\mathbb Z:M<n\le M+N\}$, and define the sequence $z:I\to\mathbb C$ by \begin{align*} z(n):=e(f(n)). \end{align*} We write $z_n:=z(n)$ for $n\in I$. The cited inequality requires a complex sequence indexed by the interval $M<n\le M+N$, and this is precisely the set $I$. The cited inequality also requires the pointwise bound $|z_n|\le 1$ for every $n\in I$. Since $f$ is real-valued on $I$ and $e(t)=\exp(2\pi i t)$ for real $t$, each $e(f(n))$ lies on the unit circle in $\mathbb C$. Hence \begin{align*} |z_n|=|e(f(n))|=1\le 1. \end{align*} Finally, the cited inequality requires an integer $H$ satisfying $1\le H\le N$, and this is exactly one of the hypotheses of the theorem. Therefore the Van Der Corput Inequality gives \begin{align*} \left|\sum_{M<n\le M+N} z_n\right|^2 \lesssim \frac{N^2}{H}+\frac{N}{H}\sum_{1\le h<H}\left|\sum_{M<n\le M+N-h}z_{n+h}\overline{z_n}\right|. \end{align*} The left-hand side is already the desired exponential sum after substituting the definition of $z_n$: \begin{align*} \left|\sum_{M<n\le M+N} z_n\right|^2=\left|\sum_{M<n\le M+N} e(f(n))\right|^2. \end{align*} The remaining task is therefore only to rewrite the correlation terms $z_{n+h}\overline{z_n}$ in terms of the finite difference $\Delta_h f(n)$. [/guided] [/step] [step:Rewrite the shifted correlations as finite-difference sums] Fix an integer $h$ with $1\le h<H$ and an integer $n$ with $M<n\le M+N-h$. Then $M<n<n+h\le M+N$, so both $f(n)$ and $f(n+h)$ are defined by the stated domain of $f$. Since $f(n)\in\mathbb R$, complex conjugation gives \begin{align*} \overline{e(f(n))}=e(-f(n)). \end{align*} Hence \begin{align*} z_{n+h}\overline{z_n}=e(f(n+h))e(-f(n)). \end{align*} Using the identity $e(a)e(b)=e(a+b)$ for [real numbers](/page/Real%20Numbers) $a,b$, this becomes \begin{align*} z_{n+h}\overline{z_n}=e(f(n+h)-f(n)). \end{align*} By the definition of the finite difference $\Delta_h f(n)=f(n+h)-f(n)$, we have \begin{align*} z_{n+h}\overline{z_n}=e(\Delta_h f(n)). \end{align*} Therefore, for each $h$ with $1\le h<H$, \begin{align*} \sum_{M<n\le M+N-h}z_{n+h}\overline{z_n}=\sum_{M<n\le M+N-h}e(\Delta_h f(n)). \end{align*} [guided] We now identify exactly what the correlation term in van der Corput's inequality means for this particular sequence. Fix an integer $h$ satisfying $1\le h<H$ and an integer $n$ satisfying $M<n\le M+N-h$. Since $h\ge 1$, we have $M<n<n+h$, and the upper bound on $n$ gives $n+h\le M+N$. Thus both indices $n$ and $n+h$ lie in the stated domain of $f$. The values $f(n)$ and $f(n+h)$ are real because $f$ is real-valued. By definition of the sequence, \begin{align*} z_{n+h}=e(f(n+h)). \end{align*} Also, since $e(t)=\exp(2\pi i t)$ and $t\in\mathbb R$ implies $\overline{\exp(2\pi i t)}=\exp(-2\pi i t)$, we have \begin{align*} \overline{z_n}=\overline{e(f(n))}=e(-f(n)). \end{align*} Multiplying these two identities gives \begin{align*} z_{n+h}\overline{z_n}=e(f(n+h))e(-f(n)). \end{align*} The exponential notation satisfies $e(a)e(b)=e(a+b)$ for real numbers $a,b$, so the right-hand side is \begin{align*} e(f(n+h)-f(n)). \end{align*} The finite difference in this theorem is oriented as \begin{align*} \Delta_h f(n)=f(n+h)-f(n). \end{align*} Thus the correlation is exactly \begin{align*} z_{n+h}\overline{z_n}=e(\Delta_h f(n)). \end{align*} Summing over all integers $n$ with $M<n\le M+N-h$ preserves this identity term by term: \begin{align*} \sum_{M<n\le M+N-h}z_{n+h}\overline{z_n}=\sum_{M<n\le M+N-h}e(\Delta_h f(n)). \end{align*} [/guided] [/step] [step:Substitute the correlation identity into the van der Corput bound] Replacing each correlation sum in the bound from the first step by the identity obtained in the second step gives \begin{align*} \left|\sum_{M<n\le M+N} e(f(n))\right|^2 \lesssim \frac{N^2}{H}+\frac{N}{H}\sum_{1\le h<H}\left|\sum_{M<n\le M+N-h}e(\Delta_h f(n))\right|. \end{align*} The implicit constant is the absolute constant from [citetheorem:9057]. This is the claimed A-process bound. [/step]