Duality in finite-dimensional normed spaces Suppose we endow $\mathbb{R}^n$ with a norm $\|\cdot\|$; call such a normed space $X$. Then, as a vector space, the dual space $X^*$ is also $\mathbb{R}^n$. Let $x\in X$ and $f\in X^*$. Consider the duality bracket $\langle x, f\rangle$. When $x$ and $f$ are thought as the elements of $\mathbb{R}^n$ is $\langle x, f\rangle$ the canonical scalar product of vectors?
 A: You are almost right. Note that for $X = \mathbb{R}^n$ the dual space $X^*$ is isomorphic to $\mathbb{R}^n$. If $\langle - , - \rangle$ denotes a scalar product on $X$, then $$\Phi_{\langle - , - \rangle}:  X \to X^*, x \mapsto (y \mapsto \langle y , x \rangle) $$
is an isomorphism and under this isomorphism the canonical pairing $(-,-) : X \times X^* \to \mathbb{R}, (x,f) \mapsto f(x)$ becomes the map 
$$ (-,-) \circ (\text{id} \times \Phi_{\langle -,- \rangle}) : X \times X \to X \times X^* \to \mathbb{R} $$
which is identical to $\langle - , - \rangle$.
But note that this really depends on the way you identify $X$ with $X^*$ (namely via $\Phi_{\langle - , - \rangle}$).
A: Yes, but the order of your presentation is somewhat imprecise.
The vector space $X^{*} = \mathrm{Hom}(\mathbb{R}^n, \mathbb{R})$ consists of linear functionals on $X = \mathbb{R}^n$ and so in general you need to specify how you identify vectors $v \in \mathbb{R}^n$ with linear functionals $\varphi \in X^{*}$. The duality bracket in general is a bilinear map $(\cdot, \cdot) \colon X \times X^{*} \rightarrow \mathbb{R}$, so you can't even write $(x,y)$ with $x,y \in X$ without specifying in advance how you identify the vector $y \in X$ with a linear functional $\varphi \in X^{*}$. 
Now, since $X$ has a canonical scalar product $\left< \cdot, \cdot \right>$, you can decide to identify a vector $v \in X$ with the linear fuctional $\varphi_v(w) = \left<v, w\right>$. The map $v \mapsto \varphi_v$ is an isomorphism providing an identification between $X$ and $X^{*}$ (this works for any real finite dimensional inner product space). If you do that, then indeed
$$ "(v,w)" := (v, \varphi_w) = \varphi_w(v) = \left< v, w \right>. $$
Alternatively, we can think of $\mathbb{R}^n$ as a vector space consisting of column vectors. Since $X = \mathbb{R}^n$, we can identify an element $\varphi \colon \mathbb{R}^n \rightarrow \mathbb{R}$ in $X^{*}$ with the $1 \times n$ matrix representing $\varphi$ with respect to the standard ordered bases of $\mathbb{R}^n$ and $\mathbb{R}$. This provides an isomorphism $X \rightarrow X^{*}$ written sloppily as $v \mapsto v^t$ (where $v^t$ is a row vector that is interpreted as a linear functional by declaring that $(v^t)(w) = v^t \cdot w$ - the expression $v^t \cdot w$ is a $1 \times 1$ matrix that we identify with a scalar). Using this identification, we also have
$$ "(v,w)" := (v, w^t) = (w^t)(v) = w^t \cdot v = \left< w, v \right>. $$
In any case, the point is that you can't ask what is $\left<x, f \right>$ without defining first how you identify $f \in X$ with an element of $X^{*}$.
