## Wednesday, April 25, 2012

### An inequality involving moments

$\newcommand{\bR}{\mathbb{R}}$  Suppose that $w:\bR\to \bR$ is a nonnegative, even smooth function  decaying fast at $\infty$, $w\in\mathscr{S}(\bR)$.

Define

$$s_m(w)= \int_{\bR^m} w(|x|) dx,\;\; d_m(w):=\int_{\bR^m} x_i^2 w(|x|) dx,\;\;\forall i$$

$$h_m(w) = \int_{\bR^m} x_i^2x_j^2 w(|x|) dx,\;\;\forall i<j.$$

Is it true that

$$d_m(w)^2\geq s_m(w) h_m(w) \tag{A}$$

for any $m\geq 2$ and any $w$  satisfying the above restrictions?

Example 1.   Observe first that

$$s_n(w)= \left(\int_{S^{m-1}} dA\right)\int_0^\infty r^{m-1} w(r) dr$$

$$d_n(w)= \left( \int_{S^{m-1}}x_1^2 dA(x)\right)\int_0^\infty r^{m+1} w(r) dr,$$

$$h_n(w)= \left( \int_{S^{m-1}}x_1^2x_2^2 dA(x) \right)\int_0^\infty r^{m+3} w(r) dr,$$

and

$$a_m:=\int_{S^{m-1}} dA = \frac{2\pi^{\frac{m}{2}}}{\Gamma(\frac{m}{2})},\;\; b_m:=\int_{S^{m-1}}x_1^2 dA(x)= \frac{\pi^{\frac{m}{2}}}{\Gamma(1+\frac{m}{2})}=\frac{a_m}{m},$$

$$c_m:= \int_{S^{m-1}}x_1^2x_2^2 dA(x)= \frac{\pi^{\frac{m}{2}}}{2\Gamma(2+\frac{m}{2})} = \frac{b_m}{m+2}.$$

Thus

$$d_m^2= b_m^2 \left(\int_0^\infty r^{m+1} w(r) dr)\right)^2=\frac{a_m^2}{m^2} \left(\int_0^\infty r^{m+1} w(r) dr)\right)^2,$$

and

$$s_m d_m= \frac{a_m^2}{m(m+2)} \left(\int_0^\infty r^{m-1} w(r) dr\right)\left(\int_0^\infty r^{m+3} w(r) dr\right),$$

so that the inequality (A)  is equivalent to

$$\left(\int_0^\infty r^{m+1} w(r) dr)\right)^2\geq \frac{m}{m+2} \left(\int_0^\infty r^{m-1} w(r) dr\right)\left(\int_0^\infty r^{m+3} w(r) dr\right). \tag{B}$$

A. Let us now choose $w(t)=t^{2k} e^{-t^2}$, $k$ nonnegative integer. Then for any $a>0$ we have

$$\int_0^\infty t^a w(t) dt=\int_0^\infty t^{a+2k} e^{-t^2} dt$$

($s=t^2$)

$$= \frac{1}{2}\int_0^\infty s^{\frac{a+2k-1}{2}} e^{-s} ds = \frac{1}{2}\Gamma\left(k+\frac{a+1}{2}\right).$$

For this choice of weight   the inequality (A) becomes

$$\Gamma(k +1+\frac{m}{2})^2\geq \frac{m}{m+2}\Gamma(k+\frac{m}{2})\Gamma(k+2+\frac{m}{2}).$$

This is equivalent to

$$k+\frac{m}{2}=\frac{\Gamma(k +1+\frac{m}{2})}{\Gamma(k+\frac{m}{2})}\geq \frac{m}{m+2} \frac{\Gamma(k+2+\frac{m}{2})}{\Gamma(k+1+\frac{m}{2})}= \frac{m}{m+2}\left(k+1+\frac{m}{2}\right) .$$

One can easily verify that the last inequality  holds for any $m\geq 2$, $k\geq 0$. It   turns into an equality when $k=0$.

B.  Suppose that $w(t)=(1+t^{2k}) e^{-t^2}$. Then

$$\int_0^\infty r^a w(t) dr= \int_0^\infty r^a e^{-r^2} dr + \int_0^\infty r^{a+2k} e^{-r^2} dr$$

$$= \frac{1}{2}\left( \Gamma\left(\frac{a+1}{2}\right)+\Gamma\left(k+\frac{a+1}{2}\right)\;\right).$$

In this case the inequality (A)  has the equivalent form

$$\frac{\Gamma(1+\frac{m}{2})+\Gamma(k +1+\frac{m}{2})}{\Gamma(\frac{m}{2})+\Gamma(k+\frac{m}{2})}\geq \frac{m}{m+2}\times \frac{\Gamma(2+\frac{m}{2})+\Gamma(k+2+\frac{m}{2})}{\Gamma(1+\frac{m}{2})+\Gamma(k+1+\frac{m}{2})},$$

or

$$\frac{\Gamma(1+\frac{m}{2})}{ \Gamma(\frac{m}{2})}\times \frac{1+\prod_{j=0}^{k-1} (1+j+\frac{m}{2})}{ 1+\prod_{j=0}^{k-1} (j+\frac{m}{2})} \geq \frac{m}{m+2}\times \frac{\Gamma(2+\frac{m}{2})}{\Gamma(1+\frac{m}{2})} \times \frac{1+ \prod_{j=0}^{k-1} (2+j+\frac{m}{2})}{ 1+\prod_{j=0}^{k-1} (1+j+\frac{m}{2})},$$

which further simplifies to

$$\frac{1+\prod_{j=0}^{k-1} (1+j+\frac{m}{2})}{ 1+\prod_{j=0}^{k-1} (j+\frac{m}{2})}\geq \frac{1+ \prod_{j=0}^{k-1} (2+j+\frac{m}{2})}{ 1+\prod_{j=0}^{k-1} (1+j+\frac{m}{2})}.$$

For $k=1$ this reduces to

$$\frac{2+\frac{m}{2}}{1+\frac{m}{2}}\geq \frac{3+\frac{m}{2}}{2+\frac{m}{2}}.$$

This is obviously true since $(m+4)^2> (m+2)(m+6)$.  $\Box$

We  can reformulate  the inequality (A)  as a convexity inequality as follows. For $a>1$ define

$$T_a(w)=\frac{\int_0^\infty t^{a-1} w(t) dt}{ \int_0^\infty t^{a-1} e^{-t^2} dt}.$$

Then the inequality (A) can be rewritten as

$$\frac{T_{m+2}}{T_m}\geq \frac{T_{m+4}}{T_{m+2}} \Longleftrightarrow T_{m+2}(w)^2\geq T_m(w) T_{m+4}(w). \tag{C}$$

Update 1. The inequality  (B) is false in  this generality. It suffices to choose $w(t)$ that approximates the  singular measure

$$\mu_n =\delta_1+\frac{1}{n^{m-2}}\delta_n,$$

where $\delta_c$ denotes the Dirac delta measure on $\bR$ concentrated at $c$. I  wonder if the inequality does hold if $(t)$ is log-concave in a rather special way

$$w(t)=e^{-U(t)}$$

where  $U(t)$ is smooth, even, convex and $\lim_{t\to\infty} U(t)=\infty$.

Also,  what are the asymptotics of

$$R_m=\frac{\left(\int_0^\infty r^{m+1} w(r) dr\right)^2}{\left(\int_0^\infty r^{m} w(r) dr\right)\left(\int_0^\infty r^{m+3} w(r) dr\right)}$$

as $m\to \infty$?

## Monday, April 16, 2012

### Live Blogging from Banff.

The conference is about to start.

## Monday, April 2, 2012

### This is only the first step

I'm starting this math blog and I hope it will work.      I hope you can see this beautiful equation

$$e^{\pi \boldsymbol{i}} +1=0,\;\;\boldsymbol{i}=\sqrt{-1},$$

or this wonderful equality

$$\int_{\mathbb{R}} e^{-\frac{x^2}{2}} dx=\sqrt{2\pi}.$$