What is the general form of linear operators on continuous functions?

Question

I was wondering if there was a representation for a set of operators dense in the space of linear operators $B$ mapping $C(a,b) \to C(c,d)$. I thought that maybe integral operators give a general representation; that is given the right $E(u,t)$, $$\int f(t)E(u,t) dt$$ approximates an arbitrary linear, bounded operator $T$ resulting in $$\int f(t)E(u,t) dt = (Tf)(u)$$

I am honestly very new to operator theory and I am really clueless as to whether or not a general representation exists. I would very much appreciate it if someone could guide me in the right direction.

Answer 1

Yes, more or less. Ordinary integrals do not give a general representation of bounded operators between spaces of continuous functions, the identity operator on $C(a,b)$ is not of this form, at least if you want $E(u,t)$ to actually be a function. However, there is the Schwartz kernel theorem that represents every operator on the space of test functions (infinitely smooth of compact support) in the integral form if the kernels $E(u,t)$ are allowed to be distributions. Since test functions are continuous any operator on continuous functions has to be representable in this form as well.

For the example above of the identity operator the kernel is then $E(u,t):=\delta(u-t)$, where $\delta$ is the Dirac's delta function. Of course, not all distributional kernels define bounded operators on continuous functions, only those that are distributional densities of signed Borel measures do. An equivalent alternative representation in terms of Stieltjes integrals $\int f(t)dK(u,t)$ is given in Tucker's paper, but it is rather abstract and technical. Linear Operators by Dunford and Schwartz may have a more accessible version.

As for density and approximation, it depends in what sense. Distributions can be approximated by continuous and even by test functions in some weak senses, so this will give you approximation of bounded operators by integral ones with "ordinary" kernels in those senses. But by norm the identity operator on $C(a,b)$ can not be approximated by integral operators with say continuous kernels, because such operators are compact, and the identity is not.

Answer 2

Every linear functional $x^{\star}$ on $C[a,b]$ has the form $$ x^{\star}(f) = \int_{a}^{b}f(t)\,d\mu(t) $$ for a unique finite complex Borel measure $\mu$ on $[a,b]$. Furthermore, $$ \|x^{\star}\| = \|\mu\|, $$ where $\|\mu\|$ is the total variation of $\mu$ on $[a,b]$. If $K : C[a,b]\rightarrow C[a,b]$ is a bounded linear operator, then, for each $t \in [a,b]$, let $\delta_{t}^{\star}$ be the evaluation functional at $t$. That is, $\delta_{t}^{\star}(f)=f(t)$. The functional $\delta_{t}^{\star}$ is bounded on $C[a,b]$. So $\delta_{t}^{\star}\circ K=K^{\star}\delta_{t}^{\star}$ is a bounded linear functional on $C[a,b]$. Hence, there is unique finite complex Borel measure $\mu_{t}$ such that $$ (Kf)(t)= \delta_{t}^{\star}(Kf) = (K^{\star}\delta_{t}^{\star})(f)=\int_{a}^{b}f(s)d\mu_{t}(s),\\ \|\mu_{t}\|=\|K^{\star}x_{t}^{\star}\|. $$ If there is a dominating positive finite meausre $\mu$ such that $\mu_{t} << \mu$ for all $t\in[a,b]$, then you can write $d\mu_{t}(s) = K_{t}(s)d\mu(s)$ where $K_{t}(s) \in L^{1}_{\mu}[a,b]$ for each $t$. That gets you close to such a representation, under the additional assumption: $$ Kf = \int_{a}^{b}f(s)K_{t}(s)\,d\mu(s) = \int_{a}^{b}f(s)K(t,s)\,d\mu(s). $$ Without the additional assumption, the integrals of continuous functions with respect to finite complex Borel measures can be expressed in terms of functions of bounded variation, which gives a representation as a Riemann-Stieltjes integral $$ Kf = \int_{a}^{b}f(s)d_{s}L_{t}(s)=\int_{a}^{b}f(s)d_{s}L(t,s). $$