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Can one think of an example of a continuous $2\pi$ periodic function whose Fourier series fails to converge on $\mathbb{R}$.

I referred this in the wikipedia page but no avail: It might be interesting to note that Jean-Pierre Kahane and Yitzhak Katznelson proved that for any given set E of measure zero, there exists a continuous function ƒ such that the Fourier series of ƒ fails to converge on any point of E.

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    Dear Chandru1, do you mean that it fails to converge at *every* point (in which case the answer is no, as Akhil Mathew points out; it will automatically converge at almost all points), or at *some* point? In the latter case examples exist, as you point out in your answer. Is it that you want an explicit example?2010-08-12
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    @Matt E: Yes, i am in need of an explicit example. In one of searches which i found they say Fejer has given one example, but i am unable to locate it.2010-08-12
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    @Hi all : I have posted one example below which is due to Fejer.2010-08-13
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    Hmm. Where is that *Wikipedian protester* when you need him...2010-10-21
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    It might be worth noting that pointwise convergence may not be what one wants: sometimes/often implicitly one wants/needs _uniform_ pointwise convergence, since, otherwise, the Fourier series of a continuous function isn't converging in $C^o$ to the continuous function.2011-06-25
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    can anyone tell me the example of a continuous function whose fourier series diverges?2011-10-12
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    @preeti Please use answers only for answers to the question - not to pose other questions on related topics. Instead, post a new question, linking to related questions when need be.2012-08-16

2 Answers 2

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All, please see this example.

Let $G_{n}$ denote the grouping of this $2n$ numbers, $$\frac{1}{2n-1},\frac{1}{2n-3},...,\frac{1}{3},1,-1,-\frac{1}{3},\cdots,-\frac{1}{2n-1}$$

We take a strictly increasing sequence of positive integers ${\lambda_n}$ and consider the groups $G_{\lambda_1},G_{\lambda_2},\cdots,$. We multiply each number of the group $G_{\lambda_n}$ by $n^{-2}$ and obtain the sequence $$\frac{1}{1^{2}(2\lambda_{1}-1)}, \cdots,-\frac{1}{1^{2}(2\lambda_{1}-1)}, \frac{1}{2^{2}(2\lambda_{2}-1)},...,-\frac{1}{2^2(2\lambda_{2}-1)},....,$$

say $\alpha_{1},\alpha_{2},\cdots$. Our aim is to show that $$\sum\limits_{n=1}^{\infty} \alpha_{n} \cos{nx}$$ is the fourier series of a continuous function. We group the terms in the following way $$\sum\limits_{n=1}^{2\lambda_{1}} \alpha_{n} \cos{nx} + \sum\limits_{n=2\lambda_{1}+1}^{2\lambda_{1}+2\lambda_{2}} \alpha_{n}\cos{nx} + \sum\limits_{n=2\lambda_{1}+2\lambda_{2}+1}^{2\lambda_{1}+2\lambda_{2}+2\lambda_{3}} \alpha_{n} \cos{nx}\cdots$$

The last series can be written as $$\sum\limits_{n=1}^{2\lambda_{1}} \alpha_{n} \cos{nx} + \sum\limits_{n=2}^{\infty} \frac{\phi(\lambda_{n},2\lambda_{1}+2\lambda_{2} + \cdots + 2\lambda_{n-1},x)}{n^2}$$

where $$\phi(n,r,x)= \frac{\cos{(r+1)x}}{2n-1} + \frac{\cos{(r+2)x}}{2n-3} + \cdots + \frac{\cos{(r+n)x}}{1} - \frac{\cos{(r+n+1)x}}{1} - \cdots - \frac{\cos{(r+2n)x}}{2n-1}$$

Now one can show that there is a constant $M$ (independent of $n,r$ and $x$) such that $|\phi(n,r,x)|\leq M$. From this it follows that the grouped series $$\sum\limits_{n=1}^{2\lambda_{1}} \alpha_{n} \cos{nx} + \sum\limits_{n=2}^{\infty} \frac{\phi(\lambda_{n},2\lambda_{1}+2\lambda_{2} + \cdots + 2\lambda_{n-1},x)}{n^2}$$ converges absolutely on $\mathbb{R}$, say to $f(x)$, and $f$ is continuous on $\mathbb{R}$. It is also easy to check that $$f(x) \sim \sum\limits_{n=1}^{\infty} \alpha_{n} \cos{nx}$$

We shall finally show that ${\lambda_n}$ can be chose so that the above series diverges at zero, that is $S_{n} = \alpha_{1} + \alpha_{2} + \cdots + \alpha_{n}$ diverges to infinity.

Since $$S_{2\lambda_{1}+2\lambda_{2}+ \cdots + 2 \lambda_{n-1} + \lambda_{n}} = \frac{1}{n^2} \Bigl( \frac{1}{2\lambda_{n}-1} + \cdots + \frac{1}{3} + 1 \Bigr)$$ behaves as $\frac{\ln{\lambda_{n}}}{{2n^{2}}}$ as $n \to \infty$, it is enough to take $\lambda_{n}=n^{n^2}$. Then the fourier series does not converge to $f$ at $x=2k\pi, \ k\in \mathbb{Z}$.

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No. Any continuous function is in $L^2$. The Carleson-Hunt theorem states that an $L^2$ function's Fourier series converges almost everywhere to the function.

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    Hey thanks for the reply. But, lets wait and see for some more time. My innate feeling is that your answer might be wrong, although i request you to give me some time to think.2010-08-12
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    Did you have a look at "comparing the reciprocals" problem. I worked on it for quite some time, and did get anything. It would better that someone discusses about that problem, which still doesn't have any comments and answers.2010-08-12
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    Why did someone vote down this answer (cancelling my upvote)?2010-08-12
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    @Chandru: I hadn't, but I can't see how that's relevant here; it's a known result in harmonic analysis. (It was a conjecture for continuous functions posed by Lusin.)2010-08-12
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    What does this have to do with the question?2010-10-21
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    If the function fails to converge on a set os measure zero, it can still converge almost everywhere. "Almost everywhere" means the complemet of a measure zero set (which is E in the questions)2016-09-30