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Question:

Let $X$ be a random variable taking values in $\mathbb{R}$ and let $F$ be its probability distribution function (that may not have a density). Denote the characteristic function as $\varphi(s) = E[e^{isX}] = \int_{\mathbb{R}}e^{isx}dF(x)$.

I am struggling to find non-trivial sufficient conditions to ensure that $|\varphi(s)|$ approaches $0$ sufficiently slowly for large $s$.

More precisely I would like to find some function $g$ such that $|\varphi(s)|^{-1}= O(g(s))$.

My Attempt:

It seems promising to put some sort of tail restriction on the distribution of $X$, like sub-gaussian or sub-exponential, but I am still stuck.

For example, for some $K,C >0$ suppose $P(|X|\geq t) \leq Ke^{-tC}$ for all $t$ sufficiently large. With this I have obtained $$ \left| \int_{\mathbb{R}}e^{isx}dF(x) \right| \geq \left| \int_{{|x|}\leq t}e^{isx}dF(x) \right| - \left| \int_{{|x|}> t}e^{isx}dF(x) \right| \geq \left| \int_{{|x|}\leq t}e^{isx}dF(x) \right| - \int_{{|x|}> t} \left|e^{isx}\right|dF(x) $$ so for $t$ sufficiently large, $$ |\varphi(s)| = \left| \int_{\mathbb{R}}e^{isx}dF(x) \right| \geq \left| \int_{{|x|}\leq t}e^{isx}dF(x) \right| - Ke^{-tC} $$ But I am not sure how show the first term is bounded away from $0$, or approaches $0$ at a certain rate as $s$ grows. (possibly letting $t$ be a function of $s$?)

Any citations/suggestions are much appreciated. Lukacs (1970) Theorem 7.3.2 seems promising but is only for purely imaginary arguments of $\varphi$.

(If its relevant, my ultimate goal is to show $\int_{a}^{b} \frac{1}{s |\varphi(s)|} ds = O(h(b))$ for some function $h$).

Chad Brown
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  • The characteristic function doesn't even vanish at infinity in general, so this question doesn't make sense...surely you want a density. – Andrew May 21 '25 at 01:59
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    On the other hand, if you look at various characteristic functions, there seems to be no reason to believe that decay of fourier transform is related to decay of tail probability – Andrew May 21 '25 at 07:58
  • Thank you for you're answer, but I'm not sure I follow. $\int_{R}dF(x) = 1$ so F must be decaying at the tails (or =0) and $|e^{isx}|\leq 1$ so shouldn't $\varphi$ be decaying? and shouldn't that decay be related to how $F$ decays? Also Lukacs 1970 gives some results related to this, I believe in Chapter 7 for sub-exponential random variables showing they decay at a certain rate. Am I miss-understanding something here? – Chad Brown May 21 '25 at 16:32
  • Also, I do not want to assume the density of F exists. My whole goal is to relax previous work that has required a density. – Chad Brown May 21 '25 at 16:36
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    The reason the fourier transform of a $L^1$ function vanishes at infinity is because of the rapid oscillations of the exponential. In particular, the fourier transform of an absolutely continuous distribution vanishes at infinity. The same cannot be said for a general borel probability measure. – Andrew May 21 '25 at 16:54
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    I honestly don't know what a sufficient condition besides having a density would make the characteristic functions $C_0$. The most natural thing to try is probably gaussian regularization, and then try to figure out when you can commute limits. – Andrew May 21 '25 at 16:57
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    The only things I know are somehow related are the Riemann–Lebesgue lemma and the Schwartz's Paley–Wiener theorem... maybe using a smooth bump function as a density? – Joako May 26 '25 at 01:14

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