I'm studying harmonic analysis by myself and I encountered the following claim about multipliers: consider a sequence of complex numbers $\{m_{n}\}_{n \in \mathbb{Z}}$ that satisfies: $$\sum_{n \in \mathbb{Z}}m_{n}m_{n1} \leq B, \ \lim_{n \rightarrow \infty}m_{n} = 0$$ where $B > 0$ above is some fixed constant. Based on the sequence, let's define a multiplier operator $T$ as follows: $$Tf(x)= \sum_{n \in \mathbb{Z}}m_{n}\hat{f}(n)e^{2\pi inx}$$ Then for any trigonometric polynomial $f$ and any $p \in (1,\infty)$, there exists some constant $C_p > 0$, such that: $$Tf_{L^p} \leq C_pBf_{L^p}$$ I haven't proved this claim yet...what I have done is to try proving the estimate for a more general $f$ in $L^p$. However, it seems that I get neither a proof nor a counterexample for the general case...any ideas on this? (For me, it's just really weird that the estimate only holds for trigonometric polynomial....I think it can be generalized to general functions in $L^p$)

$\begingroup$ Once you prove the estimate for trigonometric polynomials, it extends by continuity to any $f \in L^p$. $\endgroup$– Giorgio MetafuneApr 16, 2021 at 12:44

5$\begingroup$ This is something very classical, but it is far from trivial. For example, the case when $(m_n)_n =(1_{n <n_0})_n$ for some integer $n_0$ is equivalent to the boundedness of the Hilbert transform on $L^p$, $1<p<\infty$.There are elementary but ingenious proofs (look for Cotlar's identity). What you ask is covered by many classical theorems (eg Marcinkiewicz multiplier theorem), but more elementarily just follows from the boundedness of the Hilbert transform, as the closed convex hull of the symbols $(1_{n<n_0})_n$ is exactly your class of symbols for $B=1$. $\endgroup$– Mikael de la SalleApr 16, 2021 at 13:11

$\begingroup$ Cool, so here assuming that $f$ is trigonometric gives us that the we only need to deal with finitely many terms in $\{m_n\}$...However, when $\{m_n\}$ is not the indicator function but some general terms, how are we supposed to relate the expression to the Hilbert transform? Do we need to rewrite the expression as some convex combination? (Thanks in advance!) $\endgroup$– pureorappliedApr 16, 2021 at 15:37
1 Answer
For a trigonometric polynomial, you have that $\hat{f}(n)$ has compact support. So you can summation by parts. Let $\mu_n = m_n  m_{n1}$. Your assumption shows that $\mu_n$ is absolutely summable. The behavior at infinity means that you can write
$$ m_n = \sum_{k \leq n} \mu_k $$
This shows that
$$ Tf = \sum m_n \hat{f}(n) e^{2\pi i n x} = \sum_n \sum_{k\leq n} \mu_k \hat{f}(n) e^{2\pi i n x} = \sum_k \mu_k \sum_{n\geq k} \hat{f}(n) e^{2\pi i n x}$$
The interchange of sums is allowed since the $n$ sum is a finite sum.
Taking the $L^p$ norm you have
$$ \ Tf \_{p} \leq \sum_k \mu_k \ \sum_{n \geq k} \hat{f}(n) e^{2\pi i n \bullet} \_p \leq B \sup_k \ \sum_{n\geq k} \hat{f}(n) e^{2\pi i n \bullet} \_p $$
The boundedness of the final term by $\ f\_{L^p}$ is, as Mikael de la Salle observed, equivalent to the boundeness of the Hilbert transform.