Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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

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 $n$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.