If $|f(x)-f(y)| \leq A |x-y|^\alpha$, then $|\hat{f}(n) | \leq C(1+|n|)^{-\alpha}$

Question

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 $C>0$.

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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.

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

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Answer 1

Using $e^{i\pi}=-1$,

$$\hat{f}(n) = -\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{i\pi}f(x)e^{-inx}dx=-\frac{1}{2\pi}\int_{-\pi}^{\pi}f(y+\pi/n)e^{-iny}dy$$ hence $$2|\hat{f}(n)|\leq\frac{1}{2\pi}\int_{-\pi}^{\pi}|f(x)-f(x+\pi/n)|dx=O(\frac{1}{|n|^\alpha})~~\textrm{as}~~n\rightarrow \infty.$$

You may need $(1+|n|)^\alpha$ at $n=0.$

Answer 2

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 $|\hat{f}(n) - \hat{f}(1)| \leq C (1+ |n|^{-\alpha} )\leq C (1+|n|)^{-\alpha}$.