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.

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Perhaps you can apply the one found here: http://books.google.com/books?id=XqqNDQeLfAkC&pg=PA84

Snapshot:

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Does it follow then that if $f$ is absolutely continuous, then the Fourier series converges {\italic uniformly} to $f$? I doubt it. (Not quite sure if that word "integrable" is in Lebesgue sense.) –  TCL Nov 18 '10 at 1:45
@TCL: I am not really sure. –  Aryabhata Nov 18 '10 at 2:10
In this book, they mean Riemann Integration. –  Jesse Madnick Nov 18 '10 at 8:24
@TCL: That's an interesting question. Well, if $f$ is absolutely continuous, then it has a derivative $f'(x)$ almost everywhere. Furthermore, $f(x) - f(a) = \int_a^x f'(t)\,dt$ (using the Lebesgue Integral). Now as to whether $f'(x)$ is absolutely Riemann integrable... I'm still not sure (there's probably something small that I'm missing.) –  Jesse Madnick Nov 18 '10 at 8:34
In view of the fact that the Fourier series of every absolutely continuous function converges uniformly (to the function) (see my answer below), I am sure the "integrable" mentioned above is in Lebesgue sense. –  TCL Nov 18 '10 at 15:14

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.

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I found the following theorems from the book "Introduction to classical real analysis" by Karl R. Stromberg, 1981.

1. (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.

2. 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.

3. (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)

4. (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)

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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.

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The Holder condition is only valid for $0 \le x \le 1/n$ (if the wiki page is correct). In any case, that is the only problematic region,I suppose. So, +1. –  Aryabhata Nov 18 '10 at 1:21