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The question refers to the following paper by Jean Bourgain: http://arxiv.org/abs/math-ph/0011053

Specifically, I can't derive the following inequality in (1.20):

\begin{equation} \left|\sum_{|k|\geq1}\varphi(k)\left[\frac{1}{R}\sum_{1\leq r\leq R}e^{2\pi i r k\omega}\right]e^{2\pi i k \theta}\right|\leq C\sum_{1\leq|k|\leq K}\frac1{|k|}\frac1{R\|k\omega\|+1}+\left|\sum_{|k|>K}\varphi(k)\left[\frac{1}{R}\sum_{1\leq r\leq R}e^{2\pi i r k\omega}\right]e^{2\pi i k \theta}\right| \end{equation}

where $\varphi(k)=\mathcal O\left(\frac1{|k|}\right)$, $R<K$, $\omega\in\mathbb T$, $k\in\mathbb Z$ and $\|\cdot\|$ denotes the distance to the nearest integer.

What is certainly true is $\sum_{1\leq r\leq R}e^{2\pi i r k\omega}=\frac{\sin(\pi R k\omega)}{\sin(\pi k\omega)}e^{i\pi (R+1)k\omega}$ hence $\left|\sum_{1\leq r\leq R}e^{2\pi i r k\omega}\right|\leq\frac1{2\|k\omega\|}\leq\frac1{\|k\omega\|}$.

Either it's trivial and I don't see it or...well...

Thanks for your help!

share|improve this question
    
The exponential sum is at most the minimum of $\| k\omega\|^{-1}$ and $R$. Plugging this in, the $1/R\| k\omega\| +1$ factor can be thought of as a (constant multiple) of the maximum of $1$ and $1/R\| k\omega\|$, which is what gets from this exponential bound sum. –  Thomas Bloom Aug 22 '13 at 10:51
    
sorry, I don't get it...consider the following: \begin{align} \left|\sum_{1\leq r\leq R}e^{2\pi i r k\omega}\right|&\leq\min(R,\|k\omega\|^{-1})\\ &=\frac1{\max\left(R^{-1},\|k\omega\|\right)} \\ &\geq\frac1{R^{-1}+\|k\omega\|} \end{align} where it was used that $R$ and $\|k\omega\|$ are positive... –  Eduard Tetzlaff Aug 22 '13 at 16:08
2  
for positive $a,b$, $\max(a,b)\geq (a+b)/2$. Now reread Thomas' answer, or your reply. 2 is absorbed in C. –  ofer zeitouni Aug 24 '13 at 21:36
    
yeah I also found that answer on friday...just didn't write it! Thanks anyway! –  Eduard Tetzlaff Aug 26 '13 at 13:15

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