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?

  • 1
    $\begingroup$ 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 $\endgroup$ – Quickbeam2k1 May 29 '13 at 21:02
  • $\begingroup$ 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. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.