Example of not surjective natural map from vector space to its double dual

Question

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.

Answer 1

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.

Answer 2

Let $\dim V = \infty$ and $B = (b_1, b_2, ...)$ be a basis of V.

$(.)^{**}: V \longrightarrow V^{**}, v \mapsto v^{**}$, where $v^{**}: V^* \longrightarrow K, f \mapsto f(v)$. It easy to show that $(.)^{**}$ is a homomorphism and injective.

Consider $\psi \in V^{**}: \psi(b_i^*) = 1$ for $\forall i \in \mathbb{N}$.

Now you can see that $(.)^{**}$ is not surjective because you cannot find an element in V that maps to $\psi$ because linear combinations are finite.