In my little research project, I faced the following problem: Assume that $\rho$ is a probability density function with support $[0,\infty)$ and 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!
Edit: taking into account of the answer written by Iosif Pinelis, there is no hope for a non-trivial $f$ if we do not post further restrictions on the pdf $\rho$, I am wondering if we can hope for a non-trivial $f$ (meaning $f$ is not identically zero) if we only look at "smooth" $\rho$.


1 Answer 1


Such a function $f$ must be identically zero.

Indeed, by approximation, the problem can be restated as follows:

Suppose that $f\colon[0,\infty)\to[0,\infty)$ is a function such that for any nonnegative iid random variables (r.v.'s) $X,Y,Z$ we have $$H\ge f(G),\tag{1}$$ where $$G:=E|X-Y|,\quad H:=E|X-Y|-\tfrac12\,E|X+Y-2Z|.$$ Prove that then $f=0$.

To prove this, take any real $a\ge0$ and let each of the r.v.'s $X,Y,Z$ take each of the values $0,2a$ with probability $1/2$, if $a>0$; if $a=0$, let each of the r.v.'s $X,Y,Z$ take the only value $0$. Then $G=a$ and $H=0$, so that (1) implies $0\ge f(a)$, for each real $a\ge0$. Since $f\ge0$, we conclude that $f=0$. $\Box$

Remark 1: Here, more generally, instead of assigning probability $1/2$ to each of the values $0,2a$, we can assign probabilities $p$ and $q=1-p$ to the values $0$ and $a/(2pq)$, respectively, for any $p\in(0,1)$.

Response to a comment by the OP: The OP is saying in that comment that the mean $\mu$ was assumed to be fixed. That assumption was not stated in the OP. It was only said that $\mu>0$. However, even assuming $\mu$ is fixed, Remark 1 above gives essentially the same conclusion, that $f=0$. Indeed, in the conditions of Remark 1, for any $a\in(0,2\mu)$, take $p=a/(2\mu)\in(0,1)$. Then $EX=a/(2p)=\mu$, as needed, and $f(a)=0$. So, $f=0$ on the interval $(0,2\mu)$ and hence, in view of the condition $f(0)=0$, we have $f=0$ on the interval $[0,2\mu)$.

Note also that we only need to define $f$ on the interval $[0,2\mu)$, because $E|X-Y|\in[0,2\mu)$. Indeed, by the triangle inequality, $E|X-Y|\le EX+EY=2\mu$, and $E|X-Y|=EX+EY$ only if $|X-Y|=X+Y$ almost surely (a.s.), which may happen only if $X=0$ a.s., which would imply $\mu=0$, which would contradict the assumption $\mu>0$.

  • $\begingroup$ Thanks for your answer! But I don't expect that the relation $H\geq f(G)$ to hold for all p.d.f. $\rho$ whose support is contained in $[0,\infty)$. For instance, is it possible to show something like this for "smooth" $\rho$ (instead of being a combination of Dirac masses)? $\endgroup$
    – Fei Cao
    Jun 25, 2021 at 19:48
  • $\begingroup$ @FeiCao : As briefly noted in my answer, no degree of smoothness can possibly help here, because any distribution can be appropriately approximated by an arbitrarily smooth one. If you want to impose additional, "structural" conditions on the distribution (such as, say, the log-concavity of the pdf), that would be quite a different question, to be posted separately. For now, your question, as posted, has been fully answered. $\endgroup$ Jun 25, 2021 at 19:59
  • $\begingroup$ I agree with you. Thank you! $\endgroup$
    – Fei Cao
    Jun 25, 2021 at 20:06
  • $\begingroup$ In fact, I thought your answer is still missing something. In my OP I indeed said that $\rho$ has some fixed mean $\mu > 0$, so your example will only work when $a = \mu$ (you are forbidden to take any real $a>0$). With your argument, we can only say that $f(\mu) = 0$, am I right? $\endgroup$
    – Fei Cao
    Jun 25, 2021 at 23:24
  • $\begingroup$ @FeiCao : The assumption that the mean $\mu$ was fixed was not stated in your OP. It was only said that $\mu>0$. However, even assuming $\mu$ is fixed, Remark 1 in my answer gives essentially the same conclusion, that $f=0$. I have added details on this. $\endgroup$ Jun 27, 2021 at 2:47

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