Consider $f \in L^1(\mathbb R)$ and it's Fourier transform $\hat f : \mathbb R \rightarrow \mathbb C$ defined by$$\hat f(\xi) = \int_{-\infty}^{+\infty} e^{-ix\xi}f(x)dx.$$It is known (Riemann-Lebesgue Lemma) that$$\hat f(\xi) \xrightarrow[\xi \rightarrow \pm \infty]{} 0.$$I would like to strengthen this decay, assuming more properties on $f$. Assume that$$\operatorname{supp} f \subset [-A,+A], \quad f \in L^2(-A,+A),\quad f(-x) = -f(x),$$so that$$\hat f(\xi) = -2i \int_0^A f(x) \sin(x\xi)dx.$$I then wonder whether$$|\hat f(\xi)| \leq \frac{C}{1+|\xi|^{1/2}}.$$Observe that by definition (or as is classical) we have$$f \in H^s(\mathbb R) \Longleftrightarrow (1+|\xi|^2)^{s/2}\hat f(\xi) \in L^2(\mathbb R)$$so that, assuming that $f$ is such that$$|\hat f(\xi)| \leq \frac{C}{1+|\xi|^\alpha},$$one obtains that $f \in H^s(\mathbb R)$ for any $s < \alpha - 1/2$. The choice $\alpha = 1/2$ forces $s < 0$ and prohibits this argument to allow $f$ to be (strictly) more regular than $L^2$.
It is also interesting to have in mind that
- If $f \in BV(\mathbb R) \cap L^1(\mathbb R)$, then $\hat f (\xi) = O(1/\xi)$
- If $f \in L^\infty(\mathbb R)$, it may happen that $\hat f (\xi) \neq O(1/\xi)$, see this post