Decay of positive definite function in $L^p$ 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.
 A: 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.
