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\begin{document}
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{\bf \LARGE  Some research in analysis at UCLA }
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John Garnett, Christoph Thiele, Terence Tao\\
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{
Abstract: We give a brief taste of some active research in harmonic and 
complex analysis here at UCLA.
}
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\page

\noindent
Bounded analytic functions (John Garnett)

\bi
\item We all know Louiville's theorem: if $f$ is bounded and analytic
on all of $\C$, then $f$ must be constant.

\item You can weaken Louiville's theorem a bit.  For
instance, if $f$ is bounded and analytic on $\C \backslash \{0\}$, then
$f$ must still be constant (since the singularity at 0 must be removable).

\item In fact, there are several sets $\Gamma$ with the property that
bounded analytic functions on $\C \backslash \Gamma$ are automatically
bounded.  Such sets are said to have \emph{zero analytic capacity}.

\item Generally, the smaller the set $\Gamma$, the more likely it is to
have zero analytic capacity.  Knowing which sets have zero analytic capacity
tells you to the smallest sets for which you can cram all the unbounded
or non-analytic behaviour of a non-constant complex function.

\item If $\Gamma$ has Hausdorff dimension less than 1, then $\Gamma$
has zero analytic capacity.  When $\Gamma$ has Hausdorff dimension greater
than 1, $\Gamma$ has non-zero analytic capacity.  (Major tool: the Cauchy
integral formula).

\item (All subsets of $\R^2$ have a Hausdorff dimension, which is a number 
somewhere between 0 and 2.  Roughly speaking, this dimension measures
how efficiently one can cover the set by small balls.  The standard
middle-thirds Cantor set, for instance, has Hausdorff dimension 
$\log 2/\log 3$).

\item The intermediate case, when $\Gamma$ has Hausdorff dimension 1,
is not completely settled.  If $\Gamma$ is on the real axis, then it has
zero analytic capacity if and only if it has (1-dimensional) measure zero.
However, there exist sets with positive 1-dimensional measure which still have
zero analytic capacity, such as the \emph{Garnett set} (the product
of two middle-halves Cantor sets).

\item The finer resolution of this problem is ongoing research by John Garnett
and his students.

\ei

\page

Multilinear operators (Christoph Thiele, Terry Tao)

\bi

\item  Traditionally, real-variable harmonic analysis has concerned
itself with \emph{linear} operators $T$ and how to control them in various
norms.  In other words, we ask questions like ``If we have some sort
of control on $f$, what does that tell us about $Tf$?''

\item More recently, though, harmonic analysis has begun to move
away from the linear framework, towards genuinely non-linear problems.
As an intermediate step we consider multi-linear estimates.

\ei

\page

Example: solving an ODE

\bi

\item Suppose you wanted to solve an ODE such as
$$ \dot u_\eps(t) = \eps V(t) u_\eps^2(t) + W(t), \quad u(0) = 0,$$
where $\eps$ is a small number, and $V$, $W$ are given functions.  We 
could try to solve this by the method of power series, writing
$$ u_\eps(t) = u_0(t) + \eps u_1(t) + \eps^2 u_2(t) + \ldots$$

\item Expanding both sides and comparing co-efficients, we can solve for
$u_0, u_1, \ldots$:
$$ u_0(t) = \int_0^t W(s)\ ds$$
$$ u_1(t) = \int_0^t V(s) u_0(s)^2\ ds$$
$$ u_2(t) = 2\int_0^t V(s) u_0(s) u_1(s)\ ds$$
etc.  If you expand things out fully, you see that $u_0$ is a linear
function of $W$, $u_1$ is a trilinear function of $V$ and $W$,
$u_2$ is a quintilinear function of $V$ and $W$, and so forth.

\ei

\page

Example: Paraproducts

\bi

\item The simplest example of a bilinear estimate is
\emph{H\"older's inequality}:
$$ \| uv \|_{L^p(\R)} \leq \| u\|_{L^q(\R)} \| v\|_{L^r(\R)}
\hbox{ when } \frac{1}{p} = \frac{1}{q} + \frac{1}{r}.$$

\item We can split the pointwise product $uv$ as
$$ u(t)v(t) = \int_{-\infty}^t u(s)v'(s)\ ds + 
\int_{-\infty}^t u'(s) v(s)\ ds.$$
(Assume some vanishing at $-\infty$).

\item Each individual integral $\int_{-\infty}^t u(s) v'(s)\ ds$ is
a bilinear function of $u$ and $v$, and is an example of a
\emph{paraproduct}.  It occurs often in the study of PDE.

\item Rough theorem: Paraproducts satisfy H\"older's inequality too,
except for some endpoints.

\ei

\page

Frequency decomposition

\bi

\item The way we deal with operators like paraproducts is via frequency
decomposition: breaking up a function $u$ into low-frequency and high-frequency
components $u = u_{low} + u_{high}$:

\psfig{figure=decomp.eps,height=3in,width=6.6in}

\item Formally, this is done using the Fourier transform.  There are some
standard results about these types of decompositions; these results
are known collectively as Littlewood-Paley theory.

\item We can then split up a paraproduct into several pieces, e.g.
$$ \int_{-\infty}^t u_{high}(s)v'_{low}(s)\ ds $$
$$ \int_{-\infty}^t u_{low}(s)v'_{high}(s)\ ds. $$
The idea is that the first term is very small because $v_{low}$ is
almost constant, and the second term is almost $u_{low} v_{high}$,
because $u_{low}$ is almost constant.  In either case one can simplify
the expression to something more tractable.

\item Christoph Thiele and I study much more singular operators than
paraproducts, including the \emph{Bilinear Hilbert transform}
$$ B(f,g)(x) = p.v. \int f(x+t) g(x-t) \frac{dt}{t}.$$
This operator can also be analyzed by frequency decomposition, but on
a level more refined than the low-high analysis above.  One also needs
some spatial decomposition as well.  The enemy is then the Uncertainty
principle, which says that you can localize frequency to a range of width
$d\xi$ and space to a range of width $dx$ only when $d\xi dx \geq 1$.

\ei

\page

Oscillatory integrals, Kakeya problems, PDE (Terry Tao)

\bi

\item Solutions to PDE such as the free Schr\"odinger equation
$$ i u_t(t,x) + \Delta u(t,x) = 0; \quad u(0,x) = f(x)$$
are given by \emph{oscillatory integrals}, in this case 
$$ u(t,x) = \frac{1}{(4\pi t)^{n/2}} \int e^{i |y-x|^2/4t} f(y)\ dy.$$

\item The phase $e^{i |y-x|^2/4t}$ should lead to some cancellation in
this integral; it should usually be much smaller than, say, $\int |f(y)|\ dy$.
But how to quantify this?

\item One answer is via $L^p$ norms.  Typically one has something like $L^2$
control on $f$, and one is interested in $L^p$ spacetime control on $u$:
$$ (\int\int |u(t,x)|^p\ dx dt)^{1/p} \leq C \|f\|_2.$$
This particular estimate is known to be true exactly when $p=4$.

\item Such estimates are very handy in the perturbation theory of the
Schr\"odinger equation, and for instance can be used to obtain a rigorous
theory for the non-linear Schr\"odinger equation
$$ i u_t(t,x) + \Delta u(t,x) = |u(t,x)|^2 u(t,x).$$
In two spatial dimensions, it's known that this solution has global $L^2$ 
solutions if the initial data starts out in $L^2$ with small norm.  (The 
large norm problem is a big open problem).  Basically one uses the
method of power series mentioned earlier.

\item These are \emph{linear} estimates, because $u$ depends linearly on
$f$.  But ultimately we want to study bilinear and multi-linear estimates.
In the last ten years this has been an active area of research (and
for many equations besides Schr\"odinger).

\item The Schrodinger equation has some special ``travelling wave'' solutions, in which $u$ is concentrated on a ``tube'' in space time, something of
the form
$$ \{ (t,x): |t| \leq R^2; |x - vt - x_0| \leq R\}.$$
By the principle of superposition, one can then construct other solutions
which are supported on unions of tubes.

\item One approach to understand solutions of PDE is to first understand
the purely geometric problem of overlapping tubes.  

\psfig{figure=sticky.eps,height=4in,width=4in}

\item The very difficult \emph{Kakeya problem} asks to what extent
tubes can overlap.  It states: if $0 < \delta \ll 1$, and you have
a collection of $\delta \times 1$ tubes in $\R^3$ whose directions are all
$\delta$-separated, what is the maximum median overlap of these tubes?
The conjecture is $O(\delta^{-\eps})$ for all $\eps > 0$.  The best result
known to date is $\eps > \frac{1}{2} - 10^{-10}$.

\ei

\end{document}
