# Distribution Space over Smooth Functions is Unitary Module

## Theorem

The distribution space over smooth functions is a unitary module.

## Proof

Let $\phi \in \map \DD {\R^d}$ be a test function.

### Module Axiom $\text M 1$: Distributivity over Module Addition

 $\ds \alpha \map {\paren {T_1 + T_2} } \phi$ $=$ $\ds \map {\paren {T_1 + T_2} } {\alpha \phi}$ Definition of Multiplication of Distribution by Smooth Function $\ds$ $=$ $\ds \map {T_1} {\alpha \phi} + \map {T_2} {\alpha \phi}$ $\ds$ $=$ $\ds \alpha \map {T_1} \phi + \alpha \map {T_2} \phi$ Definition of Multiplication of Distribution by Smooth Function

Hence:

$\alpha \paren {T_1 + T_2} = \alpha T_1 + \alpha T_2$

$\Box$

### Module Axiom $\text M 2$: Distributivity over Scalar Addition

 $\ds \map {\paren {\alpha_1 + \alpha_2} T} \phi$ $=$ $\ds \map T {\paren {\alpha_1 + \alpha_2}\phi}$ Definition of Multiplication of Distribution by Smooth Function $\ds$ $=$ $\ds \map T {\alpha_1 \phi + \alpha_2 \phi}$ Real Multiplication Distributes over Addition $\ds$ $=$ $\ds \map T {\alpha_1 \phi} + \map T {\alpha_2 \phi}$ Definition of Distribution $\ds$ $=$ $\ds \alpha_1 \map T \phi + \alpha_2 \map T \phi$ Definition of Multiplication of Distribution by Smooth Function

Hence:

$\paren {\alpha_1 + \alpha_2} T = \alpha_1 T + \alpha_2 T$

$\Box$

### Module Axiom $\text M 3$: Associativity

 $\ds \paren {\alpha \beta} \map T \phi$ $=$ $\ds \map T {\paren {\alpha \beta} \phi}$ Definition of Multiplication of Distribution by Smooth Function $\ds$ $=$ $\ds \map T {\beta \paren {\alpha \phi} }$ Real Multiplication is Associative, Real Multiplication is Commutative $\ds$ $=$ $\ds \map {\paren {\beta T} } {\alpha \phi}$ Definition of Multiplication of Distribution by Smooth Function $\ds$ $=$ $\ds \map {\paren {\alpha \paren {\beta T} } } \phi$ Definition of Multiplication of Distribution by Smooth Function

Hence:

$\paren {\alpha \beta} T = \alpha \paren {\beta T}$

$\Box$

### Unitary Module Axiom $\text {UM} 4$: Unity of Scalar Ring

Let $\mathbf 1 \in \map {C^\infty} {\R^d}$ be a constant mapping such that:

$\R^d \stackrel {\mathbf 1} {\mapsto} 1$

Then:

 $\ds \mathbf 1 \cdot \map T \phi$ $=$ $\ds \map T {1 \cdot \phi}$ Definition of Multiplication of Distribution by Smooth Function $\ds$ $=$ $\ds \map T \phi$

Hence:

$1 \cdot T = T$

$\Box$

$\blacksquare$