About a theorem of dual space Thm: Suppose $V$ is a finite-dimensional vector space with an ordered basis $\beta=\{x_1,\ldots,x_n\}$. Let $f_i$, $1\leq i\leq n$, be the $i^{\text{th}}$ coordinate function with respect to $\beta$ as defined. Let $\beta^{*}=\{f_1,f_2,\ldots,f_n\}$. Then $\beta^{*}$ is an ordered basis for$V^{*}$, and, for any $f\in V^{*}$, we have $f=\sum_{i=1}^n f(x_i)f_i$.
My question here is: is $f$ an element of $V^{*}$, and is that a subspace of $V^{*}$? So for $f(x_i)$, it is a coordinate function with respect to $x_i$, and how usually how do we represent/define that function, or are they always the same?
 A: Since $\beta$ is a basis for $V$, for every $v \in V$ there exists unique scalars $a_1,a_2,\dots,a_n$ (let's say, in some field $F$) so that $v = a_1x_1 + a_2x_2 + \cdots + a_nx_n$, right? 
So, for $i \in \{1,\dots,n\}$ the function $f_i : V \to F$ is defined by $f_i(x) = a_i$. For example, in $\mathbb{R}^3$, consider the basis $\beta = \{(1,1,1),(1,1,0),(1,0,0)\}$. Observe that any $(x,y,z) \in \mathbb{R}^3$ can be written as
$$(x,y,z) = z(1,1,1) + (y-z)(1,1,0) + (x-y)(1,0,0)$$
and then we define three functions $f_1,f_2,f_3 \in (\mathbb{R}^3)^*$ by
\begin{align}
f_1(x,y,z) &= z \\ f_2(x,y,z) &= y-z \\ f_3(x,y,z) &= x-y.
\end{align}
Now, we choose an arbitrary element of $(\mathbb{R}^3)^*$, for example, $g : \mathbb{R}^3 \to \mathbb{R}$ defined by $g(x,y,z) = x+y+z$. According to the theorem, 
\begin{align}
g &= g(1,1,1)f_1 + g(1,1,0)f_2 + g(1,0,0)f_3 \\
&= 3f_1 + 2f_2 + f_3
\end{align}
and this is correct, since when we evaluate $3f_1 + 2f_2 + f_3$ in $(x,y,z)$ we obtain
\begin{align}
(3f_1 + 2f_2 + f_3)(x,y,z) &= 3f_1(x,y,z) + 2f_2(x,y,z) + f_3(x,y,z) \\
&= 3z + 2(y-z) + (x-y) = x+y+z.
\end{align}
Now, the theorem states that these coordinate functions (that depends of the choice of a basis) forms a basis for the space of all linear functions thats going from $V$ to the field, i.e. the dual space. 
Also, the theorem tell us that how to write each element of the dual space in terms of this coordinate functions, you just need to evaluate the function that you want in the basis and then pick the resulting scalars.
