# Underlying Mapping of Evaluation Linear Transformation is Element of Double Dual

## Theorem

Let $\struct {R, +, \times}$ be a commutative ring with unity.

Let $G$ be an $R$-module.

Let $G^*$ be the algebraic dual of $G$.

Let $G^{**}$ be the double dual of $G$.

For each $x \in G$, let $x^\wedge: G^* \to R$ be defined as:

$\forall t \in G^*: \map {x^\wedge} t = \map t x$

Then:

$x^\wedge \in G^{**}$

## Proof

We have that $x^\wedge$ is a mapping from $G^* \to R$.

It remains to be demonstrates that $x^\wedge$ is in fact a linear transformation.

Hence we need to show that:

$(1): \quad \forall u, v \in G^*: \map {x^\wedge} {u + v} = \map {x^\wedge} u + \map {x^\wedge} v$
$(2): \quad \forall u \in G^*: \forall \lambda \in R: \map {x^\wedge} {\lambda \times u} = \lambda \times \map {x^\wedge} u$

Hence:

 $\text {(1)}: \quad$ $\ds \map {x^\wedge} {u + v}$ $=$ $\ds \map {\paren {u + v} } t$ Definition of $x^\wedge$ $\ds$ $=$ $\ds \map u t + \map v t$ Definition of Pointwise Addition of Linear Transformations $\ds$ $=$ $\ds \map {x^\wedge} u + \map {x^\wedge} v$ Definition of $x^\wedge$

and:

 $\text {(2)}: \quad$ $\ds \map {x^\wedge} {\lambda \times u}$ $=$ $\ds \map {\paren {\lambda \times u} } t$ Definition of $x^\wedge$ $\ds$ $=$ $\ds \lambda \times \map u t$ Definition of Linear Transformation $\ds$ $=$ $\ds \lambda \times \map {x^\wedge} u$ Definition of $x^\wedge$

Hence the result.

$\blacksquare$