Question

Prove that for all x in R^n and all multi-indices a,
|x^a|<= |x|^(|a|)

Answer 1

We need to prove that $\frac{|x^a|}{|x|^{|a|}} \leq 1$. Rewrite the last expression as:$\frac{|x^{\alpha}|}{|x|^{|\alpha|}} = \frac{ \sqrt{ x_1^{2\alpha_1} + \dots + x_n^{2\alpha_n} } }{|x|^{|\alpha|} } \leq ( \frac{ x_1^{2\alpha_1} + \dots + x_n^{2\alpha_n} }{|x|^{2|\alpha|}})^{\frac{1}{2}} \leq \left(\frac{x_1^{2\alpha_1}}{|x|^{2|\alpha|}} + \dots + \frac{x_n^{2\alpha_n}}{|x|^{2|\alpha|}}\right)^{\frac{1}{2}} = \left|\left(\frac{x_1^{\alpha_1}}{|x|^{\alpha_1}}, \dots, \frac{x_n^{\alpha_n}}{|x|^{\alpha_n}}\right )\right|$ .

The right side of the last expression is equal to$|e^a|$, where $e = (\frac{x_1}{|x|}, ... \frac{x_n}{|x|})$ is the unit vector.

Using sup-norm inequality $|x^a| \leq ||x||_{\infty}^{|a|}$ and norm inequality $||x||_{\infty}^{|a|} \leq |x|^{|a|}$ , obtain $|e^a| \leq |e|^a = 1$.

Hence, $\frac{|x^a|}{|x|^{|a|}} \leq 1$.