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This is a stronger one related to the question Convergence of $\lim_{n,v \rightarrow \infty} \int_0^1 f_n (x) e^{-i2\pi v x} \mbox{d} x $.

$F_n(x) : [0,1] \rightarrow \bf R $, for $1 \leq i \leq n$, $F_n(x)= n\cdot g_{n,i}(x)$ if $x \in [\frac{i-1}{n}, \frac{i}{n})$, with $g_{n,i}$ a series of integrable functions. As $n, v \in \bf N$ goes to infinity simultaneously at the same rate, prove the convergence of
$$\lim_{n,v \rightarrow \infty} \int_0^1 F_n(x) e^{-i2\pi v x}\,\mbox{d} x $$ if $v/n$ is not an integer.

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    Consider a simple case. Suppose we let $g_{n,i}(x)=n$ for $\frac{i-1}{n}\le x<\frac{i-\frac{1}{2}}{n}$, and $g_{n,i}(x)=-n$ for $\frac{i-\frac{1}{2}}{n}\le x<\frac{i}{n}$. What would the limit be?2011-01-02
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    @TCL, That simple case has been answered by Nick Kirby in the question [Convergence of $\lim_{n,v \rightarrow \infty} \int_0^1 f_n (x) e^{-i2\pi v x} \mbox{d} x $] linked above.2011-01-02
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    With the hypotheses of the post as written now, $(F_n)$ may be basically **any** sequence of integrable functions, hence it is a logical impossibility to reach any other conclusion.2011-11-06
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    Clarification question: What is the reason for introducing $g_{n,i}$ with 2 indices? Why we cannot piece together $g$'s with the same $n$ and call it $g_n$?2012-06-09

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