In today's class my teacher used the following inequality: \begin{equation}\tag{1}\sum_{i=1}^\infty 2^{2i}\int_{\{ 2^i<|x|<2^{i+1} \}}|\hat{f}(x)|^2dx<C\|f\|_{H^1}^2,\end{equation} where $\hat{f}$ denotes the Fourier transform.
While I was looking for this result I found the question Use of the Littlewood-Paley decomposition to recover the $H^s$ norm. In particular, I have seen that \begin{equation}\tag{2}\lVert f\rVert_{H^1}^2\leq \sum_{j=0}^\infty (2^j)^{2} \lVert P_{2^j} f\rVert_2^2\end{equation} where $P_{2^i}$ is the Littlewood-Paley projectors defined as \begin{equation} (P_{2^j}f)^\wedge=\left[\phi\left( \frac{x}{2^j} \right)- \phi\left( \frac{x}{2^{j-1}}\right)\right] \widehat{f}(x), \end{equation} for $\phi\in C^{\infty}_0(\mathbb{R}^n)$ a bump function with $\operatorname{supp}(\phi)\subset\{|x|<2\}$ and $\phi(x)=1,\forall |x|<1$.
My question is: Is it possible to deduce $(1)$ from $(2)$?
Yes, but it is more complicated than a direct proof. On the set $x\in(2^i,2^{i+1})$, $2^{2i} \le x^2$. So $$LHS \le \sum_i\int_{(2^i,2^{i+1})} |x\hat f|^2 = \int |x\hat f|^2 = \int |\widehat Df|^2 = \int |Df|^2 \le RHS. $$ In order to recover (1) from (2) you need an additional property of the LP pieces, which is that $\sum_j \left[\phi\left( \frac{x}{2^j} \right)- \phi\left( \frac{x}{2^{j-1}}\right)\right] ^2\le C$. Then proceed similarly to the direct proof.