# possibility of bounding one functional by another functional

This is a natural follow-up question related to one of my previous questions at here. Assume that $$\rho$$ is a log-concave probability density function with support $$[0,\infty)$$ and fixed mean $$\mu >0$$. Let $$H[\rho] = \iiint_{y,v,w\geq 0} \rho(v)\rho(w)\rho(y)\left(\frac{|v-y|+|w-y|}{2} - \left|\frac{v+w}{2} -y\right|\right)\,\mathrm{d}v\,\mathrm{d}w\,\mathrm{d}y \,\,(\geq 0)$$ and $$G[\rho] = \iint_{x,y\geq 0} \rho(x)\rho(y)|x-y|\,\mathrm{d}x\,\mathrm{d}y.$$ I am wondering if it is possible to obtain functional inequalities of the form $$H[\rho] \geq f(G[\rho])$$ for some non-negative function $$f \colon [0,\infty) \to [0,\infty)$$ with $$f(0) = 0$$. Of course, the best scenario I can hope for is for $$f(x) = c\cdot x$$ for some $$c >0$$, but this is might be too good to be true. Thanks for your help!

Remark: taking into account of the comment made by Iosif Pinelis, there is no hope for a non-trivial $$f$$ if we do not impose any "structural" restrictions on the pdf $$\rho$$ (such as log-concavity of the density).

$$\newcommand{\R}{\mathbb R}$$Let $$p:=\rho$$, $$H:=H[p]$$, $$G:=G[p]$$. Let us show that $$\begin{equation*} f(x)\equiv kx \end{equation*}$$ with $$\begin{equation*} k=1/14334 \end{equation*}$$ will do. We will not need the restriction that the support of the distribution is $$[0,\infty)$$.

By approximation, without loss of generality (wlog), $$p(x)>0$$ for all real $$x$$ -- for instance, one may approximate $$p$$ by its convolution $$p*g$$ with the density $$g$$ of a centered normal distribution with an arbitrarily small variance. Then $$p*g>0$$ on $$\R$$ and $$p*g$$ is arbitrarily close to $$p$$ and log concave, since $$p$$ and $$g$$ are log concave -- see e.g. this Wikipedia article.

Therefore and because $$p$$ is a log-concave density, $$p$$ is continuous and attains its maximum value, say $$p_*[>0]$$, at some point $$c\in\R$$, so that $$p_*=p(c)\ge p(x)$$ for all real $$x$$. Moreover, again because $$p$$ is a log-concave density, there exist (unique) real $$a$$ and $$b$$ such that $$\begin{equation*} \text{a Again by the log-concavity of $$p$$, \begin{equation*} p(x)\le q(x):= \left\{ \begin{aligned} q_1(x):=p_*\exp\Big\{-\frac{x-c}{a-c}\Big\} &\text{ if }x As an illustration, for $$p(x)=xe^{-x}1(x>0)$$, here are the graphs $$\{(x,p(x))\colon-2\le x\le6\}$$ (blue), $$\{(x,q(x))\colon-2\le x\le6\}$$ (black), $$\{(x,q_1(x))\colon a\le x\le c\}$$ (dashed), and $$\{(x,q_2(x))\colon c\le x\le b\}$$ (dashed):

For this particular $$p$$, we have $$c=1$$, $$a=-W_0\left(-1/e^2\right)=0.15859\dots$$, and $$b=-W_{-1}\left(-1/e^2\right)=3.1461\dots$$, where $$W_j$$ is the $$j$$th branch of the Lambert $$W$$ function.

By shifting, wlog $$a=0.$$ So, \begin{align*} G&\le\iint\limits_{\R^2} q(x)q(y)|x-y|\,dx\,dy \\ &=p_*^2\frac{e^2 b^3+9 e b^3+3 b^3+3 b^2 c-12 e b^2 c-3 b c^2+12 e b c^2}{3 e^2} \\ &\le p_*^2\frac{\left(1+3 e+e^2/3\right) b^3}{e^2}, \end{align*} since $$0. Moreover, again by the log-concavity of $$p$$, we have $$p\ge p_*/e$$ on the interval $$[a,b]=[0,b]$$, so that $$1=\int_\R p\ge\int_0^b p_*/e=bp_*/e$$, whence $$p_*\le e/b$$ and $$\begin{equation*} G\le (1+3 e+e^2/3) b. \tag{1} \end{equation*}$$

On the other hand, because $$p\ge p_*/e$$ on the interval $$[a,b]=[0,b]$$ and the integrand in the definition of $$H$$ is $$\ge0$$, we have \begin{align*} H&\ge\Big(\frac{p_*}e\Big)^3\iiint\limits_{[0,b]^3} \Big(\frac{|x-z|+|y-z|}2 -\Big|\frac{x-z+y-z}2\Big|\Big)\,dx\,dy\,dz \\ &=\Big(\frac{p_*}e\Big)^3\frac{b^4}{24}. \end{align*} Also, $$1=\int_\R p\le\int_\R q=p_*b(1+1/e)$$, so that $$p_*\ge1/(b(1+1/e))$$ and hence $$\begin{equation*} H\ge\Big(\frac1{(e+1)b}\Big)^3\frac{b^4}{24}=\frac b{24(e+1)^3}. \tag{2} \end{equation*}$$

Comparing (1) and (2), we get $$\begin{equation*} H\ge\frac G{24(e+1)^3(1+3 e+e^2/3)}\ge\frac G{14334}, \end{equation*}$$ as claimed.

• Remarkable effort in writing this answer. Although I have one concern, in my problem, I really need to impose that the support of $\rho$ is contained in $[0,\infty)$ (i.e., no mass outside $\mathbb{R}_+$), so in that case, your sentence "By shifting, wlog, $a =0$" causes me some trouble... Jun 30 '21 at 5:47
• @FeiCao : Your case is just a particular case of the general one dealt with in this answer. So, I do not see any difficulty here. Also, you can do the same calculations with an arbitrary real $a$, without any shifting -- only the calculations will be a bit more complicated. Jun 30 '21 at 5:51
• Thanks! Actually I checked your profile and found that you are a very good researcher in the area of probability, I might use (not very sure) this answer for a publishable research project, may I know whether you will be interested in co-author a paper? (I am merely a Ph.D student in applied math at ASU who just finished my fourth year) If you are interested, you can email me at fcao5@asu.edu, and I am very happy to share the potential research project behind it....Best regards. Jun 30 '21 at 6:17
• @FeiCao : Concerning $G$, take $U=X-Y$ and $V_n=Z_n-W_n$, where $X,Y$ are iid random variables each with density $p$, and $Z_n,W_n$ are iid centered normal random variables each with variance $1/n$, say, such that $Z_n,W_n$ are independent of $X,Y$. Similarly, for $H$. Jun 30 '21 at 18:12
• @FeiCao : I have added the description of these $a,b,c$. Jun 30 '21 at 18:52