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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


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?

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The details depend strongly on what you take "function space" to mean. The case of $C(X)$ (scalar valued continuous functions on a compact Hausdorff space $X$) is discussed, with references, in… . The question is different but the answer points out that any bounded operator from $C(X)$ to a Banach space can be represented as an integral with respect to an operator valued measure. The Krein school of functional analysis heavily exploited this kind of idea in other function spaces. – leslie townes Nov 6 '11 at 21:45

A strong affirmative answer to one form of the question is L. Schwartz' Kernel Theorem. Perhaps the simplest case is that every continuous linear map $T$ from the (secretly "nuclear") Frechet space of smooth functions on the circle to the space of distributions on the circle is given by a distribution $K$ on a product of two circles, by $(Tf)(g)=K(f\times g)$. The distribution $K$ is the "kernel" (not in the sense of things mapped to $0$ by $T$), and is a (weak-*) limit of (classical) functions of two variables... so is an extension of integrate-against those classical two-variable functions.

On the other hand, many seemingly-more-elementary versions of the question have negative answers. The most immediate is that continuous linear maps from $L^2[a,b]$ to itself are mostly not given by integrating against kernel-functions $K(x,y)$ in $L^2([a,b]\times [a,b])$. One reason for this is that linear maps given by such kernels are (Hilbert-Schmidt, and) compact maps, and most maps on infinite-dimensional Hilbert spaces are not compact.

Similar negative answers to forms of the question eventually lead to the somewhat-more-complicated context in which Schwartz' theorem is formulated. (Some notes about this are at especially )

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