Assume $f(x), x \in \mathbb{R}$ is a function with a compact support such that its Fourier transform $\hat{f}(\xi)$ has a decay rate
$$\hat{f}(\xi) \lesssim \frac{1}{|\xi|^\gamma + 1}$$
for some $\gamma \ge 1$.
Now set $$h(x) = xf(x).$$ Since $f$ has a compact support, $h$ should have similar or better regularity than $f$. Can we now get the following decay estimate of the Fourier transform of $h$ ?
$$\hat{h}(\xi) \lesssim \frac{1}{|\xi|^\gamma + 1}$$
I know now we have $\hat{h}(\xi) = -i\partial\_\xi \hat{f}(\xi)$, but it seems hard to only use this relation to get the decay estimate.



Decay estimate of Fourier transform of a compactly supported function
The answer is positive. Since $f$ has compact support, $g:=\hat{f}$ extends to an
entire function of exponential type $\sigma$ with some $\sigma>0$.
Then your estimate on
the real line and the Phragmen - Lindelof theorem imply that
$$\log |g(z)|\leq \sigma |y|-\gamma\log|z|,\quad z=x+iy,$$
which gives that $|g(z)|=O(|z|^{-\gamma}), z\to\infty$ not only on the real line,
but also in a horizontal strip $|y|<2$. Then Cauchy inequality implies that $|g'|$
also satisfies the same estimate on the real line.


Reference for Phragmen-Lindelof theorem: B. Ya. Levin, Lectures on entire functions, lecture 6.


Added. Instead of the above inequality, one can write a more precise one,
which also follows from Phragmen-Lindelof:
$$\log|g(z)|\leq \sigma|y|+\frac{y}{\pi}\int\_{-\infty}^\infty\frac{\log|g(t)|}{(x-t)^2+y^2}dt.$$
So to obtain an estimate for complex $z$ from the estimate on the real line, one only need to estimate the Poisson integral in the right. This permits to deal with estimates other than $|x|^{-\gamma}$.


