Let $a,b\in\mathbb{Z}$, and $f\in C^2([a,b])$ such that $|f''(t)|\asymp \lambda$ for $a\le t\le b$. Prove that 
$$\sum\_{a<n\le b}\{f(n)\}=\frac{1}{2}(b-a)+O(\lambda^{1/3}(b-a)+\lambda^{-1/2}).$$
Where $\{x\}$ is the fractional part of $x$. 


I find this problem in G.Tenenbaum's book in the page 118.


Gérald Tenenbaum. Introduction to Analytic and Probabilistic Number Theory, Cambridge: Cambridge University Press, 1995.


Can you help me with this problem?



Prove that $\sum_{a<n\le b}\{f(n)\}=\frac{1}{2}(b-a)+O(\lambda^{1/3}(b-a)+\lambda^{-1/2})$
Write the function $x\mapsto\{x\}-\frac{1}{2}$ as a Fourier series, and approximate this series by a smoothed finite sum using Vaaler's lemma. You obtain something of the form
$$
\sum\_{a<n\leq b}\left(\{f(n)\}-\frac{1}{2}\right) = \sum\_{a<n\leq b}\sum\_{k\leq K} a\_k e(k f(n)) + R,
$$
where $e(t)=e^{2\pi i t}$, the $a\_k$ are the coefficients of the approximation to the Fourier series, and $R$ is some remainder depending on $K$.


Interchange summation and estimate each some of the form $\sum\_{a<n\leq b} e(k f(n))$ using the Weyl-van der Corput method. Finally, optimize $K$.


This approach is worked out in Graham-Kolesnik "Van der Corput's method of exponential sums" and in Montgomery "Ten lectures on the interface between analytic number theory and harmonic analysis".


Alternatively you can use the Erdös-Turan-inequality together with Weyl-van der Corput to bound the discrepancy of the sequence $\{f(n)\}$, and then use Koksma's inequality. Note that this is not really a different proof, the Fourier part is just hidden within the Erdös-Turan inequality.


