Decay estimate of Fourier transform of a compactly supported function 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.
 A: 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 $|x|^{-\gamma}$.
A: I came up with an answer that is more real analysis. This method works for any $h(x) = m(x)f(x)$ as long as $m(x)$ is smooth and $f(x)$ is compactly supported.
Without loss of generalization, assume the support of $f(x)$ is $[0, 1]$. Set $$g(x) = m(x)\chi(x).$$ Here $\chi(x)$ is a smooth cutoff function such that $\chi(x) = 1$ on $[0,1]$. Hence $g(x)$ is Schwartz and $h(x) = g(x)f(x)$.
We then have
\begin{equation}
\hat{h}(\xi) = \int_{-\infty}^\infty \hat{f}(\xi-\eta)\hat{g}(\eta)d\eta.
\end{equation}
For $|\xi| \gg 1$, we write
\begin{equation}
\begin{split}
\hat{h}(\xi) 
&= ( \int_{|\eta|<|\xi|/2} + \int_{|\xi|/2 < |\eta| < 2|\xi|} + \int_{|\eta|>2|\xi|})\hat{f}(\xi-\eta)\hat{g}(\eta)d\eta \\
&= T_1 + T_2 + T_3.
\end{split}
\end{equation}
For $T_1$, now $\xi - \eta \sim \xi$, we have
$$|T_1| \lesssim \frac{1}{|\xi|^\gamma+1}\int_{|\eta|<|\xi|/2}|\hat{g}(\eta)|d\eta\lesssim \frac{1}{|\xi|^\gamma+1}.$$
For $T_2$, we have
$$|T_2| \lesssim |\hat{g}(\xi)|\int_{|\xi|/2 < |\eta| < 2|\xi|}\frac{1}{1+|\xi-\eta|^\gamma}d\eta \lesssim (1+|\xi|^{1-\gamma})|\hat{g}(\xi)|$$ which decays much faster than $\hat{f}(\xi)$ since  $g$ is Schwartz.
For $T_3$, $|\xi-\eta|\sim |\eta|$, $$|T_3| \lesssim \int_{|\eta|>2|\xi|}|\hat{f}(\eta)||\hat{g}(\eta)|d\eta \lesssim \frac{1}{|\xi|^\gamma+1}.$$
In conclusion, it holds that $|\hat{h}(\xi)| \lesssim \frac{1}{|\xi|^\gamma+1}.$
