1
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ yes I have gotten this far, sorry I should have posted my workings $\endgroup$ – mathanalysis87 Feb 8 '15 at 23:41
  • $\begingroup$ It is never too late to add it in an edit :) $\endgroup$ – Ivo Terek Feb 8 '15 at 23:41
2
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ sorry I don't follow - how did you get the max part from? $\endgroup$ – mathanalysis87 Feb 8 '15 at 23:43
  • $\begingroup$ $|x_j| \le \max |x_i|$ 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. $\endgroup$ – quid Feb 8 '15 at 23:45
  • $\begingroup$ that makes sense - but how did you involve $||x||_2$? $\endgroup$ – mathanalysis87 Feb 8 '15 at 23:48
  • $\begingroup$ I bound $\max |x_i|$ by $||x||_2$ as $\sqrt{\sum x_i^2} \ge \sqrt{\max x_i^2}= \max |x_i| $. $\endgroup$ – quid Feb 8 '15 at 23:50
  • $\begingroup$ thinki get it thanks $\endgroup$ – mathanalysis87 Feb 8 '15 at 23:52
2
$\begingroup$

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 $C = \sqrt{\sum_{k = 1}^n \|e_i\|^2}$.

$\endgroup$
4
  • $\begingroup$ using this you used that $\sum ||x_i|| \leq \sum x_i^2$ could you explain how you got that $\endgroup$ – mathanalysis87 Feb 9 '15 at 0:00
  • $\begingroup$ 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 $$\sum_{k = 1}^n |x_i|\|e_i\| = v\cdot w \le \|v\|_2\|w\|_2 = C\|v\|_2 = C\|x\|_2.$$ $\endgroup$ – kobe Feb 9 '15 at 0:08
  • $\begingroup$ 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 $\endgroup$ – mathanalysis87 Feb 9 '15 at 0:14
  • $\begingroup$ Right, the actual CS inequality is $|v\cdot w| \le \|v\|_2\|w\|_2$, from which it follows that $v \cdot w \le |v\cdot w| \le \|v\|_2\|w\|_2$. $\endgroup$ – kobe Feb 9 '15 at 0:17

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.