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Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous positive-definite function with $f(0)=1$. Positive-definiteness of $f$ means $$ \sum_{i=1}^{n}\sum_{j=1}^{n}f(x_i-x_j)y_i y_j \geq 0 $$ for all $n\geq 1, x,y\in \mathbb{R}^n$.

Note that, by Bochner's theorem, $f = \widehat{\mu}$ for some Borel probability measure $\mu$ on $\mathbb{R}$.

Question. If $f \in L^{p}$ for some $p < \infty$, must we have $f(x)=O(|x|^{-c})$ for some $c>0$?

Edit: Formerly, I had the condition $p>2$ instead of $p < \infty$. Thanks to Christian Remling for pointing out that the condition $p>2$ adds nothing and the condition $p<\infty$ is needed.

Context: This is a natural extension of this question: https://math.stackexchange.com/questions/2296804/lp-implies-polynomial-decay

I posted this first to MSE a few months ago, got several up-votes, but nothing helpful: https://math.stackexchange.com/questions/2306071/decay-of-positive-definite-functions-in-lp

I am following the advice of Cross posts to Math SE regarding cross-posting.

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    $\begingroup$ $f=\widehat{\mu}$ is bounded, so if $f\in L^p$, then also $f\in L^q$ for all $q\ge p$. So the restriction that $p>2$ is not really doing anything. (However, you do have to exclude $p=\infty$.) $\endgroup$ Aug 17, 2017 at 22:35
  • $\begingroup$ @ChristianRemling I guess I meant that $f \in L^p$ for some $p > 2$, but $f \notin L^2$. When I wrote this question, I was thinking that the question becomes easy when $f=\widehat{\mu} \in L^2$ (because then $d\mu = gdx$ and $f = \widehat{g}$ for some $g \in L^2$). But I can't remember now why I thought that it becomes easy in this case. $\endgroup$
    – Linden
    Aug 17, 2017 at 23:06
  • $\begingroup$ I see, thanks. But in fact that wouldn't help because if there is a counterexample $f\in L^2$, then you can just take $f+g$, with $g$ having power decay, but $g\notin L^2$ (take a suitable Cantor type measure as $\widehat{g}$ to do this), and also produce a counterexample with $p>2$ that isn't in $L^2$. $\endgroup$ Aug 17, 2017 at 23:12
  • $\begingroup$ @ChristianRemling In that case $\mu=\hat{f}+\hat{g}$ will be non-singular (to Lebesgue measure). That naturally leads one to ask what happens with the question if we add the condition that $f=\hat{\mu}$ and $\mu$ singular. This condition implies $f \in L^p$ with $p>2$ and $f \notin L^2$. That's probably what I was thinking about with the original condition $p>2$. $\endgroup$
    – Linden
    Aug 19, 2017 at 11:00
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    $\begingroup$ I don't think that will change anything fundamentally, though it might be quite a bit more difficult to find a concrete counterexample now. I don't see any possible mechanism that could make such a statement true. Clearly, $f\in L^p$ by itself will not imply power decay, which is a smoothness condition of sorts on $\mu$, and it's hard to believe that the additional hypothesis of positivity of $\mu$ will now somehow magically produce this smoothness (at least my intuition suggests that a positive function can be just as nasty as a general one). $\endgroup$ Aug 19, 2017 at 18:20

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No, this does not follow. We can take $f=g*g$, with $g\simeq 1$ near $x_n$, with $x_n$ very rapidly increasing. We'll also choose $0\le g\le 1$ as an even continuous function from $L^1$. This will make sure that $\widehat{f}=\widehat{g}^2$ is positive, as required.

Moreover, $f\in L^1$ also, but power decay is prevented by just taking the $x_n$ large enough. More specifically, if $g(x)=\sum h(a_n(x-x_n))$, with a compactly supported $h$ and $a_n,|x_n|\to\infty$ and if also $g=1$ near zero, then $f(x_n)\ge \int h(a_nt)\, dt>0$, and this will be $\ge Cx_n^{-\alpha}$ for any given constants $C,\alpha$ if we just take $x_n$ large enough. Notice that it suffices to show that $f$ does not satisfy any of the estimates $f(x)\le N x^{-1/N}$, $x\ge N$, and for each such potential bound, we use one $x_n$ to refute it.

Finally, if $\widehat{f}$ is not in $L^1$, then we modify the argument by also multiplying $\widehat{f}$ by a smooth cut-off function $\varphi$ with $\varphi, \widehat{\varphi}\ge 0$ to fix this (as above, we can take $\varphi=\psi*\psi$ to do this). This will change $f$ itself to $\widehat{\varphi}*f$, but this will still be in $L^1$ and fail to satisfy power bounds if the $x_n$ increase rapidly.

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  • $\begingroup$ Initially, I thought $\widehat{f}\in L^1$ would just follow from $\int \widehat{f}=f(0)<\infty$ ("Fourier inversion"), and the last paragraph would become unnecessary. But that seems a bit formal and I now don't know how to make a proof out of this. $\endgroup$ Aug 18, 2017 at 7:16
  • $\begingroup$ I'm not sure what you mean. Are you saying you are not confident the last paragraph is correct? $\endgroup$
    – Linden
    Aug 19, 2017 at 1:42
  • $\begingroup$ No, I'm saying it might be unnecessary. $\endgroup$ Aug 19, 2017 at 2:33
  • $\begingroup$ Okay, I'm with you now. $\endgroup$
    – Linden
    Aug 19, 2017 at 3:03
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    $\begingroup$ Regarding $\widehat{f} \in L^1$: Since $g \in L^1$ with $0 \leq g \leq 1$, we have $g \in L^p$ for all $1 \leq p \leq \infty$. In particular, $g \in L^2$ and hence $\widehat{g} \in L^2$ by Plancherel, so that $\widehat{f} = \widehat{g}^2 \in L^1$. Now you have $f, \widehat{f} \in L^1$ and $f$ is continuous (convolution of two $L^2$ functions), so that Fourier inversion holds pointwise. $\endgroup$
    – PhoemueX
    Dec 17, 2018 at 9:38

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