Integral Transform is Mapping

Theorem
Let $\map F p$ be an integral transform:


 * $\map F p = \ds \int_a^b \map f x \map K {p, x} \rd x$

Let $T$ be the integral operator associated with $\map F p$.

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

That is, for every $\map f x$ there exists a unique $\map F p$.

Proof
Let $p$ be fixed.

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

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

We have that both $\map f x$ and $\map K {p, x}$ are integrable.

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

From Definite Integral is Unique, for a given $p$ there is exactly one real number $\map F p$ for which:
 * $\map F p = \ds \int_a^b \map f x \map K {p, x} \rd x$