[proofplan] We first reduce the real endpoint sum to a finite integer interval and show that the monotonicity and nonresonance of $f'$ force all relevant discrete first differences $f(n+1)-f(n)$ to lie in one nonresonant interval modulo $\mathbb Z$. The core estimate is a discrete Kusmin-Landau lemma: if the phase increments are monotone and stay a distance at least $\lambda$ from the integers, then the corresponding exponential sum has size $O(\lambda^{-1})$. We prove that lemma directly by discrete summation by parts, writing each term as a first difference multiplied by $(e(\delta)-1)^{-1}$ and using the bounded variation of this multiplier along a [monotone sequence](/page/Monotone%20Sequence) of increments. Applying the lemma to the sequence $\theta_n=f(n)$ gives the desired estimate. [/proofplan]

[step:Reduce the sum to consecutive integer indices] Let \begin{align*} I:=\{n\in\mathbb Z:a<n\le b\}. \end{align*} If $I=\varnothing$, then the asserted estimate is immediate. Otherwise there exist integers $M,N\in\mathbb Z$ with $M\le N$ such that \begin{align*} I=\{M,M+1,\dots,N\}. \end{align*} Define the finite sequence $\theta_M,\dots,\theta_N\in\mathbb R$ by \begin{align*} \theta_n:=f(n)\qquad(M\le n\le N). \end{align*} It remains to prove \begin{align*} \left|\sum_{n=M}^{N}e(\theta_n)\right|\lesssim \lambda^{-1}. \end{align*} [/step]

[step:Place all first differences in one nonresonant unit interval]If $M=N$, then \begin{align*} \left|\sum_{n=M}^{N}e(\theta_n)\right|=1\le \lambda^{-1}, \end{align*} because $0<\lambda\le 1/2$. Hence assume $M<N$. For $M\le n<N$, define \begin{align*} \Delta_n:=\theta_{n+1}-\theta_n=f(n+1)-f(n). \end{align*} Since $f\in C^1([a,b];\mathbb R)$ and $[n,n+1]\subset [a,b]$ for every $M\le n<N$, the [fundamental theorem of calculus](/theorems/632) gives \begin{align*} \Delta_n=\int_n^{n+1}f'(x)\,d\mathcal L^1(x). \end{align*} Because $f'$ is continuous and monotone, its image on $[a,b]$ is an interval in $\mathbb R$. Since $\|f'(x)\|_{\mathbb R/\mathbb Z}\ge\lambda$ for every $x\in[a,b]$, this interval cannot contain an integer. Therefore there is an integer $r\in\mathbb Z$ such that either \begin{align*} f'([a,b])\subset [r+\lambda,r+1-\lambda] \end{align*} or \begin{align*} f'([a,b])\subset [r-1+\lambda,r-\lambda]. \end{align*} Replacing $f$ by $-f$ if necessary only conjugates the exponential sum and preserves its absolute value, so it is enough to treat the first case. In that case each average $\Delta_n$ satisfies \begin{align*} \Delta_n\in [r+\lambda,r+1-\lambda]. \end{align*} Moreover, the sequence $(\Delta_n)_{M\le n<N}$ is monotone. If $f'$ is nondecreasing and $M\le m<n<N$, then $f'(x)\le f'(x+n-m)$ for every $x\in[m,m+1]$, so the change of variables $y=x+n-m$ gives \begin{align*} \Delta_m=\int_m^{m+1}f'(x)\,d\mathcal L^1(x)\le \int_n^{n+1}f'(y)\,d\mathcal L^1(y)=\Delta_n. \end{align*} If $f'$ is nonincreasing, the same comparison gives $f'(x)\ge f'(x+n-m)$ for every $x\in[m,m+1]$, hence $\Delta_m\ge\Delta_n$. Thus $(\Delta_n)_{M\le n<N}$ is either nondecreasing or nonincreasing.[/step]

[guided]The point of this step is to convert the hypothesis on the continuous derivative into the exact discrete hypothesis needed for the exponential sum. The relevant discrete quantities are not the values $f'(n)$, but the increments \begin{align*} \Delta_n:=f(n+1)-f(n)\qquad(M\le n<N). \end{align*} Since $f\in C^1([a,b];\mathbb R)$, the fundamental theorem of calculus applies on every interval $[n,n+1]\subset [a,b]$, and gives \begin{align*} \Delta_n=\int_n^{n+1}f'(x)\,d\mathcal L^1(x). \end{align*} We next use the nonresonance condition. The set $f'([a,b])$ is an interval because $f'$ is continuous and $[a,b]$ is connected. It is also monotone as a parametrized image, but connectedness is what prevents jumping across excluded integer neighborhoods. The condition \begin{align*} \|f'(x)\|_{\mathbb R/\mathbb Z}\ge \lambda \end{align*} means that $f'(x)$ never enters any open interval $(m-\lambda,m+\lambda)$ around an integer $m\in\mathbb Z$. Since $0<\lambda\le 1/2$, the complement of these forbidden intervals is the union of closed intervals \begin{align*} [m+\lambda,m+1-\lambda]\qquad(m\in\mathbb Z). \end{align*} A connected interval contained in this union must lie in one such component. Thus there is an integer $r\in\mathbb Z$ such that, after possibly replacing $f$ by $-f$, we have \begin{align*} f'([a,b])\subset [r+\lambda,r+1-\lambda]. \end{align*} Replacing $f$ by $-f$ changes each term $e(f(n))$ to $e(-f(n))=\overline{e(f(n))}$, so the absolute value of the whole sum is unchanged. Now each $\Delta_n$ is the average of $f'$ over $[n,n+1]$ with respect to $\mathcal L^1$. Since all values of $f'$ on that interval lie in $[r+\lambda,r+1-\lambda]$, the average also lies in the same interval: \begin{align*} \Delta_n\in [r+\lambda,r+1-\lambda]. \end{align*} Finally, we verify monotonicity of the discrete averages explicitly. Suppose first that $f'$ is nondecreasing and let $M\le m<n<N$. For every $x\in[m,m+1]$ we have $x\le x+n-m\in[n,n+1]$, and therefore \begin{align*} f'(x)\le f'(x+n-m). \end{align*} Integrating this pointwise inequality over $[m,m+1]$ with respect to $\mathcal L^1$ and then using the translation change of variables $y=x+n-m$, whose Jacobian is $1$, gives \begin{align*} \Delta_m=\int_m^{m+1}f'(x)\,d\mathcal L^1(x)\le \int_m^{m+1}f'(x+n-m)\,d\mathcal L^1(x)=\int_n^{n+1}f'(y)\,d\mathcal L^1(y)=\Delta_n. \end{align*} Thus $(\Delta_n)_{M\le n<N}$ is nondecreasing in this case. If $f'$ is nonincreasing, then for the same $m,n$ and $x\in[m,m+1]$ we have $f'(x)\ge f'(x+n-m)$, and the identical translation calculation gives $\Delta_m\ge\Delta_n$. Thus $(\Delta_n)_{M\le n<N}$ is nonincreasing in that case. In both cases the first-difference sequence is monotone.[/guided]

[step:Prove the discrete Kusmin-Landau estimate] We prove the following finite sequence estimate. Let $u_0,\dots,u_L\in\mathbb R$ be [real numbers](/page/Real%20Numbers) with $L\ge 0$, and put \begin{align*} \delta_j:=u_{j+1}-u_j\qquad(0\le j<L). \end{align*} Suppose that $(\delta_j)_{0\le j<L}$ is monotone and that, for some integer $r\in\mathbb Z$, \begin{align*} \delta_j\in [r+\lambda,r+1-\lambda]\qquad(0\le j<L). \end{align*} Then \begin{align*} \left|\sum_{j=0}^{L}e(u_j)\right|\le 4\lambda^{-1}. \end{align*} If $L=0$, then $|e(u_0)|=1\le \lambda^{-1}$. Assume $L\ge 1$. Replace $u_j$ by $u_j-rj$. Since $rj\in\mathbb Z$ for $j\in\mathbb Z$, this leaves every term $e(u_j)$ unchanged and reduces the hypotheses to \begin{align*} \delta_j\in[\lambda,1-\lambda]\qquad(0\le j<L). \end{align*} Define $z_j:=e(u_j)$ for $0\le j\le L$ and define $c_j\in\mathbb C$ by \begin{align*} c_j:=\frac{1}{e(\delta_j)-1}\qquad(0\le j<L). \end{align*} Since $\delta_j\in[\lambda,1-\lambda]$, the denominator is nonzero. Also \begin{align*} z_{j+1}-z_j=z_j(e(\delta_j)-1), \end{align*} so \begin{align*} z_j=c_j(z_{j+1}-z_j)\qquad(0\le j<L). \end{align*} Summing this identity and using discrete summation by parts gives \begin{align*} \sum_{j=0}^{L}z_j=z_L+c_{L-1}z_L-c_0z_0-\sum_{j=1}^{L-1}(c_j-c_{j-1})z_j. \end{align*} Therefore \begin{align*} \left|\sum_{j=0}^{L}z_j\right|\le 1+|c_0|+|c_{L-1}|+\sum_{j=1}^{L-1}|c_j-c_{j-1}|. \end{align*} Define the function \begin{align*} c:[\lambda,1-\lambda]&\to\mathbb C \end{align*} by \begin{align*} c(x):=(e(x)-1)^{-1}. \end{align*} The identity \begin{align*} c(x)=-\frac{1}{2}-\frac{i}{2}\cot(\pi x) \end{align*} shows that $c$ has total variation at most $\cot(\pi\lambda)$ on $[\lambda,1-\lambda]$, because $\cot(\pi x)$ is decreasing on that interval. Since $(\delta_j)$ is monotone, the points $\delta_j$ traverse the interval $[\lambda,1-\lambda]$ in one direction, so the polygonal variation $\sum_{j=1}^{L-1}|c(\delta_j)-c(\delta_{j-1})|$ is bounded by the total variation of $c$ on $[\lambda,1-\lambda]$. Moreover \begin{align*} |c_j|=\frac{1}{2|\sin(\pi\delta_j)|}\le \frac{1}{4\lambda}, \end{align*} using $|\sin(\pi x)|\ge 2\|x\|_{\mathbb R/\mathbb Z}$ for $x\in\mathbb R$. Finally $\cot(\pi\lambda)\le (\pi\lambda)^{-1}\le \lambda^{-1}$. Combining these estimates gives \begin{align*} \left|\sum_{j=0}^{L}e(u_j)\right|\le 1+\frac{1}{2\lambda}+\lambda^{-1}\le 4\lambda^{-1}, \end{align*} because $0<\lambda\le 1/2$. This proves the finite sequence estimate. [/step]

[step:Apply the discrete estimate to the phase sequence] Apply the finite sequence estimate from the previous step to \begin{align*} u_j:=\theta_{M+j}=f(M+j)\qquad(0\le j\le N-M). \end{align*} The corresponding first differences are \begin{align*} u_{j+1}-u_j=\Delta_{M+j}, \end{align*} and the preceding step verified that they are monotone and lie in one interval $[r+\lambda,r+1-\lambda]$. Hence \begin{align*} \left|\sum_{n=M}^{N}e(f(n))\right| = \left|\sum_{j=0}^{N-M}e(u_j)\right| \le 4\lambda^{-1}. \end{align*} The constant $4$ is absolute, so this is precisely \begin{align*} \left|\sum_{\substack{n\in\mathbb Z, a<n\le b}} e(f(n))\right|\lesssim \lambda^{-1}. \end{align*} This completes the proof. [/step]