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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? –  TCL Jan 2 '11 at 15:37
    
@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. –  baikal Jan 2 '11 at 20:25
    
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. –  Did Nov 6 '11 at 9:44
    
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 $g_n$? –  timur Jun 9 '12 at 15:04

1 Answer 1

Let $F_n(x)=nx$. We can modify the values of $F_n$ at finitely many points so it satisfies the maximum and minimum conditions in the post. Then

$$\int_0^1 F_n(x)e^{-i2\pi vx} dx=\int_0^1 nx e^{-i2\pi vx} dx=\frac{i}{2\pi}\frac{n}{v}$$

and clearly the limit that you are interested in does not exist.

EDIT. One can also modify $F_n(x)$ at those finitely many points so that $F_n(x)$ is continuous for all $n$.

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If $n$ and $v$ goes to infinity at the same rate (which I should include in the conditions), that limit exists. –  baikal Jan 4 '11 at 20:58
    
No. Say $n/v\to 1$, then I can change $F_n(x)=p_n x$ where $p_n$ is a sequence such that $p_n/v$ has no limit. –  TCL Jan 4 '11 at 21:44
    
Thanks TCL, I changed the condition. –  baikal Jan 9 '11 at 16:48

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