Turning a definite integral equation into a differential one. Consider the following equation where $p(u)$ is a probability distribution and where $g$ is the unknown :
\begin{equation}
 g = \int_{-\infty}^\infty p(u)f(u,g)  \mathrm{d} u
\end{equation}
Using the definition of a functional derivative :
$$ \frac{\delta F[f(x)]}{\delta f(y)}=\lim _{\epsilon \rightarrow 0} \frac{F[f(x)+\epsilon \delta(x-y)]-F[f(x)]}{\epsilon}$$
Am I allowed to say that the first equation is equivalent to :
$$ \frac{\delta g[p(u)]}{\delta p(u_0)} = f(u_0,g)$$
If not, where is my mistake ?
I am trying to understand how to invert definite integral equations. Any reference or advice is always welcome.
Thank you.
 A: If you define the functional $F$ of the function $p$ as
$$
  F[p] = \int_{-\infty}^{\infty} L(u,p(u)) \, du = \int_{-\infty}^{\infty} p(u) f(u,g) \,du
$$
then the functional derivative $\delta F / \delta p$ is $f(u,g)$.  It seems that you are viewing your integral equation as a recursively defined functional with $g=F$.  This seems to me to be over-complicating things.  It is simpler to just solve the integral equation for the unknown real number $g$ using numerical integration and Newton's method.  Define
$$
  I(g) = \int_{-\infty}^{\infty} p(u) f(u,g) \,du.
$$
Note that $I(g)$ is simply a function from $\mathbb{R}$ to $\mathbb{R}$.  Then your integral equation is
$$
  g - I(g) = 0.
$$
The Newton's method iteration for this equation is
$$
  g_{n+1} = g_n - \frac{g_n - I(g_n)}{1 - I'(g_n)}
$$
where
$$
  I'(g_n) = \int_{-\infty}^{\infty} p(u) f_y(u, g_n) \,du. 
$$
If there is no closed form for $I(g)$ or $I'(g)$, you need to use some form of numerical integration (Gaussian quadrature would probably be the most accurate and efficient) to evaluate $I(g_n)$ or $I'(g_n)$.

Consider one of the simplest probability distributions, $p(u) = \delta(u - u_0)$, where $\delta$ is the Dirac delta function.  In this case, $I(g)$ has the closed form $f(u_0,g)$.  The equation to solve is then $g - f(u_0,g) = 0$.  In general, you still need to use Newton's method to solve this.  The obvious exception is for $f$ such that there is some closed form solution.

