Suppose $F$ is the Gaussian hypergeometric function 2F1, $x\in \mathbb{N}$, $t>x \in \mathbb{N}$, $\mu \in (0,1)$. Is the function: $$f(\mu)=\mu^x F(x,-t+x,2+x,\mu),$$$$f(\mu)=\mu^{x+1} F(x,-t+x,2+x,\mu),$$ convex in $\mu$? I've done some numerical examples that seem to verify this with the following manipulate Mathematica code (apologies for the mess):
Manipulate[
Plot[
{
Subscript[\[Mu], 1]^(x + 1) b*
Hypergeometric2F1[x, -t + x, 2 + x, Subscript[\[Mu], 1]]
+ (1 - b) Subscript[\[Mu], 2]^(x + 1)*
Hypergeometric2F1[x, -t + x, 2 + x, Subscript[\[Mu], 2]],
(b*Subscript[\[Mu], 1] + (1 - b) Subscript[\[Mu], 2])^(x + 1)
Hypergeometric2F1[
x, -t + x, 2 + x, (b*Subscript[\[Mu], 1] + (1 - b) Subscript[\[Mu], 2])
]
},
{b, 0, 1}
],
{t, 1, 80, 1},
{x, 1, 80, 1},
{Subscript[\[Mu], 1], 0, 1},
{Subscript[\[Mu], 2], 0, 1}
]
EDIT: Forgot to add $x+1$ instead of $x$.
