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
- 1958: G.E.H. Reuter: Elementary Differential Equations & Operators: Chapter $2$: $\S 1.4$: The inverse of an operator
- 1968: Peter D. Robinson: Fourier and Laplace Transforms ... (previous) ... (next): $\S 1.1$. The Idea of an Integral Transform