Inequality involving Pochhammer symbols

Question

Let $m,S$ be integers satisfying $2\leq m\leq S$. I would like to show that

$$h_1\left(x\right) h_3\left(x\right) \leq h_2^2\left(x\right)$$

for all $x\geq 0$ where

$$h_k\left(x\right) \equiv {}_2F_1\left(k,k-1-m;S-m+k,-x\right).$$

${}_2F_1$ denotes the hypergeometric function.

What I have so far...

Note that \begin{multline*} h_{1}\left(y\right)h_{3}\left(y\right)=\sum_{d=0}^{\infty}\left(-y\right)^{d}\sum_{p=0}^{d}\frac{\left(1\right)^{\left(p\right)}\left(-m\right)^{\left(p\right)}}{\left(S-m+1\right)^{\left(p\right)}p!}\frac{\left(3\right)^{\left(q\right)}\left(2-m\right)^{\left(q\right)}}{\left(S-m+3\right)^{\left(q\right)}q!}\\ =\sum_{d=0}^{\infty}y^{d}\sum_{p=0}^{d}\frac{m_{p}}{\left(S-m+1\right)^{\left(p\right)}}\frac{\left(q+1\right)\left(q+2\right)\left(m-2\right)_{q}}{2\left(S-m+3\right)^{\left(q\right)}} \end{multline*} where $q\equiv d-p$ and the Pochhammer symbols are defined here: https://en.wikipedia.org/wiki/Pochhammer_symbol. Similarly, \begin{multline*} h_{2}^{2}\left(y\right)=\sum_{d=0}^{\infty}\left(-y\right)^{d}\sum_{p=0}^{d}\frac{\left(2\right)^{\left(p\right)}\left(1-m\right)^{\left(p\right)}}{\left(S-m+2\right)^{\left(p\right)}p!}\frac{\left(2\right)^{\left(q\right)}\left(1-m\right)^{\left(q\right)}}{\left(S-m+2\right)^{\left(q\right)}q!}\\ =\sum_{d=0}^{\infty}y^{d}\sum_{p=0}^{d}\frac{\left(p+1\right)\left(m-1\right)_{p}}{\left(S-m+2\right)^{\left(p\right)}}\frac{\left(q+1\right)\left(m-1\right)_{q}}{\left(S-m+2\right)^{\left(q\right)}}. \end{multline*} So it is sufficient to show $$ \sum_{p=0}^{d}\frac{\left(q+1\right)\left(q+2\right)m_{p}\left(m-2\right)_{q}}{2\left(S-m+1\right)^{\left(p\right)}\left(S-m+3\right)^{\left(q\right)}}-\frac{\left(q+1\right)\left(p+1\right)\left(m-1\right)_{p}\left(m-1\right)_{q}}{\left(S-m+2\right)^{\left(p\right)}\left(S-m+2\right)^{\left(q\right)}}\leq0, $$ for all $d\geq0$.

Note that for $d \geq 2m - 1$, the above is zero (see also @Semiclassical's comment below), so that we only have to verify the claim for $0 \leq d \leq 2\left(m -1\right)$.

Answer 1

Added 2014-09-01:

• Some mistakes corrected in the calculation of the coefficients of $h_2^2(x)-h_1(x)h_3(x)$
• Two interesting references regading log-concavity added

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Note: This is not a complete answer but hopefully a proper approach towards a solution. As I started from scratch it looks somewhat different compared to the calculations from @par. At the time I can provide two additional aspects.

1.) A nice representation of $h_k$:

Let $h_k={}_{2}F_{1}(k,k-1-m;S-m+k;-x)$ with $2\leq m\leq S$ and $k\in\{1,2,3\}$. The following is valid:

\begin{align*} h_k=\frac{1}{\binom{S}{m-k+1}}\sum_{n=k}^{m}\binom{n}{k-1}\binom{S}{m-n}x^{n-k+1} \end{align*}

This follows from

\begin{align*} h_k&={}_{2}F_{1}(k,k-1-m;S-m+k;-x)\\ &=\sum_{n\geq1}\frac{(k)_n(k-1-m)_n}{(S-m+k)_n}\frac{(-x)^n}{n!}\\ &=\sum_{n=1}^{m-k+1}\frac{(k)_n(m-k+2-n)_n}{(S-m+k)_n}\frac{x^n}{n!}\\ &=\sum_{n=1}^{m-k+1}\frac{(n+k-1)!}{(k-1)!}\frac{(m-k+1)!}{(m-n-k+1)!}\frac{(S-m+k-1)!}{(S-m+n+k-1)!}\frac{x^n}{n!}\\ &=\frac{1}{\binom{S}{m-k+1}}\sum_{n=1}^{m-k+1}\binom{n+k-1}{n}\binom{S}{m-n-k+1}x^n\\ &=\frac{1}{\binom{S}{m-k+1}}\sum_{n=k}^{m}\binom{n}{k-1}\binom{S}{m-n}x^{n-k+1} \end{align*}

So, for $k\in\{1,2,3\}$ we get a representation for $h_k$ which looks attractive:

\begin{align*} h_1(x)&=\frac{1}{\binom{S}{m}}\sum_{n=1}^{m}\binom{S}{m-n}x^n\\ h_2(x)&=\frac{1}{\binom{S}{m-1}}\sum_{n=2}^{m}n\binom{S}{m-n}x^{n-1}\tag{1}\\ h_3(x)&=\frac{1}{2\binom{S}{m-2}}\sum_{n=3}^{m}n(n-1)\binom{S}{m-n}x^{n-2} \end{align*}

We clearly see, that essentially differentiating $h_1$ generates $h_2$ and $h_3$. But at the time I couldn't find a way to use this fact in order to show the log-concavity property of $h_k$:

\begin{align*} h_2^{2}(x) \geq h_1(x)h_3(x) \qquad\qquad x>0\tag{2} \end{align*}

While playing with the $h_k$ and doing some computation another interesting aspect appeared:

2.) Crucial assumption: The polynomial $h_2^2(x)-h_1(x)h_3(x)$ has non-negative coefficients only.

If we can show that the coefficients $$[x^n](h_2^2(x)-h_1(x)h_3(x))\geq 0$$ for all $n\geq 0$ the log-concavity property (2) follows immediately.

Using following shortcuts in (1): $$\beta_m=\binom{S}{m}^{-1}\quad\text{and}\quad a_n=\binom{S}{m-n}$$ we can write

\begin{align*} h_1(x)&=\beta_m\sum_{n=1}^{m}a_nx^n\\ h_2(x)&=\beta_{m-1}\sum_{n=2}^{m}na_nx^{n-1}\\ h_3(x)&=\frac{1}{2}\beta_{m-2}\sum_{n=3}^{m}n(n-1)a_nx^{n-2} \end{align*}

In fact $a_n$ also depends on $m$ but this won't matter in the following calculation and I'd like to keep the notation simple.

Now we first calculate $h_1(x)h_3(x)$ and $h_2^2(x)$.

\begin{align*} h_1(x)&h_3(x)=\\ &=\frac{1}{2x^2}\beta_m\beta_{m-2}\sum_{j=1}^{m}a_jx^j\sum_{l=3}^{m}l(l-1)a_lx^l\\ &=\frac{1}{2x^2}\beta_m\beta_{m-2}\sum_{n=4}^{2m}\sum_{{j+l=n}\atop{{1\leq j \leq m}\atop{3\leq l \leq m}}}l(l-1)a_ja_lx^n\\ &=\frac{1}{2}\beta_m\beta_{m-2}\left(\sum_{n=4}^{m+1}\sum_{j=1}^{n-3}(n-j)(n-j-1)a_ja_{n-j}x^{n-2}\right.\\ &\qquad+\sum_{n=m+2}^{m+2}\sum_{j=n-m}^{m-1}(n-j)(n-j-1)a_ja_{n-j}x^{n-2}\\ &\qquad+\left.\sum_{n=m+3}^{2m}\sum_{j=n-m}^{m}(n-j)(n-j-1)a_ja_{n-j}x^{n-2}\right)\tag{3}\\ \end{align*}

and

\begin{align*} h_2^2(x)&=\frac{1}{x^2}\beta_{m-1}^{2}\sum_{j=2}^{m}ja_jx^j\sum_{l=2}^{m}la_lx^l\\ &=\frac{1}{x^2}\beta_{m-1}^{2}\sum_{n=4}^{2m}\sum_{{j+l=n}\atop{2\leq j,l \leq m}}jla_ja_lx^n\\ &=\beta_{m-1}^{2}\left(\sum_{n=4}^{m+2}\sum_{j=2}^{n-2}j(n-j)a_ja_{n-j}x^{n-2}\right.\\ &\qquad+\left.\sum_{n=m+3}^{2m}\sum_{j=n-m}^{m}j(n-j)a_ja_{n-j}x^{n-2}\right)\tag{4}\\ \end{align*}

Observe, that the inner sums in (3) and (4) have different index ranges which is the reason for splitting the outer sum. We have to respect this when calculating the coefficients. Therefore we split the calculation of the coefficients for $[x^n](h_2^2(x)-h_1(x)h_3(x))$ into 3 parts accordingly to \begin{align*} 4 \leq& n \leq m+1\\ &n=m+2\\ m+3 \leq &n \leq 2m \end{align*}

Here's the calculation for the first index region $4 \leq n \leq m+1$. The other calculations can be done analogously.

The coefficients of the polynomial $h_2^2(x)-h_1(x)h_3(x)$ with $n\in \{4,\dots,m+1\}$ are

\begin{align*} [x^{n-2}]&(h_2^2(x)-h_1(x)h_3(x))=\\ &=2(n-2)\beta_{m-1}^2a_2a_{n-2}\tag{5}\\ &\quad+\sum_{j=2}^{n-3}(n-j)a_ja_{n-j}\left(\beta^2_{m-1}j-\frac{1}{2}\beta_m\beta_{m-2}(n-j-1)\right)\tag{6}\\ &\quad-\frac{1}{2}\beta_m\beta_{m-2}(n-1)(n-2)a_1a_{n-1}\tag{7} \end{align*}

So, the idea is to find a proper estimate in order to show that the coefficients above are non-negative.

Computational plausbility checks:

Computation of some values of $S$ and $2\leq m \leq S$ supports the assumption and the calculations show that the middle sum (6) is non-negative whenever $m\neq S$. In case $m = S$, the difference of the first summand (5) and the third summand (7) is always greater than the middle sum (6), so that the coefficients are always non-negative.

Log-concavity:

Note: The log-concavity seems to be ubiquitiuos within this example. For every $n$ the sequence of binomial coefficients $\{\binom{n}{k}\}$ is log-concave, so that e.g.

$$\binom{S}{m-1}^2\geq\binom{S}{m-2}\binom{S}{m}$$

holds, which could be helpful for estimating the coefficients when trying to simplify (2) and (4), resp. (3).

Interesting papers regarding log-concavity:

Recently I found two interesting papers from Dmitry Karp, Sergei Sitnik and Segrei Kalmykov. I think, they both provide valuable information for the current problem. One is a Survey about log-convexity and log-concavity discussing also hypergeometric functions. The other one deals with Log-convexity and log-concavity of hypergeometric-like functions. In the second paper the section containing Theorem 6 looks promising.

Answer 2

I'll prove here some other easy cases, the first being in the comments.

My numerical experiments indicate that one should require $2 \le k \le m \le S$ to show \begin{align} q_k(x) := h_{k-1}(x)h_{k+1}(x) - h_{k}^2(x) \le 0 \end{align} Step 0, one sees immediately that $h_k(x) = h_{k,S,m}(x)$ is a polynomial of degree $(m-k+1)$.

Step 1, $m=k$, one has \begin{align} h_{k-1,S,k}(x) &= 1+\frac{(2k-2)}{(S-1)}x+\frac{k(k-1)}{S(S-1)}x^2\\ h_{k,S,k}(x) &= 1+\frac{k}{S}x\\ h_{k+1,S,k}(x) &= 1 \end{align} which gives \begin{align} q_k(x) &= h_{k-1,S,k}(x)\:h_{k+1,S,k}(x)-h_{k,S,k}(x)^2\\ &= \left(1+\frac{(2k-2)}{(S-1)}x+\frac{k(k-1)}{S(S-1)}x^2\right)\cdot 1 -\left(1+\frac{k}{S}x\right)^2\\ &= -x(2S+kx)\frac{(S-k)}{(S-1)S^2}. \end{align}

This is clearly a degree 2 polynomial in x with zeros $\{-2S/k,0\}$. So $x\ge0$ gives $q_k(x) \le 0$. $\square$

Step 2, $m=k+1$, $2 \le k \le k+1 \le S$, we have \begin{equation} (5kS+3S-k-3)\ge (5k(k+1)+3(k+1)-k-3) = 5k^2+7k+6 > 0, \end{equation} which gives \begin{align} q_k(x) = -x(2S(S^2-1)+S(5kS+3S-k-3)x +4k(k+1)Sx^2+k(k+1)^2x^3)\frac{(S-k-1)}{(S-2) (S-1)^2 S^2} \le 0. \square \end{align}

Now the easy coefficients of the general case. Step 0, gives that $\text{deg}(q_k(x)) = 2m-2k+2$.

Step 3, $r\ge 2$, $m=k+r$, $2\le k\le k+r \le S$, one has, without proving the equality, \begin{align} [x^0](q_k(x)) &= 0,\\ [x^1](q_k(x)) &= -2\cdot\frac{(S+1)(S-k-r)}{(S-r-1)(S-r+0)(S-r+1)} \le 0,\\ [x^{2r+1}](q_k(x)) &= -(2r+2)\cdot\frac{ (S-k-r) \prod\limits_{\nu=0}^{r-1}(k+\nu) \prod\limits_{\nu=1}^{r}(k+\nu) }{ \prod\limits_{\nu=0}^{r}(S-\nu) \prod\limits_{\nu=1}^{r+1}(S-\nu) } \le 0,\\ [x^{2r+2}](q_k(x)) &= -\frac{ (S-k-r) \prod\limits_{\nu=0}^{r}(k+\nu) \prod\limits_{\nu=1}^{r}(k+\nu) }{ \prod\limits_{\nu=0}^{r}(S-\nu) \prod\limits_{\nu=0}^{r+1}(S-\nu) } \le 0.\square \end{align}

Next the remaining three cases for $r=2$.

Step 4, $r=2$, $m=k+r$, $2\le k\le k+r \le S$, we have

\begin{align} (3kS+2S-4k-4) & \ge (3k(k+2)+2(k+2)-4k-4) \ge 3k^2+4k > 0,\\ (4kS+2S+k-4) & \ge (4k(k+2)+2(k+2)+k-4) \ge 4k^2+11k > 0, \\ (7kS+11S-2k-7) & \ge (7k(k+2)+11(k+2)-2k-7) \ge 7k^2+23k+15 > 0, \end{align}

which gives, without proving the equality, \begin{align} [x^2](q_k(x)) &= -3(3kS+2S-4k-4)\cdot\frac{(S-k-2)(S+1)}{(S-3)(S-2)^2(S-1)S} \le 0,\\ [x^3](q_k(x)) &= -4(4kS+2S+k-4)\cdot\frac{(S-k-2)(k+1)}{(S-3)(S-2)^2(S-1)S} \le 0,\\ [x^4](q_k(x)) &= -2(7kS+11S-2k-7)\cdot\frac{(S-k-2)k(k+1)}{(S-3)(S-2)^2(S-1)^2S} \le 0. \square \end{align}