# 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.

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.
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 $$|x|^{-\gamma}$$.
• 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? – Jacob Lu May 2 at 3:45
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 $$$$\hat{h}(\xi) = \int_{-\infty}^\infty \hat{f}(\xi-\eta)\hat{g}(\eta)d\eta.$$$$ For $$|\xi| \gg 1$$, we write $$$$\begin{split} \hat{h}(\xi) &= ( \int_{|\eta|<|\xi|/2} + \int_{|\xi|/2 < |\eta| < 2|\xi|} + \int_{|\eta|>2|\xi|})\hat{f}(\xi-\eta)\hat{g}(\eta)d\eta \\ &= T_1 + T_2 + T_3. \end{split}$$$$ For $$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}.$$