Suppose that function $f$ satisfies following properties: $f'' \in \mathrm{C}[a,b]$, $\exists A,B\geq 1$ such that $f''(x)\geq 1/A$ and $|f'(x)|<D$ for all $x\in [a,b]$. Prove that $$\displaystyle \sum\limits_{a\leq x < b}e^{2\pi i f(x)}\ll (f'(b) - f'(a))\sqrt{A} + \sqrt{A} + \ln(D(b-a))$$ where $x$ are integers from $[a,b)$ and $X\ll Y$ means $X = O(Y)$.
The problem is hard, I used some techniques from proofs of lemmas in the book, where this problem is from. Then it all boiled down to considering the following sum: $$\sum\limits_{f'(a)+\Delta\leq n\leq f'(b) - \Delta}I_{n}$$ where $$I_{n} = \displaystyle \int\limits_{[a]+1.5}^{[b]-0.5}e^{2\pi i (f(x) - nx)}dx$$ and $\Delta\in (0,1)$, $f'(b) - f'(a) > 10$. One can also consider only the case when $f'(x) > 0$ (just substituting $f(x)$ with $f(x) - [f'(a)]x$ won't change the sum at the beginning) and $f''\ll 1/A$ since $f''$ is continuous. So we are left to prove that $$\int\limits_{a}^{b}e^{2\pi i(f(x) - nx)}dx \ll \sqrt{A}$$ One can notice that since $n\in (f'(a),f'(b))$ then for every $n$ in the sum there is $c\in (a,b)$ such that $(f(x) - nx)'\Bigg\rvert_{x=c} = 0$, so we can use Taylor formula to obtain $f(x) - nx = f(c) - nc + \frac{f''(\xi)}{2}(x-c)^{2}$. So I was able to get the $\sqrt{A}$, but one integral was still remaining: $$\int\limits_{\pi(x-c)^{2}>A}\sin(\pi (x-c)^{2} f''(\tilde{\xi}))dx$$ (Set $\pi(x-c)^{2}>A$ is taken inside $[a,b]$ of course). And I don't know how to make an estimate for this integral, since I do not know what is the behavior of $f''$ (and $\tilde{\xi}$ is a function of $x$). Can you help?