Fixed point iteration involving integral $f(x)=\int_{-\infty}^{\infty}k(x,t)f(x-t)dt+g(x)$.

Question

Let $g(x)$ and $k(x,t)$ be two known functions. Is it possible to uniquely solve for $f(x)$ that is defined as:

$$f(x)=\int_{-\infty}^{\infty}k(x,t)f(x-t)dt+g(x) \tag{1}$$

I am looking for a numerical approach. One way that I was thinking of is formulating the above equation in a fixed-point interation framwork. Consider $f^{0}(x)$ is an initial guess of $f(x)$. Then,

$$f^{n+1}(x)=\int_{-\infty}^{\infty}k(x,t)f^{n}(x-t)dt+g(x) \tag{2}$$

provides an iterative approach to find $f(x)$. The question, however, I would like to get some inputs is:

Question: If this the right approach, what should be the initial guess $f^{0}(x)$?

If not, what is the right approach to address this problem?

Answer 1

You can rewrite the equation as a Fredholm integral equation. Specifically, using the change of variable $x - t = y$, the equation becomes $$ f(x) = \int_{-\infty}^{+\infty} \tilde K(x,y) f(y)\,dy + g(x), $$

where $\tilde K(x,y) = K(x,x-y)$. You'll find a lot about Fredholm equations.