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I'm having trouble proving the following inequality:

$$\forall p>1 \quad \forall m\geq 0 \quad \dfrac{m^2\Gamma(\dfrac{2m}{p})\Gamma(\dfrac{2m}{q})}{\Gamma(\dfrac{2m+2}{p})\Gamma(\dfrac{2m+2}{q})}\geq\dfrac{1}{4}p^2(p-1)^{\frac{2}{p}-2},$$ where as usual $q=\dfrac{p}{p-1}$. In fact, it seems clear from Mathematica that for a fixed $p$, the LHS is a decreasing function of $m$ (strictly unless $p=2$, in which case it's constant). The RHS can be seen to be the limit as $m\to \infty$. I actually only care about integer $m\geq 0$, but I don't find that helpful.

I have tried both a direct approach (three known inequalities that are nice enough to apply here, but lead to wrong inequalities) and working with the derivative, which naturally involves instances of the digamma function. Proving that the LHS is decreasing is equivalent to the following inequality: $$\forall p>1 \quad \forall m\geq 0 \quad \dfrac{1}{m}+\dfrac{1}{p}(\psi(\dfrac{2m}{p})-\psi(\dfrac{2m+2}{p}))+\dfrac{1}{q}(\psi(\dfrac{2m}{q})-\psi(\dfrac{2m+2}{q}))\leq0,$$ which again seems to be correct (if you're wondering, the limit as $m\to 0$ is negative for $p\neq2$). Much like before, I tried using two inequalities (for the digamma function), as well as the series representation. They seemed promising at first, but the inequalities gave me positive upper bounds, while the series converges too slowly to be useful (I suspect that any partial sum is positive for large enough $m$).

Any advice would be much appreciated. I'll be glad to explain more about the inequalities I've tried if requested.

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  • $\begingroup$ For $p=2$ this is equality, is not it? $\endgroup$ – Fedor Petrov Feb 2 at 17:04
  • $\begingroup$ Right. That's the only case where you get equality (and for all $m$ at that), I believe. I'll change the inequality sign to be precise. $\endgroup$ – Yonatan Shelah Feb 2 at 17:06
  • $\begingroup$ It looks (numerically) like $h_{a,b}(x)=\psi(a/x)-\psi(b/x)$ is convex in $x\in (0,\infty)$ whenever $a>b$. If true, applying this for $a=2m+2,b=2m$ we get by Jensen inequality with weights $p^{-1},q^{-1}$: $p^{-1}h(p)+q^{-1}h(q)\geqslant h(p^{-1}\cdot p+q^{-1}\cdot q)=h(2)=\frac1m$ as needed. $\endgroup$ – Fedor Petrov Feb 2 at 22:24
  • $\begingroup$ Nice. I'll focus on proving convexity, but that is of course an inequality in its own right. The second derivative seems quite small even when $a,b$ are far apart. $\endgroup$ – Yonatan Shelah Feb 2 at 22:53
  • $\begingroup$ it suffices (and is necessary) to prove that $\partial_a \partial_x^2 \psi(a/x)\geqslant 0$. This may be written as an integral or a sum, but still did not help me. $\endgroup$ – Fedor Petrov Feb 3 at 1:09
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By Fedor Petrov's comment, it is enough to show that \begin{equation} g(a,x):=\frac{\partial^2}{\partial^2 x}\,\psi(a/x) \end{equation} is increasing in $a>0$ for each $x>0$. We have \begin{equation} g(a,x)=f(u)/x^2, \end{equation} where $u:=a/x$ and \begin{equation} f(u)=\psi''(u)u^2+2\psi'(u)u. \end{equation} So, it is enough to show that $f'>0$ on $(0,\infty)$.

By the Gauss formula, Theorem 1.6.1
\begin{equation} \psi(u)=\int_0^\infty\Big(\frac{e^{-z}}z-\frac{e^{-zu}}{1-e^{-z}}\Big)dz \end{equation} for $u>0$, we have \begin{equation} f'(u)=\psi'''(u)u^2+4\psi''(u)u+2\psi'(u) =\int_0^\infty U(z)(2 - 4 u z + u^2 z^2)e^{-zu} dz, \tag{1} \end{equation} where \begin{equation} U(z):=\frac{z}{1-e^{-z}}. \end{equation} Integrating twice by parts, we have \begin{equation} f'(u) =\int_0^\infty U'(z)z (-2 + u z)e^{-zu} dz =\int_0^\infty U''(z)z^2 e^{-zu}dz>0, \end{equation}
as desired. Here we used the fact that \begin{equation} U''(z)=\frac{e^{-2z} (2 + e^z (-2 + z) + z)}{(1 - e^{-z})^3} =\frac{e^{-2z} }{(1 - e^{-z})^3}\,\int_0^z(z-t)e^t t\,dt>0 \end{equation} for $z>0$.

Added in response to Fedor Petrov's comment on this answer: The factor $2 - 4 u z + u^2 z^2$ in the integrand in (1) changes its sign (twice) when $z$ varies from $0$ to $\infty$. That does not allow us to obtain the desired (and, a priori, plausible) inequality $f'(u)>0$ right from (1). However, the hope was that, if we integrate the factor $(2 - 4 u z + u^2 z^2)e^{-zu}$ repeatedly in $z$ (in the sense of the indefinite integral), then the sign pattern of this factor will get sufficiently smoothed out and become a constant sign pattern. Fortunately, this worked, and we were also fortunate to have $U''>0$ on $(0,\infty)$.

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  • $\begingroup$ Thank you. This proof checks out and I'll be sure to give an acknowledgement to both of you. $\endgroup$ – Yonatan Shelah Feb 3 at 2:48
  • $\begingroup$ It should not be a miracle which it looks like. Is there any explanation without calculation why double integration by parts worked? $\endgroup$ – Fedor Petrov Feb 3 at 7:53
  • $\begingroup$ @FedorPetrov : I have added such an explanation, at the end of the answer. $\endgroup$ – Iosif Pinelis Feb 3 at 14:38
  • $\begingroup$ @IosifPinelis this explains nicely why you decided to integrate by parts, but it looks not a coincidence that after double integration we get such a nice expression as $z^2 e^{-zu}$. $\endgroup$ – Fedor Petrov Feb 3 at 15:29
  • $\begingroup$ @FedorPetrov : I also had such a no-coincidence impression, but that I cannot explain at the moment. This seems to suggest an intriguing possibility of a generalization. If you ever find such a generalization, I would certainly like to hear about it. By the way, $U'>0$ but $U'''<0$ on $(0,\infty)$. $\endgroup$ – Iosif Pinelis Feb 3 at 15:43

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