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\begin{document}
\Large
\vskip 1in
\bc
{\bf \LARGE   The almost everywhere convergence of \\ ~ \\ Bochner-Riesz multipliers in the plane
\\ ~ \\ for $p < 2$
}\\
\bigskip
\bigskip
\bigskip
\bigskip
\bigskip
Terence Tao \\
\bigskip
{\it Mathematics Department \\ \bigskip UCLA \\ }  
\bigskip
\ 
\bigskip
{\tt www.math.ucla.edu/$\sim$tao/preprints}
\ec

\page

\noindent
{\LARGE
The purpose of this talk is to:
}
\bigskip
\bi
 \item State the problem of almost-everywhere convergence of Bochner-Riesz multipliers in the plane

 \item State the classical positive and negative results

 \item Show some new positive and negative results

\ei

\page

\noindent
Classical theory of Fourier summation (one dimension)

\bi
\item Let $f$ be an $L^p(\R)$ function ($1 < p < \infty$), $t > 0$, and consider the
Dirichlet operator
$$ S_t f(x) = \int_{|\xi| < t} e^{2\pi i x \cdot \xi} \hat f(\xi)\ d\xi.$$
\item For smooth functions $S_t f$ converges to $f$ uniformly
as $t \to \infty$.
What happens if $f$ is merely in $L^p$?
\item $S_t f$ converges to $f$ in $L^p$.  (Hardy?)
\item $S_t f$ converges a.e. to $f$.  (Carleson-Hunt)
\item Can speed up the rate of convergence by taking a Riesz mean instead of
a Dirichlet operator:
$$ S_t^\delta f(x) = \int e^{2\pi i x \cdot \xi} (1-\frac{|\xi|}{t})^\delta_+ \hat f(\xi)\ d\xi.$$
When $\delta = 0$ this is the Dirichlet mean; when $\delta = 1$ this is the Fejer mean.
The higher the $\delta$, the better the convergence.
\ei
\page

\bigskip
\noindent
What happens in two dimensions?

\bi
\item Suppose we define the same operator in two dimensions:
$$ S_t f(x) = \int_{|\xi| < t} e^{2\pi i x \cdot \xi} \hat f(\xi)\ d\xi.$$
This is called the disc multiplier.
\item When $f$ is smooth, $S_t f(x)$ still converges to $f$ uniformly.
What happens if $f$ is merely in $L^p$?
\item Does $S_t f$ converge to $f$ in $L^p$ in general?
NO unless $p=2$!  (Fefferman's example)
\item Does $S_t f$ converge a.e. to $f$?  NO if $p<2$ (Fefferman's example)
UNKNOWN if $p = 2$.  ($p>2$ is equivalent to $p=2$).  This problem is very
difficult.
\item What about Bochner-Riesz means of order $\delta > 0$?
$$ S_t^\delta f(x) = \int e^{2\pi i x \cdot \xi} (1-\frac{|\xi|}{t})^\delta_+ \hat f(\xi)\ d\xi.$$
\ei

\page

The ``easy'' problem: $L^p$ convergence of Bochner-Riesz means for $\delta > 0$.

\bi 
\item  Under what conditions on $p$, $\delta$ does $S_t^\delta f$ converge to $f$ in $L^p$ for
all $f \in L^p$?
\item By some soft analysis, this is equivalent to proving the estimate
$$ \| S_1^\delta f\|_p \lesssim \|f\|_p$$
for test functions $f$.
\item Take $f$ to be a standard bump function.  Then a stationary phase (or Bessel function)
computation shows that $|S_1^\delta f(x)|$ decays like $(1+|x|)^{-\delta -\frac{1}{2}}$.
Thus a necessary condition is that $\delta + \frac{1}{2} > \frac{2}{\pp}$. (Herz)
\item By duality we also obtain the condition $\delta + \frac{1}{2} > \frac{2}{p}$. (Herz)
\item When $\delta > 0$, these necessary conditions are also suffiicent.
\item If $\delta > 1/2$ then the kernel of $S_1^\delta$ turns out to be integrable, and
the result follows from Young's inequality.  (Bochner)
\item If $p=2$ and $\delta > 0$ then the result follows trivially from Plancherel's theorem.
\item When $p=4$, $\delta > 0$, the result follows by explicitly computing the $L^4$ norm
(Carleson, Sj\"olin, Fefferman, 1972).  When $p=4/3$ the result follows by duality.
\item By interpolation between all of these results, one obtains the sharp sufficient 
conditions on $p$ and $\delta$.
\ei

\page

The ``hard'' problem: almost everywhere convergence of Bochner-Riesz means for $\delta > 0$

\bi
 \item When $p \leq 2$, this problem is equivalent to the maximal function estimate
$$ \| \sup_{t>0} |S_t^\delta f| \|_{p,\infty} \lesssim \|f\|_p$$
(Stein, 1961).  When $p > 2$ the equivalence is not perfect, but
pretty good (Garcia de Cuerva, Rubio de Francia, $\sim 1980$).
 \item The same necessary conditions $\delta + \frac{1}{2} > \frac{2}{p}, \frac{2}{\pp}$
can be shown to be true.  Are they sufficient? (Stein, 1978)
 \item When $\delta > 1/2$ the kernels $S_t^\delta$ are integrable, and the result
follows for $1 < p < \infty$ from the Hardy-Littlewood maximal inequality.
 \item When $p=2$, $\delta > 0$ one can replace the supremum by a square function
using Sobolev embedding, and then use Plancherel's theorem.  (Stein, 1961)
 \item When $p=4$, $\delta > 0$, the result follows by replacing the supremum by
a square function and explicitly computing the $L^4$ norm (Carbery, 1983).
 \item Duality doesn't work anymore!
\ei

\page
\noindent
The necessary conditions above are not sufficient when $p<2$!
\bi
 \item Stein's conjecture is true for radial functions (Kanjin, Kojima, 1983) or
lacunary approaches (Carbery, 1986), but fails in general!
 \item Theorem (T., 1997): The estimate
$$ \| \sup_{t\sim 1} |S_t^\delta f| \|_{p,\infty} \lesssim \|f\|_p$$
(and thus almost everywhere convergence) fails when $\delta < \frac{3}{2p}-1$.
 \item Consider first the Herz example, in which $f$ is a bump function.  The convolution
kernel of $S_t^\delta$ is essentially $e^{\pm 2\pi i |x|/t} |x|^{-1/2-\delta}$ when
$t \sim 1$, and so
$S_t^\delta f(x) \sim e^{\pm 2\pi i |x|} |x|^{-1/2 - \delta}$ (there is no cancellation).
This gives the Herz condition $\delta > \frac{2}{p} - \frac{3}{2}$.
 \item To improve upon the Herz example, we'd like to widen the support of $f$, however
one then runs into the cancellation in the kernel.  To improve upon this example
we have to do something non-radial.
 \item Let $f = e^{i x_1} \psi(x_1, x_2/R)$ for some large $R$ and a bump function
$\psi$.  If there were no cancellation, then $S_t^\delta * f(x)$ would be about
$R |x|^{-1/2 - \delta}$ in size for $|x| \gg R$.
 \item When $|x| \sim R^2$ then the above estimate is sharp for a special value of
$t$ (depending on $x$), thanks to a Taylor series approximation.
\ei

\psfig{figure=kernel.eps,height=5.5in,width=6.5in}
\page

What morals can we draw from this counterexample?

\bi
 \item For sake of discussion let's restrict ourselves to the unit frequency
situation $t \sim 1$.
 \item Bad things happen when $f$ has small support (say $R$) and the piece of the
kernel we are interested in has large support (say $R^2$).
 \item Bad things happen when the support of $f$ is thin (i.e. unit thickness).
 \item Hopefully: one can show that the estimates improve when the support of
$f$ is large and thick. 
\ei

\page

Positive results for $1 < p < 2$

\bi
\item From the known results when $p=1$ or $p=2$ and complex interpolation
we know that one has a.e. convergence when $\delta > \frac{1}{p} - \frac{1}{2}$.  
(Stein, 1956).  First use of complex interpolation!

\item The case $p \leq 1$ was handled by Stein, Taibleson, and Weiss in 1981.
(The best results use Hardy spaces instead of Lebesgue spaces).

\item From the above counterexample one does not have a.e. convergence
when $\delta < \frac{3}{2p} - 1$.

\item Theorem: (T., 1998) One has a.e. convergence when 
$\delta > \max(\frac{3}{4p} - \frac{3}{8}, \frac{7}{6p}-\frac{2}{3})$.

\item We will sketch the proof of this theorem in the following slides.
\ei

\psfig{figure=maxboch.eps,height=5.5in,width=6.5in}
\page

Preliminary reduction

\bi

\item For the given range of $\delta, p$ we have to prove the estimate
$$ \| \sup_{t>0} |S_t^\delta f| \|_{p,\infty} \lesssim \|f\|_p.$$

\item Note that the frequency $t$ is not localized, and neither is
the kernel of $S_t^\delta$.  However, it turns out that one can use
a combination of Littlewood-Paley theory and Calder\'on-Zygmund
theory to localize the frequency and kernel respectively, if one is
willing to lose an epsilion in the $\delta$ index (and that's not
a problem). 

\item By rescaling we can localize the frequency to $t \sim \lambda$
and the kernel to size $1$.  The estimate then becomes (ignoring epsilons)
$$ \| S f\|_{L^p_x L^\infty_t}
\lesssim \lambda^{-\frac{1}{2}+\delta} \|f\|_p$$
where
$$ S f(x,t) = \int e^{2\pi i \lambda t |x-y|} a(x,y,t) f(y)\ dy$$
and $a(x,y,t)$ is any smooth function supported on the region $t, |x-y| \sim 1$, $|x|, |y| 
\lesssim 1$.

\ei
\page

$$ S f(x,t) = \int e^{2\pi i \lambda t |x-y|} a(x,y,t) f(y)\ dy$$
\bi
\item Trivially we have the estimate
$$ \| Sf \|_{L^\infty_x L^\infty_t} \lesssim \|f\|_1.\eqno(1)$$
This estimate is sharp as long as there is no cancellation in $Sf$.
\item Also, we have
$$ \| S f\|_{L^2_x L^\infty_t}
\lesssim \lambda^{-\frac{1}{2}} \|f\|_2\eqno(2)$$
This is obtained by using Sobolev embedding to replace the $L^\infty_t$ norm
with an $L^2_t$ norm, then using some $L^2$ theory.  This estimate is also sharp,
as shown by the Knapp example $f = e^{i \lambda x_1} \psi(x_1, \lambda^{1/2} x_2)$.
\item Interpolating between these two estimates gives Stein's classical result.
\item In this rescaled problem the analogue of the improved Herz example is
$f = e^{i \lambda x_1} \psi(\lambda^{1/2} x_1, \lambda x_2)$.  This example
is sharp for (1), but not sharp for (2) (which is why there is a gap between
the negative results from this example, and the Stein estimates).  In
fact you gain a factor of $\lambda^{-\frac{1}{4}}$ in (2) for this
example.
\ei

\page

{\LARGE How to improve upon the classical estimates?}

$$ \| Sf \|_{L^\infty_x L^\infty_t} \lesssim \|f\|_1.\eqno(1)$$
$$ \| S f\|_{L^2_x L^\infty_t}
\lesssim \lambda^{-\frac{1}{2}} \|f\|_2\eqno(2)$$
\bi
\item By Marcinkeiwicz interpolation we may take $f$ to be a characteristic
function $f = \chi_E$.
\item When $E$ is very small then (1) is superior to (2), but when $E$ is very large
then (2) is superior to (1).  The only time the two are comparable is when $|E| \sim 
\lambda^{-1}$; in any other instance, one can do better than just interpolating (1) 
and (2).
\item Because the kernel of $S$ oscillates with wavelength $\lambda^{-1}$, the
uncertainty principle allows us to assume that $E$ is the union of ``dots''
of diameter $\lambda^{-1}$.  By the previous reduction there are about $\lambda$ of
these dots.
\item The main difficulty is the sup norm $L^\infty_t$ on the left-hand side.
For $p \geq 2$ one can use Sobolev embedding and replace it with an $L^2_t$ norm,
but this method is very inefficient for $p < 2$ (in fact one can show that you can't
do better than Stein's estimates this way!).
\item Since (2) is not sharp for the improved Herz example, the obvious thing to do is
to look for an improvement to (2).  Based on the Knapp and improved Herz examples,
one expects an improvement to (2) whenever the support of $f$ is small.  (For instance,
if $f$ is supported on a $\lambda^{-1}$-ball, then one gets a gain of $\lambda^{-\frac{1}{2}}$
in (2) just from (1) and H\"older's inequality.)  In fact, we have more:
\item Lemma.  If $f$ is supported in a small square $Q$ then 
$$ \| S f\|_{L^2_x L^\infty_t}
\lesssim \lambda^{-\frac{1}{2}} |Q|^{1/4} \|f\|_2.\eqno(2')$$
This is proved by rescaling $Q$ to the unit ball and then applying the argument used
to prove (2).  (The difference is that Sobolev's inequality becomes more favourable).
\item (2') is sharp with respect to all known examples, which is a good sign.
\item This estimate shows that one can improve upon Stein's estimate if $f$ is
concentrated in a small square, or perhaps a small number of small squares.
If $E$ is contained in
a collection of disjoint squares $Q$ such that $|E \cap Q| \gtrsim \alpha |Q|^{1/2}$,
then (2') and the triangle inequality give
$$ \| S \chi_E \|_{L^2_x L^\infty_t}
\lesssim \lambda^{-\frac{1}{2}} \alpha^{-1} |E|.\eqno(3)$$
This is a good estimate if $\alpha$ is large, say $\alpha \gg \lambda^{-1}$.
\item This motivates the following idea: use a Calder\'on-Zygmund decomposition
to break $E$ up into a collection of sets concentrated in small squares $Q$,
for which $|E \cap Q| \gtrsim \alpha |Q|^{1/2}$, plus a sparsely distributed remainder,
for which $|E \cap Q| \lesssim \alpha |Q|^{1/2}$ for all squares $Q$.
\ei

\page

\bi
\item If $\alpha \gg \lambda^{-1}$ then the concentrated part can be dealt with favourably
by (3).  Thus we may assume that $E$ is sparse in the sense that
$$ |E \cap Q| \lesssim \lambda^{-1} |Q|^{1/2}.$$
Heuristically, this means that $E$ is the $\lambda^{-1}$-neighbourhood of a set of
dimension at most 1.
\item This still gives a lot of possibilities as to the shape of $E$: it could be
clustered around a curve, or in a lattice, or something in between.  The two
extreme possibilities are given as follows:

\psfig{figure=possibility.eps,height=3.5in,width=6.5in}

\item If we can find an argument which improves upon the classical estimates on both
of these extreme cases, then we can probably handle the general case.

\ei

\page

The lattice case

\bi 
\item If we can show any cancellation whatsoever, then we can improve upon the $L^1$
estimate (1), and thus improve upon the Stein estimates.

\item Let's only look at the contribution of $E$ on a single $\lambda^{-1/2} \times 1$
rectangle, and look at $S \chi_E$ on a single $\lambda^{-1/2}$ square which lies in
the double of that rectangle. If there is cancellation for the typical example
of this situation, then there is significant cancellation overall.

\psfig{figure=slice.eps,height=2in,width=6in}

\item The significance of this example is that the phase has been linearized:
$\lambda t|x-y|$ essentially becomes $\lambda t (x_1 - y_1)$.  The operator is now
basically a Fourier transform evaluated at $\lambda t$.

\item However, Plancherel's theorem states that the number of frequencies in which no cancellation
occurs in the Fourier transform is proportional to the support of the function:
$$ | \{ \xi: |\hat \chi_A(\xi)| \sim |A| \} | \lesssim |A|^{-1}.$$
 So, if we wish to avoid cancellation, 
$t$ must be essentially restricted to a set of measure $\lambda^{-1/2}$.

\item If $t$ was constant then we're just studying a single Bochner-Riesz operator, and
one gets a huge amount of cancellation from the standard estimates (especially
the Carleson-Sj\"olin $L^4$ estimate), partly because $E$ is so sparse.  The cancellation 
is so great that there is still some left if $t$ is only restricted to a set of measure
$\lambda^{-1/2}$ as above.  Thus we do get some cancellation, and thus some gain over
the Stein estimates.
\ei

\page

The thin case
\bi

\item Now suppose the set $E$ is concentrated as much as possible while still obeying
the sparseness condition
$$ |E \cap Q| \lesssim \lambda^{-1} |Q|^{1/2}.$$
A good model of this occurs when $E$ is clustered near a one-dimensional set.

\item Now one restricts $E$ to a $\lambda^{-1/2}$ square $Q$ and investigates the possibility
of cancellation.  Let's say the square is centered at the origin, and contains a fair
amount of $E$ (say about $\lambda^{1/2}$ dots of diameter $\lambda^{-1/2}$).  In this case
the phase can be approximated by
$$ \lambda t |x-y| = \lambda t|x| - \langle y, \lambda t\frac{x}{|x|} \rangle + O(1)$$
Discarding the unimportant $\lambda t|x|$ term, the operator thus becomes
a Fourier transform evaluated at $\lambda t \frac{x}{|x|}$.

\psfig{figure=square.eps,height=5.5in,width=6.5in}

\item By invoking Plancherel's theorem as before,
we conclude that $t$ is again restricted to a set of measure $\lambda^{-1/2}$
for ``most'' directions $x/|x|$.  

\item But we have several squares $Q$ with this property.  An argument of C\'ordoba can then
be used to show that $t$ is restricted to sets of measure $\lambda^{-1/2}$ for most $x$.
(This is related to the fact that Besicovitch sets have full dimension in the plane).

\item Now one can use the standard $L^4$ theory as before.

\item Intermediate cases, in which $E$ is neither totally sparse or totally one-dimensional,
can be handled by a mix of the above two arguments.  
\ei

\end{document}
