<s>[INST] If $|f(x)-f(y)| \leq A |x-y|^\alpha$, then $|\hat{f}(n) | \leq C(1+|n|)^{-\alpha}$ Let $f$ be a $2\pi$ periodic continuous function and let
\begin{align}
\hat{f}(n) = \frac{1}{2\pi} \int_0^{2\pi} e^{-inx} f(x) dx
\end{align}
I want to show if
\begin{align}
|f(x)-f(y)| \leq A |x-y|^\alpha, \quad0< \alpha \leq 1
\end{align}
then  $|\hat{f}(n) | \leq C(1+|n|)^{-\alpha}$ for some $C>0$.

What I know is following
\begin{align}
  |f(x) - f(y)| \leq A |x-y|^\alpha \quad \Rightarrow \quad |\frac{f(x)}{x}| \leq A |x|^{\alpha-1}. 
\end{align}
And from MVT, $\exists c \in [0,x]$ such that $|f'(c)| \leq A |x|^{\alpha-1}$.  And from the range of $\alpha$, if $\alpha=1$, $|f'(c)| \leq 1$, and for other case $0<\alpha<1$, $f'(c)=0$.
The hints from the textbook say  $\hat{f}(n) = -\frac{1}{2\pi} \int_{0}^{2\pi} e^{-inx} f(x-\frac{\pi}{n}) dx$)
But I have
\begin{align}
  \hat{f}(n) &= \frac{1}{2\pi} \int_0^{2\pi} e^{-inx} f(x) dx  \\
  & = \frac{1}{2\pi} \int_{-\frac{\pi}{n}}^{2\pi - \frac{\pi}{n}} e^{-in(x-\frac{\pi}{n})} f(x-\frac{\pi}{n}) dx \\
&
  = - \frac{1}{2\pi} \int_{-\frac{\pi}{n}}^{2\pi - \frac{\pi}{n}} e^{-inx} f(x-\frac{\pi}{n})dx
\end{align}
which is slightly different.

Assuming hint is correct, even so I have no idea of showing $|\hat{f}(n) | \leq C(1+|n|)^{-\alpha}$

 [/INST] Comment from @Medo
\begin{align}
  |\hat{f}(n) - \hat{f}(1)| &= \frac{1}{2\pi} \left| \int_0^{2 \pi} (e^{-inx} f(x) - e^{-ix} f(x) ) dx \right|    \\
  &
  = \frac{1}{2\pi} \left| \int_0^{2\pi} \left( e^{-inx} f(x) - f(x+\frac{1}{n} ) - f(x+\frac{1}{n}) - e^{-ix} f(x) \right) dx \right|   \\
  & \leq \frac{1}{2\pi} \int_0^{2 \pi} \left( |f(x) - f(x+\frac{1}{n}) | + |f(x+\frac{1}{n}) - f(x)| \right) dx   
\end{align}
Then in the last step using Holder continuity condition, I can see $|\hat{f}(n) - \hat{f}(1)| \leq C (1+ |n|^{-\alpha} )\leq C (1+|n|)^{-\alpha}$.
</s>