Upper bound on hypergeometric function ${}_2F_1$

Question

I wonder if the following upper bound holds:

$${}_2F_1[-m, -m; -(m+l); z]\leq 1,\tag{1}$$

where ${}_2F_1[a,b;c;z]$ is the Gauss hypergeometric function, $m,l=0,1,2,\ldots$, and $0<z<1$. Clearly, this holds for $l=0$ since (1) then reduces to the binomial identity in the answer to a related question. Thus, induction seems to be a promising path to proving this upper bound. However, direct application of the expansion:

$$\begin{align}{}_2F_1[-m, -m; -(m+l); z] &= {}_2F_1[-m, -m; -(m+l-1); z] \\ &\quad + \dfrac{m^2 z}{(m+l)(m+l-1)}{}_2F_1[-(m-1), -(m-1); -(m-1+l-1); z]\end{align}$$ derived for the aforementioned solution does not seem to work here. Numerical experiments seem to confirm this bound, though perhaps there is a counterexample. I tried induction on $m$ instead of $l$ to no avail. Any ideas?

Answer 1

First show that, for $m >0$, the function is always decreasing over the interval $0<z<1$:

$$\begin{align*}\dfrac{d}{dz}{}_2F_1(-m,-m;-[m+l];z) &= \dfrac{d}{dz}\sum_{n=0}^{m} \dfrac{(-m)_{n}(-m)_{n}}{(-(m+l))_{n}}\dfrac{z^{n}}{n!}\\ \\ &= \sum_{n=1}^{m} \dfrac{(-m)_{n}(-m)_{n}}{(-(m+l))_{n}}\dfrac{z^{n-1}}{(n-1)!} \\ \\ &= \sum_{k=0}^{m-1} \dfrac{(-m)_{k+1}(-m)_{k+1}}{(-(m+l))_{k+1}}\dfrac{z^{k}}{k!} \\ &= -\dfrac{m^2}{m+l} \sum_{k=0}^{m-1} \dfrac{(-(m-1))_{k}(-(m-1))_{k}}{(-(m-1+l))_{k}}\dfrac{z^{k}}{k!}\\ \\ &= -\dfrac{m^2}{m+l} {}_2F_1(-[m-1],-[m-1];-[m-1+l];z) \\ \\ &\le -\dfrac{m^2}{m+l} (1-z)^{m-1} \\ \\ \therefore \; \dfrac{d}{dz}{}_2F_1(-m,-m;-[m+l];z) &\lt 0\\ \end{align*}$$

which used a lower bound for ${}_2F_1(-[m-1],-[m-1];-[m-1+l];z)$ that comes from this answer.

Then, for $m>0$, the function is at its maximum value in $0<z<1$ as $z\rightarrow0$:

$$\begin{align*}\lim_{z\to0}{}_2F_1(-m,-m;-[m+l];z) &= \lim_{z\to0} \sum_{n=0}^{m} \dfrac{(-m)_{n}(-m)_{n}}{(-(m+l))_{n}}\dfrac{z^{n}}{n!} \\ \\ &= \lim_{z\to0}\left[ 1 + \sum_{n=1}^{m} \dfrac{(-m)_{n}(-m)_{n}}{(-(m+l))_{n}}\dfrac{z^{n}}{n!}\right]\\ \\ \lim_{z\to0}{}_2F_1(-m,-m;-[m+l];z) &= 1\\ \end{align*}$$

One can then conclude:

$${}_2F_1(-m,-m;-[m+l];z) \le 1 \quad z\in (0,1), m > 0$$

For the special case of $m = 0$: $${}_2F_1(0,0;-l;z) = 1$$

One can then finally conclude:

$${}_2F_1(-m,-m;-[m+l];z) \le 1 \quad z\in (0,1)$$