Determine $\|f\|$ when we consider the norm $\|\cdot\|_{2}$. Let $f:\mathbb{C}^{n}\to\mathbb{C}$ a linear form, define as, $$f(x)=\sum_{i=1}^{n}{a_{i}x_{i}}\qquad, a_{i}\in\mathbb{C}$$
Determine $\|f\|$ when on $\mathbb{C}^{n}$ we consider the norm $\|\cdot\|_{2}$. 
My approach: note that $$| f(x)|=\left|\sum_{i=1}^{n}{a_{i}x_{i}}\right|\leq \sum_{i=1}^{n}{|a_{i}||x_{i}|}\leq\sum_{i=1}^{n}{|a_{i}|}$$
 A: So $f(x)=a\cdot x$, $a=(a_{1},...,a_{n})$, Cauchy-Schwartz gives $|f(x)|\leq\|a\|_{2}\|x\|_{2}$, so $\|f\|\leq\|a\|_{2}$, but with $x=\|a\|_{2}^{-1}\overline{a}$, $f(x)=\|a\|_{2}$, so $\|f\|=\|a\|_{2}$.
A: Let $\langle \cdot, \cdot \rangle$ denote the standard inner product on $\mathbb{C}^n$ and $\overline{a} = \left(\overline{a_1}, \ldots, \overline{a_n}\right)$.
Notice that $$f(x) = \sum_{i=1}^n x_i a_i = \langle x, \overline{a}\rangle = \left\langle (x_1, \ldots, x_n), \left(\overline{a_1}, \ldots, \overline{a_n}\right)\right\rangle, \text{ for } x = (x_1, \ldots, x_n) \in \mathbb{C}^n$$
We have:
$$|f(x)| = \left|\langle x, \overline{a}\rangle\right| \stackrel{CSB}{\le} \|x\|_2\left\|\overline{a}\right\|_2 = \|x\|_2\|a\|_2$$
so $\|f\| \le \|a\|_2$.
Furthermore, for $x = \overline{a}$ we have:
$$f\left(\overline{a}\right) = \left\langle \overline{a}, \overline{a}\right\rangle = \left\|\overline{a}\right\|_2^2$$
So $$\|f\| \ge \frac{\left|f(\overline{a})\right|}{\left\|\overline{a}\right\|_2} = \left\|\overline{a}\right\|_2 = \left\|a\right\|_2$$
We conclude that $\|f\| = \|a\|_2$.
This result is actually a consequence of the Riesz representation theorem, which states that all bounded linear functionals on a Hilbert space are of the form $\langle \cdot , v \rangle$ for some $v$, and that the norm of the functional is equal to the norm of the vector which represents it. In this case, $f$ is represented by $\overline{a}$, so $\|f\| = \left\|\overline{a}\right\|_2 = \left\|a\right\|_2$.
