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\def \sgn{{\rm sgn}}
\def \proof {{\it Proof.  }}

\rm
\centerline{\bigheading On the almost everywhere convergence}
\centerline{\bigheading of wavelet summation methods}
\footnote{}{\small 1991 {\it Mathematics Subject Classification.}  Primary
42C15; Secondary 40A30.}
\footnote{}{\small Keywords: hard sampling, soft sampling,
almost-everywhere convergence.}
\line{}
\centerline{Terence Tao}
\vglue 1 cm
\centerline{\it Abstract}
{
\leftskip = 80 pt
\rightskip = 80 pt
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\small
Let $\psi$ be a rapidly decreasing one-dimensional wavelet.  We show
that the wavelet expansion of any $L^p$ function converges pointwise 
almost everywhere under the wavelet projection, hard sampling, and soft
sampling summation methods, for $1 &lt; p &lt; \infty$.  In fact, the partial
sums are uniformly dominated by the Hardy-Littlewood maximal function.
}
\baselineskip = 18 pt

{\heading 0.  Introduction}

In this note we will discuss the almost everywhere convergence of various
summation methods for wavelet expansions.  We shall not aim for maximal
generality; instead, we will restrict our attention to the expansion of an 
$L^p$ function
$f$ in one dimension, for $1 &lt; p &lt; \infty$, generated by a $0$-regular 
orthogonal wavelet $\psi$.  By this we mean
that $\psi$ has integral $0$, is bounded and rapidly decreasing, and 
that the set of functions $\{\psi_{j,k}\} = \{2^{j/2} 
\psi(2^j\cdot - k)\}$ for $j,k \in \Z$, form an orthonormal 
basis for $L^2(\R)$.  The wavelet expansion of $f$ is given by
$f \sim \sum_{j,k} a_{j,k} \psi_{j,k}$, where
$a_{j,k} = \langle f, \psi_{j,k}\rangle$.  In [4] it is shown 
that this series converges unconditionally in $L^p$~norm to $f$; however, the 
pointwise behaviour of this series is more dependent on the summation method
used and is less well understood.  The summation methods that are 
used in applications almost always rely on one of the following three partial
summation operators:
$$
\leqalignno{
P_J f &amp;= \sum_{j &lt; J;k} a_{j,k} \psi_{j,k};
&amp;\hbox{the {\it wavelet projections}}\cr
T_\lambda f &amp;= \sum_{|a_{j,k}| &gt; \lambda} a_{j,k} \psi_{j,k};
&amp;\hbox{the {\it hard sampling} operators}\cr
\tilde T_\lambda f &amp;= \sum_{|a_{j,k}| &gt; \lambda}
\sgn(a_{j,k}) \bigl(|a_{j,k}| - \lambda\bigr) \psi_{j,k}.
&amp;\hbox{the {\it soft sampling} operators}\cr}
$$
Note that the sum in $P_J f$ converges absolutely since
$\psi$ is rapidly decreasing and $a_{j,k} = O(2^{-j/2} 2^{j/p} \|f\|_p)$,
by H\"older's inequality.  We shall prove the absolute
convergence of the other two sums in the next section (though 
for $p \geq 2$ one always has a finite number of terms).

\proclaim Theorem 1.  If $f \in L^p(\R)$ for some $1 &lt; p &lt; \infty$, then
for almost every $x$ one has
$$\lim_{J\to \infty} P_J f(x) = \lim_{\lambda \to 0} T_\lambda
f(x) = \lim_{\lambda \to 0} \tilde T_\lambda f(x) = f(x).$$

The almost everywhere convergence of the wavelet projection operators
was already shown (under more general hypotheses) in [3].  
The non-linear hard and soft 
sampling methods are used frequently in applications
(see [1],[2]) but this seems to be the first pointwise convergence result
for these methods.  In contrast, we have learned recently that 
T.W. K\"orner has found a counterexample to the pointwise convergence of 
the hard sampling operator for the Fourier series expansion.
We also note that if one assumes some minimal regularity 
(e.g. right-continuity) for $\psi$, then one can get the pointwise convergence 
of the operators of Theorem 1 on the entire Lebesgue set of $f$ (c.f. [3]).

Finally, we show that there are (somewhat artificial) wavelet summation 
methods which need not converge pointwise, at least for wavelets with
little regularity.

\proclaim Theorem 2.  Suppose
that $\psi$ is the Haar wavelet $\chi_{[0,1/2)} - \chi_{[1/2,1)}$.
Fix $\alpha &gt; 0$, $1 &lt; p &lt; \infty$ and consider the operator 
$T^\alpha_\lambda$ defined on $L^p(\R)$ for $\lambda&gt;0$ by
$$
T^\alpha_\lambda f = \sum_{|a_{j,k}| \geq \alpha^j \lambda}
a_{j,k} \psi_{j,k} $$
\item{(a)} If $\alpha \not = 2^{-1/2}$, then for all $f \in L^p$, 
one has $\lim_{\lambda \to 0} T^\alpha_\lambda f(x) = f(x)$ at 
every Lebesgue point $x$ of $f$.
\item{(b)} If $\alpha = 2^{-1/2}$, then there exists an $f \in L^p$ for
which $\lim \sup_{\lambda \to 0} |T^\alpha_\lambda f(x)| =
\infty$ for all $x \in \R$.

The series defining $T^\alpha_\lambda f$ need not converge
absolutely; if this is the case, we shall define the value of the series
to be the pointwise limit (where it exists) of its formal wavelet 
projections, so
$$T^\alpha_\lambda f(x) = \lim_{J \to \infty} \sum_{\eqalign{
|a_{j,k}| &amp;\geq \alpha^j \lambda\cr
j&amp;&lt; J\cr}}
a_{j,k} \psi_{j,k}(x).$$
The operator $T^\alpha_\lambda$ can be thought of as a variant of the
hard thresholding operator $T_\lambda$, but with a weighted threshold
that depends on the resolution level.  The reason that convergence fails
when $\alpha = 2^{-1/2}$ can be sketched as follows.
For this value of $\alpha$ one is
effectively applying a threshold to the ``heights'' of the summands
$a_{j,k} \psi_{j,k}$ rather than to the co-efficients $a_{j,k}$.  By
``fine-tuning'' the magnitudes of terms with comparable heights, one can
effectively sum such terms in an arbitrary order, and (with a suitable
choice of ordering) this allows one to get large non-cancellatory
contributions to the partial sums $T_\lambda f(x)$ for each $x$ (the
exact values of $\lambda$ for which non-cancellation occurs depends on $x$).
The ability to get significant non-cancellation seems to be due to the
lack of regularity (both smoothness and moment conditions) of the Haar
wavelet; it may be possible that better convergence results can be obtained
for more regular wavelets.

The work in this paper originated from a question posed by Elias Stein.
The author also thanks 
Anna Gilbert for her proofreading, the loan of several books and
articles, and for many valuable discussions.

{\heading 1. Proof of Theorem 1.}

In the following the symbol $C$ will denote various constants that
depend only on $\psi$.  In order to make the notion of a Lebesgue point
well defined, we shall not adopt the practice of identifying 
functions which agree almost everywhere.  Recall that a point $x \in \R$
is a Lebesgue point of $f$ if ${1 \over 2r} \int_{-r}^{r} |f(x+t) -
f(x)|\ dt$ converges to $0$ as $r$ tends to $0$; it is a classical
theorem of Lebesgue that, for any locally integrable function $f$,
almost every element of $\R$ is a Lebesgue point of $f$.  Closely
related to the concept of Lebesgue point is the Hardy-Littlewood maximal
operator $M$; this sublinear operator is defined by
$$Mf(x) = \sup_{r &gt; 0} {1 \over 2r} \int_{-r}^{r} |f(x+t)|\ dt.$$
Many expressions, especially ones involving approximations to the
identity, can be controlled in magnitude by the maximal function.
In particular, if $\phi(x)$ is a
function such that $|\phi(x)| \leq C_N A (1 + {|x-x_0| \over r})^{-N}$ for
all $N &gt; 0$ and some fixed $A$,$r$,$x_0$, then 
$$|\int_\R f(x) \phi(x)\ dx| \leq C A r Mf(x_0).$$
Another such estimate is the simple inequality $|f(x)| \leq Mf(x) \leq
\|f\|_\infty$,
which holds whenever $x$ is a Lebesgue point of $f$.  Proofs of these
statements can be found in [5].

Let $f$ be an arbitrary function in $L^p(\R)$.
We first show that to prove Theorem 1, it suffices to show the 
pointwise estimates
$$ \sup_J |P_J f(x)|,\ \sup_\lambda |T_\lambda f(x)|,\
\sup_\lambda |\tilde T_\lambda f(x)| \leq C Mf(x)
\eqno(1)$$
for almost every $x$.
If our operators were linear, the implication would follow from standard
arguments (see [5]), but only 
the $P_J$ are linear and so the argument requires a little care.  We
shall show the argument for the $\tilde T_\lambda$ only, as it is the
most non-linear of the three; the proof for
the other two operators is analogous (and can be simplified slightly).
Assume now 
that (1) holds for some fixed $f$ and $x$.  Since we only require convergence
almost everywhere, we may assume that $x$ is a Lebesgue point of $f$ and
of every $\psi_{j,k}$.  Choose any $\varepsilon &gt; 0$.  By the usual
method of convolving $f$ with a smooth approximation to the identity,
and using the fact that $x$ is a Lebesgue point of $f$, and that $f$ is
in an $L^p$ class for some finite $p$, one can find a
$g$ in $C^\infty_0(\R)$ for which $|M(f-g)(x)| \leq C\varepsilon$.
In turn, one can approximate
$g$ uniformly almost everywhere
by a finite linear combination $h$ of wavelets (for this
fact, see [4]).  Thus one can find
such an $h$ for which $|f(x)-h(x)| \leq |M(f-h)(x)| \leq C\varepsilon$.  To
conclude the argument, one observes that
$\tilde T_\lambda f(x) = h(x) + \tilde T_\lambda (f-h)(x) + O_h(\lambda)$
as $\lambda \to 0$.  To see this, note that the wavelet co-efficients of
$\tilde T_\lambda f$ and $h + \tilde T_\lambda(f-h)$ 
differ by $O(\lambda)$ when the wavelet occurs in the expansion of $h$, and
are identical otherwise.  Thus $|\tilde T_\lambda f(x) - f(x)| = 
O(\varepsilon)$ for sufficiently small $\lambda$, so $\lim_{\lambda \to
0} \tilde T_\lambda f(x) = f(x)$, as desired.

We now show that (1) holds for the $P_J$.
We note that this estimate was first proven in [3] under weaker decay
 hypotheses on $\psi$ and $f$; we shall give an alternative proof below.
The argument relies on two observations.  The first is the crude estimate
$$ |\psi_{j,k}(x) \overline{\psi_{j,k}(y)}| \leq C_N 2^{-j} (1 + 2^j|x-y|)^{-N}
(1 + |2^j x - k|)^{-N} \eqno(2)$$
for all $N &gt; 0$; this is a simple consequence of the rapid decrease of $\psi$
and the easy inequality $1 + 2^j|x-y| \leq (1+|2^j x-k|)(1+|2^j y -
k|)$.
The second is the fact that
$$\sum_{j,k} \psi_{j,k}(x) \overline{\psi_{j,k}(y)} = 0$$
for almost every $x,y \in \R$.
Note that this sum converges absolutely for $x
\not = y$ by (2).  The identity follows from the fact that 
the $\{\psi_{j,k}\}_{j,k \in \Z}$ form an orthonormal basis for $L^2(\R)$, so 
that the above summation converges in a distributional sense
to $\delta(x-y)$.  Since the series converges both
distributionally and absolutely almost everywhere, the two limits must 
agree almost everywhere, hence the claim.

As a consequence of these two observations, we see that the kernel
defined almost everywhere by
$$K(x,y) = \sum_{j&lt;0;k} \psi_{j,k}(x) \overline{\psi_{j,k}(y)}
= - \sum_{j\geq 0;k} \psi_{j,k}(x) \overline{\psi_{j,k}(y)}$$
converges absolutely and is bounded and rapidly decreasing away from 
the diagonal.  However, we have
$$\eqalign{P_J f(x) &amp;= \sum_{j &lt; J; k} \psi_{j,k}(x) \int_\R 
\overline{\psi_{j,k}(y)}
f(y)\ dy\cr
&amp;= \int_\R 
\sum_{j &lt; J;k} \psi_{j,k}(x)\overline{\psi_{j,k}(y)} f(y)\ dy\cr
&amp;= 
\int_\R 2^J K(2^{-J}x,2^{-J}y) f(y)\ dy\cr}$$
from Fubini's theorem; the conditions for Fubini's theorem follow from
(2), the fact that $f$ is in an $L^p$ class for some $p&gt;1$, and H\"older's 
inequality.  
Since $K(2^{-J}x,2^{-J}y)$, as a function of $y$, is rapidly decreasing
away from $x$, the desired estimate (1) now follows from the maximal
function estimates mentioned at the start of this section.

We now show that (1) holds for the $T_\lambda$.
Fix $f \in L^p(\R)$, $\lambda \in \R$, and $x
\in \R$.
Since we may assume that $Mf(x)$ is nonzero, we can choose a $J$ 
for which $\lambda \sim 2^{-J/2} Mf(x)$; one can think of $J$ as the
finest scale for which significant terms in the summation for $T_\lambda
f(x)$ can appear.  We will show that
$|T_\lambda f(x) - P_J f(x)| \leq C Mf(x),$
which will imply (1) from our results for the $P_J$.  We shall first
need some simple estimates.
For each $j,k \in \Z$, let $D(j,k)$ be the integer part of $2 + |2^j x
- k|$.  Note that for each $j$ and for each integer $d \geq 2$
there are at most two $k$ for which $D(j,k) = d$.  From the rapid
decrease of $\psi$ and the usual estimates on approximations to the
identity, we have that, for all $N &gt; 0$,
$$\eqalign{
|\psi_{j,k}(x)| &amp;\leq C_N 2^{j/2} D(j,k)^{-N}\hbox{\ and}\cr
|a_{j,k}| &amp;\leq C 2^{-j/2} D(j,k) Mf(x);\hbox{\ thus}\cr
|a_{j,k}\psi_{j,k}(x)| &amp;\leq C_N D(j,k)^{-N} Mf(x).\cr}$$
Note that the second estimate implies
that $|a_{j,k}|$ is greater than $\lambda$ only if $j &lt; J + C \log D(j,k)$.

The estimate for $|T_\lambda f(x) - P_J f(x)|$ is now straightforward. 
 Indeed, we have
$$\eqalign{
|T_\lambda f(x) - P_J f(x)| &amp;=
\bigl|\sum_{|a_{j,k}| &gt; \lambda} a_{j,k} \psi_{j,k}(x) - \sum_{j&lt;J;k}
a_{j,k} \psi_{j,k}(x)\bigr|\cr 
&amp;\leq
\bigl|\sum_{|a_{j,k}| \leq \lambda; j&lt;J} a_{j,k} \psi_{j,k}(x)\bigr|
+ \bigl| \sum_{|a_{j,k}| &gt; \lambda; J \leq j} a_{j,k} \psi_{j,k}(x)\bigr|\cr
&amp;\leq \sum_{j &lt; J;k} \lambda |\psi_{j,k}(x)| + \sum_{J \leq j &lt; J+C \log D(j,k)
} |a_{j,k}\psi_{j,k}(x)|\cr
&amp;\leq \sum_{d=2}^\infty \sum_{j &lt; J} C_N \lambda 2^{j/2} d^{-N} + 
\sum_{d=2}^\infty \sum_{J \leq j &lt; J + C \log d} C_N d^{-N} Mf(x)\cr
&amp;\leq C \lambda 2^{J/2} + \sum_{d=2}^\infty C Mf(x) d^{-N} \log d\cr
&amp;\leq C Mf(x) + C Mf(x),\cr}$$
as desired.  Note that this argument also shows that
the sum defining $T_\lambda f(x)$ is absolutely convergent, since we
know the sum for $P_J f(x)$ is.  

The proof of (1) for $\tilde T_\lambda$
is almost identical to that for $T_\lambda$, although one must use the estimate
$$\sum_{|a_{j,k}| &gt; \lambda; j&lt;J} \sgn(a_{j,k}) (|a_{j,k}| - \lambda)
\psi_{j,k}(x) = \sum_{j&lt;J;k} a_{j,k} \psi_{j,k}(x) + O\bigl(\sum_{j &lt; J;k}
\lambda |\psi_{j,k}(x)|\bigr)$$
in the above argument; we omit the details.  Thus (1) holds for all
three operators under consideration, and the theorem is proved.
\hfill$\square$

We remark that these results can be extended easily to higher dimensions,
to bi-orthogonal wavelets, and to decompositions involving more than one 
``mother wavelet''. The decay conditions 
on $\psi$ can also be relaxed; for instance, one can get the pointwise 
convergence result for the $P_J$ alone if one merely assumes that $\psi$ has 
an integrable radial
majorant.  The proofs of these assertions are simple applications of the
ideas developed in [3].

{\heading 2. Proof of Theorem 2.}

The proof of (a) is an easy generalization of the corresponding argument
for the $T_\lambda$ in the proof of Theorem 1, although if $\alpha &lt; 2^{-1/2}$ 
one has to 
use $I-P_J$ instead of $P_J$ as the approximating operator.  (Here $J$
is chosen so that $\alpha^J \lambda \sim 2^{-J/2} Mf(x)$).
We omit the details.  Instead of using the argument in the previous
section to bound $P_J$, one can compute $P_J f$ explicitly to achieve
estimate (1) for all $x \in \R$, which will ensure pointwise convergence
at every Lebesgue point, and not just almost everywhere.
Alternatively, one can use the methods in [3].

Now suppose that $\alpha = 2^{-1/2}$.  
We shall construct, for each integer $N \geq 1$, a
function $f_N = \sum_{j,k} a^N_{j,k} \psi_{j,k}$ on $[0,1)$ for
which:
\item{(i)} All but a finite number of the $a^N_{j,k}$ are $0$ and those
that are not zero satisfy $1 &lt; 2^{j/2} a^N_{j,k} &lt; 2$;
\item{(ii)}  $\|f_N\|_p \leq C_p N^{1/2}$ for every
$1 &lt; p &lt; \infty$;
\item{(iii)} $\sup_\lambda |T^\alpha_\lambda f_N(x)| &gt; N/3$
for every $x \in [0,1)$.

Once we do this, one can see that, if one chooses an increasing sequence
of natural numbers $M_i$ whose successive differences are sufficiently
large, the function
$$ F(x) = \sum_{i=1}^\infty 2^{-i}
\sum_{k=0}^{2^{M_i}-1} f_{3^i}(2^{M_i}x - k)$$
will be in $L^p([0,1))$ for every 
$1 &lt; p &lt; \infty$, but is such that
$\lim \sup_\lambda |T^\alpha_\lambda F(x)| = 
\infty$ for every $x \in [0,1)$.  The function $f(x) = \sum_{n=-\infty}^\infty
2^{-|n|} F(x-n)$ thus satisfies all the required conditions in (b).

We shall define $f_N$ explicitly:
$$f_N(x) = \sum_{j=0}^{N-1} \sum_{k=0}^{2^j-1} \bigl(1 + 2^{-j}(k+{1
\over 2})\bigr)
\psi(2^j x - k).$$
The reason for this particular choice of co-efficients is that
each $T^\alpha_\lambda f_N$ will consist only of the terms $a^N_{j,k}
\psi_{j,k}$ corresponding to wavelets whose ``centers'' are to the right
of $\lambda - 1$.  At $x = \lambda - 1$, the nature of
the Haar wavelet will cause this sum to be large and positive.

We now show that $f_N$ has the properties (i-iii) stated above.
The verification of (i) is immediate, and (ii) follows quickly from
Khinchin's inequality (see [4]).
To prove (iii), fix an $x \in [0,1)$ and observe that
all but $N$ of the factors $\psi(2^j x - k)$ in the sum defining $f_N(x)$ are 
zero, and the remainder are either $+1$ or $-1$.  Furthermore,
this factor is nonnegative if $1 + 2^{-j}(k+{1 \over 2}) 
&gt; 1 + x$ and nonpositive otherwise.  Thus, by inspection of the
definition of $T^\alpha_\lambda$ for $\lambda = 1+x$ and $\lambda = 2$,
we have
$$\eqalign{
T^\alpha_{1+x} f(x) &amp;= \sum_{\psi(2^j x - k) = +1} (1 +
2^{-j}(k+{1 \over 2})),\hbox{\ and}\cr
T^\alpha_{1+x} f(x) - T^\alpha_2 f(x) &amp;= \sum_{\psi(2^j x -
k) = -1} (1 + 2^{-j}(k+{1 \over 2}))\hbox{; so}\cr
2T^\alpha_{1+x} f(x) - T^\alpha_2 f(x) &amp;= \sum_{\psi(2^j x -
k) \not = 0} (1 + 2^{-j}(k+{1 \over 2})) &gt; N,\cr}$$
which implies (iii).  This proves the theorem.
\hfill$\square$

{\heading References}

[1] A. Averbuch, G. Beylkin, R. Coifman, M. Israeli, {\it Fast Adaptive
Algorithm for the Elliptic equations with periodic boundary conditions},
preprint.

[2] D.L. Donoho, I.M. Johnstone, {\it Ideal spatial adaptation by wavelet
shrinkage}, Biometrika {\bf 81} (1994), 425-455.

[3] S. Kelly, M.A. Kon, and L.A. Raphael, {\it Local convergence for
wavelet expansions}, J. Funct. Anal. {\bf 126} (1994), 102-138.

[4] Y. Meyer, ``Ondelettes'', Hermann, Paris, 1990.

[5] E. Stein, ``Singular Integrals and Differentiability
Properties of Functions'', Princeton University Press,
1970.

\parskip = 0 pt
\rightline{Department of Mathematics}
\rightline{Princeton University}
\rightline{Princeton, NJ 08544}
\rightline{\it tao@math.princeton.edu}
\bye 
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