Convergence of a Fourier series Let $f$ be the $2\pi$ periodic function which is the even extension of $$x^{1/n}, 0 \le x \le \pi,$$ where $n \ge 2$. 
I am looking for a general theorem that implies that the Fourier series of $f$ converges to $f$, pointwise, uniformly or absolutely. 
 A: I found the following theorems  from the book "Introduction to classical real analysis" by Karl R. Stromberg, 1981.


*

*(Zygmund) If $f$ satisfies a Hölder (also called Lipschitz) condition of order $\alpha\gt 0$ and $f$ is of bounded variation on $[0,2\pi]$, then the Fourier series of $f$ converges absolutely (and hence uniformly).  p. 521.
This applies to the example in my  question.

*If $f$ is absolutely continuous, then the Fourier series of $f$ converges uniformly  but not necessarily absolutely.  p. 519 Exercise 6(d) and p.520 Exercise 7c.

*(Bernstein) If $f$ satisfies a Holder condition of order $\alpha\gt 1/2$ , then the  Fourier series of $f$ converges absolutely (and hence uniformly).  p.520 Exercise  8 (f)

*(Hille) For each $0<\alpha\le 1/2$, there exists a function that satisfies a Holder condition of order $\alpha$ whose Fourier series converges uniformly, but not absolutely.  p.520 Exercise  8 (f)
A: Perhaps you can apply the one found here: http://books.google.com/books?id=XqqNDQeLfAkC&pg=PA84
Snapshot:

A: You can actually do your example pretty explicitly. For $f(x) = |x|^{\alpha}$ you actually have that $\hat{f}(n) = c_{\alpha} n^{-1 - \alpha}  + O(n^{-2})$, so the cosine series is absolutely convergent. If a Fourier series of a continuous function converges, it has to converge to the original function; I refer you to a Fourier analysis text for this fact though.
To get the above expression for $\hat{f}(n)$, recall that $\hat{f}(n)$ is defined as
$\hat{f}(n) = 2\int_0^1x^{\alpha} \cos(n\pi x)dx$. Integrating by parts, this is the same as ${2 \alpha \over \pi}n^{-1}\int_0^1x^{\alpha - 1} \sin(n\pi x)dx$. This in turn can be written as ${2 \alpha \over \pi}n^{-1} \int_0^\infty x^{\alpha - 1} \sin(n\pi x)dx - 
{2 \alpha \over \pi}n^{-1} \int_1^\infty x^{\alpha - 1} \sin(n\pi x)dx$. (These integrals are convergent improper integrals). 
By changing variables $x$ to $nx$ the first integral becomes the main term $c_{\alpha}n^{-1 - \alpha}$. By doing one more integration by parts and then taking absolute values, the second term is bounded by $C n^{-2}$ as needed.
A: I looked it up on Wikipedia. Assuming that the articles are correct, it seems that your function satisfies some Hölder condition and thus by the Dirichlet-Dini chriterion its Fourier series converges pointwise to f.
You could also show that your function is Lipschitz and then you'd have absolute convergence of its Fourier series.
