Vector space/Bidual/Natural mapping/Fact/Proof

Proof

Let ${}v\in V$ be fixed. First of all, we show that ${}\Psi (v)$ is a linear form on the dual space ${}{V}^{*}$. Obviously, ${}\Psi (v)$ is a mapping from ${}{V}^{*}$ to ${}K$. The additivity follows from

${}(\Psi (v))(f_{1}+f_{2})=(f_{1}+f_{2})(v)=f_{1}(v)+f_{2}(v)=(\Psi (v))(f_{1})+(\Psi (v))(f_{2})\,,$

where we have used the definition of the addition on the dual space. The compatibility with the scalar multiplication follows similarly from

${}(\Psi (v))(sf)=(sf)(v)=s(f(v))=s((\Psi (v))(f))\,.$

In order to prove the additivity of ${}\Psi$, let ${}v,w\in V$ be given. We have to show the equality

${}\Psi (v+w)=\Psi (v)+\Psi (w)\,.$

This is an equality inside of ${}{\left({V}^{*}\right)}^{*}$, in particular, it is an equality of mappings. So let ${}f\in {V}^{*}$ be given. Then, the additivity follows from

${}(\Psi (v+w))(f)=f(v+w)=f(v)+f(w)=(\Psi (v))(f)+(\Psi (w))(f)\,.$

The scalar compatibility follows from

${}(\Psi (sv))(f)=f(sv)=s(f(v))=s((\Psi (v))(f))\,.$

In order to prove injectivity, let ${}v\in V$ with ${}\Psi (v)=0$ be given. this means that for all linear forms ${}f\in {V}^{*}$, we have ${}f(v)=0$. But then, due to fact, we have

${}v=0\,.$

By the criterion for injectiviy, ${}\Psi$ is injective.

In the finite-dimensional case, the bijectivity follows from injectivity and from fact.

To fact