# 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?

• Did you just try to estimtate $|f(\alpha_1,...,\alpha_n)-f(\beta_1,...,\beta_n)|$ in terms of $C \sum |\alpha_j-\beta_j|$ for example. I think you need the triangle inequality on the normed and potential infinite dimensional space $X$ and maybe even the linear independence – Quickbeam2k1 May 29 '13 at 21:02
• You might want to break up your function $f$ as a composition $f = g \circ h$, where $h : \mathbb{R}^n \to X$ is given by $h(a_1,\dotsc,a_n) = a_1x_1 + \cdots + a_n x_n$, and $g : X \to \mathbb{R}^{\geq 0}$ is given by $g(x) = \|x\|$. Since a composition of continuous functions is continuous, you might find it easier to show that $g$ and $h$ individually are continuous. – Branimir Ćaćić May 29 '13 at 21:09

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 $|f(a)-f(b)|\leq \varepsilon$.
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.