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

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.

share|cite|improve this question
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? – Matt E Aug 12 '10 at 14:13
@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. – anonymous Aug 12 '10 at 16:46
@Hi all : I have posted one example below which is due to Fejer. – anonymous Aug 13 '10 at 14:16
Hmm. Where is that Wikipedian protester when you need him... – SamB Oct 21 '10 at 0:27
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. – paul garrett Jun 25 '11 at 18:15

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.

share|cite|improve this answer
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. – anonymous Aug 12 '10 at 12:02
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. – anonymous Aug 12 '10 at 12:12
Why did someone vote down this answer (cancelling my upvote)? – Tom Boardman Aug 12 '10 at 12:24
@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.) – Akhil Mathew Aug 12 '10 at 17:16
What does this have to do with the question? – SamB Oct 21 '10 at 0:30
up vote 0 down vote accepted

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 $x=2k\pi, \ k\in \mathbb{Z}$.

share|cite|improve this answer

Your Answer


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