globally lipschitz condition of the equation for the inverse function implies uniqueness? $g'(\theta)=-(2-\theta-2g(\theta))\frac{f(\theta)}{1-F(\theta)}\equiv G(g, \theta)$, $\theta\in[0,1]$
with assumptions
(i)$F(\cdot)$ is a continuous c.d.f. and 
(ii)$g(\cdot)$ is continuous and strictly decreasing on $[0,1]$.
I can analytically solve this equation by using integrating factor 
$L(\theta,x)=(\frac{1-F(\theta)}{1-F(x)})^2 $ and integrating from 1 to x
$g(x)=\frac{1}{[1-F(x)]^2}\left([1-F(1)]^2g(1)+\int_x^1(2-\theta)(1-F(\theta)) f(\theta)d\theta\right)$
The boundary condition is implied by that expression from left continuity at $\theta=1$,  $g(1-)=1/2$?[not sure about this] 
The problem is that at upper boundary point $\theta=1$,    $g'(1)=-\infty$, which implies $G(g, \theta)$ is not locally lipschitz in $g$?
Denote $h(\cdot)$ the inverse function of $g(\cdot)$ on $[0,1]$.
$h'(y)=- \frac{1}{2-h(y)-y}\frac{1-F(h(y))}{f(h(y))}\equiv H(h, y)$ 
Since $h'(1/2)$ is bounded at $h(1/2)=1$,  $H(h, y)$ satisfies locally Lipschitz in $h$ at $\theta=1$?  
Does this implies $g(\cdot)$ has a unique global solution? Thanks.
 A: Switching to the inverse function does not help. In the equation
$$
h'(y)=- \frac{1}{2-h(y)-2y}\frac{1-F(h(y))}{f(h(y))}
$$
the right hand side is not a Lipschitz function of $h$ when $(y,h)\approx (1/2,1)$.

Instead of trying to squeeze into a general theorem, we can prove uniqueness directly. Multiplying both sides of the equation by the integrating factor $(1-F)^2$, we get 
$$\frac{d}{d\theta}\left( (1-F(\theta))^2 g(\theta) \right) = -(2-\theta) (1-F(\theta)) f(\theta) \tag1$$ 
It follows that 
$$  (1-F(\theta))^2 g(\theta)   = - \int(2-\theta) (1-F(\theta)) f(\theta) \,d\theta \tag2$$ 
This is not just some solution; this is a statement about any solution that the equation may have.  
If $g$ is continuous, the left side of (2) is zero at $\theta=1$. This determines the antiderivative on the right of (2). That's it, the uniqueness is proved. 

This does not actually show existence of a continuous $g$ that satisfies the equation. But if the function $1-F$ has decent behavior at $\theta=1$, without crazy changes of the rate of decay, then the antiderivative in (2) will decay like $(1-F)^2$ as $\theta\to 1$. 
