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Consider the functional equation

$$ g\left(a\right) = \int_0^1 \frac{e^{c(a,h)+f(h)}}{1+e^{c(a,h)+f(h)}}dh $$

and this holds for all $a$. $g(a)$ and $c(a,h)$ are known functions on a continuous support. Assume that $\frac{\partial g(a)}{\partial a}\not=0$, $\frac{\partial c(a,h)}{\partial a}\not=0$, and $\frac{\partial c(a,h)}{\partial h}$.

How can I show that there is one unique $f(h)$ that solves this equation? The ultimate goal is to figure out under what situation will $f(h)$ be uniquely pinned down. So any additional assumptions on $g$ and $c$ that's needed can be added.

Consider the functional equation

$$ g\left(a\right) = \int_0^1 \frac{e^{c(a,h)+f(h)}}{1+e^{c(a,h)+f(h)}}dh $$

and this holds for all $a$. $g(a)$ and $c(a,h)$ are known functions on a continuous support. Assume that $\frac{\partial g(a)}{\partial a}\not=0$, $\frac{\partial c(a,h)}{\partial a}\not=0$, and $\frac{\partial c(a,h)}{\partial h}\not=0$.

How can I show that there is one unique $f(h)$ that solves this equation? The ultimate goal is to figure out under what situation will $f(h)$ be uniquely pinned down. So any additional assumptions on $g$ and $c$ that's needed can be added.

Consider the functional equation

$$ g\left(a\right) = \int_0^1 \frac{e^{c(a,h)+f(h)}}{1+e^{c(a,h)+f(h)}}dh $$

and this holds for all $a$. $g(a)$ and $c(a,h)$ are known functions on a continuous support. How can I show that there is one unique $f(h)$ that solves this equation?

Consider the functional equation

$$ g\left(a\right) = \int_0^1 \frac{e^{c(a,h)+f(h)}}{1+e^{c(a,h)+f(h)}}dh $$

and this holds for all $a$. $g(a)$ and $c(a,h)$ are known functions on a continuous support. Assume that $\frac{\partial g(a)}{\partial a}\not=0$, $\frac{\partial c(a,h)}{\partial a}\not=0$, and $\frac{\partial c(a,h)}{\partial h}$.

How can I show that there is one unique $f(h)$ that solves this equation? The ultimate goal is to figure out under what situation will $f(h)$ be uniquely pinned down. So any additional assumptions on $g$ and $c$ that's needed can be added.

Consider the functional equation

$$ g\left(a\right) = \int_0^1 \frac{e^{h+f(h)}}{1+e^{h+f(h)}}dh $$

and this holds for all $a$. $g(a)$ is a known function on a continuous support. How can I show that there is one unique $f(h)$ that solves this equation?

Consider the functional equation

$$ g\left(a\right) = \int_0^1 \frac{e^{c(a,h)+f(h)}}{1+e^{c(a,h)+f(h)}}dh $$

and this holds for all $a$. $g(a)$ and $c(a,h)$ are known functions on a continuous support. How can I show that there is one unique $f(h)$ that solves this equation?