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By plotting the function and its derivatives, one can easily be convinced that the function $$f(x):=\log\binom{x}{p x}=\log\Gamma(x+1)-\log\Gamma(px+1)-\log\Gamma((1-p)x+1),$$ defined for $x>0$ and $p \in (0,1)$, is completely monotone (i.e., for all $x$, $f(x)>0$, $f'(x)<0$, $f''(x)>0$, etc). How can this statement be proved, or is there a known proof?

By Bernstein's theorem on monotone functions, the statement is equivalent to $f$ being the Laplace transform of a non-negative function. I tried to use the Taylor expansion of log-gamma $$\log\Gamma(x+1) = -\gamma x+\sum_{j=2}^\infty \frac{\zeta(j)}{j} (-x)^j,$$ and divide the $j$th term by $j!$ (which is basically what the inverse Laplace transform does). But the result doesn't seem to be a known function (without the $\zeta(j)$ constants, though, the resulting series would essentially become the expansion of the exponential integral function.

Edit: The fact that $f'(x)<0$ is equivalent to concavity of the digamma function, which can be shown by considering its integral form (see the first answer below). However, the higher derivatives seem less obvious to work with.

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4 Answers 4

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The following is the completely monotonic claim that actually holds (also hinted by Iosif Pinelis).

Claim. Let $f(x)=\log\binom{x}{px}$; then, $f''$ is CM.

We prove this claim as a corollary of the following impressive generalization.

Theorem. (Karp and Prilepkina, 2015). Consider the ratio \begin{equation*} W(x) := \frac{\prod_{i=1}^p \Gamma(A_ix+a_i)}{\prod_{j=1}^q\Gamma(B_jx+b_j)}, \end{equation*} where $(A_i,a_i)$ and $(B_j,b_j)$ are positive scalars. Then, $(\log W(x))''$ is CM if and only if \begin{equation*} P(u) = \sum_{i=1}^p \frac{e^{-a_iu/A_i}}{1-e^{-u/A_i}} - \sum_{j=1}^q \frac{e^{-b_ju/B_j}}{1-e^{-u/B_j}} \ge 0,\quad\forall u > 0. \end{equation*} In the affirmative case, $(\log W(x))'' = \int_0^\infty e^{-xu}uP(u)du$.

Proof of the claim. Let $q:= 1-p$. We need to check positivity of \begin{equation*} k(u) := \frac{e^{-u}}{1-e^{-u}} - \frac{e^{-u/p}}{1-e^{-u/p}} - \frac{e^{-u/q}}{1-e^{-u/q}}. \end{equation*} Observe that for $p\to 0$ or $p\to 1$, $k(u)\to 0$, which is promising. Simplifying, we reduce $k\ge 0$ to showing positivity of \begin{equation*} g(u) := \frac{1}{e^u-1} - \frac{1}{e^{u/p}-1} - \frac{1}{e^{u/q}-1}, \end{equation*} which I leave to the OP to verify.

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  • $\begingroup$ NIce! What Karp and Prilepkina do (in their Lemma 1) is differentiate the Dirichlet formula I was trying to use, to get a nicer integral expression for $\psi'$. Then the rest is indeed easy. $\endgroup$ Sep 1, 2017 at 1:28
  • $\begingroup$ Nice, thanks! Positivity of $g(u)$ is easy to show. $\endgroup$
    – MCH
    Sep 1, 2017 at 2:17
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    $\begingroup$ @IosifPinelis I also started with differentiating the $\psi$ formula to get nice expressions, but then remembered that D. Karp had once sent me a paper with CM of Gamma functions, so decided to look into it for a more general result. Curiously, however, in general verifying $P(u)\ge 0$ is hard. Karp and Prilepkina provide some sufficient conditions also, and some necessary ones, but otherwise, seems not so easy to use their condition in general. $\endgroup$
    – Suvrit
    Sep 1, 2017 at 2:17
  • $\begingroup$ @Suvrit : I of course agree that in general it could be hard to verify $P(u)\ge0$. $\endgroup$ Sep 1, 2017 at 3:00
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This is a not a complete answer, but possibly a first step in it.

Using the Dirichlet formula (Theorem~1.6.1 in G. E. ANDREWS, R. ASKEY, AND R. ROY, Special functions, volume 71 of Encyclopedia of Mathematics and its Applications, Cambridge University Press, Cambridge, 1999) and then the substitution $v:=\frac1{1+z}$, one has \begin{equation} \psi(x+h)-\psi(x)=\int_0^\infty\Big(\frac1{(1+z)^x}-\frac1{(1+z)^{x+h}}\Big)\frac{d z}z =\int_0^1 v^{x-1}\frac{1-v^h}{1-v}d v \end{equation} for positive $x$ and $h$. So, \begin{equation} f'(x)=\int_0^1(pv^{px}+qv^{qx}-v^x)\frac{dv}{1-v}, \end{equation} where $q:=1-p$. In particular, it follows that $f'$ is positive (rather than negative). However, it appears that $f''$ is indeed completely monotone.

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  • $\begingroup$ I suppose one can also directly use Euler's integral representation of Harmonic numbers: $H_n=\int_0^1 (1-x^n) dx/(1-x)$, which is essentially equivalent to your proof. $\endgroup$
    – MCH
    Aug 31, 2017 at 22:29
  • $\begingroup$ I am very sorry -- but the term $-v^x$ in the parentheses under the last integral sign was missing. $\endgroup$ Aug 31, 2017 at 22:42
  • $\begingroup$ There should also be a multiplicative $\log v$ factor resulting from the derivative of $v^x$, right? I think that's what makes the derivative negative. The idea is clear, though. Thanks again! $\endgroup$
    – MCH
    Aug 31, 2017 at 22:53
  • $\begingroup$ How do you argue positivity of the second derivative, though? Inside the integral we'll get a $-v^x+p^2 v^{p x}+(1-p)^2 v^{x(1-p)}$ which seems to be sometimes positive and sometimes negative depending on $v$... $\endgroup$
    – MCH
    Aug 31, 2017 at 23:11
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    $\begingroup$ $f'>0$ also follows because $v^{px}>v^x$ and $v^{qx}>v^x$ for $x>0$ and $p$ and $v$ in $(0,1)$. $\endgroup$ Sep 1, 2017 at 0:13
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This is a footnote to Suvrit's answer: He reduces the problem to verifying the condition $$\frac{1}{e^u-1} \geq \frac{1}{e^{u/p}-1} + \frac{1}{e^{u/q}-1}$$ for $u>0$, $p+q=1$ and $p,q > 0$. Put $g(p) = \tfrac{1}{e^{u/p}-1}$ and extend continuously to $g(0)=0$. It is sufficient to show that $g$ is convex on $[0,1]$.

We have $$\frac{d^2 g}{(dp)^2} = \frac{e^{u/p} u}{p^3 (e^{u/p}-1)^2} \left( \frac{u}{p} \frac{e^{u/p}+1}{e^{u/p}-1}-2 \right)$$ so we just need to show the quantity in parentheses is positive. Putting $v=u/p$, we need to show $\tfrac{v (e^v+1)}{e^v-1} \geq 2$. We can rewrite this as $\tfrac{v}{2} \geq \tfrac{e^{v/2}-e^{-v/2}}{e^{v/2}+e^{-v/2}}= \tanh \tfrac{v}{2}$. Since the derivative of $\tanh x$ is $\tfrac{1}{\cosh^2 x}<1$, we have $x \geq \tanh x$. $\square$

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  • $\begingroup$ Thanks for filling in the missing details! Indeed, convexity here suffices to prove the desired superadditivity $g(p+q)\ge g(p)+g(q)$. $\endgroup$
    – Suvrit
    Sep 1, 2017 at 2:14
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    $\begingroup$ Another way to see positivity of $g$ is to prove $p/(e^u-1)\geq 1/(e^{u/p}-1)$, which is immediate by rearranging and taking the derivative in $u$. $\endgroup$
    – MCH
    Sep 1, 2017 at 2:20
  • $\begingroup$ @MahdiCheraghchi Oh, that's much better. Maybe the nicest way to see it is, since $e^x$ is convex, $\tfrac{e^{u/p}-1}{u/p} \geq \tfrac{e^u-1}{u}$. $\endgroup$ Sep 1, 2017 at 11:30
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As done in the above answers and comments, the positivity of the function \begin{equation*} g(u)= \frac{1}{e^u-1}-\frac{1}{e^{u/p}-1} - \frac{1}{e^{u/q}-1} \end{equation*} is equivalent to the inequality \begin{equation} \frac{1}{e^u-1} \geq \frac{1}{e^{u/p}-1} + \frac{1}{e^{u/q}-1}\tag{1}\label{1} \end{equation} for $u>0$ and $p,q>0$ such that $p+q=1$.

The inequality \eqref{1} was recovered in [1, Lemma 1] by Alzer in the form of \begin{equation} \frac{1}{y-1}-\frac{1}{y^{1/\lambda}-1}-\frac{1}{y^{1/(1-\lambda)}-1}>0, \quad y>1, \quad \lambda\in(0,1).\tag{2}\label{2} \end{equation} In [2, Lemma 1.4], Ouimet generalized the inequality \eqref{2} to \begin{equation} \frac1{y-1}-\sum_{k=1}^n\frac{1}{y^{1/\lambda_k}-1}>0,\tag{3}\label{3} \end{equation} where $y>1$ and $\lambda_1,\lambda_2,\dotsc,\lambda_n\in(0,1)$ such that $\sum_{k=1}^{n}\lambda_k=1$.

In the paper [3], the inequality \begin{equation} \Biggl(\sum_{k=1}^n\lambda_k\Biggr) H\biggl(\frac{x}{\sum_{k=1}^n\lambda_k}\biggr) \ge\sum_{k=1}^{n}\lambda_k H\biggl(\frac{x}{\lambda_k}\biggr)\tag{4}\label{4} \end{equation} for $\lambda_k>0$ and $x>0$ was proved to be true, where $$ H(t)=\frac{t}{e^t-1}, \quad t\in\mathbb{R} $$ is the generating function of the Bernoulli numbers.

In the preprint [4], the inequality \begin{equation} \Biggl(\sum_{k=1}^n\lambda_k\Biggr)^3 H\biggl(\frac{x}{\sum_{k=1}^n\lambda_k}\biggr) \ge\sum_{k=1}^{n}\lambda_k^3 H\biggl(\frac{x}{\lambda_k}\biggr)\tag{5}\label{5} \end{equation} was proved to be true for $\lambda_k>0$ and $x>0$.

In the preprint [5], Ouimet proved the inequality \begin{equation} \sum_{i=1}^m\frac1{y^{1/\nu_i}-1}+\sum_{j=1}^n\frac1{y^{1/\tau_j}-1} >\sum_{i=1}^m\sum_{j=1}^n\frac1{y^{1/\lambda_{ij}}-1}\tag{6}\label{6} \end{equation} for $y>1$ and $0<\lambda_{ij}\le1$, where $\nu_i=\sum_{j=1}^n\lambda_{ij}$ and $\tau_j=\sum_{i=1}^m\lambda_{ij}$ satisfying $\sum_{i=1}^m\nu_i=\sum_{j=1}^n\tau_j=1$.

In the paper [6, Theorem 3.1], Ouimet's inequality \eqref{6} was refined as \begin{equation} \sum_{i=1}^m\frac1{e^{x/\nu_i}-1}+\sum_{j=1}^n\frac1{e^{x/\tau_j}-1} \ge2\sum_{i=1}^m\sum_{j=1}^n\frac1{e^{x/\lambda_{ij}}-1}\tag{7}\label{7} \end{equation} for $x>0$ and $\lambda_{ij}>0$, where $\nu_i=\sum_{j=1}^n\lambda_{ij}$ and $\tau_j=\sum_{i=1}^m\lambda_{ij}$.

The inequality \eqref{7} established in [6, Theorem 3.1] extends, generalizes, and refines all of the above inequalities other than \eqref{5}.

In the paper [7], the inequalities \eqref{5} and \eqref{7} were generalized as follows.

  • For $\alpha\ge1$, $x>0$, and $\lambda_{ij}>0$ for $1\le i\le m$ and $1\le j\le n$, denote $\nu_i=\sum_{j=1}^n\lambda_{ij}$ and $\tau_j=\sum_{i=1}^m\lambda_{ij}$. Then \begin{equation}\label{alpha>=1-inequal}\tag{8} \sum_{i=1}^m\frac{\nu_i^{\alpha-1}}{e^{x/\nu_i}-1}+ \sum_{j=1}^n\frac{\tau_j^{\alpha-1}}{e^{x/\tau_j}-1} \ge2\sum_{i=1}^m\sum_{j=1}^n\frac{\lambda_{ij}^{\alpha-1}}{e^{x/\lambda_{ij}}-1}. \end{equation}
  • Let $\alpha\ge1$, $x>0$, and $\lambda_{ijk}>0$ for $1\le i\le\ell$, $1\le j\le m$, and $1\le k\le n$. Then \begin{multline}\label{2:1Sum-Ineq} \sum_{k=1}^n\sum_{j=1}^m\frac{\bigl(\sum_{i=1}^\ell\lambda_{ijk}\bigr)^{\alpha-1}}{e^{x/\sum_{i=1}^\ell\lambda_{ijk}}-1} +\sum_{i=1}^\ell\sum_{k=1}^n\frac{\bigl(\sum_{j=1}^m\lambda_{ijk}\bigr)^{\alpha-1}}{e^{x/\sum_{j=1}^m\lambda_{ijk}}-1}\\ +\sum_{j=1}^m\sum_{i=1}^\ell\frac{\bigl(\sum_{k=1}^n\lambda_{ijk}\bigr)^{\alpha-1}}{e^{x/\sum_{k=1}^n\lambda_{ijk}}-1} \ge3\sum_{k=1}^n\sum_{j=1}^m\sum_{i=1}^\ell\frac{\lambda_{ijk}^{\alpha-1}}{e^{x/\lambda_{ijk}}-1} \end{multline} and \begin{multline}\label{1:2Sum-Ineq} \sum_{k=1}^n\frac{\bigl(\sum_{j=1}^m\sum_{i=1}^\ell\lambda_{ijk}\bigr)^{\alpha-1}}{e^{x/\sum_{j=1}^m\sum_{i=1}^\ell\lambda_{ijk}}-1} +\sum_{i=1}^\ell\frac{\bigl(\sum_{k=1}^n\sum_{j=1}^m\lambda_{ijk}\bigr)^{\alpha-1}}{e^{x/\sum_{k=1}^n\sum_{j=1}^m\lambda_{ijk}}-1}\\ +\sum_{j=1}^{m} \frac{\bigl(\sum_{i=1}^\ell\sum_{k=1}^n\lambda_{ijk}\bigr)^{\alpha-1}}{e^{x/\sum_{i=1}^\ell\sum_{k=1}^n\lambda_{ijk}}-1} \ge3\sum_{k=1}^n\sum_{j=1}^m\sum_{i=1}^\ell\frac{\lambda_{ijk}^{\alpha-1}}{e^{x/\lambda_{ijk}}-1}. \end{multline}

The inequality \eqref{alpha>=1-inequal} was also employed in the paper [8].

In the preprint [9], whose revised version has been accepted by Acta Mathematica Scientia, the inequality \begin{equation} \Biggl(\sum_{k=1}^{n}w_k\lambda_k\Biggr)^{\alpha+1}H\Biggl(\frac{x}{\sum_{k=1}^{n}w_k\lambda_k}\Biggr) \le\sum_{k=1}^{n}w_k\lambda_k^{\alpha+1}H\biggr(\frac{x}{\lambda_k}\biggr) \end{equation} was proved, where $\alpha\ge0$, $n\ge2$, $\lambda_k\in(0,\infty)$ and $w_k\in(0,1)$ for $1\le k\le n$, and $\sum_{k=1}^{n}w_k=1$.

Finally, a closely-related or similar inequality was proved in the preprint [10, Lemma 2.1].

References

  1. H. Alzer, Complete monotonicity of a function related to the binomial probability, J. Math. Anal. Appl. 459 (2018), no. 1, 10--15; available online at https://doi.org/10.1016/j.jmaa.2017.10.077.
  2. F. Ouimet, Complete monotonicity of multinomial probabilities and its application to Bernstein estimators on the simplex, J. Math. Anal. Appl. 466 (2018), no. 2, 1609--1617; available online at https://doi.org/10.1016/j.jmaa.2018.06.049.
  3. F. Qi, D.-W. Niu, D. Lim, and B.-N. Guo, Some logarithmically completely monotonic functions and inequalities for multinomial coefficients and multivariate beta functions, Appl. Anal. Discrete Math. 14 (2020), no. 2, 512--527; available online at https://doi.org/10.2298/AADM191111033Q.
  4. F. Qi, A logarithmically completely monotonic function involving the $q$-gamma function, HAL preprint (2018), available online at https://hal.archives-ouvertes.fr/hal-01803352v1.
  5. F. Ouimet, Complete monotonicity of a ratio of gamma functions and some combinatorial inequalities for multinomial coefficients, arXiv preprint (2019), available online at https://arxiv.org/abs/1907.05262v2.
  6. F. Qi, W.-H. Li, S.-B. Yu, X.-Y. Du, and B.-N. Guo, A ratio of finitely many gamma functions and its properties with applications, Rev. R. Acad. Cienc. Exactas Fis. Nat. Ser. A Math. RACSAM. 115 (2021), no. 2, Paper No. 39, 14 pages; available online at https://doi.org/10.1007/s13398-020-00988-z.
  7. Feng Qi and Bai-Ni Guo, From inequalities involving exponential functions and sums to logarithmically complete monotonicity of ratios of gamma functions, Journal of Mathematical Analysis and Applications 493 (2021), no. 1, Paper No. 124478, 19 pages; available online at https://doi.org/10.1016/j.jmaa.2020.124478.
  8. Feng Qi and Dongkyu Lim, Monotonicity properties for a ratio of finite many gamma functions, Advances in Difference Equations 2020, Paper No. 193, 9 pages; available online at https://doi.org/10.1186/s13662-020-02655-4.
  9. Feng Qi, Complete monotonicity for a new ratio of finitely many gamma functions, Acta Mathematica Scientia Series B English Edition 42B (2022), no. 2, 511--520; available online at https://doi.org/10.1007/s10473-022-0206-9.
  10. Frederic Ouimet and Feng Qi, Logarithmically complete monotonicity of a matrix-parametrized analogue of the multinomial distribution, arXiv (2021), available online at https://arxiv.org/abs/2105.01494.
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