MathOverflow is a question and answer site for professional mathematicians. 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

I came across these inequalities while learning about Schwartz functions (Classical Fourier Analysis, Grafakos) and I have no idea how to prove this:

For $x \in \mathbb{R}^{n}$ and $\alpha = (\alpha_{1}, \ldots, \alpha_{n}) \in \mathbb{N}^{n}$, we set

$$ x^{\alpha} = x_{1}^{\alpha_{1}}\cdots x_{n}^{\alpha_{n}}.$$

Then prove that there exists a constant $c_{n,\alpha}$ such that

$$\left| x^{\alpha}\right| \leq c_{n,\alpha}|x|^{|\alpha|}$$

where $|\alpha| = \alpha_{1} + \cdots + \alpha_{n}$.

Conversely, for every $k \in \mathbb{N}$, there exists a $C_{n,k}$ such that

$$|x|^{k} \leq C_{n,k}\sum\limits_{|\beta| = k}|x^{\beta}|$$

Any help would be appreciated.

P.S. Please let me know if the question is too elementary for this forum.

share|cite|improve this question
What is $|x|$? The Euclidean or $l_2$ norm? – Todd Trimble Jul 28 '13 at 16:04
$|x| = \sqrt{x_{1}^{2} + \cdots + x_{n}^{2}}$ – Vishal Gupta Jul 29 '13 at 2:50
up vote 2 down vote accepted

The first inequality with constant 1 follows from $ \vert x^\alpha\vert\le\Vert x\Vert_{\infty}^{\vert \alpha \vert},\quad\text{where $\Vert x\Vert_{\infty}$ is the sup-norm.} $

The second equality, also with constant 1, is due to $ \Vert x\Vert_{\infty}^k=\max_{1\le j\le n} \vert x_j\vert^k\le \max_{\vert \alpha\vert=k} \vert x^\alpha\vert. $

share|cite|improve this answer
Although I meant the Euclidean norm, but this is good enough as I can use the inequalities between these two norms. – Vishal Gupta Jul 27 '13 at 7:28

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.