Why most of the continuous transforms are based on integral operator? I am curious about why most of the continuous transforms, for instance fourier/ Laplace are based on integral operator?
OR,
In other words, why integral operator provides the "best" correlation between two functions?
Thanks
 A: From the point of Functional Analysis, we consider the operator's space(or domain & range). Taking Fourier Transform as an example, we can see it like this: $T:C(R^n)\rightarrow{C(R^n)}$ or $T:L^2(R^n)\rightarrow{L^2(R^n)}$. So we enlarge the space(or domain & range) through Lebesgue Integral and we will find more beautiful and smooth in $L^2(R^n)$ than $C(R^n)$. Actually, we can make a larger(or more suitable) space(or domain & range) by using integral operator.
We can also see this question from the Representation Theory. If we look this question carefully, we will find that $T:L^2(R^n)\rightarrow{L^2(R^n)}$ is a transform from a space into its dual space. It is a way that studying a space by studying its dual space. So we can see this question from the Representation Theory:
(1)Fourier Transform: $T:L^2(R^n)\rightarrow{L^2(R^n)}$. It is a Representation Theory about group $(R^n,+)$.
(2)Fourier Series: It is about the transform of period functions to series and converse. First, we can see that period functions is defined on $R^n/Z^n$ or $n$ dimensional torus $T^n$ which is still a group by seen as the product $S$. Second, series can be seen a function defined on group $Z^n$. So transform is $T:L^2(T^n)\rightarrow{L^2(Z^n)}$ and $T:L^2(Z^n)\rightarrow{L^2(T^n)}$. It is a Representation Theory about $T^n$ and $Z^n$.
From above, we know the integral operator may be related to the Topological Group Representation Theory and this will use some integral or precisely, Haar Measure.
The book about this: "A First Course in Harmonic Analysis" #http://en.bookfi.org/book/443200# 
