Several questions about integral operators. I have been fumbling with expressions of the form
\begin{equation}
A\{f\}(s) = \int A(s,t)f(t)\operatorname{dt}
\tag{$\star$}
\end{equation}
as a generalization of the matrix product. When looking these up online I found them under the name of "integral operators".
By formal manipulations I have found that the analogue of the identity matrix $I$ should be the Dirac $\delta$, since:
$$
I\{f\}(s) = \int I(s,t)f(t) \operatorname{dt} 
$$
and this should intuitively equal $f(s)$, so $I(s,t) = \delta(s-t)$. However, I have not found any other obvious interesting "functions".
Here are some questions I have about these kind of "functions":


*

*Is there no "real" identity element? I know the Dirac $\delta$ is not a function, but in my analysis class we have approximated it with a sequence of Dirichlet "kernels".

*What is the analogous notion to the trace of a matrix? Intuitively, I guessed that $tr(A) = \int A(t,t) \mathop{dt}$ should be the right notion but then $$tr(\delta) = \int \delta(0) \operatorname{dt} = \int \delta(0)\mathbb{1} \operatorname{dt} = 1$$ whereas $\operatorname{trace}(I) = n$ the dimension of the vector space. EDIT: as pointed out in the comments, we have $tr(\delta) = \infty$. Seeing as the space of functions is in some sense infinite, should there be a definition of trace?

*We have the Fourier transform $$\mathfrak{F}\{f\}(s) = \int e^{-2\pi i s t} f(t) \operatorname{dt}$$ and the Laplace transform $$\mathfrak{L}\{f\}(s) = \int e^{-s t} f(t) \operatorname{dt},$$ so can one view these as integral operators? If so, what are the analogous matrices in the finite-dimensional case? The answer $A_{j,k} = e^{-2 \pi i j k}$ seems disappointing. Going back to question 2, what is the trace of the Fourier/Laplace transform, if it exists?

*The formal considerations lack technical context. What is the natural setting (space) for the "function" $A(s,t)$? I should think it would be $L^1(\mathbb{R} \times \mathbb{R})$, yet the Fourier transform is complex... What should the bounds of the integral in $(\star)$ be?

*Is it possible to develop the concept of eigenvalues in this manner?
Thank you very much for your answers in advance. For context, I have had courses in Analysis up until the Fourier series and transform, one semester of Complex Analysis, a course covering most of Munkres' Topology and passing familiarity with Lebesgue integration.
 A: A nice framework for this kind of questions is $L^2$, if you take $a\in L^2(\Omega^2)$ and $f\in L^2(\Omega)$ then for a.e.  $s\in \Omega$ you have $a(s,\cdot)\in L^2(\Omega)$ and so $a(s,\cdot)f\in L^1(\Omega)$ a.e. and thus $Af(s)=\int_{\Omega}a(s,t)f(t)\text{d}t$ has a meaning almost everywhere.
Under some assumptions on $a$ you can show that $A$ is a continuous linear operator of $L^2(\Omega)$. For example if $M_1=\text{sup}_{x\in \Omega} \int_{\Omega}|a(x,y)|\text{d}y <\infty$ and $M_2=\text{sup}_{y\in \Omega} \int_{\Omega}|a(x,y)|\text{d}x <\infty$ then $||A||_{\mathcal{L}(L^2(\Omega))}\leq \sqrt{M_1 M_2}$ (this is mostly an application of Fubini's theorem).
Making further assumption we can define an important class of integral operators : Suppose that there exist an orthonormal basis $(e_n)$ of $L^2(\Omega)$ such that $\sum ||Ae_n||_2<\infty$ then $A$ is called a Hilbert-schmidt operator and $\sqrt{\sum ||Ae_n||_2}=||A||_\star$ his Hilbert-Schmidt Norm. One can then show that the HS norm doesn't depend on the basis, that the HS norm is inferior to the operator norm... An other important fact is that every HS operator can be written in the integral operator form.
Hilbert-Schmidt operators have the property to be compact (every bounded set is maped to a relatively compact set, note that closed bounded set are not automatically compact because we're in infinite dimension). If you also suppose that $a$ is hermitic, ie if $a(x,y)=\bar{a}(y,x)$ (where $\bar{a}$ is the complex conjugate of $a$) then your operator $A$ will be selfadjoint (with the same definition given sometimes for finite dimensional self adjoint operator : $(Af,g)=(f,Ag)$ where $(\cdot,\cdot)$ is the inner product). 
An important theorem is that a compact operator is diagonalisable : there exists a base $(e_n)$ such that $Ae_n = \lambda_n$. Moreover $\lambda_n$ can be choosed decreasing and always have limit $0$. Just like in the finite dimension case, if $A$ is self adjoint then the eigenvalues will be reals. Reciprocaly if you have a diagonalisable operator such that the eigenvalues tends to $0$ then it is compact. For a HS operator you even have $\sum |\lambda_n|^2=||A||_\star^2=||a||_{L(\Omega^2)}^2$.
Of course there is lot of generalisations to this things and you should search for "spectral theory" if you want more informations.
back to the questions :
1)Of course there is no identity if you want $a$ to be a classical function, $Af(s)=f(s)$ for every function mean that $a(s,\cdot)=0$ away from every neightborhood of $s$, so it's $0$ a.e. and thus $Af(s)=0$ for every $f$, which is absurd. If you only restrain to suitable measure instead of function then you already have found what would be the identity.
2)in what i've written you can take diagonalisable operator and if the sum of their eigenvalue is absolutely convergent then it is a reasonable definition for the trace. More generaly you can define a trace for a larger class of operators, search for trace class.
3)You can of course see them as integral operators. unfortunately the function $e^-i\pi st$ is not $L^2$ on $\mathbf{R}^d$. But if you take a $1$-periodic function on $\mathbf{R}$, which is $L^2$ over its period, then its Fourier coefficients will be in $l^2(\mathbf{Z})$ and in fact the application that map the function to its fourier coefficients is an isometric isomorphism from $L^2([0,1])$ to $l^2(\mathbf{Z})$ and you can have interesting results in that way. For the finite dimensional case search for discrete fourier transform, it's often used in numerical analysis.
For the case of Fourier transform it is known that $\mathcal{F}$ is a unitary operator of $L^2$ and you can try to see if it's in the trace class but i really doubt it since $\mathcal{F}$ is a bijection of $\mathcal{S}(\mathbf{R}^d)$ (the schwatrz space) onto itself and it is an infinite dimensional space.
4) see what i have already written.
5) see what i have written upway in my post. integral operators are operators, an eigenvalue will be exactly what you would expect : a complex number $\lambda$ such that it exists a vector $v\neq 0$ with $Av=\lambda v$. Once again if you want to know more about this or generalizations of the concept of eigenvalue search for spectral theory.
As a side note I would say that the analogy with matrices is sometime relevant : taking a hilbert basis of $L^2(\Omega)$ will give you an isomorphism of hilbert spaces between $L^2(\Omega)$ and $l^2(\mathbf{N})$, and in this space expressing a continuous operator as an "infinite matrix" is pretty easy. Moreover if your operator is compact it'll behave kind of like a finite dimentional linear operator, for more info search "fredholm alternative".
