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Let $f$ satisfies $$|f(x+u) - f(x)|\leq L|u|^{\alpha}$$

for some constants $L$ and $\alpha$. If $\alpha = 1$ then $f$ is called Lipschitz continuous, and if $0 < \alpha < 1$ then $f$ is Hölder continuous. Show that if $f$ is $2\pi$-periodic and Lipschitz or Hölder continuous, then its Fourier series converges to $f(x)$ for each $x$.

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2 Answers 2

Let $$ D_N(t)=\frac{1}{2\,\pi}\,\frac{\sin(N+1/2)t}{\sin(t/2)} $$ be the Dirichlet kernel, and let $$ S_N(f;x)=\sum_{k=-N}^N\hat f(k)e^{ikx}=\int_{-\pi}^{\pi}D_N(t)f(x-t)\,dt $$ be the $N$-th partial sum of the Fourier series of $f$. Then, for all $x\in[-\pi,\pi]$ $$ S_N(f,x)-f(x)=\int_{-\pi}^{\pi}D_N(t)(f(x-t)-f(x))\,dt=\int_{-\pi}^{\pi}\sin\frac{(2\,N+1)t}{2}h(t)\,dt $$ where $$ h(t)=\frac{f(x-t)-f(x)}{\sin(t/2)}\;. $$ Since $f$ is Hölder continuous of order $\alpha$, it follows that $|h(t)|\le Ct^{-1+\alpha}$ for some constant $C>0$. In particular, $h$ is integrable on $[-\pi,\pi]$. The proof is finished applying the Riemann-Lebesgue lemma.

It should be noted that your problem is a particular case of Dini's convergence criterion.

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+1 Cute :) ((Misprints: 1. On Line -3 you should make a $Lip(\alpha)$ assumption. 2. That function is not Lipschitz - it is Hölder continuous. 3. On line -3 it is $h$ that is integrable.)) –  AD. Oct 5 '11 at 18:27
    
@AD Than kyou. I have edited my answer. –  Julián Aguirre Oct 6 '11 at 9:39

A classical theorem says that if $f\in Lip(\alpha)$, for some $\alpha$ with $\alpha>1/2$, then the Fourier series of $f$ belongs to the Wiener algebra (that is $\sum|\hat{f}(n)|< \infty$).

The proof can be found in Katznelson's book An Introduction to Harmonic Analysis. See also Hölder Condition, Convergence of Fourier series and Wiener algebra.

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