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Prove that:

$\quad\quad\quad\quad f(x) = \log\big( {}_2F_1(a,\,b\,;\,c\,;\,x^{-1})\big),\;\;a,b,c>0$

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:

$\quad\quad\quad\quad g(x) = \frac{\Large _2F_1(a,\,b\,;\,c\,;\,\alpha x)}{\Large _2F_1(a,\,b\,;\,c\,;\,x)},\;\; 0<\alpha<1,\;\,a,b,c>0$

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.

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.

Prove that:

$\quad\quad\quad\quad f(x) = \log\big( {}_2F_1(a,\,b\,;\,c\,;\,x^{-1})\big),\;\;a,b,c>0$

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:

$\quad\quad\quad\quad g(x) = \frac{\Large _2F_1(a,\,b\,;\,c\,;\,\alpha x)}{\Large _2F_1(a,\,b\,;\,c\,;\,x)},\;\; 0<\alpha<1,\;\,a,b,c>0$

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

Prove that:

$\quad\quad\quad\quad f(x) = \log\big( {}_2F_1(a,\,b\,;\,c\,;\,x^{-1})\big),\;\;a,b,c>0$

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:

$\quad\quad\quad\quad g(x) = \frac{\Large _2F_1(a,\,b\,;\,c\,;\,\alpha x)}{\Large _2F_1(a,\,b\,;\,c\,;\,x)},\;\; 0<\alpha<1,\;\,a,b,c>0$

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.

Prove that:

$\quad\quad\quad\quad f(x) = \log\big( {}_2F_1(a,\,b\,;\,c\,;\,x^{-1})\big),\;\;a,b,c>0$

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:

$\quad\quad\quad\quad h(x) = \frac{\Large _2F_1(a,\,b\,;\,c\,;\,\alpha x)}{\Large _2F_1(a,\,b\,;\,c\,;\,x)},\;\; 0<\alpha<1,\;\,a,b,c>0$

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

Prove that:

$\quad\quad\quad\quad f(x) = \log\big( {}_2F_1(a,\,b\,;\,c\,;\,x^{-1})\big),\;\;a,b,c>0$

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:

$\quad\quad\quad\quad g(x) = \frac{\Large _2F_1(a,\,b\,;\,c\,;\,\alpha x)}{\Large _2F_1(a,\,b\,;\,c\,;\,x)},\;\; 0<\alpha<1,\;\,a,b,c>0$

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