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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.

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2 Answers 2

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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.

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  • $\begingroup$ 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? $\endgroup$ Commented Feb 13 at 16:54
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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.

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  • $\begingroup$ What do you mean by "In our set?" What is $p$? Did you mean "sinc function" instead of "since function?" $\endgroup$
    – Mark Viola
    Commented Feb 15 at 16:49
  • $\begingroup$ It should be "setting", I have corrected the answer. Thanks for the comment. $\endgroup$ Commented Feb 16 at 20:30

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