Sobolev spaces fourier norm equivalence

You have $\xi=(\xi_1,\ldots,\xi_n)$ and $|\xi|^2=\xi_1^2+\ldots+\xi_n^2.$

If you expand $F(\xi)=(1+|\xi|^2)^k$ by binomial expansion, you will observe that the expression has precisely the same terms as $S(\xi)$. Only their coefficients differ. Therefore, you can replace the coefficients of $F(\xi)$ by a suitable number to bound $S(\xi)$ from either side.

More precisely, if $a_{\alpha}$ are the coefficients of $F(\xi)$, then a choice of the form $C_1a_{\alpha}\leq 1$ and $C_2a_{\alpha}\geq 1$ for all $\alpha$ will produce the inequalities $$C_1(1+|\xi|^2)^k\leq\sum_{|\beta|\leq k}|\xi^{\beta}|^2\leq C_2(1+|\xi|^2)^k$$

$C_1$ and $C_2$ depend on the coefficients alone and they are simply binomial coefficients which depend on $k$ and $n$.

There's a couple of simplifications that make it easier.

• There is a finite number of terms in the sum $S$, so we need only bound each one individually. Ie, show that for each $\alpha \le k$, $|\xi^\alpha|^2 \le C (1+|\xi|^2)^k$.
• We only need to consider the "tails" of the functions where $\xi$ is sufficiently large. This follows since the ratio $F/S$ is continuous in $\xi$ and therefore attains it's max and min in any compact set.
• Since $|\xi|$ is large, we likewise need only consider the case $|\alpha|=k$.

In this regime, $$(1+|\xi|^2)^k \approx (|\xi|^2)^k = \left( \sum_i \xi_i^2 \right)^k$$

whereas $$|\xi^\alpha|^2 = \sum_i \xi_i^{2\alpha_i}$$

with $\sum_i \alpha_i = k$. Applying Cauchy Schwarz with the "1-trick" yields, $$\left( \sum_i \xi_i^{\alpha_i} \right)^2 \le \left( \sum_i \xi_i^{2 \alpha_i}\right) \left(\sum_i 1^2 \right) = C \sum_i \xi_i^{2 \alpha_i}.$$

Then since $\alpha_i \le k$ and $\xi$ is sufficiently large, we have: $$\sum_i \xi_i^{2 \alpha_i} \lesssim \sum_i \xi_i^{2 k} \le \left( \sum_i \xi_i^2 \right)^k.$$

Altogether this implies $|\xi^\alpha|^2 \lesssim (1+|\xi|^2)^k$ as required.