In the paper "A Convolution Inequality Concerning Cantor-Lebesgue Measures", the Littlewood-Paley theory is used to estimate the norm of multiplier operator in Lemma 1. It is claimed that Lemma 2 is an elementary variant and the proof involves the boundedness of Hilbert transform, but the proof is not given.

  1. How can we prove Lemma 2? How is it related to Hilbert transform?
  2. Is there any version of Littlewood-Paley theorem for decomposition into finitely many pieces? Will this help proving Lemma 2?
  3. It is claimed that he constants involved in Lemma 1 and 2 will tend to $1$ as $p,q$ tend to $2$. Why is this true? Do these follow from Riesz-Thorin interpolation theorem?

I will be grateful for any help and hints~

  • 1
    $\begingroup$ All your questions are quite easy to answer with a bit of work and a standard Harmonic Analysis textbook. Just some hints. Question 2 is irrelevant. 1 is a direct consequence of the L^p-bounds for the Hilbert transform H, and the fact that any 1-D frequency projection can be written as a linear combination of modulations/translations of H. About the constants: the Hilbert transform is an isometry on L^2 (thus has norm 1) and the L-P inequalities are identities in L^2 with constant 1. $\endgroup$ May 12, 2013 at 9:59
  • $\begingroup$ Thanks for your hints. Is Hilbert transform used here only for boundedness of each $\| m_j \|$ instead of proving the inequality itself? Then can we just prove in the following way? $$\| m \| = \| \sum m_j \| \leq \sum \| m_j \| \leq N \max \| m_j \|$$ Then is $A_2$ simply $N$? $\endgroup$ May 13, 2013 at 21:04

1 Answer 1


I think it should work like that: One proves the $(2,2)$ inequality with constant 1, since the frequency intervals are disjoint, by using the $(2,2)$ type of the Hilbert transform. Then for general $p\leq q$, you can prove a bound which is linear in $N$, more or less as you do above. This however doesn't tend to $1$ as $p,q\to 2$. But now for any $p\leq q$, you can find $ 1 < p_1,q_1<\infty $ so that $p$ is between $p_1$ and $2$ (the exact order depends on whether $p>2$ or $p<2$ ) and, likewise, $q$ is between $2$ and $q_1$. To complete the proof interpolate between the $(2,2)$ bound and the $(p_1,q_1)$ bound.

  • $\begingroup$ Thanks, I have tried writing it out, it seems there is still some problem. $$\| T_m \|_{2,2} = \| m \|_\infty = \max_j \| m_j \|_\infty \leq \max_j \| T_{m_j} \|_{p,q} $$ for any $p \leq q$. Take $s$ and $t$ such that $p$ is between $s$ and $2$, $q$ is between $t$ and $2$. $$\| T_m f \|_2 \leq \max_j \|T_{m_j}\|_{s,t} \|f\|_2$$ $$\| T_m f \|_t \leq N \max_j \| T_{m_j} \|_{s,t} \| f \|_s$$ Let $M = \max_j \| T_{m_j} \|_{s,t}$. Applying interpolation we can get $$\| T_m f \|_q \leq M^{1- \theta} (N M)^{\theta} \| f \|_p = N^\theta M \| f \|_p$$ $\endgroup$ May 14, 2013 at 23:06
  • $\begingroup$ That is, $$ \| T_m \|_{p,q} \leq N^\theta \max \| T_{m_j} \|_{s,t}.$$ I cannot get the $(p,q)$ norm instead of $(s,t)$ norm. Is there any way to fix this? $\endgroup$ May 14, 2013 at 23:09

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