Prove that:


$$ f(x) = \log\big( 
{}\_2F\_1(a,\,b\,;\,c\,;\,x^{-1})\big),\;\;a,b,c>0
$$


is convex (and decreasing) on $(1,\infty)$.


It actually seems that the stronger result that $f\big((x+1)^{\beta}\big)$, $\beta>0$, is completely monotonic, is true. I saw a post on here proving a similar result using continued fractions to show that all the Taylor series coefficients are positive when $c\ge a+b$. I wonder if such an approach could used to show that the coefficients of a series expansion of $f\big((x+1)^{\beta}\big)$ have alternating signs for $x>0$ ( for arbitrary $a,b,c>0$ ). Or compute the inverse Laplace transform?


This result would imply that $f(\!\sqrt{x})$ is convex on $(1,\infty)$, which is equivalent to the function:


$$ g(x) = \frac{ \_2F\_1(a,\,b\,;\,c\,;\,\alpha x)}{ \_2F\_1(a,\,b\,;\,c\,;\,x)},\;\; 0<\alpha<1,\;\,a,b,c>0$$


being decreasing on $[0,1]$, which is what I originally wanted to show.


This result is important to show UMP properties of multiple determination coefficient tests.



Log convexity of hypergeometric function for $a,b,c>0$
Using the series representation of the hypergeometric function, we see that $\_2F\_1(a,b\,;c\,;(1+x)^{-1})$, $x>0$, is the pointwise convergent limit of positive sums of completely monotonic functions for $a,b,c>0$, and is thus completely monotonic.


Since completely monotonic functions are log convex, it follows that $\log \big({}\_2F\_1(a,b\,;c\,;(x+1)^{-1}\big)$ is convex on $(0,\infty)$.


