Integral Operator is Linear

Theorem

Let $T$ be an integral operator.

Let $f$ and $g$ be integrable real functions on a domain appropriate to $T$.

Then $T$ is a linear operator:

$\forall \alpha, \beta \in \R: \map T {\alpha f + \beta g} = \alpha \map T f + \beta \map T g$

Corollary 1

$\forall \alpha, \beta \in \R: \map T {f + g} = \map T f + \map T g$

Corollary 2

$\forall \alpha \in \R: \map T {\alpha f} = \alpha \map T f$

Proof

Let $T$ be expressed in its full form as an integral fransform:

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

for some integrable function $\map K {p, x}$.

Then:

\(\ds \map T {\alpha f + \beta g}\)

\(=\)

\(\ds \int_a^b \paren {\alpha \map f x + \beta \map g x} \map K {p, x} \rd x\)

\(\ds \)

\(=\)

\(\ds \int_a^b \paren {\alpha \map f x \map K {p, x} + \beta \map g x \map K {p, x} } \rd x\)

Real Multiplication Distributes over Addition

\(\ds \)

\(=\)

\(\ds \alpha \int_a^b \map f x \map K {p, x} \rd x + \beta \int_a^b \map g x \map K {p, x} \rd x\)

Linear Combination of Definite Integrals

\(\ds \)

\(=\)

\(\ds \alpha \map T f + \beta \, \map T g\)

Definition of Integral Operator

$\blacksquare$