A linear functional $\omega:v\mapsto\omega(v)$ on a finite dimensional vector space $X$ of dimension $N$ with an inner product (·\ ,·) is an element of the dual vector space X', and a couple of isomorphisms later I'll find $\omega(v)=(\omega,v).$
Now the Riesz representation theorem for $L^p(\Omega)$, $1\le p<\infty$, says that for some \omega\in L^p(\Omega)', there is an $u\in L^q(\Omega)$ with
$\omega(v)=\int_\Omega u(x)v(x) \mathbb{d}x.$
Secondly, a continuous endomorphism $A: v\mapsto A(v)$ of a finite dimensional vector space can naturally be represented by a matrix with $N^2$ coefficients $A_n^{p}$, which acts on the vectors as
$(A(v))_n=\sum_{p=1}^NA_n^{p}v_p.$
My question now is if every linear map $A: f\mapsto A(f)$ on function spaces can represented similar to the matrixes operation, i.e. as
$(A(f))(x) = \int_\Omega A(x,p)f(p) \mathbb{d}p,$
even if this might only be true in a distributional sense.
Clearly $f(x) \mapsto \int A(x,p)f(p) \mathbb{d}p$ is a linear map and besides the regular examples where $A(x,p)$ is really just a nice function itself (like for the Fourier transformation with $A(x,p)=e^{\mathbb{i}xp}$), I am aware of distributions like the Dirac delta ``A(x,p)=\delta(x-p)´´, which for example represents the identity $f(x) \mapsto \int \delta(x-p)f(p) \mathbb{d}p = f(x)$. But do i get everything this way? Can I write all operators using an integral which a priori gathers information of the function $f(x)$ on its whole domain?
The question arose when I ask for an ``matrix-like´´ representation of common linear operators like D\equiv\frac{\partial}{\partial x}: f(x)\mapsto f'(x). I guess ``D=-\delta'\ ´´ works since f(x)\mapsto(D(f))(x)=\int (-\delta'(x-p))f(p) \mathbb{d}p=\int \delta(x-p)f'(p) \mathbb{d}p=f'(x). But are there even integral representations for, say, $\Delta=\sum_i\frac{\partial^i}{\partial x^i}$ or even the Laplace–Beltrami operator? Maybe using nesting of delta-distributions and metric-coefficient functions? Is this possible, is it useful in any sense, and are there more difficult linear operators than that? And if not, why? Is it advantageous to view the linear operators on function spaces like that?