# A challenging inequality that involves the digamma function and polygamma functions

Let $$f(x)=x \psi(x+1)$$, where $$\psi$$ is the digamma function. Define $$g(x)=(f(ax)+f((1-a)x)-f(x))-(f(ax+by)+f((1-a)x+(1-b)y)-f(x+y)),$$ where $$0\le a,b\le 1$$ and $$x,y\ge 0$$.

How to show that $$g(x)$$ is increasing in $$x$$ on $$[0,\infty)$$? Thank you!

Here is what I have tried. Let $$h(x,y)=f(ax+by)+f((1-a)x+(1-b)y)-f(x+y)$$. Then we have $$g(x)=h(x,0)-h(x,y)$$. To show that $$g(x)$$ is increasing, it suffices to show that $$\frac{\partial}{\partial x}(h(x,0)-h(x,y))\ge 0.$$ Then it is sufficient to show that $$\frac{\partial^2 h(x,y)}{\partial y \partial x}\le 0,$$ which is equivalent to $$f''(x+y)\ge abf''(ax+by)+(1-a)(1-b)f''((1-a)x+(1-b)y),$$ where $$f''(x)=2\psi_1(1+x)+x\psi_2(1+x)$$, where $$\psi_1$$ and $$\psi_2$$ are the first and second derivatives of the digamma function $$\psi$$. Then if I use the series representation of polygamma functions $$\psi_m(z)=(-1)^{m+1}m!\sum_{k\ge 0}\frac{1}{(z+k)^{m+1}},$$ the above inequality is equivalent to $$\sum_{k\ge 1}k\left[ \frac{1}{(x+y+k)^3}-\frac{ab}{(ax+by+k)^3}-\frac{(1-a)(1-b)}{((1-a)x+(1-b)y+k)^3} \right]\ge 0.$$ I find that if $$a,b,x,y$$ are fixed, $$\frac{1}{(x+y+k)^3}-\frac{ab}{(ax+by+k)^3}-\frac{(1-a)(1-b)}{((1-a)x+(1-b)y+k)^3}$$ is negative for small $$k$$ and positive when $$k$$ is sufficiently large. So it seems that the inequality cannot be addressed term by term. I am thinking about if it can be proved by re-arranging the terms in the series and grouping them into positive groups. I also tried the similar strategy by using the integral representation of polygamma functions $$\psi_m(z)=(-1)^{m+1}\int_0^\infty \frac{t^m e^{-zt}}{1-e^{-t}}dt,$$ but it did not work either.

Max reduced the problem to proving that $$(xf'')'\geqslant 0$$. We have $$f=x\psi(x+1)=1+x\psi(x)$$, so $$f''=(x\psi(x))''=x\psi''+2\psi'$$, $$f'''=x\psi'''+3\psi''$$,$$(xf'')'=xf'''+f''=x^2\psi'''+4x\psi''+2\psi'$$. Exactly this guy is proved to be non-negative in the recent clever MO answer by Iosif Pinelis to a different question. I wonder whether this is coincidence.

• Good catch, and curious coincidence. It may be the case that topic starters work on related problems. – Max Alekseyev Mar 28 at 0:11

Numerical evidence suggests that for any $$a\in(0,1]$$ and $$x\geq 0$$, we have $$(\star)\qquad f''(ax)\leq \frac{f''(x)}{a}.$$

The inequality $$(\star)$$ allows to easily prove that $$f''(x+y)\ge abf''(ax+by)+(1-a)(1-b)f''((1-a)x+(1-b)y).$$

Indeed, we have $$f''(ax+by) = f''(\frac{ax+by}{x+y}(x+y)) \leq \frac{x+y}{ax+by} f''(x+y)$$ and similarly $$f''((1-a)x+(1-b)y) \leq \frac{x+y}{(1-a)x+(1-b)y}f''(x+y)$$. Since $$f''(x+y)>0$$, it remains to notice that $$1 - ab\frac{x+y}{ax+by} - (1-a)(1-b)\frac{x+y}{(1-a)x+(1-b)y} = \frac{(a-b)^2xy}{(ax+by)((1-a)x+(1-b)y)} \geq 0.$$

I do not yet have a proof of $$(\star)$$.

• For what it worth, your conjecture is equivalent to $g(x) =xf''$ being increasing function (it is $g(ax) \le g(x)$.) – Fedor Petrov Mar 26 at 0:39
• @FedorPetrov: Yes, indeed. – Max Alekseyev Mar 26 at 0:51
• Thanks! I can show that the limit $\lim_{x\to +\infty} f''(x)-af''(ax)=0$ using the two-sided approximations presented in arxiv.org/pdf/0903.1984.pdf. It remains to show that $f''(x)-af''(ax)$ is decreasing in $x$. Numerical results show that $f''(x)-af''(ax)$ is totally monotone. – Heptafluoride Mar 26 at 3:46
• I find it quite remarkable that exactly this inequality appeared recently here on MO (see my answer and the reference therein). – Fedor Petrov Mar 27 at 0:14

Inspired by Max's answer, I finally have a proof for the inequality $$h(x)\triangleq f''(x)-a f''(ax)\ge 0.$$

Recall that $$\psi_k(z)\triangleq \frac{d^k}{dz^k}\psi(z)$$. Using the two-sided approximation presented in https://arxiv.org/abs/0903.1984 $$\frac{(k-1)!}{\Bigl\{x+\Bigl[\frac{(k-1)!}{|\psi_k(1)|}\Bigr]^{1/k}\Bigr\}^k} +\frac{k!}{x^{k+1}}<\bigl|\psi_k(x)\bigr|<\frac{(k-1)!}{\bigl(x+\frac12\bigr)^k}+\frac{k!}{x^{k+1}},$$ we have $$h(x)\le x \left(a^2 \left(\frac{4}{(2 a x+3)^2}+\frac{2}{(a x+1)^3}\right)-\frac{2}{(x+1)^3}-\frac{1}{\left(x+1+\sqrt{-\frac{1}{\psi ^{(2)}(1)}}\right)^2}\right)+2 \left(-\frac{a}{(a x+1)^2}-\frac{a}{a x+\frac{6}{\pi ^2}+1}+\frac{1}{(x+1)^2}+\frac{1}{x+\frac{3}{2}}\right)$$ and $$h(x) \ge x \left(a^2 \left(\frac{2}{(a x+1)^3}+\frac{1}{\left(a x+1+\sqrt{-\frac{1}{\psi ^{(2)}(1)}}\right)^2}\right)-\frac{2}{(x+1)^3}-\frac{4}{(2 x+3)^2}\right)+2 \left(-\frac{a}{(a x+1)^2}-\frac{2 a}{2 a x+3}+\frac{1}{(x+1)^2}+\frac{1}{x+\frac{6}{\pi ^2}+1}\right).$$ By the squeeze theorem, we know that $$\lim_{x\to \infty} h(x)=0.$$ Now it remains to show that $$h(x)$$ is decreasing in $$x$$. Let us show an even stronger result that $$h(x)$$ is completely monotone. By Bernstein's theorem on monotone functions, it suffices to show that $$h(x)=\int_0^\infty e^{-tx} dg(t),$$ where $$g(x)$$ is the cdf of a non-negative finite Borel measure on $$[0,\infty)$$.

Note that $$h(x)=\int_0^\infty\frac{\left(t^2 \left(a^2 \text{csch}^2\left(\frac{t}{2}\right)-\text{csch}^2\left(\frac{t}{2 a}\right)\right)\right) \exp (-t x)}{4 a^2}dt,$$ which completes the proof. This integral representation is obtained by using the integral representation of polygamma functions $$\psi_m(z)=(-1)^{m+1}\int_0^\infty \frac{t^m e^{-zt}}{1-e^{-t}}dt.$$ Plugging in the above integral representation and change the variable $$at$$ to $$t$$, we have $$h(x)= \int_0^{\infty } t (2-t x) \left(\frac{1}{\exp (t)-1}-\frac{1}{a \left(\exp \left(\frac{t}{a}\right)-1\right)}\right) \exp (-t x) \, dt.$$ Then we split it into two parts $$h(x)=h_1(x)+h_2(x)$$, where $$h_1(x)=\int_0^{\infty } 2t \left(\frac{1}{\exp (t)-1}-\frac{1}{a \left(\exp \left(\frac{t}{a}\right)-1\right)}\right) \exp (-t x) \, dt$$ and $$h_2(x)=\int_0^{\infty } t (-t x) \left(\frac{1}{\exp (t)-1}-\frac{1}{a \left(\exp \left(\frac{t}{a}\right)-1\right)}\right) \exp (-t x) \, dt\\=\int_0^{\infty } t^2 \left(\frac{1}{\exp (t)-1}-\frac{1}{a \left(\exp \left(\frac{t}{a}\right)-1\right)}\right) d\exp (-t x)\\ =\int_0^\infty \exp (-t x) d\left(t^2 \left(\frac{1}{\exp (t)-1}-\frac{1}{a \left(\exp \left(\frac{t}{a}\right)-1\right)}\right)\right)\\=\int_0^\infty \exp(-tx) t \left(\frac{t}{a^2 \left(e^{t/a}-1\right)^2}+\frac{t-2 a}{a^2 \left(e^{t/a}-1\right)}-\frac{e^t (t-2)+2}{\left(e^t-1\right)^2}\right) dt\,.$$ Putting them together again and simplifying it yields the desired result.