Let $f:[−\pi, \pi] \to \mathbb R$ be continuous with $f(−\pi) = f(\pi)$ extend $f$ to $\mathbb R$ by periodicity conditions. Assume that for some $\alpha > \frac{1}{2}$ we have
$$\displaystyle\int_{-\pi}^{\pi}\big|\frac{f(x+h) - f(x)}{h^\alpha}\big|^2 < \infty.$$ Show that $\displaystyle\sum_{n = -\infty}^{\infty} |\hat f(n)| < \infty$
Define $ g_h(x)= f(x+h) -f(x), ~ x \in [-\pi, \pi] $ and $ h \in \mathbb R $. Observe that $$ \hat g_h(n) = (e^{inh}-1)\hat f(n), ~~~h \in \mathbb R ~~\mbox{and}~~|\hat g_h(n)|= 2|\sin \dfrac{nh}{2}||\hat f(n)| $$ By hypothesis, there exists $ M\geq 0$ such that
$$ \displaystyle\int_{-\pi}^{\pi} |g_h(x)|^2 \,dx = M h^{2 \alpha}. $$ Now Parseval's identity yields $$ 2 \pi \displaystyle\sum_{-\infty}^{\infty} |\hat g_h(n)|^2 = M h^{2 \alpha}. $$ Then $$ 2 \pi \displaystyle\sum_{-\infty}^{\infty} 4|\sin \dfrac{nh}{2}|^2|\hat f(n)|^2 = M h^{2 \alpha}. $$ Choose $ h= \dfrac{\pi}{2^k} $ for arbitrary but fix $ k \in \mathbb N $ and $ 2^{k-1} \leq |n| < 2^k $ which implies $ \dfrac{\pi}{4} \leq |n|\dfrac{h}{2} < \dfrac{\pi}{2}$. Then $ |\sin (\dfrac{|n|h}{2})|^2 = |\sin (\dfrac{nh}{2})|^2 \geq \dfrac{1}{2}$ and [ $\displaystyle\sum_{-\infty}^{\infty} |\hat f(n)|^2 \leq \dfrac{M}{4\pi} (\dfrac{\pi}{2^k})^{2 \alpha}$. ] By Cauchy Schwarz inequality, \begin{align*} \displaystyle\sum_{2^{k-1} \leq |n| < 2^k} |\hat f(n)| &\leq (\displaystyle\sum_{2^{k-1} \leq |n| < 2^k} 1^2)^\frac{1}{2} (\displaystyle\sum_{2^{k-1} \leq |n| < 2^k} |\hat f(n)|^2)^\frac{1}{2} \\ & \leq (2^k - 2^{k-1} )^\frac{1}{2} \big[\dfrac{M}{4\pi} (\dfrac{\pi}{2^k})^{2 \alpha}\big]^\frac{1}{2}\\ &= \dfrac{\sqrt{M}\pi^{\alpha-\frac{1}{2}}}{2^\frac{3}{2}}\dfrac{1}{2^{(\alpha-\frac{1}{2})k}} \end{align*} By taking summation over $ k \in \mathbb N $ it follows that (in addition use the fact that $ |\hat f(0)|^2 < \infty \Rightarrow |\hat f(0)| < \infty$), $$ \displaystyle\sum_{n = -\infty}^{\infty} |\hat f(n)| < \infty$$