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I know this may be a stupid question, but still hope someone can help me.

$\alpha \in \mathbf { N } ^ { N }$, $p \in \mathbf { N } $ and $\xi \in \mathbf { R } ^ { N }$ Why :

$\max _{|\alpha| \leq p}\left|\xi^{\alpha}\right| \leq \left(\frac{1+|\xi|^{2}}{2}\right)^{p}$

cmk
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Almendrof66
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1 Answers1

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This is false if $|\xi|<1$. Indeed for $\xi =(1/2,1/2)$ and $p=2$, the LHS is $1/2$ whereas the RHS is $$\left(\frac{1+(1/2)^2}{2}\right)^2=\frac {25} {64}<\frac1 2.$$ If $|\xi|\geq 1$, then $$|\xi^\alpha|\leq |\xi|^p\leq \left(\frac{1+|\xi|^2}{2}\right)^p,$$ here we used, in the first inequality, the fact that the map $\mathrm{R}^+\to \mathrm{R}^+: a\mapsto a^p$ is non-decreasing, and the second inequality follows from $$|a|\leq \frac{1+a^2}{2}.$$ Now varying the $\alpha$ we get the result.

jijijojo
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  • Thank you for your help, I applied your indication and I found that (if my calculation are correct!) $ \left|\xi^{\alpha}\right| \leq \left(\frac{1+|\xi|^{2}}{2}\right)^{\alpha}$ now, to get the result it sufficed to apply max on $\alpha$ but $\left(\frac{1+|\xi|^{2}}{2}\right)$ must be $>1$ to concerve the ordre !! in other words the inequality is not general – Almendrof66 Jan 05 '20 at 19:36
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    @Almendrof66 no you don't need any restriction. Please see the edited answer. – jijijojo Jan 05 '20 at 19:57
  • sorry, but the map $\mathrm{R}^+\to \mathrm{R}^+: a\mapsto a^p$ is decreasing in $]0,1[$ – Almendrof66 Jan 05 '20 at 20:30
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    @Almendrof66 No you're confused. That map is non-decreasing. This map $\mathrm{R}^+\to \mathrm{R}^+,, x\mapsto a^x$ is decreasing, for $a\in (0,1)$. – jijijojo Jan 05 '20 at 21:39
  • I've been wrestling with the proof of the first inequality for a number of hours using your indication and the euclidian norm but I didn't find the result, after searching again I found this comment, and I would like to share it with you https://math.stackexchange.com/a/446824/730650 – Almendrof66 Jan 09 '20 at 16:34