I have a question because when I search something on mathstackexchange, I've read something that I'm not sure it's true, and I would like to know if I'm right or I'm wrong.

Actually, we consider $h \in \mathbb{R}^n$, and the author of the answer of the topic I talk about claims that, for $h = (h_1, \ldots, h_n)$ such that $\|h\| \ll 0$, we have :

$$\|h\|^6 \leq \left( \sum_{i_1, i_2, i_3 \in \{0,\ldots, n \}} h_{i_1} h_{i_2} h_{i_3}\right)^2$$


$$\|h\|^6 \geq n^6 \left(\sum_{i_1, i_2, i_3} h_{i_1}h_{i_2}h_{i_3}\right)^2$$

For the second inequality, it's okay, but I'm not okay with the first one. Actually, If I consider the euclidean norm on $\mathbb{R^n}$, then we have

$$\|h\|^6 \leq \left(\sum_{i_1, i_2, i_3} h_{i_1}h_{i_2}h_{i_3}\right)^2$$

and as we have : $\sum_{i_1, i_2, i_3} h_{i_1}h_{i_2}h_{i_3} = (h_1+h_2+\cdots+h_n)^3$, we have

$$\iff \sqrt{(h_1^2+h_2^2+\cdots+h_n^2)}^6 \leq (h_1+h_2+\cdots+h_n)^6 $$

$$\iff (h_1^2+h_2^2+\cdots+h_n^2)^3 \leq (h_1+h_2+\cdots+h_n)^6 $$

But if I consider, for $\varepsilon > 0$, $(\varepsilon, -\varepsilon, 0, \ldots, 0)$, then I have, if the equality is true : $8\varepsilon^6 \leq 0$, which is absurd. Then, the equality is wrong. Have I right ?

In general, I have the impression that we have $$\sum_{i_1, \ldots, i_k} h_{i_1}\cdots h_{i_k} = O(\|h\|^k),$$ but not necessarily $\|h\|^k = O\left(\sum_{i_1, \ldots, i_k} h_{i_1}\cdots h_{i_k}\right)$.

Have I right ?

Thank you for your help !

  • $\begingroup$ That norm might not be Euclidean. BTW, what does $\|h\|\ll 0$ mean? $\endgroup$ – xbh Jul 26 '18 at 1:29
  • $\begingroup$ Yes but as all the norm are equivalent in finite dimension, I can suppose it's the euclidean norm? We can still consider the inequality and multiplied one of the term by a constant, and it doesn't change anything, so it's not a problem, no? For the notation, it means for $h$ close to zero (as close as we want). $\endgroup$ – ChocoSavour Jul 26 '18 at 9:31
  • $\begingroup$ Your thinking is fine. The first inequality is false. $\endgroup$ – Kavi Rama Murthy Jul 27 '18 at 6:41
  • $\begingroup$ Okay, thank you! :) $\endgroup$ – ChocoSavour Jul 27 '18 at 8:40

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