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Are there any (at least mildly) explicit counterexamples to the statement $$ \sum_{m \in \mathbb{Z}} \|P_m f\|_p \lesssim \|f\|_p? $$ (Or some good reason to expect this to be false?).

Here $P_m$ is the $m$-th Littlewood-Paley projection, that is $$ \widehat{P_m f} = \psi_m \widehat{f} $$ with $\psi_m(\xi) = \psi(\xi/2^m)$, $\psi(\xi) = \phi(\xi)-\phi(2\xi)$ and $\phi$ is a real radial Schwartz function supported on the closed centered ball of radius $2$ and which equals $1$ on the closed centered ball of radius $1$. $P_m$ being (initially at least) defined for Schwartz functions.

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    $\begingroup$ hint: reduce this to a scaling problem and you are down to constructing a sequence that converges in $\ell^p$ but not in $\ell^1$. $\endgroup$ – Willie Wong Apr 19 '11 at 12:03
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    $\begingroup$ If you do not mind dealing with Fourier series instead of integrals, just take $\sum_k a_k z^{2^k}$. For every $p<\infty$, this is in $L^p$ on the circle if and only if $a_k\in\ell^2$ and your LP projections are just individual monomials. $\endgroup$ – fedja Mar 13 '12 at 21:35
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If that inequality was true, for all $f\in L^p$, that would imply $$ L^p\subset \dot B^0_{p,1}\quad\text{with continuous injection}, $$ with $\dot B^0_{p,1}$ the homogeneous Besov space whose norm is precisely given by the left-hand-side of your inequality. On the other hand, the reverse inequality $$ \Vert{f}\Vert_{L^p}=\Vert{\sum_{m\in \mathbb Z}P_m f}\Vert_{L^p}\le \sum_{m\in \mathbb Z}\Vert{P_m f }\Vert_{L^p} $$ is true for all $f\in \dot B^0_{p,1}\cap L^p$. We would have the topological equality $L^p= \dot B^0_{p,1}$. But it is classical that for $1< p<\infty$, $$ L^p=F^0_{p,2} $$ where $F^0_{p,2}$ is a Triebel-Lizorkin space. Also classical is the fact that a Triebel-Lizorkin space is never a Besov space, except for $p=2$. Even in the case $p=2$ that inequality is false since it would imply $$ \dot B^0_{2,2}=L^2=\dot B^0_{2,1}, $$ which is incompatible with the strict inclusion $ \ell^1(\mathbb Z)\subset\ell^{2}(\mathbb Z). $

Bazin.

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