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Suppose that $f$ is $2 \pi$ periodic and Hölder continuous of order $\alpha > 1/2$. Show that the Fourier series of $f$ converges absolutely.

So we know that $f(x+2 \pi t) = f(x)$ for all $t \in \mathbb{Z}$. Also there exists a constant $C$ such that $$ |f(x+h)-f(x)| \leq C|h|^{\alpha}$$ for all $x$ and $h$.

Thus we want to show that $\sum_{n=-\infty}^{\infty} |c_{n}e^{inx}| = L$. What are some hints to get started?

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I posted the answer in modular form, so you can decide to read some and not others. –  40 votes Jul 12 '13 at 17:44
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1 Answer

This is a theorem of Bernstein, found, for example, in Katznelson's Introduction to Harmonic Analysis.

A simple-minded approach would be to shift the variable in $$c_n=\frac{1}{2\pi}\int_0^{2\pi} f(x)e^{-inx}\,dx\tag1$$ to $y=x+\pi/n$, thus obtaining $$c_n=\frac{1}{2\pi}\int_{0}^{2\pi} f_{\pi/n}(y-\pi/n)e^{-iny+\pi i}\,dy\tag2$$ where I write $f_h(x)=f(x-h)$. Then add (1) and (2): $$2c_n=\frac{1}{2\pi}\int_0^{2\pi} (f(x)-f_{\pi/n}(x))e^{-inx}\,dx\tag3$$ and weep in despair: the estimate is only $|c_n|=O(n^{-\alpha})$...

The thing is, estimating $|c_n|$ by maximum of integrand in (3) and then summing the estimates is just too crude. Parseval's identity is more precise.

So, let's try Parseval. Since the $n$th Fourier coefficient of $f_h-f$ is $(e^{-inh}-1)\widehat{f}(n)$, we have $$ \sum_n |e^{-inh}-1|^2|\widehat{f}(n)|^2 \le Ch^{2\alpha} \tag4$$

The inequality (4) isn't worth much unless we can bound $ |e^{-inh}-1|$ from below. There is no uniform bound for all $n$, but if we focus on some dyadic range $2^k\le |n|< 2^{k+1}$, then choosing $h=\frac{2\pi}{3}\cdot 2^{-k}$ gives good result: $\frac{2\pi}{3}\le |nh|< \frac{4\pi}{3}$, which keeps $e^{-inh}$ far away from $1$.

So, $$ \sum_{n=2^k}^{2^{k+1}-1} |\widehat{f}(n)|^2 \le \widetilde{C} 2^{-2k\alpha} \tag5$$ and the rest is a mopping-up operation: Cauchy-Schwarz turns (5) into $$ \sum_{n=2^k}^{2^{k+1}-1} |\widehat{f}(n)| \le 2^{k/2}\cdot\widetilde{C} 2^{-k\alpha} \tag6$$ which sums up.

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