Canonical Isomorphism Between $\mathbf{V}$ and $(\mathbf{V}^*)^*$

Let $V$ be any vector space over a field $k$. Let $\{e_i\}_{i \in I}$ be a basis for $V$. For each $i \in I$, there is a unique linear functional $f_i: V \rightarrow k$ such that $f_i(e_j) = \delta_{ij}$: that is, $f_i(e_i) = 1$ and for every other basis element $e_j$, $f_i(e_j) = 0$.

CLAIM: The set $\{f_i\}_{i \in I}$ is linearly independent in $V^{\vee}$, and thus $\dim V \leq \dim V^{\vee}$. It is a basis if and only if $V$ is finite-dimensional (if and only if $V^{\vee}$ is finite-dimensional).

The linear independence is easy: if $a_1 f_{i_1} + \ldots + a_n f_{i_n} = 0$, then just by evaluating at $e_{i_1},\ldots,e_{i_n}$ we find $a_1 = \ldots = a_n = 0$.

In the finite-dimensional case -- say $I = \{1,\ldots,n\}$ -- we may write any linear $g: V \rightarrow k$ as

$g = g(e_1) f_1 + \ldots + g(e_n) f_n$,

which shows that $f_1,\ldots, f_n$ is a basis for $V^{\vee}$ (and implies $\dim V = \dim V^{\vee}$).

However, in the infinite-dimensional case the $\{f_i\}_{i \in I}$ do not form a basis..essentially because in an abstract vector space all our sums must be finite sums rather than infinite sums! Indeed the subspace spanned by the $f_i$'s is precisely the set of linear functionals which are zero at all but finitely many basis elements $e_i$, whereas to give a linear functional $f$ on $V$ the values $f(e_i)$ can be absolutely arbitrary. Concretely, the functional $f$ with $f(e_i) = 1$ for all $i \in I$ does not lie in the span of the $f_i$'s.

(Remark: In fact whenever $\dim V$ is infinite, we have $\dim V^{\vee} > \dim V$. That is, not only is $\{f_i\}_{i \in I}$ not a basis, there is no basis for the dual space of cardinality equal to that of $I$. This is actually not so easy to prove, and it is not needed to answer the question.)

Now we come back to the canonical map $I: V \rightarrow V^{\vee \vee}$ given by

$I(x): f \mapsto f(x)$.

CLAIM: a) $I$ is always injective.
b) $I$ is surjective if and only if $V$ is finite-dimensional.

To prove a), let $x$ be a nonzero element of $V$ and choose a basis $\{e_i\}_{i \in I}$ for $V$ in which $x$ is one of the basis elements, say $x = e_1$. Then $f_1$ is a linear functional which does not vanish at $x$, so $I(x)$ is a nonzero element of $V^{\vee \vee}$.

To prove b) we first use the fact that if $V$ is finite-dimensional, $\dim V = \dim V^{\vee} = \dim V^{\vee \vee}$. Thus $I: V \rightarrow V^{\vee \vee}$ is an injection between two vector spaces of the same finite dimension, so it must be an isomorphism.

Finally, if $I$ is infinite-dimensional, then one can see by choosing bases, dual sets and dual dual sets as above that $I$ is not surjective. (A good first step here is to confirm that in the finite-dimensional case, if we choose a basis $e_1,\ldots,e_n$ for $V$, a dual base $f_1,\ldots,f_n$ for $V^{\vee}$ and then a dual dual base $g_1,\ldots,g_n$ for $V^{\vee \vee}$, then the map $I$ is precisely the one which maps $e_i$ to $g_i$ for all $i$.) I can supply more details upon request. Note also that if we are willing to make use of the above parenthetical fact that $\dim V^{\vee} > \dim V$ when $\dim V$ is infinite, then we see that $\dim V^{\vee \vee} > \dim V^{\vee} > \dim V$, and thus not only is $I: V \rightarrow V^{\vee \vee}$ not an isomorphism, but moreover there is no isomorphism of vector spaces from $V$ to $V^{\vee \vee}$. Again though, this lies significantly deeper.

Added: Let me say a bit about the more ambitious approach of showing $\operatorname{dim}_k V^{\vee} > \operatorname{dim}_k V$ for any infinite-dimensional vector space $V$. Let $I$ be a basis for $V$, so $V \cong \bigoplus_{I} k$. As mentioned above, to give a linear functional on $V$ it is necessary and sufficient to assign to each basis element an arbitrary value in $k$, whence an isomorphism $V^{\vee} \cong k^I = \prod_{I} k$. Thus dualization replaces a direct sum over $I$ with a direct product over $I$. When $I$ is finite there is no difference, so we recover $V \cong V^{\vee}$. However, when $I$ is infinite I claim that

$$ \operatorname{dim}_k V^{\vee} = \operatorname{dim}_k k^I = \# k^{\# I} \geq 2^{\# I} > \# I = \operatorname{dim}_k V.$$

I know of almost no standard texts which include a proof of this result, and indeed some cleverness / real ideas seem to be required (unlike the above discussion of the non-surjectivity of $I$ in the infinite-dimensional case which is, while somewhat lengthy to write out in detail, really very straightforward). However, by coincidence I just found on the web a very nice proof of this result which deduces it from Dedekind's Linear Independence of Characters. Please see this note of France Dacar.

The 'double dual' can be much 'larger' than the original space.

Here is a concrete example: Take $c_0$ the space of sequences that converge to zero with the $\sup$ norm. Then it turns out that $c_0^* = l_1$, the space of summable sequences, and furthermore, $l_1^* = l_{\infty}$, the space of bounded sequences. Hence $c_0^{**} = l_{\infty}$, and $l_{\infty}$ is vastly larger than $c_0$.