Characterization of $H^k$ by Fourier transform

Question

Let $u\in H^k(\mathbb{R}^n)$ a function with complex-valued.

Question 1: If $u\in H^k(\mathbb{R}^n)$ and $D^\alpha u\in L_2(\mathbb{R}^n)$, by definition for each $|\alpha|\le k,$ then we have

$$\|D^\alpha u\|_{L_2}=\|\widehat{D^\alpha u}\|_{L_2}=\|(iy)^\alpha\widehat{u}\|_{L_2},$$

by properties of Fourier transform.

So is correct to afirm, for multiindex $\alpha=k e_i,$ $\{e_j\}$ basis of $\mathbb{R}^n$, that

$$\int_{\mathbb{R}^n}|y|^{2k}|\widehat{u}|^2dy=\int_{\mathbb{R}^n}|D^k u|^2 dx?$$ Cause in Evans book I found a inequality for this setup (chapter 5, section 8).

Question 2: How can I show that previous equation implies that

$$\int_{\mathbb{R}^n}(1+|y|^{k})^2|\widehat{u}|^2dy\le C\|u\|^2_{H^k} ?$$

Someone has a hint for me?

Thanks.

Answer 1

Well for the full norm you need all derivatives up to order $k$ not just the highest one, so apply the rule for general $|\alpha|\le k$, and sum the result. For example you need the $L^2$ norm in there, this will give you the $1$ since $\|u\|_2^2=\|\hat u\|_2^2$.

Answer 2

By definition we have $|D^ku|^2=\sum_{|\alpha|=k}|D^{\alpha}u|^2$ then

$$\int |y|^{2k}|\hat{u}|^2dx\leq C\int |D^ku|^2dx.$$

Note that if $a,b>0$ then $a+b\leq 2\max\{a,b\}$ and thus $(a+b)^s\leq 2^s(a^s+b^s)$, $\forall s>0$. It follows that $(1+|y|)^{2k}\leq C(1+|y|^{2k})$ and this implies

$$\int (1+|y|)^{2k}|\hat{u}|^2dx\leq C\int (1+|y|^{2k})|\hat{u}|^2dx\leq C(||u||_2+||D^ku||_2)=C||u||_{H^k}.$$