Distributional Solution to x T = 0

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

Let $\delta$ be the Dirac delta distribution.

Let $\mathbf 0$ be the zero distribution.

Suppose $T$ satisfies the following equation in the distributional sense:


 * $x T = \mathbf 0$

Then $T = \alpha \delta$ where $c \in \C$.

Proof
Let $\phi \in \map \DD \R$ be a test function.

Let $c \in \C$.

Suppose:


 * $T = c \delta$

Then:

Direct inclusion
Suppose:


 * $T \in \map {\DD'} \R : x T = 0$

Then:

So:


 * $\set {x\phi : \phi \in \map \DD \R} \subseteq \ker T$

where $\ker$ denotes the kernel of $T$.

Let $\psi = x \phi$.

Suppose $\phi \in \map \DD \R$.

Then $\psi \in \map \DD \R$.

Furthermore:


 * $\paren {\map \phi 0 = 0} \implies \paren {\map \psi 0 = 0}$

Hence:


 * $\set {x \phi : \phi \in \map \DD \R} \subseteq \set {\psi \in \map \DD \R : \map \psi 0 = 0}$

Reverse inclusion
Let $\psi \in \map \DD \R$ such that $\map \psi 0 = 0$.

Then:

Hence:


 * $\ker \delta = \set {\psi \in \map \DD \R : \map \psi 0 = 0}$

By the fundamental theorem of calculus:

Let $\ds \map \phi x := \int_0^1 \map {\psi'} {tx} \rd t$.

Then $\map \psi x = x \map \phi x$

By assumption, $\psi$ is a test function.

Hence, $\psi$ is smooth:


 * $\psi \in \CC^\infty$

By differentiating under integral sign:


 * $\phi \in \CC^\infty$

By definition, $\psi$ has a compact support.

Let $a \in \R_{\mathop > 0}$.

Suppose:


 * $\forall x \notin \closedint {-a} a : \map \psi x = 0$

Then:


 * $\ds \forall x \notin \closedint {-a} a : \map \phi x = \frac {\map \psi x}x = 0$

We have that $\phi$ is smooth and compactly supported.

By definition, $\phi \in \map \DD \R$.

Altogether:


 * $\set {\psi \in \map \DD \R : \map \psi 0 = 0} \subset \set {x \phi : \phi \in \map \DD \R}$

Thus:


 * $\ker \delta \subseteq \set {x \phi : \phi \in \map \DD \R} \subseteq \ker T$

By Vectors whose Kernels are Subsets of Each Other:


 * $\exists c \in \C : T = c \delta$