I am looking for some sufficient conditions for an even, continuous, nonnegative, non-increasing, non-convex function to be non-negative definite. In other words $$ \int_0^\infty f(x)\cos(x\omega) \, dx\ge 0, \quad \omega \in \mathbb{R}. $$ The function $f(x)=\exp(-x^{\alpha})[x^{\alpha} \log(x) + \delta/\alpha ]$, $\alpha\in(1,2)$, $\delta\ge 2$. Note that for $\alpha\in(0,1)$, $f(x)$ is convex. I have tried complex detour numerically (as in Tuck, E. O.: On positivity of Fourier transforms, Bull. Austral. Math. Soc., 74 (2006) 133– 138) – did not work out. It seems the imaginary part always increases around $0$. I have tried Polya type conditions () - no luck either. For $\alpha\in(0,1)$ everything works out as the function is convex. Thank you in advance for any hints or references!

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    $\begingroup$ You can find a list of necessary conditions in On the positivity of Fourier transforms. $\endgroup$ Jul 15, 2020 at 17:42
  • $\begingroup$ @CarloBeenakker, thank you, good read, I am seeking sufficient conditions though as I believe the function I am interested in is positive definite - cannot prove it rigorously but have some hand waving arguments $\endgroup$ Jul 15, 2020 at 17:51
  • $\begingroup$ positive definite for all $\delta$ ? $\endgroup$ Jul 15, 2020 at 18:45
  • $\begingroup$ @CarloBeenakker, $\delta\ge 2$, I have edited the question, thank you $\endgroup$ Jul 15, 2020 at 19:12
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    $\begingroup$ You did 11 edits in 2 days, please don't permanently edit your post. Edit 7 was changing "Sufficient" into "sufficient" and Edit 12 was the reverse edit, so it looks like artificially putting the post in the front list. $\endgroup$
    – YCor
    Jul 16, 2020 at 21:36

3 Answers 3


please show me how to prove for $|\omega|>20$

With great pleasure. We shall just show that $F(y)=\int_0^\infty e^{-x^a}x^a\log x\cos(yx)\,dx>0$ for large enough $y>0$. Note that $\cos(yx)=\Re e^{iyx}$, so we have the real part of a contour integral from $0$ to $+\infty$ of $e^{-z^a}z^a\log z e^{iyz}\,dz$. The integrand oscillates like crazy on the line, so we would like to move the contour up to get less oscillation. Ideally we would like to have a curve $\Gamma$ parameterized by $z(t)$ so that $\Re \left[e^{-z^a}z^a\log z e^{iyz}z'(t) \right] \ge 0$ everywhere on $\Gamma$. If we can do it, it would be the end of the story. However, for the integral in question it is, clearly, impossible (without the extra $1$ or $2$, the Fourier transform has zero integral, so it cannot be positive everywhere), so we'll settle for less: the integral over the "head part" of the contour will have relatively large positive real part and the tail will be small.

We will use the same curve $\Gamma$ that is normally used to prove that the Fourier transform of $e^{-|x|^a}$ is non-negative, namely, the curve $z(x)=x+i\xi(x)$, $x>0$, where $\xi(x)$ satisfies $(x+i\xi)^a=g+iyx$, $\xi,g\in\mathbb R$. This will make $e^{-z^a}e^{iyz}$ real positive on $\Gamma$. Note that $\xi$ is well-defined and continuous in $x$ for $1\le a<2$ and $\xi(x)\asymp x^{2-a}$ at $+\infty$, so this change of the contour is legitimate.

On $\Gamma$, we have $e^{-z^a}e^{iyz}=e^{-H}$ where $H=g+y\xi$. Differentiating the identity defining $\Gamma$, we get $az^{a-1}(1+i\xi_x)=g_x+iy$, so $$ H_x=g_x+y\xi_x=\Re [(g_x+iy)(1-i\xi_x)]=\Re[az^{a-1}|1+\xi_x|^2]>0 $$ for $z$ in the first quadrant, so $H$ is strictly increasing.

We now want to evaluate $$ \Re\int_{\Gamma} e^{-H}z^a\log z\,dz=\int_{\Gamma} \Re\left[\frac{z^{a+1}}{a+1}\left(\log z-\frac 1{a+1}\right)\right]d(-e^{-H})\,. $$ To do it, we will switch to polar coordinates $z=Re^{i\theta}$. Notice that the equation for $\Gamma$ becomes $R^a\sin(a\theta)=yR\cos\theta$. Since $\theta\mapsto \frac{\sin(a\theta)}{\cos\theta}$ increases from $0$ to $+\infty$ as $\theta$ runs from $0$ to $\pi/2$, the curve $\Gamma$ intersects every circumference centered at $0$ only once, so the radius $R$ is a legitimate parameter on $\Gamma$.

We also have $\Re [z^a\bar z]=\Re[(x-i\xi)(g+iyx)]=x(g+y\xi)=xH$, so $$ H=R^a\frac{\cos((a-1)\theta)}{\cos\theta} $$ Now come a couple of observations. The first one is that $$ \sin(a\theta)=\sin((a-1)\theta)\cos\theta+\cos((a-1)\theta)\sin\theta $$ and $$ -\sin((a-1)\theta)\cos\theta+\cos((a-1)\theta)\sin\theta=\sin((2-a)\theta)>0 $$ for $\theta\in[0.\frac{\pi}2]$. Hence $$ R^a\cos((a-1)\theta)\sin\theta\le R^a\sin(a\theta)=yR\cos\theta\le 2R^a\cos((a-1)\theta)\sin\theta\,. $$ Juxtaposing this with the polar formula for $H$, we see that $H\sin\theta\le yR\le 2H\sin\theta$ on $\Gamma$.

The second observation is that $\sin(a\theta)\le 1$, so for $R\in[0,1]$, we have $\cos\theta\le\frac 1y$, i.e., $\theta\in[\theta_0,\frac\pi 2]$ where $\theta_0=\arccos\frac 1y$ is quite close to $\frac\pi 2$ for large $y$.

Now we are ready to look at the real part of $\frac{z^{a+1}}{a+1}\left(\log z-\frac 1{a+1}\right)$ when $R\le 1$. It is $$ \frac{R^{a+1}}{a+1}\left[\cos((a+1)\theta)\left(\log R-\frac 1{a+1}\right)-\theta\sin((a+1)\theta)\right]\,. $$ Note that $(a+1)\theta\in [2\theta_0,\frac{3\pi}2)$, so as soon as $\theta_0\ge\pi/4$ ($y>\sqrt 2$), the cosine is negative. Thus we can ignore $\log R<0$, which leaves us with the expression $$ -\cos((a+1)\theta)\frac 1{a+1}-\theta\sin((a+1)\theta) $$ on $[\theta_0,\frac {\pi}2]$. Taking the derivative, we see that it is increasing in $\theta$, so $\theta=\theta_0$ is the worst case. Again, it can easily be verified that this expression is greater than, say, $c(a)=\frac 1{2(a+1)}$ if $\theta_0$ is close enough to $\frac\pi 2$. Taking into account the bound $H\le yR/\sin\theta_0=:H_1R$ on $[0,1]$ and noting that for $R=1$, we have $H=\Re[z^a]+y\xi\ge -1+y\sin\theta_0=:H_0$, we immediately conclude that $$ \int_{\Gamma:R<1}\dots d(-e^{-H})\ge c(a)\int_0^{H_0/H_1}\frac {R^{a+1}}{a+1}H_1e^{-H_1R}\,dR $$ For large $y$, we have $H_0\approx H_1\approx y$ (see the accurate estimates above), so the whole integral is about $$ \frac{c(a)y}{a+1}\int_0^1 R^{a+1}e^{-yR}\,dR\approx \frac{c(a)}{a+1}y^{-a-1}\int_0^\infty r^{a+1}e^{-r}\,dr $$ i.e., we have just power decay in $y$ here.

On the other hand, since $R\le 2H/y=R(H)$ all the way through, we have $$ \left|\int_{\Gamma: R>1}\dots d(-e^{-H})\right|\le\frac 1{a+1}\int_{H_0}^\infty R(H)^{a+1}\sqrt{\left(\log R(H)+\frac 1{a+1}\right)^2+\frac{\pi^2}4} \,e^{-H}\,dH\,, $$ which decays exponentially in $y$, so the whole integral is positive for large $y$. I leave the accurate estimates of the minimal $y$ for which it works to you. $y>20$ is certainly enough, but I won't be suprised if one can push it down to $y>10$.

I'm still struggling with small $y$... :-( Meanwhile, feel free to ask questions if something is unclear.

  • $\begingroup$ the proof is awesome, I am struggling to find a way to evaluate $y$ numerically. I believe it should be $y$ which makes the second integral positive. What is $H_0$? $\endgroup$ Jul 17, 2020 at 16:05
  • $\begingroup$ @TanyaVladi Sorry, forgot to write: $H_0=-1+y\sin\theta_0$ (the lower bound for $H$ at the moment of escape from the unit circle). $\endgroup$
    – fedja
    Jul 17, 2020 at 17:20
  • $\begingroup$ got it, thank you. Just a follow up question, it seems using your method one can get a lower bound on $F(y)$, it is possible to estimate when the global minimum occur? the number of zeros of $F(y)$ $\endgroup$ Jul 17, 2020 at 17:55
  • $\begingroup$ I have run numerics. Basically I find points on the $\Gamma$ by solving $R^{\alpha-1}\sin(\alpha\theta)=\cos(\theta)y$ for fixed $y$ and $\alpha$- get pairs of $(x,\xi)$. As you can imagine $\xi$ is an increasing function of $x$. I have noticed that large $y$ do not work as real part of ${z^{\alpha+1}\over \alpha+1}$ $\{\log(z)-{1\over \alpha+1}\}$ is negative for large $\xi$. On the other hand tiny $y$s work perfectly as the real part is positive, What am I missing? $\endgroup$ Jul 17, 2020 at 21:07
  • $\begingroup$ @TanyaVladi You are probably missing the logic of this proof: The real part for the large $y$ is negative once the curve leaves the unit disk and the second integral estimates the negative contribution, so it has to be subtracted. The positive contribution comes from the part of the curve in the unit disk, which is bounded from below by the first integral. The point is that for large $y$ the first (positive) integral decays in a power (in $y$) fashion while the second one decays exponentially, so the positive contribution wins. I'm getting a cutoff at around $y=8.5$ uniformly in $a$. $\endgroup$
    – fedja
    Jul 17, 2020 at 22:14

This can be very complicated, so I will only give a reference which contains many highly non-trivial results of this sort:

MR0428382 J. V. Linnik and I. V. Ostrovskiĭ, Decomposition of random variables and vectors. Translated from the Russian. Translations of Mathematical Monographs, Vol. 48. American Mathematical Society, Providence, R. I., 1977.

Look especially in the chapter which is called Necessary conditions for $I_0$, I think it is Chapter IV.

  • $\begingroup$ I have the book and checked it. Are you implying that my function is of Levy Hinchin representation? I do not see it, can you please drop another hint? $\endgroup$ Jul 14, 2020 at 23:05
  • $\begingroup$ I only answered the first sentence in your message, not about this specific function. I do not claim that your function has a Levy Hinchin representation. $\endgroup$ Jul 14, 2020 at 23:20
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    $\begingroup$ Change of contour should work (as I commented in your other post, it does work if you care only about $|\omega|>20$) but it seems rather delicate because the function has a zero in the first quadrant and treating it properly in the full range requires a lot of extra care. On the side note, asking slight variations of the same question several times in a row is usually frowned upon, though you seem lucky so far :-) $\endgroup$
    – fedja
    Jul 15, 2020 at 1:51
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    $\begingroup$ @TanyaVladi It isn't easy if you want to get a proof in the full range and do not want to settle for anything less, as it seems to be the case (otherwise I would post my $|\omega|>20$ solution long ago). Your question has been crystal clear all the time. Just give people some time to figure it out. :-) Of course, there is no explicit formula for anything here but it is not what makes it tricky. The real issue is the unpleasant behavior of the argument of $1+z\log z$ in the upper half-plane and I'm still struggling with it in the case of small $\omega$. $\endgroup$
    – fedja
    Jul 15, 2020 at 19:36
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    $\begingroup$ @fedja, please show me how to prove for $|\omega|>20$. I can show it is positive for small $\omega$ $\endgroup$ Jul 15, 2020 at 20:54

How about the following reference T. Gneiting, Criteria of Po´lya type for radial positive definite functions, Proc. Am. Math. Soc. 129 (2001), 2309–2318. I have plotted the results for 6th derivative all numeric but it seems working,

  • $\begingroup$ Looks very interesting. So what particular $k,l,\alpha$ do you suggest? $\endgroup$
    – fedja
    Jul 26, 2020 at 18:16

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