Let $m\geq 2$ and $B^{m}\subset \mathbb{R}^{m}$ be the unit OPEN ball . For any fixed multi-index $\alpha\in\mathbb{N}^{m}$ with $|\alpha|=n$ large and $x\in B^{m}$

$$|x^{\alpha}|^{2}\leq \int_{B^{m}}|y^{\alpha}|^{2}dy\,??$$

  • $\begingroup$ It seems to be true because the right side behaves as a power of $|\alpha|^{-p}$ (for some $p>0$) and the left-one is like a power of a positive number $<1$...but i dont have a formal proof...not yet.. $\endgroup$ – GoldSoundz Aug 3 '12 at 22:45
  • $\begingroup$ So the inequality may be true for almost every $x\in S^{m}$? $\endgroup$ – GoldSoundz Aug 4 '12 at 2:56
  • $\begingroup$ I included an answer to your updated question in my old answer. I suggest you to open a new question instead of modifying the old one, because it makes the old answers nonsense. $\endgroup$ – timur Aug 10 '12 at 0:46
  • $\begingroup$ Sorry the update. But in the unit open ball is it possible to have $$\max_{B^{m}}|x^{\alpha}|=1$$ when $\alpha=(n,0,...,0)$, $n>1$? $\endgroup$ – GoldSoundz Aug 10 '12 at 0:56
  • $\begingroup$ You have to use supremum. This means that for any given $\varepsilon>0$, you can choose $x$ in the open unit ball and $n$ large, so that the left hand side of your inequality is larger than $1-\varepsilon$, while the right hand side is smaller than $\varepsilon$. $\endgroup$ – timur Aug 10 '12 at 1:04

No. For a counterexample, take $\alpha=(n,0,\ldots,0)$. Obviously, $\max_{S^m}|x^\alpha|=1$, but an easy calculation shows $$ \int_{S^m}|y^\alpha|^2{\mathrm{d}}\sigma(y) \to 0, $$ as $n\to\infty$.

For the updated question, that involves the open unit ball, the answer is the same. With the same counterexample, we have $$ \int_{B^m}|y^\alpha|^2{\mathrm{d}}y \to 0, $$ as $n\to\infty$.


Using the Bergman inequality, for each $K \subset B^{m}$ compact there exists $M_{K}>0$ such that $$|x^{\alpha}|\leq M_{K}||p_{\alpha}||_{2},\quad \alpha\in\mathbb{N}^{m+1},\,x\in K.$$


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