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$f_n(x) : [0,1] \rightarrow \bf R $, and $f_n$ is $\frac{1}{n}$-periodic, $\max f_n(x) = n$, $\min f_n(x) = -n$.

As $n$ and $v$ goes to infinity simultaneously, prove the convergence of
$$\lim_{n,v \rightarrow \infty} \int_0^1 f_n(x) e^{-i2\pi v x}\,dx $$ if $v,n$ are integers and $v/n$ is not an integer.

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Is there a particular order of the limits, or do we have to prove convergence in the two variables simultaneously? – Ross Millikan Dec 30 '10 at 23:15
It is the convergence in the two variables simultaneously. – baikal Dec 31 '10 at 0:35
@user5181. I added the phrase $v,n$ are integers. – TCL Jan 1 '11 at 15:59
up vote 3 down vote accepted

I'll assume you mean to take a limit in some sense as over double sequences indexed by natural numbers $n, \nu$ such that $\frac{\nu}{n}$ is not an integer. For $f_n$ an $\frac{1}{n}$-periodic function such that $\max f_n=-\min f_n =n$, let $g_n$ be defined by $f_n(x)=ng_n(nx)$. Then $g_n$ is $1$-periodic and has max and min equal to 1 and -1. Let $r=\frac{\nu}{n}$. Then you require that $r$ is not an integer. Then writing the integral of interest in terms of $g_n$ and $r$ and $n$, you'll be very tempted to change variables and integrate from 0 to n. Do so.

Then using the periodicity of $g_n$, you'll want to break the integral into a sum of $n$ integrals. Finally, switch the order of summation and integration, be happy that you can look up or re-derive the formula for geometric sums, and be shocked (or not, but I was pleasantly surprised).

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My feeling is, if $v/n$ is not integer, integration of the product $f_n(x)e^{-i2\pi v x}$ cannot be broken into a sum of $n$ integrals. Can you give more hints? – baikal Dec 31 '10 at 0:51
An integral from t=0 to n can always be broken into a sum of n integrals. The only part you have to be careful about is the fact that $e^{-2\pi i r t}$ won't be $1$-periodic. – AppliedSide Dec 31 '10 at 1:45
Yes I see your point, just mis-understood you. – baikal Dec 31 '10 at 1:54

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