Use of the Littlewood-Paley decomposition to recover the $H^s$ norm Let $\phi\in C^{\infty}_0(\mathbb{R}^n)$ be such that 
$$\{\lvert \xi\rvert \le 1\} \prec \phi \prec \{\lvert \xi \rvert < 2\}^{[1]} $$
and define the Littlewood-Paley projectors as
\begin{equation}
(P_{2^j}f)^\wedge=\left[\phi\left( \frac{\xi}{2^j} \right)- \phi\left( \frac{\xi}{2^{j-1}}\right)\right] \widehat{f}(\xi) 
\end{equation}
where $j\in \mathbb{Z}$ and $f$ is Schwartz. ($\,^\wedge$ denotes Fourier transform.)

Question. Suppose that $f$ is a Schwartz function such that $\widehat{f}$ is supported in $\{\lvert \xi\rvert >2\}$, and let $s>0$ be fixed. Does there exist an absolute constant $C$ such that the following inequality is true?
  $$\sum_{j=1}^\infty \left( 2^j\right)^s \lVert P_{2^j} f\rVert_2\le C \lVert f \rVert_{H^s}.$$ EDIT I now believe that the answer is negative, see comments.

Some motivation(you can safely omit reading this): 
I want to prove a maximal estimate for the Schrödinger group $S_t=e^{it\Delta}$, namely
$$\tag{1}\lVert S^\star f\rVert_{L^p(B_1)}\le C \lVert f\rVert_{H^s},$$
where $S^\star f=\sup_{t\in (0, 1)}\lvert S_tf\rvert$ and $B_1$ is the unit ball. All the articles I am consulting claim without further explanation that it is enough to prove this apparently weaker fact: 
$$\tag{2}\lVert S^\star f\rVert_{L^p(B_1)}\le C \lambda^s\lVert f\rVert_2,\qquad \operatorname{Spt}\widehat{f}\subset \{\lvert \xi\rvert \sim \lambda\}$$
(where $\lvert \xi\rvert \sim \lambda$ means that $\lambda\le \lvert \xi \rvert \le 2\lambda$). I would like to prove the implication $(2)\Rightarrow (1)$. 
Now if $f$ is a Schwartz function that is Fourier supported in $\{\lvert \xi \rvert> 2\}$, then 
$$f=\sum_{j=1}^\infty P_{2^j}f.$$
Applying the sublinearity of $S^\star$ and (2) we arrive at 
\begin{equation}
\begin{split}
\lVert S^\star f\rVert_{L^p(B_1)} &\le\sum_{j=1}^\infty \lVert S^\star P_{2^j} f\rVert_{L^p(B_1)} \\
&\le C \sum_{j=1}^\infty \left(2^j\right)^s\lVert P_{2^j}f\rVert_2,
\end{split}
\end{equation}
and this is where the Question comes in.

$\,^{[1]}$ Meaning that $\phi\ge 0$ everywhere, that $\phi(\xi)=1$ for all $\xi \in \{\lvert \xi\rvert \le 1\}$ and that the support of $\phi$ is contained in $\{\lvert \xi \rvert < 2\}$.
 A: After a talk with my advisor I believe that I have a clear view of the matter. 
Suppose that $f$ is a Schwartz function whose Fourier transform is supported away from the origin${}^{[1]}$, so that 
$$ f=\sum_{j=0}^\infty P_{2^j}f.$$
The series converges in the sense that the sum is locally finite at Fourier side. By a simple version of the Littlewood-Paley's inequality we have ${}^{[2]}$:
$$\tag{LP}\lVert f\rVert_{H^s}^2\sim \sum_{j=0}^\infty (2^j)^{2s} \lVert P_{2^j} f\rVert_2^2$$
that is, the $H^s$-norm of $f$ is comparable with the $\ell^2$-norm of the sequence 
$$\mathbf{F}=\left\{ (2^j)^s\lVert P_{2^j}f\rVert_2\ :\ j=0,1,2\ldots\right\}.$$
With this newly introduced notation we can rephrase the Question above: 

Question (rephrased): Is it true that $\lVert \mathbf{F}\rVert_{\ell^1}\le C \lVert f\rVert_{H^s}?$

That this needs not be true can be easily seen now by inserting relation (LP) into the question: 
$$\lVert \mathbf{F}\rVert_{\ell^1}\le C \lVert \mathbf{F}\rVert_{\ell^2} \tag{?!}$$
and this is clearly false as inequalities between $\ell^p$ spaces go the opposite way. For a concrete example just take a function $f$ such that $\mathbf{F}=\{\frac{1}{j}\}$.
However, not everything is lost. If we are ready to lose an $\varepsilon$ in the exponent $s$ we can apply Cauchy-Schwarz's inequality as follows:
$$
\begin{split}
\sum_{j=0}^\infty (2^j)^s\lVert P_{2^j}f\rVert_2 & = \sum_{j=0}^\infty (2^j)^{s+\varepsilon}\lVert P_{2^j}f\rVert_2 (2^j)^{-\varepsilon} \\
&\le \left( \sum_{j=0}^\infty (2^j)^{2(s+\varepsilon)}\lVert P_{2^j}f\rVert_2^2\right)^{\frac{1}{2}}\left(\sum_{j=0}^\infty (2^{-\varepsilon})^j\right)^{\frac{1}{2}} \\
&=C_\varepsilon \left( \sum_{j=0}^\infty (2^j)^{2(s+\varepsilon)}\lVert P_{2^j}f\rVert_2^2\right)^{\frac{1}{2}} \le C_\varepsilon \lVert f\rVert_{H^{s+\varepsilon}}.
\end{split}
$$
This way we get an estimate not into the desired space $H^s$ but into a slightly worse space $H^{s+\varepsilon}$. 

${}^{[1]}$ This is just a simplifying assumption and can be removed at the cost of some minor technicality. 
${}^{[2]}$ Here $A\sim B$ means that $cB\le A\le CB$ where $c, C$ are absolute constants. For this particular inequality the constants can be taken to be $c=\frac{1}{2}$ and $C=1$.
