Showing $f$ is continuous Let $x_1, x_2, ... , x_n$ be linearly independent vectors of a normed space $X$, so for any $ x \in X$ set $x = \alpha_1 x_1 + \alpha_2 x_2 + ... + \alpha_n x_n$ for real numbers $\alpha ... \alpha_n$. Prove that the function $f: \mathbb{R}^n \rightarrow \mathbb{R}^{\geq 0}$ defined by
$$f(\alpha_1, \alpha_2, ... \alpha_n) = || \alpha_1 x_1 + \alpha_2 x_2 + ... + \alpha_n x_n ||$$ is continuous. 
Do I have to show that this is a linear operator and then show that is is bounded to imply continuity? The confusing part is this is a general normed space-- I don't have an explicit norm definition. How do I proceed?
 A: You just need to look at the axioms of a norm (for instance here).
With these axioms, you can show that if the input of the function gets close to $a$, for any norm of $\mathbb{R}^n$, like euclidean or maximum norm, then the output gets close to $f(a)$.
More formally, for any $a\in\mathbb{R}^n$ and $\varepsilon>0$, there is $h>0$ such that for all $b\in\mathbb{R}^n$, if $||a-b||\leq h$ then $|f(a)-f(b)|\leq \varepsilon$.
A: It is true that $f$ is bounded since, for $M = \max\{||x_j||; 1 \leq j \leq n\}$,
$$
     |f(\alpha_1,\ldots,\alpha_n)|
\leq \sum_{i=1}^n |\alpha_i| ||x_i||
\leq M\sum_{i=1}^n |\alpha_i|, 
$$
so 
$$
     \sup\{|f(\alpha_1,\ldots,\alpha_n)|: ||(\alpha_1,\ldots,\alpha_n)||_n = 1 \}
\leq M.
$$
It is not linear, however, since
$$
     f(-(\alpha_1,\ldots,\alpha_n))
\neq -f(\alpha_1,\ldots,\alpha_n).
$$
So, strictly speaking, you cannot use your approach. Nevertheless, the function 
$g:\mathbb{R}^n \to X$ given by $(\alpha_1,\ldots,\alpha_n) \mapsto \sum_{i=1}^n \alpha_i x_i$ is clearly linear, bounded (as we just saw), and therefore continuous. Then just show that the function $||\cdot||: X \to \mathbb{R}$ is continuous, and your result will follow by composition.
