How do we know elements of a vector space can be embedded into its double dual?

Question

Consider a vector space $V$ and its double dual $V^{**}$. In general, the space $V^{**}$ is far too large and so we wish to find a coarse topology on it. So we instead only consider those maps that are generated by $V$, that is we only consider maps $\chi \in V^{**}$ such that $$\chi_v(f) = f(v) \quad\forall f \in V^*$$ and require them to be continuous, which is of course the weak* topology on $V^*$.

In most textbooks it is simply stated that we may do such an embedding that defines $\chi_v$ as above, but why is this? How do we know such $\chi_v$ even exist? What do maps look like that are not of this form?

Second, in defining the weak* topology why was the choice made to make these maps in particular continuous? Do they give us some nice property?

Answer 1

For double dual elements not of the form $\chi_v$ for $v\in V$, see https://en.wikipedia.org/wiki/Dual_space#Infinite-dimensional_case. The $\chi_v$ exist simply because we can verify that $\chi_v$ is a well-defined function that is linear, so it lies in $V^{\ast\ast}$. More generally, you can verify that $\Phi\colon V\to V^{\ast\ast}$ defined by $\Phi(v) := \chi_v$ (alternatively written $\Phi(v)(f) := f(v)$) is an injective linear map that preserves the norm on $V$ and $V^{\ast\ast}$ (if they have one), and $\Phi$ is surjective if and only if $V$ is finite-dimensional. You can see https://en.wikipedia.org/wiki/Dual_space#Double_dual for a mention of the weak topology, if that helps.

(Hit character limit in comments so changed to an answer).

Answer 2

The dual space $V^{\ast}$ by definition consists of linear functionals $V \to K$. This gives a bilinear pairing $V^{\ast} \times V \to K$ given by just evaluating a linear functional $f \in V^{\ast}$ on a vector $v \in V$. Now we can turn this bilinear pairing around and think of vectors $v \in V$ as defining linear functionals on $V^{\ast}$ itself, via $f \mapsto f(v)$. This produces a natural map

$$V \ni v \mapsto (f \mapsto f(v)) \in (V^{\ast})^{\ast}.$$

Now here is a statement that requires some caution: you will see it stated that this map is an embedding, but actually this requires the axiom of choice! In the absence of choice it's consistent with ZF that there exist infinite-dimensional vector spaces $V$ such that $V^{\ast} = 0$; see this MO discussion, for example. My favorite example is $\mathbb{R}$ as a vector space over $\mathbb{Q}$.

In the presence of choice we just pick a basis of $V$ and define linear functionals on it by picking their values arbitrarily on each basis vector; this lets us show that linear functionals on $V$ separate points, meaning that if $v \neq w$ in $V$ then there exists a linear functional $f$ such that $f(v) \neq f(w)$. This is equivalent to the condition that the double dual map is an embedding (exercise).

Second, in defining the weak* topology why was the choice made to make these maps in particular continuous? Do they give us some nice property?

These are the "canonical" maps; if all you know about $V^{\ast}$ is that it's the dual of $V$, then these are all the maps that you know you can write down "canonically" on $V^{\ast}$ given only that information and nothing else. So it's a natural question to see what topology makes them continuous and then what continuous functions with respect to that topology are.