This is just about the clearest definition of double dual space I have come across:
Let $V$ be a vector space over $\mathbb R$. We know that a
finite-dimensional vector space $V$ is isomorphic to its dual space
$V^*$. This isomorphism is not ‘very nice’ because it depends on the
choice of basis. There is however one example of a ‘nice’ or natural
isomorphism between two vector spaces. This is an isomorphism between
a finite-dimensional vector space $V$ and the dual space $V^{**}$ of
its dual $V^*.$ The space $V^{**}$ is called the double (or second)
dual of $V$. The elements of $V^{**}$ are linear functionals of linear
functionals.Let $v_0$ be a vector in $V$. For each $f \in V^*$, we write
$$v_0^*(f)=f(v_0)$$
In this formula, $v_0$ is fixed (in $V$) and $f$ is allowed to vary
(in $V^*$), so that the formula defines a function on $V^*$.
But I still don't get it. In part it might be the notation:
What is $v_0^*$ and how do you verbalize the notation? The quote defines $v_0$ as a vector in the original $V$ vector space. That is clear, although the need for the subindex $0$ is either just annoying or its meaning obscure. But there is never a definition of $v_0^*$. I presume it is something along the lines of the "vector being fed to $V^*$ from $V$" (?), but of course then comes the $(f)$, which is defined as a function… and the house of cards inside $v_0^*(f)=f(v_0)$ seems to collapse again.
Is this thing, $a_0^*(f)\in V^{**}?$
Asking for an example may be too much, but a deciphering of the notation would be a great starting point to start thinking about the concept.
CONCLUSIONS (after the accepted answer, comments and additional
studying):
I initially asked
how to "verbalize" $v_0^*(f)=f(v_0)$
meaning how to "unpack" this cluster… OK "Charlie Foxtrot" of super and sub-scripted notation in English. This is what I was thinking of as the ideal answer:
It means: Any given vector in $V$, for example, $v_0 \in V$, can be uniquely mapped to an element in $V^{**}$, corresponding to a functional on $V^*$ (that is a functional that eats functions in the space of $V^*$). This functional on $V^{*}$ can be called $v_0^*$.
That is to say, our new friend, $v_0^*$, belongs to $V^{**}$, but carries one single star because the star makes reference to what it "eats" $(V^{*})$, not where it lives $(V^{**}).$ The author kindly throws the subscript $0$ in $a_0^*$ just in case there was anyone still following at that point – poetic license like in Edgar Allan Poe's “All that we see or seem is but a dream within a dream.” But I digress. In any event, $a_0^*$ lives in the dual of the dual, and as any dual, it is a linear functional. So a function from a real vector space to $\mathbb R.$
The parenthesis, $(f)$ part $v_0^*(f)$, again is telling us that what goes into the functional $a_0^*(\text{here})$ is the functionals $(f)$ that live in $V^*$, sending $V\rightarrow \mathbb R.$ Redundant to have this $(f)$ when we already indicated the "diet" of $a_0^*(f)$ with the asterisk in the superscript? You bet. This part makes sense, though, much like $f(x)=\log x$ makes sense. What is annoying is not having suppressed the superscript $*$ in $a_0^*$. It could have been notated like $L_v(f)$, you know… $L$ for linear functional, $v\in V,$… very simple, but risking the cabalistic aroma of $a_0^*(f).$
OK. So on to the RHS of the equation, which is more of a definition, $:=f(v_0)$
is only uninspired at first glance, being deeply tricky at heart: you look at it and you may assume it the function of… $f(\color{red}{v_0}),$ but it's much more devilish… Here $v_0$ is fixed, and the actual variable is $\color{red}{f(\cdot)}.$
$v_0^*(f)=f(v_0)$ evaluates the functionals in $V^*$ at $v_0$, which remains fixed. So the functions of $V^*$ (dual space) are zipped through by the $a_0^*$ functional in $V^{**}$ one at a time, as it where, while $v_0\in V$ stays the same.
It is in this regard and evaluation map. So if can leave aside the cryptographic notation in the quoted paragraph, and call the function in $V^{**}$ something less obscure, $L_v(f)=f(v)$, where $f\in V^*$ and $v\in V$, we can look at it as a map $L_v: V^*\rightarrow \mathbb R$. Once $v$ is fixed, (let's bring out now the subscripts) to say, $v_0$, the function $L_{v_0}=f(v_0)$ is the evaluation function.
So how to verbalize this? Clearly ‘vee-nought-star' on $X^*.$ As in the answer.
It might be more straightforward to look at it as a dual pair or bracket notation:
If $V$ is finite dimensional there is a natural “bracketing map” $\phi
> : V^∗ \times V \rightarrow \mathbb R$ given by $\phi:( f,v) \mapsto
> ⟨f,v⟩.$ The expression $\langle f,v \rangle$ is linear in each
variable when the other is held fixed. If $f$ is fixed we get a linear
functional $v \mapsto f(v)$ on $V,$ but if we fix $v$ the map $f
> \mapsto \langle f,v \rangle$ is a linear map from $V^∗ \rightarrow
> \mathbb R,$ and hence is an element $j(v) \in V^{∗∗} =( V^∗)^∗,$ the
“double dual” of $V.$
This last concept is nicely explained here.
Best Answer
The displayed line is the definition of $v_0^*$. Each vector $v_0$ in $V$ determines a linear functional $v_0^*$ (which I read ‘vee-nought-star’) on $X^*$, i.e., an element $v_0^*$ of $V^{**}$. This $v_0^*$ is therefore a linear function from $V^*$ to $\Bbb R$, and it’s defined by
$$v_0^*(f)=f(v_0)\tag{1}$$
for each $f\in V^*$. That is, its value at the linear functional $f\in V^*$ is simply the value of $f$ at $v_0$.
One does of course have to verify that the function from $V^*$ to $\Bbb R$ defined by $(1)$ actually is linear, but this is quite straightforward.