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Let $f$ be a $C^2$ real-valued function on the interval $[a,b]$. Suppose that $f'(x)$ is monotone on $[a,b]$ and there is $\lambda>0$ such that $$ \min_{x\in [a,b]} |f'(x)|>\lambda $$

It is not hard to see that there exists $C>0$ such that $$ \left|\int_a^b e^{2\pi i f(x)} dx \right|< \frac{C}{\lambda}. $$ The proof could be found, for instance, in the book uniform distribution of sequences Lemma 2.1.

I want to know whether we can find some upper bound respect to the length of $[a,b]$. More precisely, for the interval $[a,b]$ small enough (for example, small enough such that $|f(a)-f(b)|<1$), I wonder whether there exists a constant $C>0$ which is independent of $a, b$ and $f$ such that $$ \left|\int_a^b e^{2\pi i f(x)} dx \right|< \frac{C|a-b|}{\lambda}. $$

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    $\begingroup$ This can't work because the LHS is $\simeq b-a$ when you send $b-a\to 0$, no matter how large $f'$ is. $\endgroup$ Commented Jul 21, 2018 at 1:55
  • $\begingroup$ An obvious upper bound is $|a-b|$. As Christian Remling pointed out, this cannot be improved asymptotically for $b-a$ very small. $\endgroup$
    – GH from MO
    Commented Jul 21, 2018 at 2:06
  • $\begingroup$ You may be interested in Chapter 8 of Stein's Harmonic Analysis or these notes by Tao math.ucla.edu/~tao/247b.1.07w/notes8.pdf $\endgroup$ Commented Jul 21, 2018 at 2:52
  • $\begingroup$ Don’t you get that \int_a^b e(f(x)) dx = e(f(b))/[2\pi i f’(b)] - e(f(a))/[2\pi i f’(a)] + O(|b-a| \sup_x |f’’(x)| / |f’(x)|^2)? I’ve just written the integrand as [2\pi i f’(x) e(f(x))] / [2\pi i f’(x)] and integrated by parts (forgive me if this is nonsense! It’s late and I’m writing from my phone.). Dunno if this is useful to you, but at least it shows you the boundary term (basically) preventing you from having the desired bound. $\endgroup$
    – alpoge
    Commented Jul 21, 2018 at 3:00
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    $\begingroup$ @alpoge, plus formatting: $$\int_a^be\bigl(f(x)\bigr)\,dx={e\bigl(f(b)\bigr)\over2\pi if'(b)}-{e\bigl(f(a)\bigr)\over2\pi if'(a)}+O\left(|b-a|\sup_x{|f''(x)|\over|f'(x)|^2}\right)$$ $\endgroup$ Commented Jul 21, 2018 at 4:34

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