Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $x\in B^{m}$

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

share|cite|improve this question
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.. – Dubglass Aug 3 '12 at 22:45
So the inequality may be true for almost every $x\in S^{m}$? – Dubglass Aug 4 '12 at 2:56
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. – timur Aug 10 '12 at 0:46
Sorry the update. But in the unit open ball is it possible to have $$\max_{B^{m}}|x^{\alpha}|=1$$ when $\alpha=(n,0,...,0)$, $n>1$? – Dubglass Aug 10 '12 at 0:56
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 $\varepsilon$. – timur Aug 10 '12 at 1:04
up vote 2 down vote accepted

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 $n\to\infty$.

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.