I am trying to solve the following exercise of Stein's book on Fourier Analysis:
Suppose $f$ is continuous of moderate decrese on $R$ whose fourier transform $\hat f$ is continuous and satisfies $ \hat f(\xi) = O(1/|\xi|^{1+\alpha})$ as $|\xi |\rightarrow \infty, 0<\alpha<1.$ Prove that $|f(x+h)-f(x)|\leq M|h|^\alpha $ for all $x,h\in R$.
The hint given is to express $f(x+h)-f(x)$ using the inversion formula and then estimate the integral for $\xi$ ranging $|\xi|\leq 1/|h|$ and $|\xi|\geq 1/|h|.$
I guess that the condition $\hat f(\xi) = O(1/|\xi|^{1+\alpha})$ will guarantee that $\hat f$ is also of moderate decrease, so then I can freely use the inversion formula. But I actually cannot prove that. I'll show what I've done:
Since $ \hat f(\xi) = O(1/|\xi|^{1+\alpha})$, there exists $\xi_0$ such that $|\xi|\geq \xi_0$ implies $|f(\xi)|\leq C/|\xi|^{1+\alpha} \leq C$ ( this last inequality only if $\xi>1).$ Then $|f(\xi)||\xi|^{1+\alpha}\leq C$ and adding $|f(\xi)|$ to this inequality, holds $$|f(\xi)|(1+|\xi|^{1+\alpha})\leq C+C, \mbox{if we suppose $\xi>1$}.$$
This actually proves that $\hat f$ is of moderate decrease, for $|\xi|>\xi_0$ and provided that $\xi_0>1$. Now, I cannot prove that for $|\xi_0|\leq 1$ and not also I could prove that $\hat f$ is of moderate decrease on the finite interval $[-\xi_0,\xi_0].$
Any help on how to continue?
Just a simple step is missing.
Let $\xi_0$ and $C$ be such that $|\hat{f}(\xi)|\leq C/|\xi|^{1+\alpha}$ when $|\xi|\geq \xi_0$, and let $$\color{blue}{\xi_1=\max\{\xi_0,1\}},\quad g(\xi)=|\hat{f}(\xi)|(1+|\xi|^{1+\alpha}),\quad \color{blue}{M=\sup_{|\xi|\leq\xi_1}g(\xi)}$$ ($M$ exists because $g(\xi)$ is continuous).
Now $|\xi|\geq\xi_1$ implies $g(\xi)\leq 2|\hat{f}(\xi)||\xi|^{1+\alpha}\leq 2C$ (like you did).
Combined, we have $g(\xi)\leq\max\{M,2C\}$ for any $\xi\in\Bbb{R}$.