0

I have this problem but I don't know how to solve it.

Considering the Euclidean norm |x|, in $\mathbb{R}^n$. If $a>0$, prove there exist $c>0$ depending only of $n$ and $a$ such that: $$c^{-1}(|x_1|^a+\dots+|x_n|^a) \leq |x|^a \leq c(|x_1|^a+\dots+|x_n|^a). $$ I tried to use the AM-GM inequality but I didn't go anywhere. Also, this exercise is after the section of Fubini's Theorem in my book of Lebesgue Measure, but I don't know how to use that. Any ideas? Thanks.

TeemoJg
  • 1,069
  • Take everything and rise it to the power $1/\alpha$. Then use that all norms are equivalent on finite dimensional spaces. Or look up a proof for that statement. – Stan Tendijck Apr 04 '19 at 00:41
  • I really want a way that doesn't use that all norms are equivalent if it is possible. – TeemoJg Apr 04 '19 at 00:42
  • Essentially proving this statement is proving that the $L_{\alpha}$ norm is equivalent to $L_2$ norm. So I suggest you try to find a prove which does this. If I am correct one inequality is probably easy and one is more difficult but that is just me trying to remember it. – Stan Tendijck Apr 04 '19 at 00:45
  • For example, first part of this question: https://math.stackexchange.com/questions/2890009/all-norms-of-mathbb-rn-are-equivalent – Stan Tendijck Apr 04 '19 at 00:47
  • @StanTendijck You may notice that, for $a\in(0,1)$, this is not a norm you're getting. – Clement C. Apr 04 '19 at 01:00
  • Ah yes you're right – Stan Tendijck Apr 04 '19 at 07:46

0 Answers0