Inverse Integral Operator is Linear if Unique

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 $p, q \in \R$:

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.