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

Define

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

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

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?

Where do the quantities $s_m,d_m,h_m$ come from?

Consider a smooth compact $m$-dimensional Riemann manifold $(M,g)$. Fix an orthonormal basis of $L^2(M)$ consisting of eigenfunctions $\Psi_k$

$$\Delta \Psi_k=\lambda_k \Psi_k, $$

$$0=\lambda_0<\lambda_1\leq \lambda_2 \leq \cdots $$

$\newcommand{\ve}{{\varepsilon}}$ For $\ve>0$ and $w$ as above we set $w_\ve(t):=w(\ve t)$. Consider random functions on $M$ of the type

$$ u_\ve =\sum_{k\geq 0} u_k \sqrt{w_\ve(\lambda_k^{1/2})} \Psi_k, $$

where the $u_k$ are independent normal r.v. with mean $0$ and variance $1$. Note that thhe variance of $u_k \sqrt{w_\ve(\lambda_k^{1/2})}$ is $w_\ve(\lambda_k^{1/2})$ which goes to zero very fast due to the fast decay of $w$. This guarantees that $u_\ve$ is almost surely smooth. $\newcommand{\bp}{\boldsymbol{p}}$ Fix a point $\bp\in M$ and normal coordinates $(x^i)$ at $\bp$. $\newcommand{\pa}{\partial}$ The numbers $s_m(w)$ $d_m(w)$, $h_m(w)$, are related to the behavior of the random variables $u_\ve(\bp)$, $\pa_{x^i}u_\ve(\bp)$ and $\pa^2_{x^ix^j}u_\ve(\bp)$ as $\ve\to 0$. More precisely the rescaled variables $\ve^mu_\ve(\bp)$, $\ve^{m+1}\pa_{x^i}u_\ve(\bp)$ and $\ve^{m+2}\pa^2_{x^ix^j}u_\ve(\bp)$, $i\neq j$, converge as $\ve \to 0$ to mean zero normal variables of variances $s_m(w)$, $d_m(w)$ and respectively $h_m(w)$.

An Example (showing that Jochen Wengenroth's example is not a counterexample.) 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

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

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

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

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

so that the inequality (A) is equivalent to

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

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 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 $m\geq 2$, $k\geq 0$.

Update 1. The inequality (A) is true for weights $w$ of the form $w(t)=(1+t^2)e^{-t^2}$ and $w(t)=t^{2k}e^{-t^2}$.

Update 2. As Mikael de la Salle indicated the inequality (A) is not valid in the stated generality. I want to rephrase the question: for which weights $w$ one has

$$\lim_{m\to\infty} \frac{d_m^2}{s_mh_m}=1,$$

i.e.,

$$ \lim_{m\to\infty}\frac{\left(\int_0^\infty r^{m+1}w(r) dr\right)^2}{\left(\int_0^\infty r^{m-1}w(r) dr\right)\left(\int_0^\infty r^{m+3}w(r) dr\right)}=1.\tag{C} $$

I could not find weights $w$ violating (C). That does not mean that there aren't any.

Here is a geometric interpretation of (C). Denote by $(-,-)_w$ the $L^2$-inner product with respect to the measure $w(r)dr$ on $(0,\infty)$. We denote by $\Vert-\Vert_w$ the associated norm. If we set

$$\mu_k(r):=r^k,\;\;\nu_k(r):=\frac{1}{\Vert \mu_k \Vert_w} \mu_k(r), $$

then the inequality (C) takes the form

$$ \lim_{m\to\infty} \bigl(\nu_{(m-1)/2}\;, \;\nu_{(m+3)/2}\bigr)_w=1.\;\;\tag{D} $$

This implies that, as $m\to\infty$, the distance between the lines spanned by $\mu_{(m-1)/2}$ and $\mu_{(m+3)/2}$ goes to zero.

I'll set

$$I_k(w):=\int_0^\infty r^kw(r) dr. $$

Update 3. Mikael de la Salle strikes again! Following his suggestion consider a weight $w$ such that

$$ w(r)= \exp(-(\log r)\log(\log r)),\;\;\forall r\geq 1.$$

Then

$$ I_k(w)\sim J_k =\int_0^\infty t^k \exp(-(\log r)\log(\log r)) dr,\;\;\mbox{as $t\to \infty$}. $$

The last integral can be estimated using the Laplace method and yields

$$J_k\sim\sqrt{2\pi\tau _k} e^{\tau_k}, \;\;\tau_k=e^{k+1}$$

In particular this shows that for this particular weight one has

$$\lim_{m\to\infty} \frac{I_{m+1}(w)^2}{I_{m-1}(w)I_{m+3}(w)}=0. $$

Mikael's suspicions were right.

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If you let $\omega\sim \frac{1}{x}$ near zero, then you can make $s$ really large while keeping $h$ and $d$ bounded. On the other hand I think $2s\cdot h\geq d^2$ holds... –  Gjergji Zaimi Apr 25 '12 at 13:31
    
The function $w$ is smooth, so it does not explode near $0$. I know that the inequality is true for certain weights $w$ such that $w(t)=e^{-t^2}$. Note if it holds for $w$, it also holds for $w_\ve$. –  Liviu Nicolaescu Apr 25 '12 at 13:37
    
How about a sequence of smooth functions which converge to $|x|^{-1}$ near zero? –  Gjergji Zaimi Apr 25 '12 at 13:48
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I have the feeling that if $w$ has heavy tail, you can have a limit zero in (C). For example if $w(x)=\exp( - (\log x )(\log \log x))$ for $x>1$. –  Mikael de la Salle May 4 '12 at 15:56
    
Thank you. That would be interesting. I'll think about it. I know at least one family of weights for which (C) happens. If $w$ has compact support and it is weakly decreasing in a neighborhood of the largest point of the support. –  Liviu Nicolaescu May 4 '12 at 17:08
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2 Answers

up vote 3 down vote accepted

As explained by Jochen Wengenroth, the answer is no in general.

Indeed, as you notice, the inequality $(A)$ is equivalent to an inequality about the moments $B_k=\int_0^\infty t^k w(t) dt$, namely $B_{m+1}^2 \geq C(m) B_{m-1} B_{m+3}$ for some $C(m)$ which is bounded above and below (my rapid computations give $C(m)=(1+2/m)^{-1}$ and do not agree with yours, so I prefer to be vague with the exact value of $C(m)$).

Anyway, for any $m$ you can find $w$ such that $B_{m+1}^2/(B_{m-1} B_{m+3})$ is arbitrarily close to zero. Indeed, if take $\mu_n$ the measure on $\mathbb R$ giving mass $1$ to $1$ and mass $1/n^{m+2}$ to $n$, then $ \int t^k d\mu_n = 1+n^{k-m-2}$, so that the ratio $B_{m+1}^2/(B_{m-1} B_{m+3})$ is equivalent to $1/n$, which goes to $0$. Of course, $\mu_n$ is not of the form $w(r)dr$, but it can be approximated by such measures.

But the reverse inequality is almost true, namely by Hoelder's inequality $B_{m+1}^2\leq(B_{m-1} B_{m+3})$, so that $d_m(w)^2/(s_m(w) h_m(w))$ is bounded above by $C(m)$.

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The inequality (B) to be proved requires $\frac{B_{m+1}^2}{B_{m-1}B_{m+3}}\geq \frac{m}{2m+1}$. For the $\mu_n$ you indicated we have $$ \frac{B_{m+1}^2}{B_{m-1}B_{m+3}}\frac{(1+n^3)^2}{(1+n)(1+n^5)}\to 1>\frac{1}{2}$$ as $n\to \infty$$ –  Liviu Nicolaescu Apr 26 '12 at 9:26
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I do not follow your comment: for the $\mu_n$ I indicate, $\frac{B_{m+1}^2}{B_{m-1} B_{m+3}} =\frac{(1+1/n)^2}{(1+1/n^3)(1+n)} \to 0$. –  Mikael de la Salle Apr 26 '12 at 11:37
    
My apologies. You are right. I was having my first coffee when I wrote the comment. In the mean time I found several counterexamples to (A) and I also realized that for my particular problem I need to impose additional constraints on $w(t)$. More precisely, I want $(t)$ to be log-concave in a rather strong way $$w(t)= e^{-U(t)}$$ $U(t)$, smooth, even, convex $\lim_{t\to\infty} U(t)=\infty$. –  Liviu Nicolaescu Apr 26 '12 at 12:22
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I don't think so: To write things probabilistically let us assume $m=2$ and $s_2=1$ (the inequality is homogeneous w.r.t. the weight $w$). Then $w(|x|)$ is the Lebesgue-density of a probability measure on $\mathbb R^2$ which is invariant under rotation (since it only depends on the norm) and for the two coordinate functions $X,Y$ you are asking for $$ E(X^2 Y^2) \le E(X^2) E(Y^2).$$ (For the special weight $w_0(t)=c\exp(-t^2/2)$ the coordinates are independent and one has even equality.) In the general case, the random vector $2^{-1/2}(X+Y,X-Y)$ is a rotation of $(X,Y)$ and has thus the same distribution as $(X,Y)$ which leads to $$ E(X^2 Y^2)= \frac{1}{4} E((X+Y)^2(X-Y)^2) = \frac{1}{4} E((X^2-Y^2)^2) = \frac{1}{2} (E(X^4)-E(X^2Y^2))$$ hence $E(X^2Y^2)= \frac{1}{3} E(X^4)$. Now, if the "tails" of $w$ are "heavier" than those of $w_0$ one gets that the inequality might be wrong.

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Thanks! This is very helpful. It opens up an intriguing can of worms. How about a weaker question: what happens to the ratio $\frac{d_m^2}{s_mh_m}$ as $m\to \infty$? –  Liviu Nicolaescu Apr 25 '12 at 14:08
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@Liviu: I believe that $\frac{d_m^2}{s_m h_m}=B_{m+1}^2/((1+2/m)B_{m-1} B_{m+3})$ where $B_k=\int t^k w(t) dt$. –  Mikael de la Salle Apr 25 '12 at 14:27
    
@ Jochen. I think that there is an error in your computations. I have included detalied computations to a class of examples including the weight $t^2e^{-t^2}$ you suggested and the inequality seems to hold. –  Liviu Nicolaescu Apr 25 '12 at 16:04
    
@ Mikael: see the new example I have just added. –  Liviu Nicolaescu Apr 25 '12 at 16:07
    
I have deleted the wrong part of the "answer" (I was confused about the marginal distribution of $X$) –  Jochen Wengenroth Apr 26 '12 at 11:29
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