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.



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.

  • $\begingroup$ Thanks. This is precisely what I was asking. For the K I'm using, it ends up being linear, but I'm interested in the general case. $\endgroup$ – Alexander Sibelius Apr 14 '14 at 3:43
  • $\begingroup$ You know if the space $L^2(\Omega,C[a,b])$ is a Hilbert space also under the norm $\mathbb{E}sup\mid X \mid ^{2}$? Also dosnt it also need a measure? $\endgroup$ – Speaker Oct 29 '18 at 18:56
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    $\begingroup$ It's not a Hilbert space, the parallelogram law fails. (Think about when $\Omega$ is a single point, in which case this space just becomes $C([a,b])$ with the sup norm, which is definitely not a Hilbert space). And your "norm" isn't a norm unless you take a square root. But in that case it will still be a Banach space. I did note that in order for any of this to make sense, you have to specify a measure on $\Omega$; I called it $P$. $\endgroup$ – Nate Eldredge Oct 29 '18 at 20:09
  • $\begingroup$ And I suppose this does not change if we change the sigma algebra to optional or predictale, right? $\endgroup$ – Speaker Oct 29 '18 at 21:16
  • $\begingroup$ @NathanChart: It'd be good if you can ask these as new questions, so that other people besides me can see them and possibly answer. But it'd also be helpful if you can give more context. $\endgroup$ – Nate Eldredge Oct 29 '18 at 21:54

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