Dx - k is Hypoelliptic

Theorem
Let $f \in \map {C^\infty} \R$ be a smooth real function.

Let $T \in \map {\DD'} \R$ be a distribution.

Let $T_f$ be the distribution associated with $f$.

Let $k \in \R$.

Suppose in the distributional sense it holds that:


 * $\paren {\dfrac \d {\d x} - k} T = T_f \quad (1)$

Then there is a $c \in \R$ such that:


 * $T = T_{F + c \map \exp {k x} }$

where $F \in \map {C^\infty} \R$ is a solution to:


 * $\paren {\dfrac \d {\d x} - k} F = f$

In other words, $\dfrac \d {\d x} - k$ is a hypoelliptic operator.

Proof
By assumption $f \in \map {C^\infty} \R$.

By Solution to Linear First Order ODE with Constant Coefficients:


 * $\exists F \in \map {C^\infty} \R : \paren {\dfrac \d {\d x} - k} F = f$

By Differentiable Function as Distribution and multiplication of distribution by a smooth function:


 * $\exists F \in \map {C^\infty} \R : \paren {\dfrac \d {\d x} - k} T_F = T_f \quad (2)$

Subtract $(2)$ from $(1)$:

where $\mathbf 0$ is the zero distribution.

This is solved by:


 * $T - T_F = T_{c \map \exp {k x} }$

or:


 * $T = T_{F + c \map \exp {k x} }$

where $c \in \C$.

Finally:


 * $\ds \paren {\dfrac \d {\d x} - k} \paren {F + c \map \exp {k x} } = f$