This is an exercise problem. I got stuck here and would like to get a hint. The problem is
Suppose $f$ is continuous and $2\pi$-periodic, and $|\hat{f}(n)|\leq |n|^{-3/2}$ for all non-zero $n\in\mathbb{Z}$. Prove that $f$ satisfies for any $x,y$:
$$|f(x)-f(y)|\leq 100|x-y|^{1/2}$$
I'm trying to proceed in the following way: since the Fourier coefficients are absolutely summable, I will expand $f$ with the Fourier series:
$$
\begin{aligned}
|f(x) – f(y) | &= \left| \lim_{N\rightarrow \infty} (S_Nf(x) – S_Nf(y)) \right| \\
&\leq \lim_{N\rightarrow \infty} \sum_{n=-N}^N \left|\hat{f}(n)\right|\left|e^{inx} – e^{iny}\right| \\
&\leq \lim_{N\rightarrow \infty} \sum_{n=-N,n\neq 0}^N |n|^{-3/2} \left|e^{inx} – e^{iny}\right| \\
\end{aligned}
$$
Next I want to get some Holder continuity property for $e^{inx}$ in order to proceed:
$$
\begin{aligned}
\sup_{x,y}\frac{\left| e^{inx} – e^{iny}\right|}{|x-y|^{1/2}} &=
\sup_{x,y}\left( \frac{\left| e^{inx} – e^{iny}\right|}{|x-y|}
\right)^{1/2} \left| e^{inx} – e^{iny}\right|^{1/2} \\
&\leq \sup_{x,y}\left( \sup_{\xi\in[x,y]} \left|ine^{in\xi}\right|\right)^{1/2}
\sqrt{2} \\
&\leq \sqrt{2n}
\end{aligned}
$$
However, if I simply plug into the original inequality, I will get an infinite sum over $1/n$, which does not converge. I think I either need to get better Holder continuity estimation for $e^{inx}$ or need to proceed in a completely different way. But I'm currently stuck here. Any help or hint would be appreciated! Thank you very much!
Best Answer
We have: $$\left|\frac{e^{nix}-e^{niy}}{x-y}\right|\leq\min\left(\frac{2}{|x-y|},2n\right)$$ hence: $$\begin{eqnarray*}\sum_{n=1}^{N}n^{-3/2}|e^{nix}-e^{niy}|&\leq& |x-y\,|\!\!\sum_{1\leq n\leq\frac{1}{|x-y|}}\frac{2}{\sqrt{n}}+2\sum_{n\geq\frac{1}{|x-y|}}\frac{1}{n\sqrt{n}}\\&\leq&4\sqrt{|x-y|}+6\sqrt{|x-y|}\end{eqnarray*}$$ and $$|f(x)-f(y)|\leq 20\sqrt{|x-y|}.$$