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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.

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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 $|x|^{-\gamma}$.

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  • $\begingroup$ Many thanks for your answer! It is very helpful! May I ask you one more question? Does the Phragmen-Lindelof theorem hold for more general functions? For example, if $f(z)$ is an entire function of exponential type and $|f(x)| \le log(2+|x|) / (1+|x|^\gamma)$ on the real axis, can we expect $|f(z)| \le Ce^{\sigma |y|}\log(2+|z|)/(1+|z|^\gamma)$? Now the estimate has a log in the numerator, will this affect the answer? $\endgroup$ – Jacob Lu May 2 at 3:45
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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 $|\hat{h}(\xi)| \lesssim \frac{1}{|\xi|^\gamma+1}.$

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