Let $f \in L^2(\mathbb R)$ be a function such that 


$$\vert f \vert\_{\alpha}:=\sup\_{h>0}h^{-\alpha}\Vert f(\bullet+h)-f \Vert\_{L^2}< \infty$$


for some $\alpha \in (0,1).$


I would like to know whether there exists $\beta \in (0,1)$ such that $\vert f \vert\_{\alpha}$ satisfies for some constant $C\_{\alpha,\beta}>0$


$$ \vert f \vert\_{\alpha} \le C\_{\alpha,\beta} \left\lVert \langle \bullet \rangle^{\beta} \widehat{f} \right\Vert\_{L^2}$$


for all $f$ for which the right-hand side is finite.


where $\langle x \rangle:=\sqrt{1+\vert x \vert^2}.$



$L^2$ bound and Sobolev spaces
This works for $\beta\ge\alpha$. In terms of the Fourier transform $g=\widehat{f}$, what you're trying to establish becomes
$$
h^{-2\alpha}\int |g(t)|^2 \left| e^{ith}-1\right|^2 \, dt \lesssim
\int |g(t)|^2 (1+t^2)^{\beta}\, dt .
$$
We can see that this holds by considering separately $|t|\ge 1/h$ and $|t|<1/h$ in the first integral. In the first contribution, estimate $|e^{ith}-1|\le 2$, and in the second one, use $h^{-\alpha}|e^{ith}-1|\lesssim |t|h^{1-\alpha}\le |t|^{\alpha}$.


