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.

  • 1
    $\begingroup$ $f=\widehat{\mu}$ 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
  • 1
    $\begingroup$ 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

1 Answer 1


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.