I recognize the fact that the natural map from an infinite dimension vector space $V$ to it's double dual space $V^{**}$ need not necessarily to be surjective because we don't have that the $\dim V$ = $\dim V^{*}$.
However, what is a simple example to show that it is not surjective in this case?
My first instinct was to pick the vector space $V = (\mathbb{Z}/\mathbb{2Z})^{\mathbb{N}}$ just a simple vector space.
However, when picking up a functional in the dual space which wouldn't be isomorphic to the natural map for any $v \in V$ I wasn't able to write one down.
Can someone help me come up with an example / some intuition as to why this happens?
Thanks.
Best Answer
You can pick any infinite dimensional vector space. Say, take $V=k[x]$ to be just polynomials in one variable. It is countable-dimensional. If you take its dual, you get $V^*=k[[x]]$, which is already uncountable-dimensional. When taking double dual $V^{**}$ it will be even bigger. So there is no way $V$ will be isomorphic to $V^{**}$, you don't even have to worry about how exactly the map $V\to V^{**}$ looks like.
And this is true in general for infinite-dimensional vector spaces. Their linear duals are getting too big.