# Khabibullin's conjecture on integral inequalities

In mathematics, Khabibullin's conjecture, named after B. N. Khabibullin, is related to Paley's problem[1] for plurisubharmonic functions and to various extremal problems in the theory of entire functions of several variables.

## The first statement in terms of logarithmically convex functions

Khabibullin's conjecture (version 1, 1992). Let ${\displaystyle \displaystyle S}$ be a non-negative increasing function on the half-line ${\displaystyle [0,+\infty )}$ such that ${\displaystyle \displaystyle S(0)=0}$. Assume that ${\displaystyle \displaystyle S(e^{x})}$ is a convex function of ${\displaystyle x\in [-\infty ,+\infty )}$. Let ${\displaystyle \lambda \geq 1/2}$, ${\displaystyle n\geq 2}$, and ${\displaystyle n\in \mathbb {N} }$. If

${\displaystyle \int _{0}^{1}S(tx)\,(1-x^{2})^{n-2}\,x\,dx\leq t^{\lambda }{\text{ for all }}t\in [0,+\infty ),}$

(1)

then

${\displaystyle \int _{0}^{+\infty }S(t)\,{\frac {t^{2\lambda -1}}{(1+t^{2\lambda })^{2}}}\,dt\leq {\frac {\pi \,(n-1)}{2\lambda }}\prod _{k=1}^{n-1}{\Bigl (}1+{\frac {\lambda }{2k}}{\Bigr )}.}$

(2)

This statement of the Khabibullin's conjecture completes his survey.[2]

## Relation to Euler's Beta function

Note that the product in the right hand side of the inequality (2) is related to the Euler's Beta function ${\displaystyle \mathrm {B} }$:

${\displaystyle {\frac {\pi \,(n-1)}{2\lambda }}\prod _{k=1}^{n-1}{\Bigl (}1+{\frac {\lambda }{2k}}{\Bigr )}={\frac {\pi \,(n-1)}{\lambda ^{2}}}\cdot {\frac {1}{\mathrm {B} (\lambda /2,n)}}}$

## Discussion

For each fixed ${\displaystyle \lambda \geq 1/2}$ the function

${\displaystyle S(t)=2(n-1)\prod _{k=1}^{n-1}{\Bigl (}1+{\frac {\lambda }{2k}}{\Bigr )}\,t^{\lambda },}$

turns the inequalities (1) and (2) to equalities.

The Khabibullin's conjecture is valid for ${\displaystyle \lambda \leq 1}$ without the assumption of convexity of ${\displaystyle S(e^{x})}$. Meanwhile, one can show that this conjecture is not valid without some convexity conditions for ${\displaystyle S}$. In 2010, R. A. Sharipov showed that the conjecture fails in the case ${\displaystyle n=2}$ and for ${\displaystyle \lambda =2}$.[3]

## The second statement in terms of increasing functions

Khabibullin's conjecture (version 2). Let ${\displaystyle \displaystyle h}$ be a non-negative increasing function on the half-line ${\displaystyle [0,+\infty )}$ and ${\displaystyle \alpha >1/2}$. If

${\displaystyle \int _{0}^{1}{\frac {h(tx)}{x}}\,(1-x)^{n-1}\,dx\leq t^{\alpha }{\text{ for all }}t\in [0,+\infty ),}$

then

${\displaystyle \int _{0}^{+\infty }{\frac {h(t)}{t}}\,{\frac {dt}{1+t^{2\alpha }}}\leq {\frac {\pi }{2}}\prod _{k=1}^{n-1}{\Bigl (}1+{\frac {\alpha }{k}}{\Bigr )}={\frac {\pi }{2\alpha }}\cdot {\frac {1}{\mathrm {B} (\alpha ,n)}}.\,}$

## The third statement in terms of non-negative functions

Khabibullin's conjecture (version 3). Let ${\displaystyle \displaystyle q}$ be a non-negative continuous function on the half-line ${\displaystyle [0,+\infty )}$ and ${\displaystyle \alpha >1/2}$. If

${\displaystyle \int _{0}^{1}{\Bigl (}\,\int _{x}^{1}(1-y)^{n-1}{\frac {dy}{y}}{\Bigr )}q(tx)\,dx\leq t^{\alpha -1}{\text{ for all }}t\in [0,+\infty ),}$

then

${\displaystyle \int _{0}^{+\infty }q(t)\log {\Bigl (}1+{\frac {1}{t^{2\alpha }}}{\Bigr )}\,dt\leq \pi \alpha \prod _{k=1}^{n-1}{\Bigl (}1+{\frac {\alpha }{k}}{\Bigr )}={\frac {\pi }{\mathrm {B} (\alpha ,n)}}.\,}$

## References

1. ^ Khabibullin B.N. (1999). "Paley problem for plurisubharmonic functions of finite lower order". Sbornik: Mathematics. 190 (2): 309–321.
2. ^ Khabibullin BN (2002). "The representation of a meromorphic function as the quotient of entire functions and Paley problem in ${\displaystyle \displaystyle \mathbb {C} ^{n}}$: a survey of some results". Mat. Fizika, analiz, geometria. 9 (2): 146–167. arXiv:math.CV/0502433.
3. ^ Sharipov, R. A. (2010). "A Counterexample to Khabibullin's Conjecture for Integral Inequalities". Ufa Mathematical Journal,. 2 (4): 99–107. arXiv:1008.2738.