Integral Transform is Mapping

Theorem
Let $F \left({p}\right)$ be an integral transform:


 * $F \left({p}\right) = \displaystyle \int_a^b f \left({x}\right) K \left({p, x}\right) \, \mathrm d x$

Let $T$ be the integral operator associated with $F \left({p}\right)$.

Then $T$ is a mapping from the domain of $T$ to its image.

That is, for every $f \left({x}\right)$ there exists a unique $F \left({p}\right)$.

Proof
Let $p$ be fixed.

In this context, $f \left({x}\right) K \left({p, x}\right)$ is the pointwise product of the functions $f \left({x}\right)$ and $K \left({p, x}\right)$.

From Pointwise Operation is Well-Defined, it follows that $f \left({x}\right) K \left({p, x}\right)$ is a real function on $x$.

We have that both $f \left({x}\right)$ and $K \left({p, x}\right)$ are integrable.

It follows from Pointwise Product of Integrable Real Functions is Integrable that $f \left({x}\right) K \left({p, x}\right)$ is an integrable function.

From Definite Integral is Unique, for a given $p$ there is exactly one real number $F \left({p}\right)$ for which:
 * $F \left({p}\right) = \displaystyle \int_a^b f \left({x}\right) K \left({p, x}\right) \, \mathrm d x$