Why are integral operators defined the way they are?

Question

I'm reading a book on functional analysis, in which the following example is given after the definition of an operator.

Let $-\infty<a<b<\infty$, and let the function $F:[a,b]\times[a,b]\times\mathbb{R}\to\mathbb{R}$ be continuous. We set

$$(Au)(x) = \int_a^x F(x,y,u(y))dy~~~\forall~x\in[a,b]$$

$$(Bu)(x) = \int_a^b F(x,y,u(y))dy~~~\forall~x\in[a,b]$$

Then, we obtain the two operators $A:C[a,b] \to C[a,b]$ and $B:C[a,b]\to C[a,b]$. The operators $A$ and $B$ are called integral operators.

1. How did the author jump to the fact that if $u$ is continuous, then so are $Au$ and $Bu$? There's no justification given in any of the prior sections.

2. Why is this example important at all? The way $F$ is defined seems so arbitrary. Why not instead define $F:[a,b]\times[a,b]\to\mathbb{R}$ and include $F(x,y)$ in the integral? Wouldn't that be simpler? Why take the limits $a$ to $b$ in the expression for $Bu$? What's the motivation/justification of defining this stuff the way it is? I'm asking because these operators seem important enough to have a name.

Answer 1

The first point is just a simple fact about the Riemann integral. If you know the dominated convergence theorem you can prove it very easily, but here is a proof that avoids that machinery.

Since $u$ is continuous, $u([a,b]) \subseteq [c,d]$ for some $c,d \in \mathbb{R}$. Now, $F$ is continuous on the compact set $[a,b]\times[a,b]\times[c,d]$ and hence is bounded there, say $F([a,b]\times[a,b]\times[c,d]) \subseteq [-C,C]$. So $$|Au(x_1) - Au(x_2)| \leq \bigg | \int_{x_1}^{x_2} |F(x,y,u(y))| dy \bigg | \leq C |x_1 - x_2|.$$

Continuity of $B$ can be established similarly, using uniform continuity of $F$ in the first variable on $[a,b]\times[a,b]\times[c,d]$.

These operators are actually incredibly important (although often we want a larger domain than $C[a,b]$, but that should give motivation to study the nice case of continuous functions).

For example, the definition hides it a bit, but you're quite possibly actually already familiar with very useful examples of such operators. The usual integral transforms (Fourier, Laplace etc.) are examples of integral operators.

Even if you don't care about these specific integral operators, you'll often find such operators cropping up as energy functionals in Calculus of Variations problems which alone is a good enough reason to care about them.