I am reading about Sobolev spaces and I have a question regarding Sobolev spaces and the Fourier transform.
So by defining the Fourier transform, $F(\cdot)$ as an isometry we get $||f||_{L^2}=||\widehat{f}||_{L^2}$ and $\widehat{D^\alpha f}=\xi^\alpha\widehat{f}$. So one can define the norm in $H^k$, $$||f||_{H^k}=\left(\int(1+|\xi|^2)^k|\widehat{f(\xi)}|^2d\xi\right)^{\frac{1}{2}}$$
then he goes on and says that the proof that this is an equivalent norm is an easy exercise and is ommited. So I tried to prove it but I got stuck very quickly. The proof boils down to having to show that the functions
$$F(\xi)=(1+|\xi|^2)^k$$
and
$$S(\xi)=\sum_{|\alpha|\le k}|\xi^\alpha|^2$$
with $\xi\in \mathbb{R}^n$ bound each other. The one directions is more or less clear since $S$ includes the terms of $F$, but I cannot see why the reverse is true.
Best Answer
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 $n$.