Explicit inverse of $x\mapsto\frac{1}{\alpha} (x+1)^{\frac{\alpha - 1}{\alpha}} | x - 1 |^{\frac{1}{\alpha}} \frac{\alpha(x - 1) + 2}{(x- 1)(x + 1)}$

Question

In the very same spirit as my previous question I am trying to find an explicit representation of the convex conjugate $f^*$ of the function $$ f \colon \mathbb R \to [0, \infty], \qquad x \mapsto | 1 - x |^{\frac{1}{a}} (1 + x)^{\frac{\alpha - 1}{\alpha}} + \iota_{[0, \infty)}(x), $$ where for any set $A$, the convex indicator function $\iota_{A}(x)$ is equal to $0$ for $x \in A$ and equal to $\infty$ else and $\alpha \in (0, 1)$.

Up to edge cases, this first boils down to finding the inverse of $$ f'(x) = \frac{1}{\alpha} (x+1)^{\frac{\alpha - 1}{\alpha}} | x - 1 |^{\frac{1}{\alpha}} \frac{\alpha(x - 1) + 2}{(x- 1)(x + 1)} = \frac{1}{\alpha} \left| \frac{x-1}{x+1}\right|^{\frac{1}{\alpha}} \left(\alpha + \frac{2}{(x- 1)}\right) $$ (to obtain $f^*(y) = y (f')^{-1}(y) - f((f')^{-1}(y))$.)

Due to the last term in parenthesis, I do not know how to invert $f'$, like it was done in this answer. We have that $f'$ is strictly monotonically increasing and continuous, so that it is bijective onto its range and hence invertible.

For $\alpha = \frac{2}{3}$, WolframAlpha gives an explicit inverse, which as a quite complicated expression, whereas for $\alpha = \frac{1}{\sqrt{2}}$, it can't find an answer in terms of standard mathematical functions.

Answer 1

$\def\a{\alpha}\DeclareMathOperator\I{I}$

Substitute $x=\frac{w+1}{w-1}$ (i.e. $w = \frac{x + 1}{x - 1}$):

$$\frac1\a\left|\frac{x+1}{x-1}\right|^\frac1\a\left(\a+\frac2{x-1}\right)=y\iff |w|^\frac1a-\frac w{\a y}-\frac{\a-1}{\a y}=0$$

For $y < 0$ there is a closed Wolfram language form via this post in section 1

$$x^r+ax+b=0\implies x=\frac{b r}{a(1-r)\text I^{-1}_\frac{b^{r-1}(r-1)}{a^r\left(\frac1r-1\right)^r}(r-1,2)}$$

where inverse beta regularized appears. Therefore:

$$\boxed{f(x)=|1-x|^\frac1\a(1+x)^\frac{\a-1}\a\implies (f’)^{-1}(x)=\frac2{\I^{-1}_{1-x}\left(2,\frac1\alpha-1\right)}-1,0\le x\le1}$$

testable here. Wolfram Alpha automatically reduces the solution to a radical form if possible.

Using $(f’)^{-1}(y)=\frac2u-1$, then $f^*(y)=2\frac{y-|1-u|^\frac1a}u-y$ using algebra and software. $1-u$ from section 7 can be used again and the absolute value removed as $u\ge0$ to get $$\boxed{f^*(y)=2\frac{y-\I^{-1}_y\left(\frac1\a-1,2\right)^\frac1\a}{\I^{-1}_{1-y}\left(2,\frac1\a-1\right)}-y,0\le y\le 1}$$

There is probably also a series solution, with closed forms for possibly the entire domain of $f(x)$ which will be added later.