$\ell^q$ analog of square function

It is a classical result in harmonic analysis that

$$\|\|P_kf\|_{\ell^2_k}\|_{L^p_x}\approx\|f\|_{L^p}$$

for $p\in(1,\infty)$, where $P_k$ is the Littlewood-Paley decomposition onto frquency $\approx 2^k$.

What if I replace the $\ell^2$ norm in $k$ by the $\ell^q$ norm? Is it possible that in addition to the trivial estimate derived from the embedding of $\ell^q$ in $\ell^2$ (or vice versa), there are some nontrivial value of $(p,q)$ for this to hold?

• What is $P_kf$? Nov 14, 2017 at 9:36
• It seems that you are asking whether the quasi norm $$\|f\|_{\dot{F}_{p,q}^{0}} := \| \|P_{k}f\|_{\ell_{k}^{q}}\|_{L_{x}^{p}}$$ for the homogeneous Triebel-Lizorkin space $\dot{F}_{p,q}^{0}$ is equivalent to the Lebesgue $L^{p}$ norm. I do not know have a copy of H. Triebel's Theory of Function Spaces in front of me, but I would suggest looking in there. Nov 14, 2017 at 16:00

1 Answer

The following counterexamples are taken from examples 6.1.10 and 6.1.11 in the textbook L. Grafakos, Classical Fourier Analysis (Third Edition).

Claim 1. Fix $1<p<\infty$ and $q<2$. Then the inequality $$\| (\sum_{j\in\mathbb{Z}} |P_{j}(f)|^{q})^{1/q}\|_{L^{p}} \lesssim_{p,q} \|f\|_{L^{p}}$$ cannot hold.

Proof. Let $\psi$ be a $C^{\infty}$ function on $\mathbb{R}$ such $\hat{\psi}\geq 0$,

$$\hat{\psi}(\xi) = \begin{cases} 1 & {\xi \in [9/8,15/8]} \\ 0 & {\xi\notin [7/8,17/8]} \end{cases},$$ and $$\sum_{j\in\mathbb{Z}} \hat{\psi}(2^{-j}\xi)^{2}=1, \qquad \xi>0.$$ Extend $\hat{\psi}$ to be an even function on $\mathbb{R}$. Let $P_{j}:=\psi(2^{-j}D)$.

Now let $\varphi$ be a Schwartz function such that $\hat{\varphi}\geq 0$ and $\mathrm{supp}(\hat{\varphi}) \subset [11/8,13/8]$. Fix $N\gg 1$, and define

$$f_{j}(x) := e^{2\pi\frac{12}{8}2^{j}x}\varphi(x), \qquad j=1,\ldots,N.$$

Since $\widehat{f}_{j}(\xi) = \hat{\varphi}(\xi-\frac{12}{8}2^{j})$ is supported on the set $$[\frac{11}{8} + \frac{12}{8}2^{j}, \frac{13}{8}+\frac{12}{8}2^{j}]\subset [\frac{9}{8}2^{j}, \frac{15}{8}2^{j}], j\gg 1,$$ $\hat{\psi}(2^{-j}\xi)=1$ on the support of $\hat{f}_{j}$.

Now define $$f := \sum_{j=3}^{N}f_{j}.$$ We have the lower bound $$\|(\sum_{j}|P_{j}(f)|^{q})^{1/q}\|_{L^{p}}\geq \|(\sum_{j=3}^{N} |f_{j}|^{q})^{1/q}\|_{L^{p}} = (N-3)^{1/q}\|\varphi\|_{L^{p}}.$$ By the Littlewood-Paley inequality, we have the upper bound $$\|f\|_{L^{p}} \lesssim_{p} \|(\sum_{j}|P_{j}f|^{2})^{1/2}\|_{L^{p}} = (N-3)^{1/2}\|\varphi\|_{L^{p}}.$$

For $N\gg 1$, we obtain a contradiction. $\Box$

Claim 2. Fix $1<p<\infty$ and $2<q<\infty$. Then the inequality $$\|g\|_{L^{p}} \lesssim_{p,q} \|(\sum_{j} |P_{j}g|^{q})^{1/q}\|_{L^{p}}$$ cannot hold.

Proof. This follows from duality and Claim 1. $\Box$.