# showing any norm on $\mathbb{R}^n$ is bounded by the standard norm

Let V be a real finite-dimensional vector space. Let $$\|\cdot\|$$ be an arbitrary norm on $\mathbb{R}^n$ Write $x = \sum_{i=1}^k x_i e_i$ where $e_i$ is the standard basis. I am trying to show that $$\|x\| \leq C\|x\|_2$$ for some constant $C>0$. Does anyone have any advice?

This is for a bigger question, but I feel as though if I solve this I can solve the whole problem

## 3 Answers

Hint: $$\|{\bf x}\| = \left\|\sum_i x_i {\bf e}_i\right\| \leq \sum_i \|x_i{\bf e}_i\| = \sum_i |x_i|\|{\bf e}_i\|.$$

• yes I have gotten this far, sorry I should have posted my workings – mathanalysis87 Feb 8 '15 at 23:41
• It is never too late to add it in an edit :) – Ivo Terek Feb 8 '15 at 23:41

It is a well-known result that all norms on a finite dimensional space are equivalent, of which yours is a special case.

Just note by the properties of a norm:

$$||x|| \le \sum_{i=1}^k |x_i| ||e_i|| \le k \max_i\{|x_i| \} \max_i\{||e_i||\} \le k \max_i\{||e_i|| \} ||x||_2$$ Setting $C = k \max_i\{||e_i|| \}$ the claim is established.

• sorry I don't follow - how did you get the max part from? – mathanalysis87 Feb 8 '15 at 23:43
• $|x_j| \le \max |x_i|$ and $||e_j|| \le \max ||e_i||$ so the product is bounded by the product. So each summand is bounded by $\max |x_i| \max ||e_i||$ and you have $k$ terms. – quid Feb 8 '15 at 23:45
• that makes sense - but how did you involve $||x||_2$? – mathanalysis87 Feb 8 '15 at 23:48
• I bound $\max |x_i|$ by $||x||_2$ as $\sqrt{\sum x_i^2} \ge \sqrt{\max x_i^2}= \max |x_i|$. – quid Feb 8 '15 at 23:50
• thinki get it thanks – mathanalysis87 Feb 8 '15 at 23:52

By the triangle inequality, $\|x\| \le \sum_{k = 1}^n |x_i|\|e_i\|$, and by the Cauchy-Schwarz inequality, $\sum_{k = 1}^n |x_i|\|e_i\| \le C\|x\|_2$ with $C = \sqrt{\sum_{k = 1}^n \|e_i\|^2}$.

• using this you used that $\sum ||x_i|| \leq \sum x_i^2$ could you explain how you got that – mathanalysis87 Feb 9 '15 at 0:00
• No, I didn't use that. Specifically, I used the fact that for any pair of vectors $v, w\in \Bbb R^n$, $v\cdot w \le \|v\|_2\|w\|_2$. Letting $v = (|x_1|,\ldots, |x_n|)$ and $w = (\|e_1\|,\ldots, \|e_n\|)$, we have $$\sum_{k = 1}^n |x_i|\|e_i\| = v\cdot w \le \|v\|_2\|w\|_2 = C\|v\|_2 = C\|x\|_2.$$ – kobe Feb 9 '15 at 0:08
• where are you getting that from? I am reading the wikipedia page here: en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality under special cases – mathanalysis87 Feb 9 '15 at 0:14
• Right, the actual CS inequality is $|v\cdot w| \le \|v\|_2\|w\|_2$, from which it follows that $v \cdot w \le |v\cdot w| \le \|v\|_2\|w\|_2$. – kobe Feb 9 '15 at 0:17