Given $a>b>0$, is there any upper bound of the following ratio of hypergeometric function? $$\frac{_2F_1(a,1b;a+1;x)}{_2F_1(a,1b;a+1;y)}$$ for $1>x>y>0$ ideally in the form like some powers of $x/y$？
1 Answer
If $b$ is not an integer, this ratio is bounded by a constant.
Indeed, Remark 1.2.4 pp 3435 of "From Gauss to Painlevé" by IwasakiKimuraShimomuraYoshida can be applied to your case, in which $\gamma\alpha\beta=b$. It yields that the function $f={}_2F_1(\alpha,\beta;\gamma,\cdot)$ you consider is continuous on [0,1]. the exact value at $1$ is given on page 73 of the same book at as $\frac{\Gamma(a+1)\Gamma(b)}{\Gamma(1)\Gamma(a+b)}$, in particular it is not $0$. As f(0)=1, using the local uniqueness of solutions in $(0,1)$, one may conclude the existence of $K>0$ such that $f(x)>K$, for every $x\in[0,1]$.
Then the function $F:[0,1]\times [0,1]\rightarrow \mathbb R, (x,y)\mapsto f(x)/f(y)$ is continuous on a compact, whence the conclusion.
Edit: Actually, the derivative $f'$ is $\left(\frac{a+1}{a(1b)}\right){}_2F_1(\alpha+1,\beta+1;\gamma+1,\cdot)$ and must have the sign of $\frac{a+1}{a(1b)}$. So an explicit upper bound can be given:
if $1b>0$ then $F$ is bounded by $f(1)/f(0)=\frac{\Gamma(a+1)\Gamma(b)}{\Gamma(1)\Gamma(a+b)}$,
if $1b<0$ then $F$ is bounded by $f(0)/f(1)=\left(\frac{\Gamma(a+1)\Gamma(b)}{\Gamma(1)\Gamma(a+b)}\right)^{1}$.

$\begingroup$ If you read the details in the mentioned book, I think you can remove the condition of b integral, replacing the remark by the preceding Lemma. For the quantitative part, if $b=1$, then $f$ is constant, so the optimal bound is $1$. $\endgroup$– gcousinMay 7, 2019 at 3:40