Suppose that $E $ is an infinite dimensional vector space. Show that there exists a dual space $E^*$ such that the natural injection $i:E^* \rightarrow L (E)$ defined by $i(e^*) = \langle - , e^* \rangle : E \rightarrow \Gamma$ is not surjective, where $L (E) $ is the set of linear mappings $\varphi : E \rightarrow \Gamma $. $ (\langle -,- \rangle$ takes values in $\Gamma )$
(In my context all vector spaces are defined over a fixed, but arbitrarily chosen field $ \Gamma $ of characteristic $ 0 $)
I know that if $E, E^*$ is a pair of vector space and if a fixed non-degenerate bilinear function, $\langle , \rangle$, in $E^* \times E$ is defined. Then $E$ and $E^*$ will be called dual with respect to the bilinear function $\langle , \rangle $.
I don't know how to use the fact that $ E $ is an infinite dimensional vector space. Could you give me any suggestion please?
A very important theorem in linear algebra that is rarely taught is:
Theorem: A vector space has the same dimension as its dual if and only if it is finite dimensional.
Your $L(E)$ is the “full dual”. Take a basis $B$ of $E$ and, for $v\in B$, define $v^*(v)=1$, $v^*(w)=0$, for $w\in B$, $w\ne v$ and extend by linearity.
Take as $E^*$ the subspace of $L(E)$ spanned by $\{v^*:v\in B\}$.