$L^2$ bound and Sobolev spaces

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

• Do you really mean to put weights in the physical space instead of the frequency space? Then no such bound is possible. – Fan Zheng Aug 23 at 11:51
• Does the following help? en.wikipedia.org/wiki/… (It is stated for Fourier series of functions on the circle, but my memory is that natural modifications work for the Fourier transform of a function on the line) The idea is that if you have suitable decay of the Fourier transform at infinity, as witnessed by an appropriate Sobolev norm, then you get the desired tail estimates needed to apply Bernstein's theorem(s) – Yemon Choi Aug 23 at 12:53
• @YemonChoi I am not entirely sure I understand the connection between your comment and the OP's question fully to be honest. Should this link imply that the answer is yes or no? – Sascha Aug 23 at 13:07
• @Sascha Sorry, I did not notice the condition $\beta<1$. Therefore my comment does not help, at least not in any direct way – Yemon Choi Aug 23 at 13:16

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 $$h^{-\alpha}|e^{ith}-1|\lesssim |t|h^{1-\alpha}\le |t|^{\alpha}$$.