Let $a>0$, then produce an $f \in L^1(\Bbb R^n)$ such that $f$ is not identically 0, $f$ has compact support, $\int_{\Bbb R^n}f(x)dx=0$ and $|\hat{f}(\xi)| \le C|\xi|^{-a}, \forall \xi \ne 0 $ and for some $C>0$
My attempt:
For $n=1$, let us consider $f=\chi_{(0,1]}-\chi_{[-1,0]}$ . Then clearly, $f \in L^1(\Bbb R^n)$, has compact support $[-1,1]$ and $\int_{\Bbb R}f(x)dx= \int_{0}^1 dx - \int_{-1}^0 dx=0$.
Now for the Fourier transform,$$\hat{f}(\xi)=\int_{0}^1 e^{-i\xi x}dx-\int_{-1}^0 e^{-i\xi x}dx$$ $$=\int_{0}^1 (\cos(\xi x) -i \sin(\xi x))dx-\int_{-1}^0 (\cos(\xi x) -i \sin(\xi x))dx$$ $$=\Big[\frac{\sin(\xi x)}{\xi}+i\frac{\cos(\xi x)}{\xi}\Big]_{0}^1-\Big[\frac{\sin(\xi x)}{\xi}+i\frac{\cos(\xi x)}{\xi}\Big]_{-1}^0$$ $$=\frac{2i}{\xi}(\cos \xi -1)$$ Thus $$\forall \xi \ne 0, |\hat{f}(\xi)| \le \frac{2}{|\xi|}|1-\cos(\xi)| \le \frac{4}{|\xi|}$$
So in the 1-dimensional case, for $a=1$, I have an explicit function. Was thinking of generalizing this to higher dimensions by looking at suitable radial functions, but haven't done any computation yet. And also what about the general $a>0$ in $\Bbb R^n$ ?
Thanks in advance for help!