Example of a function whose Fourier Series fails to converge at One point 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.
 A: 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 $x=2k\pi, \ k\in \mathbb{Z}$.
A: 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.
