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$f(x, y) : (0,1)^2 \rightarrow \mathbb R $, is a continuous function, and its partial derivatives with respect to $x$ and $y$ are all integrable. if $y=\mbox{mod}(n^2 x,1)$, i.e., $y$ is $\frac{1}{n^2}$-periodic over $x$, we can write $$ g_n(x) = f(x, \mbox{mod}(n^2 x,1))$$ and let $g_n'(x)$ be its derivative, as $n, v \in \mathbb N$ goes to infinity simultaneously with the ratio $n/v$ remain a constant, I'm looking for a proof of the convergence of
$$\lim_{n,v \rightarrow \infty} \int_0^1 g_n'(x) e^{-i2\pi v x}\mbox{d} x $$

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  • $\begingroup$ Could you please give a bit more background on where you've run into this question, what you need the answer for, and so on? The way your question is worded ("prove that...") makes it sound like an exercise that's been assigned, but if this isn't the case then giving more context may help people to give more useful answers $\endgroup$
    – Yemon Choi
    Jan 12, 2011 at 19:59
  • $\begingroup$ Yemon, this is a problem I encountered when tried to find out the convergence rate of the Fourier coefficients of $g_n(x)$ as $n$ and $v$ increase simultaneously. $\endgroup$
    – june06
    Jan 12, 2011 at 20:35
  • $\begingroup$ If this isn't homework, it should be... $\endgroup$
    – Igor Rivin
    Jan 12, 2011 at 20:37
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    $\begingroup$ You haven't said anything about proper conditions. You didn't assume $f$ to be differentiable at the start, but you later talk about $g_n'$. As Yemon says, it's also very unclear what your limiting process is. I think your question is almost meaningless as stated, and needs to be far more precise. $\endgroup$
    – Zen Harper
    Jan 14, 2011 at 9:04
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    $\begingroup$ Hi, Yemon, Zen and Nick, thank you all for helping me to clarify the question. I've made some effort to improve the question. $\endgroup$
    – june06
    Jan 14, 2011 at 15:05

1 Answer 1

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If we take $f(x,y)=2\sqrt x$, so that $g_n'(x)=1/\sqrt x$, then your integrals do not exist at all.

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  • $\begingroup$ No, $(1-e^{-2\pi iv})=0$ as $v \in \mathbb N \/$. It is converged. $\endgroup$
    – june06
    Jan 13, 2011 at 18:06
  • $\begingroup$ Wadim, thanks for your remind. So I exclude 0 from the domain. $\endgroup$
    – june06
    Jan 14, 2011 at 6:37
  • $\begingroup$ Maple recognizes the real and imaginary part of the proposed integral in terms of Fresnel C functions, and tells me that the limit exists as v tends to infinity. Knowing nothing about Fresnel C functions, I wouldn't bet on Maple's correctness. I do agree with the others that this problem should be stated more precisely. $\endgroup$ Jan 14, 2011 at 13:24

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