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.


3 Answers 3


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.

  • $\begingroup$ Good catch, and curious coincidence. It may be the case that topic starters work on related problems. $\endgroup$ Commented Mar 28, 2019 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 $(\star)$.

  • 1
    $\begingroup$ For what it worth, your conjecture is equivalent to $g(x) =xf''$ being increasing function (it is $g(ax) \le g(x) $.) $\endgroup$ Commented Mar 26, 2019 at 0:39
  • $\begingroup$ @FedorPetrov: Yes, indeed. $\endgroup$ Commented Mar 26, 2019 at 0:51
  • $\begingroup$ 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. $\endgroup$
    – Alex
    Commented Mar 26, 2019 at 3:46
  • $\begingroup$ I find it quite remarkable that exactly this inequality appeared recently here on MO (see my answer and the reference therein). $\endgroup$ Commented Mar 27, 2019 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.


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