The continuous dual of the reals I just have a few questions involving the continuous dual of $\mathbb{R}^{N}$. We know that the dual $(\mathbb{R}^{N})^{*}$ of $\mathbb{R}^{N}$ is the space of all linear forms $$a: \mathbb{R}^{N} \rightarrow \mathbb{R} \text{  } \text{ such that } x \mapsto \langle a,x \rangle$$   Consider the mapping $\pi: (\mathbb{R}^{N})^{*} \rightarrow \mathbb{R}^{N}$ defined as
$$\langle a,x \rangle = (\pi a,x ) \text{  }\text{ where }a \in (\mathbb{R}^{N})^{*}, x \in \mathbb{R}^{N}$$
where $(\cdot,\cdot)$ is the scalar product.
If we have are given a function $$F: \mathbb{R}^{N} \rightarrow (\mathbb{R}^{N})^{*}$$
then it follows that $$\langle F(x),y \rangle = (\pi F(x),y) = \sum_{j}F_{j}(x)y_{j}$$


*

*I want to show that if the functions $F_{1}(x),...,F_{N}(x)$ are continuous then $F$ is continuous. Is the following proof fine:


Take $x_{n} \rightarrow x$ in $\mathbb{R}^{N}$. Then since each of $F_{1}(x_{n}) \rightarrow F_{1}(x)...F_{N}(x_{n}) \rightarrow F_{n}(x)$ it follows that $\langle F(x_{N}),y \rangle \rightarrow \langle F(x),y \rangle$ for any $y \in \mathbb{R}^{N}$. Therefore we have that $F(x_{n}) \rightharpoonup^{*} F(x)$(weak star convergence) and since $(R^{N})^{*}$ is finite-dimensional, this is the same as $F(x_{n}) \rightarrow F(x)$. This shows that $F$ is continuous. Firstly is this proof fine and secondly is there a better way to show continuity?
2.What are the properties that defines the mapping $\pi: (\mathbb{R}^{N})^{*} \rightarrow \mathbb{R}^{N}$? I suspect that it is at least linear and bijective but I think by the Ries-representation theorem it follows that it is even an isometry?
Thanks for any assistance.   
 A: If $F$ is as defined in the OP, let $M=\max{\{\|F_j\|\}}$. Since all norms on $\Bbb R$ are equivalent, take $\|x\|:=\sum\limits_j|x_j|$. Thus:
\begin{align*}
\|F\|&=\sup{\{\|F(x)\|\,\mid\,0\leq \|x\|\leq 1\}}=\sup{\left\{\sup{\{|\left<F(x),y\right>|\,\mid\,0\leq \|y\|\leq 1\}}\,\mid\,0\leq \|x\|\leq 1\right\}}\\
&=\sup{\left\{\sup{\left\{\left|\sum_jF_j(x)y_j\right|\,\mid\,0\leq \|y\|\leq 1\right\}}\,\mid\,0\leq \|x\|\leq 1\right\}}\\
&\leq\sup{\left\{\sup{\left\{\sum_j\left|F_j(x)y_j\right|\,\mid\,0\leq \|y\|\leq 1\right\}}\,\mid\,0\leq \|x\|\leq 1\right\}}\\
&\leq\sup{\left\{\sup{\left\{\sum_j(\|F_j\|)(\|x\|)\left|y_j\right|\,\mid\,0\leq \|y\|\leq 1\right\}}\,\mid\,0\leq \|x\|\leq 1\right\}}\\
&\leq\sup{\left\{\sup{\left\{M\|x\|\sum_j\left|y_j\right|\,\mid\,0\leq \|y\|\leq 1\right\}}\,\mid\,0\leq \|x\|\leq 1\right\}}\\
&=\sup{\left\{M\|x\|\cdot\sup{\left\{\|y\|\,\mid\,0\leq \|y\|\leq 1\right\}}\,\mid\,0\leq \|x\|\leq 1\right\}}\\
&=M\sup{\{\|x\|\,\mid\,0\leq \|x\|\leq 1\}}=M.
\end{align*}
So we have that $F$ is bounded and hence continuous.
