Show that if $|\hat{f}(n)|\le\frac{C}{|n|^{1+\alpha}}(n\ne0),\alpha\in(0,1)$ and $f\in C(T)$

Question

Show that if $|\hat{f}(n)|\le\frac{C}{|n|^{1+\alpha}}(n\ne0),\alpha\in(0,1)$ and $f\in C(T)$ then $f$ is holder continuous of order $\alpha$ on $T$.

Answer 1

Writing $\hat f(n)=c_n$ for convenience:

First, since $\sum|c_n|<\infty$, $f$ is the sum of its Fourier series: $$f(t)=\sum c_ne^{int}.$$

We need to show there exists $c$ so $$|f(t+h)-f(t)|\le c|h|^\alpha.$$Since $t=(t-h)+h$ we can assume that $h>0$. If $h\ge1$ then $$|f(t+h)-f(t)|\le2||f||_\infty\le 2|||f||_\infty h^\alpha,$$so in fact we can assume $$0<h<1.$$

Choose an integer $N\ge0$ with $$2^{-(N+1)}<h\le 2^{-N}.$$Write $$f(t+h)-f(t)=\sum_{-2^N}^{2^N}c_n(e^{in(t+h)}-e^{int})+\sum_{|n|>2^N}c_n(e^{in(t+h)}-e^{int}) =I+II.$$Now $$II\le 4\sum_{n=2^N}^\infty n^{-1-\alpha}=c(2^N)^{-\alpha}\le c h^\alpha.$$For $I$ we use the fact that $|e^{it}-e^{is}|\le|t-s|$: since the $n=0$ term vanishes, $$I\le c\sum_{n=1}^{2^N}n^{-1-\alpha}(nh)=ch\sum_{n=1}^{2^N}n^{-\alpha}\le ch(2^N)^{1-\alpha}\le ch h^{\alpha-1}=ch^\alpha.$$

I left a few details to you. The answer to various questions you might have is "geometric series". Edit: or in one case "compare the sum to an integral".

Exercise: (i) Explain why the above is wrong for $\alpha\le0$. (ii) Similarly for $\alpha\ge1$.