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$\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\ii}{\boldsymbol{i}}$ $\newcommand{\eW}{\mathscr{W}}$

While investigating the distribution of critical points of random funtions on tori I was lead to the following analytical questions. Suppose that $w :\bR\to [0,\infty)$ is a Schwartz function. Define

$$ f: \bR\to \bR,\;\; f(t)=w(t^2). $$

We can use the function $f$ to define for any positive integer $m$ a Schwartz function

$$ F_m:\bR^m\to \bR,\;\;F_m(\vec{x})=f(|\vec{x}|)=w(|\vec{x}|^2). $$

We denote by $\widehat{F}_m(\vec{\xi})$ its Fourier transform and by $H_m(w, \vec{\xi})$ the Hessian of $\widehat{F}_m$ at $\vec{\xi}$.

Problem 1.

Fix a positive integer $m$. Describe the set $\eW_m$ of weights $w$ such that

$$ H_m(w,\xi) > H_m(w, 0), \;\; > \forall \vec{\xi}\neq 0. $$

Above, for two symmetric matrices $A$, $B$ the inequality $A< B$ signifies that $B-A$ is positive definite. Hence $w\in\eW_m $ if an only if the origin is a strict, absolute minimum for the Hessian map $$ \vec{\xi}\to H_m (w,\vec{\xi}). $$

Problem 2.

Describe the set

$$\eW:=\bigcap_{m>0}\eW_m. $$

Here is some information about $\eW$.

A. For any positive Schwartz function and any $m>0$ we have

$$ H_m(w,\vec{x})> H_m(w,0) $$

if $|\vec{\xi}|$ is sufficiently small, or sufficiently large. In other words, the origin is always a local minimum of the Hessian map. Thus the problem is about what happens in between.

B. For any $c>0$ we have $w(s)=e^{-c s/2}\in\eW$. Indeed $f(t)=e^{-ct^2/2}$

$$\widehat{F}_m(\vec{\xi})=const_m e^{-\frac{|\vec{\xi}|^2}{2c}}. $$

E.g., for $m=1$ we have $$ \frac{d^2}{dt^2} e^{-t^2/2}=(t^2-1) e^{-t^2/2} > -1 = \frac{d^2}{dt^2} e^{-t^2/2}|_{t=0}.$$

C. $\eW$ is a convex cone. In particular, any linear combination

$$ w(t)=\sum_i A_i e^{-c_it/2},\;\;A_i>0, $$

belongs to $\eW$. This implies that if the Schwartz function $w(t)$ is the Laplace transform of a positive finite measure $\mu$ on $[0,\infty)$, then $w(t)\in\eW$. The classical Hausdorff-Bernstein theorem shows that this happens if and only if $w$ is completely monotone, i.e.,

$$(-1)^k w^{(k)}(t)\geq 0,\;;\forall t>0, \;\;\forall k\in\mathbb{Z}_{\geq 0}. $$

Thus $\eW$ contains all the completely monotone Schwartz functions.

Here is a plausible

Conjecture (a) The weight $w$ belongs to $\eW$ if and only if $w$ is completely monotone.(Compare this with Schoenberg's theorem characterising completely monotone functions.)

(b) $\eW_m \neq \eW$, $\forall m$.

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Does this mean that the Hessian itself is strictly positive definite? –  Suvrit Jul 30 '12 at 8:45
    
Thanks for your comment. I had the inequalities in reverse order. I have edited the post to reflect this.. The Hessian at $0$ is negative definite. –  Liviu Nicolaescu Jul 30 '12 at 9:40
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