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:
\(\ds x \map T {\map \phi x}\) | \(=\) | \(\ds x \paren{c \map \delta {\map \phi x} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \delta {c x \map \phi x}\) | 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:
- $\ker \delta \subseteq \set {x \phi : \phi \in \map \DD \R} \subseteq \ker T$
- $\exists c \in \C : T = c \delta$
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 6.4$: A glimpse of distribution theory. Multiplication by $C^\infty$ functions