# Norms in R^m

• Feb 19th 2011, 10:24 PM
ragnar
Norms in R^m
Define $\displaystyle ||x||=\sqrt{(x1)^2 +...+(xm)^2}$ in R^m, and $\displaystyle ||x||_{p} = \sqrt[p]{|x1|^{p} +...+|x|^{p}}$. I need to show that for every m and p there are scalars c and C such that, for every x in R^m, $\displaystyle c||x|| \leq ||x||_{p} \leq C||x||$.

My only thought is to try to find the limit of $\displaystyle C||x||-||x||_{p}$ as each coordinate goes to infinity. If it converges the I think I'm done, no? But I don't see how to do it.
• Feb 19th 2011, 10:33 PM
ragnar
However, being in the first part of an analysis course, the notion of a limit is not yet defined so this is probably not allowed.
• Feb 20th 2011, 06:41 AM
ragnar
Let me write this a little better (now that I can do it on my computer rather than an iPhone).

Define $\displaystyle ||\text{\bf x}|| = \sqrt{(x_{1})^{2} + ... + (x_{m})^{2}}$ for $\displaystyle \text{\bf x} \in \mathbb{R}^{m}$, and $\displaystyle ||\text{\bf x}||_{p} = \sqrt[p]{|x_{1}|^{p} + ... + |x_{m}|^{p}}$. Then prove what was said above: for any $\displaystyle m, p \geq 1$, there are scalars c, C > 0 such that for any $\displaystyle \text{\bf x} \in \mathbb{R}^{m}$ we have $\displaystyle c||\text{\bf x}|| \leq ||\text{\bf x}||_{p} \leq C||\text{\bf x}||$.
• Feb 20th 2011, 09:02 AM
Opalg
Quote:

Originally Posted by ragnar
Define $\displaystyle ||\text{\bf x}|| = \sqrt{(x_{1})^{2} + ... + (x_{m})^{2}}$ for $\displaystyle \text{\bf x} \in \mathbb{R}^{m}$, and $\displaystyle ||\text{\bf x}||_{p} = \sqrt[p]{|x_{1}|^{p} + ... + |x_{m}|^{p}}$. Then prove what was said above: for any $\displaystyle m, p \geq 1$, there are scalars c, C > 0 such that for any $\displaystyle \text{\bf x} \in \mathbb{R}^{m}$ we have $\displaystyle c||\text{\bf x}|| \leq ||\text{\bf x}||_{p} \leq C||\text{\bf x}||$.

Hint: Let $\displaystyle \|{\bf x}\|_\infty = \max\{|x_1|,|x_2|,\ldots,|x_m|\}$. Show that $\displaystyle \|{\bf x}\|_\infty\leqslant\|{\bf x}\|_p\leqslant\|{\bf x}\|_1\leqslant m\|{\bf x}\|_\infty.$ Notice also that $\displaystyle \|{\bf x}\|=\|{\bf x}\|_2$.
• Feb 20th 2011, 07:33 PM
ragnar
Okay, I see how to prove the inequality of your hint. What I don't see is how this contributes to the assignment. I also see that $\displaystyle ||\text{\bf x}||_{\infty} \leq ||\text{\bf x}||_{2} \leq ||\text{\bf x}||_{1}$, but I'm still stuck.
• Feb 20th 2011, 08:25 PM
ragnar
I think I may have part of it. If I multiply $\displaystyle ||\text{\bf x}||$ by $\displaystyle \sqrt{m}$ then for whichever term is the max, I'll have $\displaystyle m$ many of it, thus meaning that the stuff under the radical will be closer to being demonstrably larger than $\displaystyle ||\text{\bf x}||_{p}$.
• Feb 20th 2011, 11:19 PM
Opalg
Quote:

Originally Posted by ragnar
Okay, I see how to prove the inequality of your hint. What I don't see is how this contributes to the assignment. I also see that $\displaystyle ||\text{\bf x}||_{\infty} \leq ||\text{\bf x}||_{2} \leq ||\text{\bf x}||_{1}$, but I'm still stuck.

Once you have proved those inequalities, you're practically there:

$\displaystyle \|{\bf x}\|_p \leqslant m\|{\bf x}\|_\infty \leqslant m\|{\bf x}\|_2,$

$\displaystyle \|{\bf x}\|_2\leqslant m\|{\bf x}\|_\infty \leqslant m\|{\bf x}\|_p$ and hence $\displaystyle m^{-1}\|{\bf x}\|_2\leqslant \|{\bf x}\|_p.$

Thus $\displaystyle m^{-1}\|{\bf x}\| \leqslant \|{\bf x}\|_p \leqslant m\|{\bf x}\|.$
• Feb 26th 2011, 08:51 PM
ragnar
Ah, got it... But WOW I'm amazed I was expected to have figured this all out without even the hint you gave.

Thank you very much for your help! That was very good.