Let $f,g$ be smooth even log-concave functions on $\mathbb{R}^{n}$, i.e.,$f=e^{-F(x)}, g=e^{-G(x)}$ for some even convex functions $F(x),G(x)$. Is it true that:
$$ \int_{\mathbb{R}^{n}} \langle \nabla f(x), x\rangle\; g dx \cdot \int_{\mathbb{R}^{n}} \langle \nabla g(x), x\rangle\; f dx \leq \int_{\mathbb{R}^{n}}\langle \nabla f(x), x\rangle\ \langle \nabla g(x), x\rangle\ dx \cdot \int_{\mathbb{R}^{n}}fg\,dx\; \quad (*) $$ The answer seems to be positive.

Motivation: There is an interesting question which asks whether the function $\varphi(t)=\mu(e^{t} K)$ is log-concave on $\mathbb{R}$ where $K$ is a symmetric convex body in $\mathbb{R}^{n}$ and $d\mu$ is an even log-concave probability measure. Now if you consider the expression $(\ln \varphi(t))''_{t=0}$ and ``do integration by parts'' several times and simplify few expressions you will arrive to an inequality which follows from (*) but definitely is weaker than (*). So I was wondering if there is a simple counterexample to (*).

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    $\begingroup$ Could you add some background / motivation to this inequality. It reminds me of some related inequalities... $\endgroup$
    – Suvrit
    Oct 31, 2015 at 21:01
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    $\begingroup$ Sure, I updated the question. Which related inequality does it remind you? Maybe you mean Pitt's inequality.. $\endgroup$ Nov 1, 2015 at 1:14
  • $\begingroup$ Perhaps as you say, it reminds me of some work after Pitt --- somehow I am having a deja vu of having seen exactly your inequality in some paper, but cannot recall where, sorry! $\endgroup$
    – Suvrit
    Nov 1, 2015 at 17:44
  • $\begingroup$ Does $\langle\nabla f,x\rangle$ mean the same as $\langle x,(\nabla f)(x)\rangle$ (also denoted as $f'(x)(x)$ or $\langle x,f'(x)\rangle$)? $\endgroup$ Nov 8, 2015 at 4:27
  • $\begingroup$ @IosifPinelis yes it does: gradient of f at point x times (dot product) with x. In one dimensional case it just xf'(x) $\endgroup$ Nov 8, 2015 at 4:32

2 Answers 2


Not sure whether it is still of interest after three years, but anyway, I suggest the following:
ensure that your assumptions on $F$ and $G$ imply
1. $x\mapsto\langle \nabla F,x\rangle$ (same for $G$) is increasing with respect to some partial order $\le$ on $\mathbb{R}^{d}$
2. $(\mathbb{R}^{d}, \mathcal{B}(\mathbb{R}^{d}), Q, \le)$ with $Q(dx)=\frac{fg}{\langle f,g\rangle}dx$ is an FKG-space, c.f. https://projecteuclid.org/download/pdf_1/euclid.ecp/1465058067.
Hope that helps. Regards


edit: I overlooked the word "even" in the question, so my attempted counter-example isn't valid. Sorry about this.

I don't think the proposed inequality holds. To see this, consider the following (one-dimensional) situation: enter image description here This represents the case where $F$ and $G$ are equal $0$ on an interval each, and grow quickly outside these intervals.

For definiteness, let $$ F(x) = \begin{cases} c(A-x)^2 & \mbox{if $x<A$,} \\ c(x-B)^2 & \mbox{if $x>B$, and} \\ 0 & \mbox{otherwise,} \end{cases} $$ for constants $A$ and $B$, and $G$ be defined similarly, for different constants $A$ and $B$. Now consider what happens when we let the constant $c$ go to infinity: In this case, the functions $f$ and $g$ will converge to step functions, while the slope at the edges of the steps goes to $\pm\infty$.

Left hand side: Since $\int_B^{B+\sqrt{c}} \exp\bigl(-(x-B)^2\bigr) \propto -\sqrt{c}$ the downward slope of $f$ makes a contribution which is asymptotically proportional to $-\sqrt{c}$, and the contribution of the upward slope is negligible in comparison. By symmetry, the second integral makes the same contribution, and thus the left-hand side is asymptotically proportional to $c$.

Right-hand side: The integral $\int f(x)g(x) \,dx$ converges to a constant and since the derivatives of $f$ and $g$ are never both large at the same time, the first integral on the right-hand side will decay as $c$ increases.

Thus, for this example, as $c\to\infty$, the left-hand side goes to infinity while the right-hand side goes to zero, and thus the proposed bound cannot hold for this example.

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    $\begingroup$ In dimension $n=1$ the inequality holds trivially because you can rewrite it as follows $\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}(\nabla F(x)\cdot x - \nabla F(y) \cdot y)(\nabla G(x)\cdot x - \nabla G(y) \cdot y)e^{-F(x)-F(y)-G(x)-G(y)}dxdy \geq 0$. Now when $n=1$ notice that since $h(x):=F'(x) x$ is even and increasing for $x \geq 0$,it easily follows that sign of $F'(x)x-F'(y)y$ always equals to the sign of $G'(x)x-G'(y)y$ $\endgroup$ May 2, 2018 at 19:32
  • $\begingroup$ @Paata I don't understand why $F'(x) x$ would be even and increasing? For example $F(x) = (x+2)^2$ seems to be a counter-example to both even and increasing. $\endgroup$ May 2, 2018 at 19:54
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    $\begingroup$ The original problem asks to prove the inequality for even convex functions $F$ and $G$. So $F(x)=(x+2)^{2}$ is not even. $\endgroup$ May 2, 2018 at 19:56
  • $\begingroup$ @Paata Thanks! I somehow overlooked the word "even" in the question. $\endgroup$ May 2, 2018 at 19:59

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