Existence of Transformations whose Commutator equals Identity

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Theorem

Let $\map {C^\infty} \R$ be the space of smooth real functions.

Let $\circ$ denote the composition of mappings.

Let $A : \map {C^\infty} \R \to \map {C^\infty} \R$ be the mapping such that:

$\forall \phi \in \map {C^\infty} \R : \forall x \in \R : \map {\paren{ A \circ \phi } } x := \map {\dfrac {d \phi} {d x} } x$

Let $B : \map {C^\infty} \R \to \map {C^\infty} \R$ be the mapping such that

$\forall \phi \in \map {C^\infty} \R : \forall x \in \R : \map {\paren{ B \circ \phi } } x := x \map \phi x$

Let $I : \map {C^\infty} \R \to \map {C^\infty} \R$ be the identity such that:

$\forall \phi \in \map {C^\infty} \R : I \circ \phi = \phi$


Then:

$\forall \Psi \in \map {C^\infty} \R : \paren {A \circ B - B \circ A} \circ \Psi = I \circ \Psi$

or with slight abuse of notation:

$A \circ B - B \circ A = I$


Proof

Let $\Psi \in \map {C^\infty} \R$.

Then:

\(\ds \paren {A \circ B - B \circ A} \circ \Psi\) \(=\) \(\ds \map {\paren {A \circ B } } \Psi - \map{ \paren{B \circ A } } \Psi\) Derivative Operator is Linear Mapping
\(\ds \) \(=\) \(\ds \map A {\map B \Psi} - \map B {\map A \Psi}\) Definition of Composition of Mappings
\(\ds \) \(=\) \(\ds \map A {x \Psi} - \map B {\dfrac d {d x} \Psi}\) by hypothesis
\(\ds \) \(=\) \(\ds \dfrac d {d x} \paren {x \Psi} - x \dfrac d {d x} \Psi\) by hypothesis
\(\ds \) \(=\) \(\ds \paren{ \Psi + x \dfrac d {d x} \Psi } - x \dfrac d {d x} \Psi\) Product Rule for Derivatives
\(\ds \) \(=\) \(\ds \Psi\) simplification
\(\ds \) \(=\) \(\ds I \circ \Psi\) Definition of Identity Mapping

$\blacksquare$

Sources

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