Commutative and Associative Product on Space of Distributions does not Exist

From ProofWiki
Jump to navigation Jump to search


Let $\map {\DD'} \R$ be the distribution space.

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

Let $\circ$ be a product operation on $\map {\DD'} \R$.


$\forall T \in \map {\DD'} \R : \forall \alpha \in \map {C^\infty} \R : \alpha \circ T := \alpha \cdot T$

where $\cdot$ stands for multiplication of distribution by a smooth function.

Suppose $\circ$ is commutative and associative.

Then $\circ$ does not exist.


Aiming for a contradiction, suppose there is such a product.

Let $\delta \in \map {\DD'} \R$ be the Dirac delta distribution.

We also have that $\ds \PV \frac 1 x$ is a distribution.

Let $\phi, \psi \in \map \DD \R$ be test functions.


\(\ds x \cdot \map {\paren { \PV \frac 1 x} } \phi\) \(=\) \(\ds \map {\paren {\PV \frac 1 x} } {x \phi}\)
\(\ds \) \(=\) \(\ds \lim_{\epsilon \mathop \to 0} \int_{\size x \mathop > \epsilon} \frac {x \map \phi x} x \rd x\)
\(\ds \) \(=\) \(\ds \lim_{\epsilon \mathop \to 0} \int_{\size x \mathop > \epsilon} \map \phi x \rd x\)
\(\ds \) \(=\) \(\ds \int_{- \infty}^\infty \map \phi x \rd x\) Definition of Test Function
\(\ds \) \(=\) \(\ds \map {\bf 1} \phi\)
\(\ds \) \(=\) \(\ds c\) $c \in \R$

Here $\mathbf 1$ is the distribution associated to the real function defined by:

$\mathbf 1 : \R \to 1$

Consider the product $\ds \delta \cdot x \cdot \PV \frac 1 x$ in the distributional sense, where $\delta$ and $\PV \frac 1 x$ are distributions, and $x$ is a smooth function.

\(\ds \paren{\map \delta \psi \circ x} \circ \map {\paren{\PV \frac 1 x} } \phi\) \(=\) \(\ds \paren{x \circ \map \delta \psi} \circ \map {\paren{\PV \frac 1 x} } \phi\) Commutativity in the first pair
\(\ds \) \(=\) \(\ds \paren{x \cdot \map \delta \psi} \circ \map {\paren{\PV \frac 1 x} } \phi\)
\(\ds \) \(=\) \(\ds \map \delta {x \psi} \circ \map {\paren{\PV \frac 1 x} } \phi\) Definition of Multiplication of Distribution by Smooth Function
\(\ds \) \(=\) \(\ds 0 \circ \map {\paren{\PV \frac 1 x} } \phi\) Definition of Dirac Delta Distribution
\(\ds \) \(=\) \(\ds 0 \cdot \map {\paren{\PV \frac 1 x} } \phi\)
\(\ds \) \(=\) \(\ds 0\)
\(\ds \map \delta \psi \circ \paren {x \circ \map {\paren{\PV \frac 1 x} } \phi}\) \(=\) \(\ds \map \delta \psi \circ \paren {x \cdot \map {\paren{\PV \frac 1 x} } \phi}\)
\(\ds \) \(=\) \(\ds \map \delta \psi \circ \map {\mathbf 1} \phi\)
\(\ds \) \(=\) \(\ds \map {\mathbf 1} \phi \circ \map \delta \psi\) Commutativity
\(\ds \) \(=\) \(\ds c \circ \map \delta \psi\)
\(\ds \) \(=\) \(\ds c \cdot \map \delta \psi\)
\(\ds \) \(=\) \(\ds \map \delta {c \psi}\) Definition of Multiplication of Distribution by Smooth Function
\(\ds \) \(\ne\) \(\ds 0\)

Altogether, we have that:

$\ds \paren {\delta \circ x} \circ \PV \frac 1 x \ne \delta \circ \paren {x \circ \PV \frac 1 x}$

Hence, the associativity is violated, and we have reached a contradiction.