# Can the canonical linear function of a vector space to his double dual space be surjective? [duplicate]

I am having problems to prove the following:

Let V be a K-Vectorspace and $dim(V)=\infty$. Prove that the canonical linear transformation $e: V \rightarrow V^{**}$ is not surjective.

I think I get the idea of the proof, since if $(v_{i})_{i \in I}$ is a basis of $V$ and $(v_{i}^{*})_{i \in I}$ is the dual family in $V^{*}$ then it might - or even can't - span the whole space $V^{*}$. But I am neither sure how I can show this, nor the actual proof.
Help would be much appreciated. Thanks in advance!

• Are you interested in $V^{**}$ or $V^{*}$ ? You comments are about the later and require a choice of basis? – Rene Schipperus May 4 '17 at 20:53
• I am interested in $V^{**}$ the comments were just my simplified thoughts – get rekt m8 May 4 '17 at 21:08

We start with a basis $\{v_i\}_{i\in I}$. From this we obtain the duals $\{v_i^*\}_{i\in I}$, defined by $v_i(v_j)=\delta_{ij}$. Now observ that for any $v\in V$, the element $e(v)\in V^{**}$ suffers from some severe restrictions: As $v$ is a linear combination of finitely many of the $v_i$, we see that $\phi_i(v)\ne 0$ for at most finitely many $i\in I$. So if we exhibit $\Phi\in V^{**}$ such that $\Phi(\phi_i)\ne 0$ for infinitely many $i\in I$, this $\Phi$ cannot be of the form $e(v)$ with $v\in V$, thus showing that $e$ is not onto.
Exhibiting such $\Phi$ is easy: Let $W=\operatorname{span}(\phi_i\mid i\in I)$ and define $\Phi_0\colon W\to K$ by $\Phi(\phi_i)=1$ for all $i\in I$. Use an arbitrary complement to $W$ in $V^{**}$ to extend $\Phi_0$ to $\Phi\in V^{**}$ as desired.
• how is $\phi _{i}$ defined? – get rekt m8 May 4 '17 at 21:25