Norm of a functional in finite-dimensional space

Question

I'd like some help to solve the following exercise:

Let $X$ be a finite-dimensional vector space over $\mathbb K \in \{\mathbb C, \mathbb R\}$. Let $\{x_1,x_2,...,x_n\}$ be a basis. If $x=\sum_{i=1}^na_ix_i$, define $\|x\|:=\max_{1\leq i\leq n} |a_i|$. If $f$ is a (bounded) linear functional on $X$, find $\|f\|$.

My attempt. Taking $x=\sum_{i=1}^na_ix_i \in X$, by linearity of $f$ and triangle inequality, we have that $$|f(x)|=\left|\sum_{i=1}^na_if(x_i)\right|\leq \sum_{i=1}^n|a_i||f(x_i)| \leq \|x\|\sum_{i=1}^n |f(x_i)|.$$ Then, $\sum_{i=1}^n |f(x_i)|$ is a candidate to be $\|f\|$. Believing it, I've been trying to find $x_0\in X$ such that $\|x_0\|=1$ and $|f(x_0)|=\sum_{i=1}^n |f(x_i)|$. However, I couldn't get it. If someone could give me a hint, I'd be grateful. Thanks in advance!

Answer 1

If $f \ne 0$, consider $v = \sum_{i=1}^n a_ix_i$ where $$a_i = \begin{cases} \frac{\overline{f(x_i)}}{|f(x_i)|}, &\text{ if $f(x_i) \ne 0$}\\ 0, &\text{otherwise}\end{cases}$$

We have $\|v\| = \max_{1 \le i \le n}|a_i| = 1$ and $$f(v) = f\left(\sum_{i=1}^n a_ix_i\right) = \sum_{i=1}^n a_if(x_i) = \sum_{i=1}^n |f(x_i)|$$

Therefore

$$\|f\| \ge \frac{|f(v)|}{\|v\|} = \sum_{i=1}^n |f(x_i)|$$

We conclude $\|f\| = \sum_{i=1}^n |f(x_i)|$.

Answer 2

How about $\Sigma_{i=1}^nr_ix_i$, wherein $r_i\in \{1,-1\}$ and $f(r_ix_i)\geq 0$?