Absolute convergence of Fourier series of a Hölder continuous function

Solution 1:

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 $$c_n=\frac{1}{2\pi}\int_{0}^{2\pi} f_{\pi/n}(y)e^{-iny+\pi i}\,dy\tag2$$ 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.