# On a differential inequality

The question has probabilistic origins, but it would take too long to elaborate. $\newcommand{\ii}{\boldsymbol{i}}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\eS}{\mathscr{S}}$

Fix a nonnegative Schwartz function $w:\bR\to \bR$. For any positive integer $m$ let $V_m:\bR^m\to\bR$ denote the Fourier transform of

$$w_m:\bR^m\to\bR,\;\; w_m(x)=w(|x|^2/2).$$

More precisely

$$V_m(\xi)=\int_{\bR^m} e^{-\ii (x,\xi)} w(|x|^2/2) dx.$$

$V_m$ is an $O(m)$-invariant Schwartz function on $\bR^m$ so that it has the form

$$V_m(\xi)= f_m\bigl(|\xi|^2/2\bigr),$$

where $f_m$ is a smooth one-variable function, $[0,\infty)\ni r\mapsto f_m(r)\in\bR$.

The function $f_m=f_{m,w}$ depends on the initial $w$. The dependence $w\mapsto f_{m,w}$ is linear and can be explicitly described in terms of the Hankel transforms.

Denote $\newcommand{\eF}{\mathscr{F}}$ by $\eF$ the class of $C^2$-functions $f:[0,\infty)\to \bR$ such that

$$f'(0)< f'(r)+2rf''(r)<-f'(0),\;\;\forall r>0. \tag{1}$$

If we set $r=t^2/2$, $g(t)= f(r)=f(t^2/2)$, then $$g''(t)=f'(t^2/2)+t^2f''(t^2/2)=f'(r)+2rf''(r),$$

and we can rephrase the above inequality as

$$|g''(t)|<|g''(0)|,\;\;\forall t>0. \tag{2}$$

Denote by $\newcommand{\eW}{\mathscr{W}}$ $\eW$ the collection of nonnegative Schwartz functions $w :\bR\to \bR$ such that

$$f_{m,w}\in\eF, \;\;\forall m>0.$$

The description (2) and the $O(m)$ invariance of $V_m(\xi)$ lead to the following equivalent description of $\eW$. More precisely $w\in\eW$ iff for any $m>0$

$$|\Delta V_m(\xi)|<|\Delta V_m(0)|,\;\;\forall \xi\in\bR^m\setminus 0, \tag{3}$$

i.e.,

$$\left\vert\int_{\bR^m} e^{-\ii(x,\xi)}|x|^2w\Biggl(\frac{|x|^2}{2}\Biggr) |dx|\right\vert <\int_{\bR^m} |x|^2w\Biggl(\frac{|x|^2}{2}\Biggr)|dx|,\;\;\forall\xi\in\bR^m\setminus 0.$$

Remark 1. The class $\eF$ contains all the completely monotone functions $f:[0,\infty)\to \bR$ such that $f''(0)>0$. (Recall that a function $f:[0,\infty)\to \bR$ is completely monotone if it is smooth and $(-1)^kf^{(k)}(t)\geq 0$, $\forall t\geq 0$, $\forall k\geq 0$.)

This follows from the following observations.

• The functions $r\mapsto g_s(r)= e^{-sr}$, belong to $\eF$ for any $s>0$.
• The set $\eF$ is a convex cone.
• By Bernstein theorem, any completely monotone function $f$ can be written as an infinite superposition of nonnegative multiples of the functions $g_s$. More precisely, there exists a finite positive Borel measure $\mu(|ds|)$ on $[0,\infty)$ such that

$$f(r)=\int_0^\infty e^{-sr} \mu(|ds|).$$

Remark 2. If $w:\bR\to\bR$ is a nonnegative Schwartz function whose restriction to $[0,\infty)$ is completely monotone, then $w\in \eW$.

Indeed, we can find a positive, finite Borel measure $\mu(|ds|)$ on $[0,\infty)$ such that

$$w(t)= \int_0^\infty e^{-st}\mu(|ds|).$$

Then

$$w_m(x)= \int_0^\infty e^{-s|x|^2/2}\mu(|ds|),$$

$$V_m(\xi)= \int_0^\infty\left(\int_{\bR^m} e^{-\ii(x,\xi)} e^{-s|x|^2/2} dx\right) \mu(|ds|)$$

$$= \int_0^\infty\left(\int_{\bR^m} e^{-\ii(y,\xi)/\sqrt{s}} e^{-|y|^2/2} dy\right) s^{-\frac{m}{2}}\mu(|ds|) = (2\pi)^{\frac{m}{2}}\int_0^\infty e^{-\frac{|\xi|^2}{2s}} s^{-\frac{m}{2}}\mu(|ds|).$$

Hence

$$f_m(r)= (2\pi)^{\frac{m}{2}}\int_0^\infty e^{-\frac{r}{s}} s^{-\frac{m}{2}}\mu(|ds|).$$

This proves that $f_m(r)$ is completely monotone. A simple computation shows $f_m''(0)>0$ and from Remark 1 we conclude $f_m\in\eF$.

Now comes the question.

Does the class $\eW$ contain examples of functions $w$ not covered by Remark 2?

Remark 3. I should perhaps mention here a theorem of Schoenberg which states that a function $w:[0,\infty)\to \bR$ is completely monotone if and only if, for any $m>0$ the function

$$\bR^m\ni x\mapsto w_n(x)=w(|x|^2/2) \in\bR$$

is positive definite. According to Bochner's theorem, this means that $w_m$ is the Fourier transform of a positive measure on $\bR^m$. In view of the above discussion we see we can rephrase the question as follows.

Is it true that $w\in \eW \Rightarrow f_{m,w}\geq 0$, $\forall m>0$?

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