\documentstyle[amssymb,amsfonts,psfig,12pt]{article} \input MathMacro.tex \newcommand{\eqn}[1]{(\ref{#1})} \newcommand{\is}{$I^{s}\;$} \newcommand{\mbf}[1]{\mbox{\boldmath ${#1}$}} \newcommand{\bc}{\begin{center}} \newcommand{\ec}{ \end{center}} \newcommand{\page}{\vfil \break} \newcommand{\bi}{\begin{itemize}} \newcommand{\ei}{ \end{itemize}} \newcommand{\benum}{\begin{enumerate}} \newcommand{\eenum}{ \end{enumerate}} \newcommand{\bd}{\begin{description}} \newcommand{\ed}{ \end{description}} \def\rp{{r^\prime}} \def\rpt{{\tilde r^\prime}} \def\rtil{{\tilde r}} \def\qtil{{\tilde q}} \def\kp{{k^\prime}} \def\kt{{\tilde{k}}} \def\qp{{q^\prime}} \def\pp{{p^\prime}} \def\qpt{{\tilde q^\prime}} \def\R{{\hbox{\bf R}}} \def\C{{\hbox{\bf C}}} \def\rn{{\R^n}} \def\Z{{\hbox{\bf Z}}} \def\dimsup{{\overline{\rm dim}}} \def\eps{{\varepsilon}} \def\implies{{\Rightarrow}} \def\RR{{\frak R}} \def\emph#1{{\it #1}} % \newcommand{\lesssim}{\mbox{$\,\stackrel{\textstyle{<}}{\sim}\,$}} \renewcommand {\theequation}{\thesection.\arabic{equation}} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}{Proposition}[section] \newtheorem{lemma}{Lemma}[section] \newtheorem{corollary}{Corollary}[section] \newtheorem{remark}{Remark}[section] \begin{document} \Large \vskip 1in \bc {\bf \LARGE Rough convolution operators on \\ homogeneous groups}\\ \bigskip \bigskip \bigskip \bigskip \bigskip \bigskip Terence Tao (UCLA)\\ \bigskip \ \bigskip {\tt www.math.ucla.edu/$\sim$tao/preprints } (in preparation) \ec \page \noindent {\LARGE The purpose of this talk is to: } \bigskip \bi \item Describe the known weak-type $(1,1)$ results for rough convolution operators in $\R^n$ \item Extend these results to general homogeneous groups \ei \page \noindent Convolution operators on $\R^n$ \bi \item Let $K(x)$ be a function on $\R^n$ which is homogeneous of degree $-n$: $$ K(\lambda x) = \lambda^{-n} K(x).$$ We are interested in the singular integral operator $T(f) = f * K$. \item In order for $T$ to be bounded even on $L^2$, it is necessary that $K$ is $L^1$ and has mean zero on the unit annulus. \item Conversely, if $K$ is $L^1$ and odd on the unit annulus, or if $K$ is $L \log L$, even, and mean zero on the unit annulus then $T(f)$ is bounded on $L^p$ for every $1 < p < \infty$. (Calder\'on-Zygmund). Thus if $K \in L \log L$ and has mean zero, then $T$ is bounded on every $L^p$. \item The $L \log L$ on $K$ was weakened to $H^1$ by Weiss, and recently weakened in a different manner by Grafakos and Stefanov. (The $L^2$ question is completely settled). \item We consider the problem of weak-type $(1,1)$ bounds. \ei \page \noindent Known results \bi \item $K$ will always be assumed to have mean zero. \item If $K$ has some smoothness (e.g. H\"older continuity) then the weak-type $(1,1)$ bound is a result of Calder\"on-Zygmund theory. \item Work by Hofmann, Christ, Rubio de Francia, and Seeger generalized this to obtain weak-type $(1,1)$ results whenever $K$ is merely in $L \log L$. There is still an open gap between $L \log L$ and $L^1$. \item The work of the first three authors is based on geometric $TT^*$ arguments, and was only shown for low dimensions. The method of Seeger relies on Fourier analysis, and covered all dimensions. \item We show that Christ and Rubio de Francia's arguments actually can be extended to all dimensions. Furthermore, they apply to homogeneous groups other than $\R^n$, whereas the Fourier methods of Seeger do not seem to extend easily. \ei \page \bigskip $L^2$ theory \bi \item Of course, to get weak(1,1) boundedness one must first obtain $L^2$ boundedness. \item When the underlying group is $\R^n$ we get $L^2$ boundedness for $L \log L$ kernels from the work of Calder\'on and Zygmund. \item If $K$ is \emph{odd} then we can extend this to homogeneous groups (and for $L^1$ kernels) by a variant of the method of rotations. (Of course, instead of rotating the Hilbert transform one must use something more curved, but these objects have been well studied, e.g. by Phong and Stein). \item But if $K$ is even and in $L \log L$ we must use a different approach. The idea is to introduce pseudo-Littlewood-Paley operators $f \mapsto f * \phi_j$, where $\phi_j$ is a rescaling of $\phi$ at scale $2^j$, and $\phi$ is Schwarz and satisfies a lot of moment conditions. To prove $L^2$ boundedness of $f \mapsto f*K$ it suffices to do prove the boundedness of the individual Littlewood-Paley projections $f \mapsto f*K*\phi_j$. By dilation invariance we may take $j=0$. \item The idea is to break up the kernel $K$ into dyadic pieces $K = \sum_K K_k$ and obtain bounds for the pieces $f \mapsto f * K_k * \phi_0$ which have a decay factor $2^{-\eps |k|}$. For $k < 0$ this follows from the mean zero condition on $K$, while for $k > 0$ the idea is to use the radial regularity of $K_k$ to eventually interact with the moment conditions in $\phi_0$. To exploit this we have to use the $TT^*$ method several times, and look at the iterated operator $$ f \mapsto f * K_k * \phi_0 * \ldots * K_k * \phi_0$$ to obtain the cancellation. One needs to convolve $n$ copies of $K_k$ because each $K_k$ is smooth in only one direction. \psfig{figure=convolute.eps,height=5in,width=5in} \item There is a bad contribution in which the n directions of smoothing have a small Jacobian (which means that they do not combine to form something with enough isotropic regularity to interact with the moment conditions of $\phi_0$), but this term is "small" and can be dealt with by crude, non-cancellatory methods. \ei \page \bigskip \noindent Weak (1,1): Preliminary reductions \bi \item We'll work in $\R^n$ for simplicity, but the arguments extend, usually without difficulty, to more general groups such as the Heisenberg group. \item Fix $f \in L^1$. By linearity, it suffices to show that $$ \{ x: |f * K(x)| \gtrsim 1 \} \lesssim \|f\|_1.$$ \item Apply a Calder\'on-Zygmund decomposition at height $1$, decomposing $f$ into a good function $g$ (which we can throw away by $L^2$ theory) and a sum of bad functions $\sum_I b_I$, where the $I$ are an essentially disjoint collection of balls, $b_I$ is mean zero and has $L^1$ norm $\lesssim |I|$ on $I$, and $\sum_I |I| \lesssim \|f\|_1$. So we only need show that $$ \{ x: | \sum_I b_I * K(x)| \gtrsim 1\} \lesssim \sum_I |I|.$$ \item Now we break up $K$. Let $K_0$ be the restriction of $K$ to the unit annulus. For each dyadic scale $j$, let $\tilde K_j$ be the smoothed-out dilate of $K_0$ at scale $2^j$: $$ \tilde K_j(x) = 2^{-nj} \int_{t \sim 1} K_0(2^{-j} tx) \phi(2^{-j} tx) \phi(t)\ \frac{dt}{t}$$ where $\phi$ is an appropriate cutoff function. Then we can write $K = \sum_j \tilde K_j$. \item Let $2^i$ denote the radius of $I$. We can rewrite our expression as $$ \sum_I b_I * K(x) = \sum_s \sum_I b_I * \tilde K_{i+s}(x).$$ \item If $s \leq 0$ then this set is supported in an exceptional set of measure $O(\sum_I |I|)$, so it suffices to estimate the contribution for $s > 0$. \item The idea is to control this contribution in $L^2$, under the stronger assumption that $K_0$ is in $L^\infty$ (not just $L \log L$). In fact, we will be able to get an exponential gain in $s$: $$ \| \sum_I b_I * \tilde K_{i+s}(x) \|_2 \lesssim 2^{-\eps s} \| K_0 \|_\infty \sum_I |I|.$$ A standard argument allows one to convert this exponential decay into a weak-type estimate for the $L \log L$ kernels: $$ \{ |\sum_s \sum_I b_I * \tilde K_{i+s}(x)| \gtrsim 1 \} \lesssim \| K_0 \|_{L \log L} \sum_I |I|.$$ \ei \page The Christ-Rubio de Francia trick \bi \item Fix $s \gg 1$. By dilation invariance we can make $\sum_I |I| \lesssim 1$. We wish to prove $$ \| \sum_I b_I * \tilde K_{i+s} \|_2 \lesssim 2^{-\eps s} \| K_0 \|_\infty.$$ \item The functions $b_I * \tilde K_{i+s}$ are supported on the annuli $2^s I_\Delta$ with radii $\sim 2^{i+s}$ centered at $I$. We emphasize this by rewriting the desired estimate as $$ \| \sum_I \psi_I (b_I * \tilde K_{i+s}) \|_2 \lesssim 2^{-\eps s} \| K_0 \|_\infty$$ where $\psi_I$ is a smooth cutoff to $2^s I_\Delta$. \item Since the $I$ are disjoint, we expect the $2^s I$ to have multiplicity at most $O(2^{ns})$, i.e. that $$ \| \sum_I \chi_{2^s I} \|_\infty \lesssim 2^{ns}.$$ It turns out that this estimate is not strictly speaking true (the dilates can pile up logarithmically), but we can refine the collection $I$ by a small amount to make it true. The idea is to first prove the bound $$ \| \sum_I \tilde \chi_{2^s I} \|_{BMO} \lesssim 2^{ns},$$ which is valid for any collection of disjoint balls $I$. \item Having made this reduction, it turns out that we can relax the $L^\infty$ control on $K_0$ to just $L^2$ control. I.e. we'll show need to show $$ \| \sum_I \psi_I (b_I * \tilde K_{i+s}) \|_2 \lesssim 2^{-\eps s} \| K_0 \|_2.$$ \item Now we change our viewpoint. Usually, we keep $K$ fixed and think of these types of estimates as an operator estimate on $f$ (or on related functions such as $b_I$). But as Christ and Rubio de Francia showed, it is often profitable to reverse this perspective, keep the $b_I$ fixed, and think of this estimate as an operator estimate on $K_0$! Note that we have converted a weak $(1,1)$ result to a $(2,2)$ result. \ei \page Next step: $TT^*$ \bi \item The Christ-Rubio de Francia trick has the non-trivial consequence of allowing one to apply the $TT^*$ method in the $K$ variable. \item Using $TT^*$, the estimate $$ \| \sum_I \psi_I(b_I * \tilde K_{i+s}) \|_2 \lesssim 2^{-\eps s} \| K_0 \|_2$$ gets transformed to $$ \| 2^{-ns} \sum_I \psi_I \tau_I D \tau_I^* \psi_I F \|_2 \lesssim 2^{-2 \eps s} \|F\|_2$$ where $\psi_I$ is a cutoff to the annulus $2^s I_\Delta$, $\tau_I$ is convolution with the signed measure $|I|^{-1} b_I$, and $D$ is the averaged dilation operator $$ DF(x) = \int_{t \sim 1} f(tx) \phi(t)\ dt.$$ \item There is a probabilistic interpretation of this operator, if one is willing to be very liberal with the notion of "probability", namely allow total probability to become less than 1, and allow negative probability. \psfig{figure=smooth.eps,height=5in,width=5in} \item For each $x$, randomly pick an $I$ such that $x \in 2^s I_\Delta$, each with probability $2^{-ns}$. (Note that the total probability here can be less than 1). Randomly pick a point in the support of $b_I$, with (signed!) probability distribution $b_I/|I|$. Dilate $F$ around this point by a random $t \sim 1$. The expected value of this operation is the expression above. \psfig{figure=smooth_homog.eps,height=5in,width=5in} \item It's easy to prove this bound without the $2^{-2\eps s}$, even without exploiting any cancellation in the $b_I$ or any smoothing in the $D$ operator. Basically one uses the fact that the $\psi_I$ overlap with multiplicity at most $2^{ns}$. So the hope is to use the cancellation in the $b_I$ and the regularity in the $D$ to obtain some additional gain. \item Unfortunately $D$ only smooths in one direction (the direction oriented toward the randomly chosen point in $b_I$). To get around this, we need to iterate the above operator $n$ times to get smoothing in every direction, which will then interact with the mean zero condition of the $b_I$ to get a gain of $2^{-\eps s}$. \ei \page A model case \bi \item Iterating the desired operator $n$ times, we find ourselves having to estimate quantities like $$ \prod_{i=1}^n \psi_{I_i} \tau_{I_i} D \tau_{I_i}^* \psi_{I_i} F. $$ To win the $2^{-\eps s}$ gain, we need to play the cancellation in the $\tau_{I_i}$ against the $n$ smoothing operators $D$. \item Let us consider a model case when all the $I_i$ are the same size. Then the cutoff functions $\psi_{I_i}$ are fairly harmless. Also, for randomly placed $I_i$, the operators $D$ smooth in very different directions. So in most cases one gets smoothing in every direction, and so one can obtain cancellation using any one of the $\tau_{I_i}$. \psfig{figure=smoothout.eps,height=5in,width=5in} \item There is an exceptional set of $n$-tuples of $I_i$ which are arranged so that the smoothing effects are in a set of directions with small Jacobian. Fortunately, these $n$-tuples are rare, and their contribution can be easily controlled. \item Now let's consider a slightly less special case in which the $I_i$ are non-decreasing in size. In the Euclidean setting this case can be handled in the same manner as previously. For general homogeneous groups there is a possible difficulty due to the non-isotropic dilation structure, in that the smoothing effects corresponding to the large $I_i$ may be in very parallel directions. However, these directions will also be very long, and so there is still more than enough isotropic smoothing to interact with $\tau_{I_1}$, which is the convolution operator with the best cancellation properties (since $I_1$ is smaller than all the other balls). \item By self-adjointness one also can handle the case when the $I_i$ are non-increasing in size. \item However, we encounter a difficulty when the balls $I_i$ alternate between being very large and very small. The effect of the large balls is to create smoothing over very long, very parallel arcs, but this is then truncated by the cutoff functions associated with the small balls so that one only obtains smoothing over short, parallel arcs. This is not enough regularity to obtain the desired $2^{-\eps s}$ gain even if one uses the $\tau_{I_i}$ associated with the smallest ball in the $I_i$. \item To get around this problem we will take a much larger iteration of our operator - instead of iterating it $n$ times, we shall iterate it $2^{2n-3}$ times! The reason for this is the following combinatorial lemma, which allows us to extract from any sequence $I_1, \ldots, I_{2^{2n-3}}$ of balls, a subsequence of $I_{i_1}, \ldots, I_{i_n}$ which is either non-decreasing or non-increasing in size, and such that the smoothing effects of the associated operators remain untruncated by any intervening operators. \item More precisely: Let $r_1, \ldots, r_{2^{2n-3}}$ be any collection of positive reals. Then there exists a subsequence $r_{i_1}, \ldots, r_{i_n}$ which is monotone, and such that $r_j \geq \min(r_{i_k}, r_{i_l})$ whenever $i_k < j < i_l$. (In other words, the balls between $I_{i_k}$ and $I_{i_l}$ do not cause any truncation problems.) \ei \end{document}