Inverse Integral Operator is Linear if Unique

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Theorem

Let $T$ be an integral operator.

Let $f$ be an integrable real function on a domain appropriate to $T$.

Let $F = \map T f$ and $G = \map T g$.

Let $T$ have a unique inverse $T^{-1}$.


Then $T^{-1}$ is a linear operator:

$\forall p, q \in \R: \map {T^{-1} } {p F + q G} = p \map {T^{-1} } F + q \map {T^{-1} } G$


Proof

Let:

$x_1 = \map {T^{-1} } F$
$x_2 = \map {T^{-1} } G$

Thus:

$F = \map T {x_1}$
$G = \map T {x_2}$


Then for all $p, q \in \R$:

\(\ds \map T {p x_1 + q x_2}\) \(=\) \(\ds p \map T {x_1} + q \map T {x_2}\) Integral Operator is Linear
\(\ds \) \(=\) \(\ds p F + q G\)

and so $x = p F + q G$ is a solution to the equation:

$\map T x = p F + q G$

But this equation has only one solution:

$x = \map {T^{-1} } {p F + q G}$

Thus $p F + q G$ must coincide with the above:

$p \map {T^{-1} } F + q \map {T^{-1} } G = \map {T^{-1} } {p F + q G}$

which proves that $T^{-1}$ is a linear operator.

$\blacksquare$


Sources