Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
1  
hint: reduce this to a scaling problem and you are down to constructing a sequence that converges in $\ell^p$ but not in $\ell^1$. –  Willie Wong Apr 19 '11 at 12:03
1  
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. –  fedja Mar 13 '12 at 21:35
add comment

1 Answer 1

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.