A maximal function for families of Hilbert transforms along homogeneous curves

Abstract

Let $H^{(u)}$ be the Hilbert transform along the parabola $(t, ut^2)$ where $u\in \mathbb {R}$. For a set U of positive numbers consider the maximal function ${\mathcal {H}}^U \,f= \sup \{|H^{(u)}\, f|: u\in U\}$. We obtain an (essentially) optimal result for the $L^p$ operator norm of ${\mathcal {H}}^U$ when $2<p<\infty$. The results are proved for families of Hilbert transforms along more general nonflat homogeneous curves.

Appendix A: Proof of Proposition 3.4

The proof is a modification of the argument for the standard Cotlar inequality regarding truncations of singular integrals, cf. [26, §I.7].

Let $m_j(\xi ) =\eta (2^{-j}\xi )m(\xi )$ and let $a_j(\xi ) = m_j(2^j\xi )$. We pick $0<\varepsilon <\min \{\alpha -d,1\}$. Then by assumption

$$\begin{aligned} \sup _{j\in {\mathbb {Z}}}\Vert a_j\Vert _{\mathscr {L}^1_\alpha }\le B<\infty \end{aligned}$$

(A.1)

which implies that $|{\mathcal {F}}^{-1} [a_j](x)|\le CB(1+|x|)^{-d-\varepsilon }$, and thus, with $K_j= {\mathcal {F}}^{-1} [m_j]$,

$$\begin{aligned} |K_j(x)|+ 2^{-j} |\nabla K_j(x)| \le C B 2^{jd} (1+2^j|x|)^{-d-\varepsilon }. \end{aligned}$$

For Schwartz functions f we have $S f=\sum _{j\in {\mathbb {Z}}} K_j* f$ and $S_nf=\sum _{j\le n} K_j*f$.

Lemma A.1

Fix $\tilde{x}\in {\mathbb {R}}^d$ and $n\in {\mathbb {Z}}$, and let $g(y)= f(y){\mathbb {1}}_{B(\tilde{x}, 2^{-n})}(y)$ and $h=f-g$. Then

• (i) $|S_n g(\tilde{x})|\lesssim B \, M[f](\tilde{x})$.

• (ii) $|S_nh(\tilde{x})-S h(\tilde{x})|\lesssim B\, M[f](\tilde{x})$.

• (iii) For $|w-\tilde{x}|\le 2^{-n-1}$ we have $|Sh(\tilde{x})-Sh(w)|\lesssim B \, M[f](\tilde{x}).$

Proof

By appropriate normalization of the multiplier we may assume $B=1$.

• (i) is immediate since for $j\le n$

  $$\begin{aligned} |K_j*g(\tilde{x})|\lesssim 2^{jd}\int _{|\tilde{x}-y|\le 2^{-n}}|g(y)| dy \lesssim 2^{(j-n)d} M[g](\tilde{x}) \end{aligned}$$

  and the assertion follows since $|g|\le |f|$.

For (ii) notice that $|S_n h(\tilde{x})-S h(\tilde{x})|\le \sum _{j>n}|K_j* h(\tilde{x})|$. For $j>n$ we estimate

where the slashed integral denotes the average. Thus we get

$$\begin{aligned} \sum _{j\ge n} |K_j*h(\tilde{x})|\lesssim M[h](\tilde{x}) \end{aligned}$$

and, since $|h|\le |f|$, the assertion follows.

Concerning (iii) we consider the terms $K_j*h(\tilde{x})-K_j*h(w)$ separately for $j\le n$ and $j>n$. The term $\sum _{j>n} |K_j*h(\tilde{x})|$ was already dealt with in (ii). Since $|w-\tilde{x}|\le 2^{-n-1}$ we have $|w-y|\approx |\tilde{x}-y|$ for $|\tilde{x}-y|\ge 2^{-n}$ and thus the previous calculation also yields

$$\begin{aligned} \sum _{j> n} |K_j*h(w)|\lesssim M[h](\tilde{x}) \lesssim Mf(\tilde{x}). \end{aligned}$$

It remains to consider the terms for $j\le n$. In that range we write

$$\begin{aligned} K_j*h(\tilde{x})-K_j*h(w) = \int _0^1\int \limits _{|\tilde{x}-y|\ge 2^{-n} }\langle \tilde{x}-w,\nabla K_j (w+s(\tilde{x}-w)-y)\rangle h(y) dy\, ds. \end{aligned}$$

Since $|w-\tilde{x}| \le 2^{-n-1}$ we can replace $|w+s(\tilde{x}-w)-y|$ in the integrand with $|\tilde{x}-y|$ and estimate the displayed expression by $C\sum _{l\ge 0} A_{l,j,n}$ where

Summing in $l > 0$ and then $j\le n$ yields

$$\begin{aligned} \sum _{j\le n} |K_j*h(\tilde{x})-K_j*h(w)| \lesssim Mh(\tilde{x}) \lesssim Mf(\tilde{x}). \end{aligned}$$

(A.2)

Proof of (3.7)

We proceed arguing as in [26, §I.7]. Fix $\tilde{x}\in {\mathbb {R}}^d$ and $n\in {\mathbb {Z}}$ and define g and h as in the lemma. For (suitable) w with $|w-\tilde{x}|\le 2^{-n-1}$ we write

$$\begin{aligned} S_n f(\tilde{x})&= S_n g(\tilde{x})+ (S_n-S)h(\tilde{x}) + Sh(\tilde{x})\nonumber \\&=S_n g(\tilde{x})+ (S_n-S)h(\tilde{x}) + Sh(\tilde{x})- Sh(w) + Sf(w)- Sg(w). \end{aligned}$$

(A.3)

By Lemma A.1

$$\begin{aligned} |S_n g(\tilde{x})|+ |(S_n-S)h(\tilde{x})| + |Sh(\tilde{x})- Sh(w)| \lesssim B \,M[f](\tilde{x}) \end{aligned}$$

and it remains to consider the term $Sf(w)-Sg(w)$ for a substantial set of w with $|w-\tilde{x}|\le 2^{-n-1}$.

By the Mikhlin-Hörmander theorem we have for all $f\in L^1({\mathbb {R}}^d)$ and all $\lambda >0$

$$\begin{aligned} {\mathrm{meas}}( \{x: |Sf(x)|>\lambda \}) \le A \lambda ^{-1} \Vert f\Vert _1 \end{aligned}$$

where $A \le C_{\alpha ,d}B$.

Now let $\delta \in (0,1/2)$ and consider the set

$$\begin{aligned} \Omega _n(\tilde{x}, \delta )=\big \{ w: |w-\tilde{x}|< 2^{-n-1}, \quad |Sg(w)|> 2^d\delta ^{-1} A \,M[f](\tilde{x})\big \}. \end{aligned}$$

In (A.3) we can estimate the term |Sg(w)| by $2^d\delta ^{-1} A \,M[f](\tilde{x})$ when $w\in B(\tilde{x}, 2^{-n-1}) {\setminus } \Omega _n(\tilde{x},\delta )$. Hence we obtain

$$\begin{aligned} |S_n f(\tilde{x})| \le \inf _{w\in B(\tilde{x}, 2^{-n-1}) {\setminus } \Omega _n(\tilde{x},\delta )} |Sf(w)| + C(\alpha , d) B (1+\delta ^{-1}) M[f](\tilde{x}). \end{aligned}$$

(A.4)

By the weak type inequality for S we have

$$\begin{aligned} {\mathrm{meas}}(\Omega _n(\tilde{x},\delta ))&\le \frac{A\Vert g\Vert _1}{2^d\delta ^{-1} A M[f](\tilde{x})} =\frac{\delta }{2^d M[f](\tilde{x})}\int _{|\tilde{x}-y|\le 2^{-n}}|f(y)| dy\\&\le \delta \, 2^{-d}\,{\mathrm{meas}}(B(\tilde{x}, 2^{-n})) =\delta \, {\mathrm{meas}}(B(\tilde{x}, 2^{-n-1})). \end{aligned}$$

Hence ${\mathrm{meas}}( B(\tilde{x}, 2^{-n-1}) {\setminus } \Omega _n(\tilde{x},\delta ))\ge (1-\delta ) {\mathrm{meas}}(B(\tilde{x}, 2^{-n-1}))$ and thus for all $r>0$

$$\begin{aligned}&\inf _{w\in B(\tilde{x}, 2^{-n-1}) {\setminus } \Omega _n(\tilde{x},\delta )} |Sf(w)| \\&\quad \le \left( \frac{1}{{\mathrm{meas}}(B(\tilde{x}, 2^{-n-1}) {\setminus } \Omega _n(\tilde{x},\delta ))} \int _{B(\tilde{x}, 2^{-n-1})} |Sf (w)|^r dw\right) ^{1/r}\\&\quad \le \left( \frac{1}{(1-\delta )|B(\tilde{x}, 2^{-n-1})|} \int _{B(\tilde{x}, 2^{-n-1})} |Sf (w)|^r dw\right) ^{1/r}. \end{aligned}$$

We obtain

$$\begin{aligned} |S_n f(\tilde{x})| \le (1-\delta )^{-1/r} (M[|Sw|^r](\tilde{x}))^{1/r} + C(\alpha ,d) (1+\delta ^{-1}) B\,M[f](\tilde{x}) \end{aligned}$$

uniformly in n. This implies (3.7). $\square$