Jtdylktuy

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

$$|x|^{k} leq C_{n,k}sumlimits_{|beta| = k}|x^{beta}|$$

Any help would be appreciated.

The first inequality is clear at $x=0$ with any constant . Now note that it suffices to prove it for $|x|=1$. This is because if it is true on the unit sphere then since $xneq0$ then $y=(frac{x_{1}}{|x|},frac{x_{2}}{|x|},...,frac{x_{n}}{|x|})=frac{x}{|x|}in S^{n-1}$ and then we have

$$begin{align} frac{|x^{alpha}|}{|x|^{|alpha|}}
& = frac{ sqrt{ x_1^{2alpha_1} + dots + x_n^{2alpha_n} } }{|x|^{|alpha|} } \
& = bigg( frac{ x_1^{2alpha_1} + dots + x_n^{2alpha_n} }{|x|^{2|alpha|}} bigg)^{frac{1}{2}}
\
& = bigg( frac{x_1^{2alpha_1}}{|x|^{2|alpha|}} + dots + frac{x_n^{2alpha_n}}{|x|^{2|alpha|}} bigg)^{frac{1}{2}}
\
& leq bigg[ bigg(frac{x_1}{|x|}bigg)^{2alpha_1} + dots + bigg(frac{x_n}{|x|}bigg)^{2alpha_n}bigg]^{frac{1}{2}}
\
& = bigg| bigg( frac{x_1^{alpha_1}}{|x|^{alpha_1}}, dots, frac{x_n^{alpha_n}}{|x|^{alpha_n}} bigg) bigg|
\
& = | y^{alpha} |
\
& leq c_{n,alpha} |y|^{|alpha|}
\
& = c_{n,alpha}bigg|bigg(frac{x_{1}}{|x|},frac{x_{2}}{|x|},...frac{x_{n}}{|x|}bigg)bigg|^{|alpha|}
\
& = c_{n,alpha}frac{1}{|x|^{|alpha|}}|(x_{1},x_{2},...x_{n})|^{|alpha|}
\
& = c_{n,alpha}frac{|x|^{|alpha|}}{|x|^{|alpha|}}
= c_{n,alpha}
end{align}$$

So $|x^{alpha}|le c_{n,alpha}|x|^{|alpha|}$.

Now we prove this inequality on the unit sphere.

$vert x^{alpha}vert=|x_1^{alpha_1}||x_2^{alpha_{2}}|...|x_n^{alpha_n}|lefrac{1}{n}sum_{k=1}^{n}|x_{k}|^{nalpha_{k}}lefrac{1}{n}big(sum_{k=1}^{n}|x_{k}|^{alpha_{k}}big)^{n}$

$lefrac{1}{n}big(sum_{k=1}^{n}(1+|x_{k}|)^{alpha_{k}}big)^{n}lefrac{1}{n}big(sum_{k=1}^{n}(1+|x|)^{alpha_k}big)^{n}lefrac{1}{n}big(n(1+|x|)^{|alpha|})^{n}=n^{n-1}2^{n|alpha|}$.

Note that I have used the Arithmetic Geometric Inequality to get the first inequality in the above proof. The second inequality follows from the inequality: $(x_{1}+x_{2}+...+x_{n})^{k}=sum_{alpha_{1}+...+alpha_{n}=k}binom{k}{alpha_{1}, alpha_{2},...,alpha_{n}}x^{alpha_{1}}...x^{alpha_{n}}ge x_{1}^{k}+...+x_{n}^{k}$ which is true when $x_{1},...x_{n}$ are all non-negative. Also note that $|x_{k}|le1+|x_{k}|$ and $|x_{k}|le sqrt{x_{1}^{2}+...+x_{n}^{2}}=|x|$ for all $k=1,2,...,n$.

For the second inequality we attack as follows:

$|x|^{k}=big(sum_{k=1}^{n}|x_{k}|^{2}big)^{frac{k}{2}}lebig(sum_{k=1}^{n}|x_{k}|big)^{frac{2k}{2}}=big(sum_{k=1}^{n}|x_{k}|big)^{k}=sum_{|beta|=k}frac{k!}{beta!}|x^{beta}|lebig(sum_{|beta|=k}big(frac{k!}{beta!}big)big)big(sum_{|beta|=k}|x^{beta}|big)=n^{k}big(sum_{|beta|=k}|x^{beta}|big)$.