Vector space/Bidual/Natural mapping/Fact/Proof

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

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

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

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

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

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

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

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

The scalar compatibility follows from

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

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

${\displaystyle {}v=0\,.}$

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

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