On the existence of a unique solution I would like to show that there exists a unique solution in $a$ for the below given equation
$$g(a)=\frac{1}{ea^{r}}\Bigg(\int^{y_l}_{-\infty} \ln\left(\frac{1}{ea^{r}}\right)f_0(y)dy$$
$$+\int^{y_u}_{y_l} e^{\frac{\ln^2a}{\ln(ab)}}{f_1}^{\frac{\ln a}{\ln(ab)}}(y){f_0}^{1-\frac{\ln a}{\ln(ab)}}(y)\ln\left( \frac{1}{ea^{r}}e^{\frac{\ln^2a}{\ln(ab)}}{f_1}^{\frac{\ln a}{\ln(ab)}}(y){f_0}^{-\frac{\ln a}{\ln(ab)}}(y)\right)dy+\int^\infty_{y_u} a\ln\left( \frac{a}{ea^r}\right)f_0(y)dy \Bigg)=\epsilon 
$$
if additionally we have
$$ea^{r}=\int^{y_l}_{-\infty} f_0(y)dy+\int^{y_u}_{y_l} e^{\frac{\ln^2a}{ln(ab)}}{f_1}^{\frac{lna}{ln(ab)}}(y){f_0}^{1-\frac{lna}{ln(ab)}}(y)dy+\int^\infty_{y_u}a f_0(y)dy 
$$
I also 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 $y_l=l^{-1}(1/a)$
$5$- $0<\epsilon<1$

Any help would be appreciated. Thanks in advance.
 A: I was able to simplify the equation
$$g(a)=\frac{1}{ea^{r}}\Bigg(\int^{y_l}_{-\infty} \ln\left(\frac{1}{ea^{r}}\right)f_0(y)dy$$
$$+\int^{y_u}_{y_l} e^{\frac{\ln^2a}{\ln(ab)}}{f_1}^{\frac{\ln a}{\ln(ab)}}(y){f_0}^{1-\frac{\ln a}{\ln(ab)}}(y)\ln\left( \frac{1}{ea^{r}}e^{\frac{\ln^2a}{\ln(ab)}}{f_1}^{\frac{\ln a}{\ln(ab)}}(y){f_0}^{-\frac{\ln a}{\ln(ab)}}(y)\right)dy+\int^\infty_{y_u} a\ln\left( \frac{a}{ea^r}\right)f_0(y)dy \Bigg)=\epsilon 
$$
using
$$z=ea^{r}=\left(\int^{y_l}_{-\infty} f_0(y)dy+\int^{y_u}_{y_l} e^{\frac{\ln^2a}{ln(ab)}}{f_1}^{\frac{lna}{ln(ab)}}(y){f_0}^{1-\frac{lna}{ln(ab)}}(y)dy+\int^\infty_{y_u}a f_0(y)dy \right) $$
we have 
$$g(a)=\frac{1}{z}\Bigg(\int^{y_l}_{-\infty} -\ln\left(z\right)f_0(y)dy$$
$$+\int^{y_u}_{y_l}-\ln{z} e^{\frac{\ln^2a}{\ln(ab)}}{f_1}^{\frac{\ln a}{\ln(ab)}}(y){f_0}^{1-\frac{\ln a}{\ln(ab)}}(y)dy$$
$$+\int^{y_u}_{y_l} e^{\frac{\ln^2a}{\ln(ab)}}{l}^{\frac{\ln a}{\ln(ab)}}(y)\ln\left( e^{\frac{\ln^2a}{\ln(ab)}}{l}^{\frac{\ln a}{\ln(ab)}}(y)\right){f_0}(y)dy$$
$$+\int^\infty_{y_u} -\ln{z}\,af_0(y)dy+ \int^\infty_{y_u} a\ln af_0(y)dy 
 \Bigg) 
$$
collecting the terms with $-\ln z$ in common parenthesis we have
$$g(a)=\frac{1}{z}\Bigg(-\ln\left(z\right)\Bigg(\int^{y_l}_{-\infty} f_0(y)dy$$
$$+\int^{y_u}_{y_l} e^{\frac{\ln^2a}{\ln(ab)}}{f_1}^{\frac{\ln a}{\ln(ab)}}(y){f_0}^{1-\frac{\ln a}{\ln(ab)}}(y)dy+\int^\infty_{y_u} \,af_0(y)dy\Bigg) $$
$$+\int^{y_u}_{y_l} e^{\frac{\ln^2a}{\ln(ab)}}{l}^{\frac{\ln a}{\ln(ab)}}(y)\ln\left( e^{\frac{\ln^2a}{\ln(ab)}}{l}^{\frac{\ln a}{\ln(ab)}}(y)\right){f_0}(y)dy$$
$$+ \int^\infty_{y_u} a\ln af_0(y)dy 
 \Bigg) 
$$
The inner parenthesis is also $z$ accordingly we can write
$$g(a)=\frac{1}{z}\Bigg(-\ln\left(z\right)z $$
$$+\int^{y_u}_{y_l} e^{\frac{\ln^2a}{\ln(ab)}}{l}^{\frac{\ln a}{\ln(ab)}}(y)\ln\left( e^{\frac{\ln^2a}{\ln(ab)}}{l}^{\frac{\ln a}{\ln(ab)}}(y)\right){f_0}(y)dy$$
$$+ \int^\infty_{y_u} a\ln af_0(y)dy 
 \Bigg) 
$$
which is
$$-\ln\left(z\right)+\frac{1}{z}\Bigg(\int^{y_u}_{y_l} e^{\frac{\ln^2a}{\ln(ab)}}{l}^{\frac{\ln a}{\ln(ab)}}(y)\ln\left( e^{\frac{\ln^2a}{\ln(ab)}}{l}^{\frac{\ln a}{\ln(ab)}}(y)\right){f_0}(y)dy+ \int^\infty_{y_u} a\ln af_0(y)dy \Bigg)$$
After this point, what I know is that, the derivative of $z$ with respect to $a$ is positive and the derivative of $l$ with respect to $y$ is also positive. What I need to show is that $g(a)$ is monotone! however the last expression that I derived depends on $y_l$ at the limits of the integral and I dont know how to deal with this case, especially  since $y_l=l^{-1}(1/a)$
