Distributional Solution to x T = 0

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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:

\(\ds x \map T {\map \phi x}\) \(=\) \(\ds x \paren{c \map \delta {\map \phi x} }\)
\(\ds \) \(=\) \(\ds \map \delta {c x \map \phi x}\) Definition of Multiplication of Distribution by Smooth Function
\(\ds \) \(=\) \(\ds 0\) Definition of Dirac Delta Distribution
\(\ds \) \(=\) \(\ds \map {\mathbf 0} \phi\)

Direct inclusion

Suppose:

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

Then:

\(\ds \forall \phi \in \map \DD \R: \, \) \(\ds 0\) \(=\) \(\ds \map {xT} \phi\)
\(\ds \) \(=\) \(\ds \map T {x \phi}\) Definition of Multiplication of Distribution by Smooth Function

So:

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

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

$\Box$


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:

\(\ds \map \delta \psi\) \(=\) \(\ds \map \psi 0\)
\(\ds \) \(=\) \(\ds 0\)

Hence:

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

By the fundamental theorem of calculus:

\(\ds \map \psi x\) \(=\) \(\ds \map \psi 0 + \int_0^x \map {\psi'} \xi \rd \xi\)
\(\ds \) \(=\) \(\ds 0 + x \int_0^1 \map {\psi'} {tx} \rd t\) $\xi = t x$
\(\ds \) \(=\) \(\ds x \int_0^1 \map {\psi'} {tx} \rd t\)

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:

$\psi \in \map {C^\infty} \R$

By differentiating under integral sign:

$\phi \in \map {C^\infty} \R$

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 Kernel of Linear Transformation contained in Kernel of different Linear Transformation implies Transformations are Proportional:

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


$\blacksquare$


Sources