# Decay property of the Fourier transform

Let $$f,g:\mathbb{R} \to (0,\infty)$$ be nonnegative functions on $$\mathbb{R}$$. If $$\lim_{|x|\to \infty} \frac{g(x)}{f(x)} = 0.$$ Can we conclude anything about the decay property of their respective Fourier transforms $$\hat f,\hat g$$? They are defined by $$\hat f(y) = \int_{\mathbb{R}} e^{-ixy} f(x)dx,$$ similar for $$\hat g$$. In particular, could we conclude anything such as $$\lim_{|y| \to \infty} \frac{|\hat g(y)|}{|\hat f(y)|} = 0,$$ provided that the division is well-defined?

A related question is Decay of Fourier Transform of a Schwartz Function.

Hint: Take $$g(x)=e^{-|x|}, f(x)=e^{-|x|/2}$$. Fourier transforms of these function can be written down explitly. [Refer to Cauchy distribution in Wikipedia]. The ratio of the FT's actually tends to a positive constant.
• Thanks for your answer. The reason that I am asking is that I would like to compare the decaying property of the Fourier transform of $\exp(-|t|^p)$ for $p>2$ with the one of $\exp(-|t|^2)$. Is there any result in that case? In other words, what happens if the decay rates of the original functions are different? Commented Feb 13 at 16:54
As a follow-up comment to the answer above, I have some new findings regarding the question. As a general principle in Fourier transform, also known as Heisenberg's Inequality, see, e.g., Theorem 1.1 in The Uncertainty Principle: A Mathematical Survey by Folland, we would expect that if the original function $$f$$ decays faster, then its Fourier transform $$\hat f$$ decays slower. In fact, the theorem says if $$f\in L^2(\mathbb{R})$$ and $$\|f\|_{L^2} =1$$, then $$\left(\int_{\mathbb{R}} x^2 |f(x)|^2 dx \right)\left(\int_{\mathbb{R}} x^2 |\hat f(x)|^2 dx\right)\geq \frac{1}{16\pi^2},$$ here $$\hat f$$ denotes the standard $$2\pi$$-Fourier transform of $$f$$. In our setting, i.e., the function $$g$$ takes the form $$t\mapsto \exp(-|t|^p)$$ and $$f$$ is the Gaussian function, if $$p=\infty$$, then $$g$$ is compactly supported and $$\hat g$$ is the Sinc function, which decays much slower than the Gaussian function.
• What do you mean by "In our set?" What is $p$? Did you mean "sinc function" instead of "since function?" Commented Feb 15 at 16:49