I'm learning the Littlewood-Paley theory by myself and I encounter the following claim:

Pick a smooth function $\chi$ such that:
$$\chi(\xi) = \begin{cases}
1 \ (|\xi| \leq \frac{1}{2})\\
0 \ (|\xi| \geq 1)
\end{cases}$$
Based on the function $\xi$ picked above, let's define an operator $S_{k} \ (\forall \ k \in \mathbb{Z})$ acting on an arbitrary function $f$ as follows:
$$S_{k}f(x) = (\chi(\frac{\xi}{2^{k}})\hat{f}(\xi))^{\vee}(x)$$
where $\hat{}$ and $\vee{}$ above denotes Fourier transform and inverse Fourier transform, respectively. Moreover, based on $\xi$, we can also define the Littlewood-Paley projection operators $P_{l} \ (\forall \ l \in \mathbb{Z})$.

Now for any $s \in \mathbb{R}$ and any $k \in \mathbb{Z}^{+} \cup \{0\}$, we have the following inequality (equivalence of norms) for any function $f$ in the Sobolev space $H^s$:
$$c^{-1}||f-S_{k}f||_{H^{s}}^2 \leq \sum_{l \geq k}2^{2ls}||P_{l}f||_{L^2}^2 \leq c||f-S_{k}f||_{H^{s}}^2$$
Above $c=c(s) \geq 1$ is some constant dependent on $s$. From the inequality above (equivalence of norms), we can further deduce that $\cap_{t \in \mathbb{R}}H^{t}$ is dense in $H^{s}$.

I roughly get the idea of deducing the dense property from the inequality listed above. However, I'm bit stuck with proving the inequality...it feels that $S_{k}f$ is some other form of "projection", so I guess probably "almost orthogonality" will be useful here? Any help/hint will be appreciated!