I'm interested in solving a particular integral equation: $$g(X) = \int_0^1 K(X,p)f(p) \ dp$$ where $f(p)\in L^1([0,1])$ and $X$ is a stochastic process of finite length; i.e. a collection of random variables. $K$ is a continuous kernel.
This looks like a Fredholm integral of the first kind, although it's not clear to me that the function $g$ lives in a Banach space. $g$ is a linear functional collections $X$ of measurable functions, so I want to show that $X$ is in some Banach space. But, I don't know much about stochastic processes. I'm not even exactly sure what a norm would look like.
Any guidance on this problem would be very helpful.
Thanks
This should be a comment, but it's too long.
I'm not sure I understand what you are defining. Typically a stochastic process is something like a jointly measurable function $X : \Omega \times (a,b) \to \mathbb{R}$, where $(\Omega, \mathcal{F}, P)$ is a probability space and $(a,b)$ is some interval (which could be replaced by a closed interval, an unbounded interval, or some other parameter space). I am guessing that by "$K$ is a continuous kernel" you mean it's a continuous function $K : \mathbb{R} \times [0,1] \to \mathbb{R}$, and so I'm inclined to read your integral as
$$g(X) = \int_0^1 K(X(\omega, t),p) f(p)\,dp$$
but then the right side has a dependence on $\omega$ and $t$ while the right side is supposed to be a number. I would guess you want to integrate these variables out somehow, but I don't see it in your question. I also don't see why you call g a "linear" functional when, for general $K$, it seems likely to be nonlinear.
Anyway, since $X$ is nothing but a measurable function on $\Omega \times (a,b)$, which can naturally be equipped with the product measure $P \times m$ (where $m$ is Lebesgue measure on $(a,b)$), under appropriate integrability conditions, you could naturally think of $X$ as an element of the Banach space $L^p(\Omega \times (a,b), P \times m)$, or possibly $L^p(\Omega; L^q((a,b)).$ Or if the process is continuous in $t$, perhaps $L^p(\Omega; C([a,b]))$. In case that helps.