# Commutative and Associative Product on Space of Distributions does not Exist

## Theorem

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 $\map {\DD'} \R$.

Suppose:

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

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

Suppose $\circ$ is commutative and associative.

Then $\circ$ does not exist.

## Proof

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.

Then:

 $\ds x \cdot \map {\paren { \PV \frac 1 x} } \phi$ $=$ $\ds \map {\paren {\PV \frac 1 x} } {x \phi}$ $\ds$ $=$ $\ds \lim_{\epsilon \mathop \to 0} \int_{\size x \mathop > \epsilon} \frac {x \map \phi x} x \rd x$ $\ds$ $=$ $\ds \lim_{\epsilon \mathop \to 0} \int_{\size x \mathop > \epsilon} \map \phi x \rd x$ $\ds$ $=$ $\ds \int_{- \infty}^\infty \map \phi x \rd x$ 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.

$\blacksquare$