Definition talk:Integral Transform/Operator

Thoughts on incorporating this
I am looking for a way to define the integral operator as follows:

Let $\CC \closedint a b$ be the space of real-valued functions continuous on closed interval.

Let $x \in \CC \closedint a b$

Let $\R$ be the set of real numbers.

The integral operator, denoted by $I$, is the mapping $I : \CC \closedint a b \to \R$ such that:


 * $\ds \map I x := \int_a^b \map x t \rd t$

In some sense this has a trivial kernel, i.e. 1. However, the point of this is to emphasize that the input, or the variable, is $\map x t$. Can we include $\map x t$ or $\map f x$ as one of variables of an integral transform? It may happen that I will later need to consider continuity or differentiation wrt $f$ or $x$.

In other words, I need to replace $\map F p = \displaystyle \int_a^b \map f x \map K {p, x} \rd x$ with something like $\map F {f, p} = \displaystyle \int_a^b \map f x \map K {p, x} \rd x$. Any thoughts?--Julius (talk) 13:31, 21 February 2021 (UTC)


 * Surely it's just a different notation? Most of the literature I've seen on the subject uses the lowercase letter for the original function, and then the uppercase for the transform of it. Like with Laplace transforms: $\map f t \to \map F s$, for example. Yes of course $F$ is a "function" of $f$, but in this notational style, the $f$ is implicit in the fact that its transform is its uppercase letter.


 * If you are going to do a considerable amount of work on integral operators, then feel free to restyle the notation accordingly if it makes it easier for you, but it would be appreciated if you keep a record of the original notation in an "also known as" or "also denoted as" section (you can find instances of the form) so as to allow us who have grown up with the existing notation to have some continuity. --prime mover (talk) 14:20, 21 February 2021 (UTC)


 * I see that integral transforms were treated as analogous to integral operators. However, integral functionals are also integral operators, and integrands are not always proportional to the first power of any single function. So now we are in a situation where a few theorems (mostly involving integral transforms) have to be renamed.--Julius (talk) 16:25, 22 February 2021 (UTC)