I have an equation for which I need to show that the solution is unique. Here is my problem:
$\frac{1}{a^{r}}\left(\int^{y_l}_{-\infty} \frac{1}{e}f_0(y)dy+\int^{y_u}_{y_l} e^{\frac{a^2}{ln(ab)}}{f_1}^{\frac{lna}{ln(ab)}}(y){f_0}^{1-\frac{lna}{ln(ab)}}(y)dy+\int^\infty_{y_u} \frac{a}{e}f_0(y)dy \right)=1 $
I have the following information about the variables in the equation:
$1$- $f_0$ and $f_1$ are some density functions where $l=f_1/f_0(y)$ is monotone increasing
$2$- $a,b,r,y_l,y_u\in\mathbb{R}$ and $a,b,r>0$,and obviously $y_u>y_l$
$3$- $ab>1$
$4$- $y_u=l^{-1}(b)$ and $y_l=l^{-1}(1/a)$
Question:
How to show that the equation has a unique solution in r for any two densities satisfying the above conditions, and any $a$ and $b$.
What I've thought about:
As $r$ can be chosen how we want such that the equation will be satisfied, there are two cases of importance $0 or $a>1$.
if $a>1$ then I have $\infty>a^r>1$ and if $0 then I have $1>a^r>0$
If I look at the equation when $a$ is very small ($0), then $y_l=l^{-1}(1/a)$ implies that $y_l$ should be very big. Accordingly we roughly have
$\frac{1}{a^{r}}\left(\int^{y_l}_{-\infty} \frac{1}{e}f_0(y)dy\right)=1$
which is
$\frac{1}{a^{r}}=e$
As $1>a^r>0$ in this case, it is obvious that the equation is satisfied for some $r$.
Another idea could be to take the derivative of the function with respect to $r$ but I dont know how to proceed with this idea.
Any help would be appreciated.