An inequality involving multi-index [closed]

Question

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.

Answer 1

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. $

Answer 2

I think for the first inequality the constant is $1$ actually. I note $|\boldsymbol\alpha|=\alpha_1+\cdots+\alpha_n$ and $\mathbf{x}^{\boldsymbol\alpha}=x_1^{\alpha_1}\cdots x_n^{\alpha_n}$

The geometric mean inequality states that if $\beta_i\geq 0$ and $\sum_{i=1}^{n} \beta_i=1$ then for positive numbers $x_i\geq 0$ and $q>0$: \begin{equation} x_1^{\beta_1} \cdots x_d^{\beta_n} \leq (\sum_{i=1}^{n}\beta_i x_i^{q})^{1/q} \end{equation} Applying this for $|x_i|$, and $\beta_i=\alpha_i/|\boldsymbol\alpha|$ gives: \begin{equation} (|x_1|^{\alpha_1} \cdots |x_n|^{\alpha_n})^{1/|\boldsymbol\alpha|}\leq \frac{1}{|\boldsymbol\alpha|^{1/q}} (\sum_{i=1}^{n}\alpha_i |x_i|^{q})^{1/q} \end{equation} So: \begin{equation} |\mathbf{x}^{\boldsymbol\alpha}|\leq \frac{1}{|\boldsymbol\alpha|^{|\boldsymbol\alpha|/q}}(\sum_{i=1}^{d}\alpha_i |x_i|^{q})^{|\boldsymbol\alpha|/q} \end{equation} But $\alpha_i\leq |\boldsymbol\alpha|$ so: \begin{equation} \frac{1}{|\boldsymbol\alpha|^{|\boldsymbol\alpha|/q}}(\sum_{i=1}^{n}\alpha_i |x_i|^{q})^{|\boldsymbol\alpha|/q} \leq \frac{|\boldsymbol\alpha|^{|\boldsymbol\alpha|/q}}{|\boldsymbol\alpha|^{|\boldsymbol\alpha|/q}}(\sum_{i=1}^{n} |x_i|^{q})^{|\boldsymbol\alpha|/q} \end{equation} Hence: \begin{equation} |\mathbf{x}^{\boldsymbol\alpha}|\leq (\sum_{i=1}^{n}|x_i|^{q})^{|\boldsymbol\alpha|/q} \end{equation} Taking $q=2$ gives: \begin{equation} |\mathbf{x}^{\boldsymbol\alpha}|\leq \|\mathbf{x}\|^{|\boldsymbol\alpha|} \end{equation}