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\begin{document}
\Large
\vskip 1in
\bc
{\bf \LARGE  Rough convolution operators on \\
homogeneous groups}\\
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Terence Tao (UCLA)\\
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\ 
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{\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}
