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:
- $\map F p = \ds \int_a^b \map f x \map K {p, x} \rd x$
$\blacksquare$
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Sources
- 1968: Peter D. Robinson: Fourier and Laplace Transforms ... (previous) ... (next): $\S 1.1$. The Idea of an Integral Transform