— 1. The Hardy-Littlewood maximal inequality —
We work in Euclidean space
with Lebesgue measure; we write
instead of
for the Lebesgue measure of a set
. For any
and
let
denote the open ball of radius
centred at
. Thus for instance
. For any
, we use
to denote the dilate of
around its centre by
.
For any
, we define the averaging operators
on
for any locally integrable
by

It is not hard to see that these averages

are well-defined, and are even continuous functions, for locally integrable

.
One can view
as an averaging operator

From
Schur’s test or
Young’s inequality (or
Minkowski’s inequality) we know that these

are contractions on every

,

:

Thus the averages

are uniformly bounded in
size as

varies. The fundamental
Hardy-Littlewood maximal inequality asserts, roughly speaking, that they are also uniformly bounded in
shape:
Proposition 1 (Hardy-Littlewood maximal inequality) We have the the strong-type
inequality 
for all
and any
, and also the weak-type
inequality 
for any
.
The sublinear operator

is known as the
Hardy-Littlewood maximal operator. It is easy to see that the above proposition is equivalent to the assertion that the Hardy-Littlewood maximal operator is weak-type

and strong-type

for all

. Note that it is not strong-type

; indeed, if

is any non-trivial function, then we easily verify the pointwise bound

, which ensures that

is not in

. (Here we use the “Japanese bracket” notation

, as a smoothed out version of the absolute value function

.)
Exercise 2 Explain why
is not of strong or weak type
for any
.
As this proposition is so fundamental we shall give several proofs of it.
— 2. First proof —
We begin with the classical proof, starting with some standard qualitative reductions. Firstly we may easily reduce to
being non-negative. A monotone convergence argument also lets us restrict to functions
which are bounded and have compact support. (At this point one may object that
might not be measurable, but one can use dominated convergence to restrict
to be (say) rational, at which point measurability is clear.) Also,
is continuous in
, so we may restrict
to a countable dense set (such as the positive rationals); another monotone convergence argument then lets us restrict
to a finite set. Of course, our bounds need to be uniform in this set, as well as being uniform in the boundedness and support of
.
It is obvious that
is bounded on
(indeed, it is a contraction on this space). So it suffices by Marcinkiewicz interpolation to prove the weak-type
inequality; by homogeneity (and the preceding reductions) it thus suffices to show that

for any non-negative bounded compactly supported

, where we are implicitly restricting

to a finite set.
Let us denote the set on the left-hand side by
; our hypotheses on
and
easily ensure that
is a compact set (alternatively, one can work with the uncountable sup
, but instead replace
by an arbitrary compact subset
of itself, and then take suprema in
at the end, noting that Lebesgue measure is a
-finite Radon measure). By construction, we thus see that for any
there exists a radius
such that
is locally large compared to the ball
:

On the other hand, what we want to show is that

is globally large compared to

:

Since the compact set

is covered by the balls

, and hence by finitely many of these balls, things look quite promising. However, there is one remaining issue, which is that these balls could overlap quite heavily, preventing us from summing
(1) to get
(2). Fortunately there is a very simple algorithm which extracts out from any collection of overlapping balls, a collection of non-overlapping balls which manages to capture a significant fraction of the original collection in measure:
Lemma 3 (Wiener’s Vitali-type covering lemma) Let
be a finite collection of balls. Then there is a subcollection
of disjoint balls such that 
Proof: We can order
in decreasing order of size. Now we select the disjoint balls
by the greedy algorithm, picking the largest balls we can at each stage. Namely, for
we choose
to be the first ball which is disjoint from all previously selected balls
(thus for instance
must equal
), until we run out of balls. Clearly this gives us a family of disjoint balls. Now observe from construction that each ball
in the original collection is either a ball
in the subcollection, or else intersects a ball
in the subcollection of equal or larger radius. In either case we see from the triangle inequality that
is contained in
. In other words,

and so

and the claim follows.
From the covering lemma it is easy to conclude (2). Indeed, since
is covered by finitely many of the balls
, the covering lemma gives us finitely many disjoint balls
,
such that

and then on summing
(1) we get
(2) (with an explicit constant of

).
Remark 4 Under mild assumptions one can generalise the covering lemma to infinite families of balls without difficulty. One can also replace balls by similar objects, such as cubes; the main property that one needs is that if two such objects overlap, then the smaller one is contained in some dilate of the larger. This is a fairly general property, and for instance holds for metric balls on a measure space with some doubling property
, but it fails for very thin or eccentric sets such as long tubes, rectangles, annuli, etc. Indeed, understanding the maximal operator for these more geometrically complicated objects is still a major challenge in harmonic analysis, leading to important conjectures such as the Kakeya conjecture, which remains open in higher dimensions despite recent progress.
Exercise 5 (Baby Besicovitch covering lemma) Let
be a collection of intervals on the real line. Show that there exist a subcollection
such that
, and such that every point
belongs to at most
of the intervals
. What is the best explicit bound for
you can get?
Exercise 6 Show that if
is supported on
, then ![\displaystyle \| Mf \|_{L^1([0,1])} \lesssim \|f\|_{L \log L([0,1])}.](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5C%7C+Mf+%5C%7C_%7BL%5E1%28%5B0%2C1%5D%29%7D+%5Clesssim+%5C%7Cf%5C%7C_%7BL+%5Clog+L%28%5B0%2C1%5D%29%7D.&bg=ffffff&fg=000000&s=0&c=20201002)
Exercise 7 For any locally integrable
, let
denote the rectangular maximal function 
where
ranges over all rectangles with sides parallel to the coordinate axes which contain
. Show that
is bounded on
for all
. (Hint: prove by induction, controlling the
-dimensional rectangular maximal function by the
-dimensional “horizontal” rectangular maximal function, applied to a one-dimensional “vertical” maximal function. A certain amount of application of the Fubini-Tonelli theorem may be needed.) Show by example that
is not of weak-type
in dimensions
.
Exercise 8 Let
,
, and let
be locally integrable on
.
Exercise 9 Let
, and let
be a ball such that
at every point of
. Show that
at every point of
.
— 3. Second proof —
Let us now give a slightly different proof of the above inequality, replacing balls by the slightly simpler structure of dyadic cubes.
Definition 10 (Dyadic cube) A dyadic cube in
of generation
is a set of the form 
where
is an integer and
.
The crucial property of dyadic cubes is the nesting property: if two dyadic cubes overlap, then one must contain the other. This leads to
Lemma 11 (Dyadic Vitali-type covering lemma) Let
be a finite collection of dyadic cubes. Then there is a subcollection
of disjoint cubes such that 
Proof: Take the
to be the maximal dyadic cubes in
– the cubes which are not contained in any other cubes in this collection. The nesting property then ensures that they are disjoint and cover all of
between them.
If we then define the dyadic maximal function

where

ranges over the dyadic cubes which contain

, then the same argument as before then gives the
dyadic Hardy-Littlewood maximal inequality 
(with no constant loss whatsoever!) which then leads via Marcinkiewicz interpolation to

for

.
We can rewrite the dyadic maximal inequality in another way. Let
be the
-algebra generated by the dyadic cubes of generation
, then

where

is the unique dyadic cube of generation

which contains

. The dyadic Hardy-Littlewood maximal inequality is then equivalent to the assertion that

and thus

for

.
Observe that if
, then there is a ball
centred at
which contains
of comparable volume:
. Because of this, one easily obtains the pointwise inequality

and so the dyadic inequality follows (up to constants) from the non-dyadic one. The converse pointwise inequality is not true (test it with

and
![{f = 1_{[0,1]}}](https://s0.wp.com/latex.php?latex=%7Bf+%3D+1_%7B%5B0%2C1%5D%7D%7D&bg=ffffff&fg=000000&s=0&c=20201002)
, for instance). However, a slightly modification of this inequality is true, thanks to the
-translation trick of Michael Christ. We first explain this trick in the context of the unit interval
![{[0,1]}](https://s0.wp.com/latex.php?latex=%7B%5B0%2C1%5D%7D&bg=ffffff&fg=000000&s=0&c=20201002)
.
Lemma 12 Let
be a (non-dyadic) interval. Then there exists an interval
which is either a dyadic interval, or a dyadic interval translated by
, such that
and
.
The only significance of
is that its binary digit expansion oscillates between
and
. Note that the claim is false without the
shifts; consider for instance the interval
for some very small
, which straddles a certain “discontinuity” in the standard dyadic mesh. The point is that the dyadic mesh and the
-translate of the dyadic mesh do not share any discontinuities.
Exercise 13 Prove the above lemma.
For intervals larger than
, a shift by
is not enough; consider for instance what happens to the interval
. Instead, we have to shift by
, which of course does not make sense as a real number. However, it does make sense in some formal
-adic sense (as the doubly infinite binary string
) which is good enough to define shifted dyadic meshes.
Definition 14 (Dyadic meshes) We define
to be the collection of all dyadic intervals in
. We define
to be the collection of all intervals of the form
, where
is a dyadic interval at some generation
and
is any integer greater than or equal to
(note that the exact choice of
is irrelevant). If
, we let
be the collection of cubes formed by the Cartesian product of intervals from
.
By modifying the above lemma one then quickly deduces
Lemma 15 Let
be a ball. Then there exists
and a shifted dyadic cube
such that
and
.
This in turn leads to the pointwise inequality bounding the dyadic maximal function by the ordinary one:

where

is the shifted dyadic maximal function

A routine modification of the proof of the dyadic maximal inequality (or translating this inequality by

and taking limits as

) shows that each of the

are individually of weak-type

, and bounded on

for

. Since there are only

many choices of

, we can then deduce the usual Hardy-Littlewood maximal inequality from the dyadic one.
Remark 16 What is going on here is that there are two ways to view the real line
. One is the “Euclidean” way, with the usual group structure and metric. The other is the “Walsh” or “dyadic” way, in which we identify
with the Cantor group 
via the binary representation,
(identifying
with
, and ignoring the measure zero sets of terminating decimals where the binary representation is not unique). The group structure is now the one inherited from the Cantor group; in the binary representation, the Cantor-Walsh addition law
is the same as ordinary addition
but where we neglect to carry bits. The usual archimedean metric
is replaced by the non-archimedean metric
, defined by 
With this metric, the dyadic intervals become the metric balls.
Exercise 17 (Hardy-Littlewood maximal inequality for filtrations) Let
be a measure space, and let
be an increasing sequence of
-finite
-algebras in
(thus
for all
). Show that 
and hence 
for all
and all
-measurable
for which the right-hand side is finite. (Hint: use monotone convergence to reduce to finitely many
. Reduce further to the case when the
are countably generated (by using the level sets of the
for rational intervals). Reduce further still to the case where the
are finitely generated, i.e. finite. Now adapt the dyadic argument. There are also simpler arguments which do not require all of these reductions.) This inequality is also known as Doob’s inequality, and implies in particular that
converges pointwise a.e. to
whenever
.
Exercise 18 (Relationship between dyadic and non-dyadic Hardy-Littlewood maximal inequalities) Let
be locally integrable. Establish the pointwise bound 
for some
depending only on
.
— 4. Third proof —
In the above arguments we obtained
bounds by first proving weak
bounds and then interpolating. It is natural to ask whether such bounds can be obtained directly. The answer is yes, but it is surprisingly more difficult to do so. Let us give two such approaches, a Bellman function approach and a
method approach, which are themselves powerful methods which apply to many other problems as well.
We begin with the Bellman function approach. This method works primarily for dyadic model operators, such as
, though it can also work for very geometric operators as well (using geometric averaging operators such as heat kernels in place of the dyadic averaging operators). For simplicity let us just work in one dimension (though it is possible to use rearrangement and space-filling curves to deduce the higher-dimensional case from the one-dimensional case), and consider the task of establishing

for some fixed

(such as

).
The idea is to work by induction on scales – in other words, to induct on the number of generations. To do this we need a “base case”, so we perform some qualitative reductions. Fix
(the case
being trivial). By a monotone convergence argument we may restrict attention only to those intervals
of length larger than
, so long as our estimates are uniform in
. By rescaling (replacing
by
) we can reduce to the case
. Let us write
for the dyadic maximal function restricted to intervals of length at least
. By a monotone convergence argument (there is a slight problem because we cannot represent
as the monotone limit of dyadic intervals; however we can do this for
and
separately, and then add up, noting that the dyadic maximal function is localised to each of these half-lines, e.g. if
is supported on
then so is
) we can also assume that
is supported on a dyadic interval
of some length
, and also we may restrict
to that interval. We can also take
to be non-negative. Our task is now to show that there exists a constant
such that

whenever

. (We make the constant

explicit here because of the induction that we shall shortly use.) Of course the point is that

is independent of

,

, and

.
We make a small but useful remark: once
and
are both restricted to
, the only dyadic intervals
which are relevant in the definition of
are those which are contained in
(including
itself). Intervals which are disjoint from
play no role, and intervals which contain
give a worse average than that arising from
itself.
The idea is to prove this by induction on the generation
of
. Our first approach will not quite work, but a subtle modification of it will.
When
the claim is trivial (as long as
), because for
a dyadic interval of generation
and
we have

Now let

, and let us see whether we can deduce the

case from the

case
without causing any deterioration in the constant 
. (Using various applications of the triangle inequality it is not hard to get from

to

, replacing

with something worse like

, but this is not going to iterate into something independent of

.)
Let us split the dyadic interval
of generation
into two “children”
of generation
(
and
stand for left and right). This also causes a split
. By induction hypothesis we have

and we also trivially have

Now if we were lucky enough to have the pointwise estimates

and similarly for

, then we could simply average the two induction hypotheses and be done. However, this is not quite the case: the correct relationship between the maximal function of

and of

is that

and similarly when

. This causes a problem. If we estimate the max by a sum and use triangle type inequalities, we will eventually get a bound such as
(4) but with

replaced by

, which is not acceptable for iteration purposes. So we have to somehow keep the max with the constant

with us in the induction argument. This eventually forces us to change the induction hypothesis
(4), replacing the left-hand side

by the more general

for some arbitrary

. Given our knowledge that max and addition are comparable up to constants, we know that
(4) is equivalent to the estimate

up to changes in the constant

. But perhaps this estimate has a better chance of being proven by induction. The key recursive inequality is now that

and similarly for

. One can try the induction strategy again, but one sees that the inability to efficiently control

in terms of

and

is a serious problem. (Hölder’s inequality of course gives

bounded by

, but this turns out to be insufficient.) Because of this, we have no choice but to also throw in the average

into the induction hypothesis somehow.
Let us formalise this as follows. Given any parameters
, let
denote the cost function

where

is understood to be nonnegative and supported on a dyadic interval

of generation

. Note that Hölder’s inequality shows that

when

, since in this case the supremum is over an empty set. Our task is thus to show that

uniformly in

and in

. As we said earlier, the

parameter is not obviously necessary yet, but will become so when we try to perform the induction, as it tracks a certain finer property of the function

which needs to be managed in order to prevent the constants from blowing up. The base case when

is again trivial; the issue is to pass from fine scales to coarse scales without destroying the boundedness of implicit constant.
Now the recursive inequality can be turned into an inequality for
. Suppose that
attains the supremum (or comes within an epsilon of it). We have

for some

. Similarly

for some

. Then we have

for

, and similarly for

; by construction we thus have

Taking suprema, we obtain the recursive inequality

(In fact, this is an equality – why?) On the other hand,

can be computed directly as

when

, and

otherwise. In principle, this gives us a complete description of

, which should allow one to determine the truth or falsity of
(5). Note that we have reduced the problem from one involving an unknown function

(which has infinitely many degrees of freedom) to one involving just three scalar parameters

, except that we have a different cost function at every scale. However, suppose that we can devise a
Bellman function 
with the property that

for all

, and such that

obeys the inequality

whenever

and

. Then an induction will show that

for all

(and note that the implied constants here are
uniform in

), thus proving
(5).
Thus the whole task is reduced to a freshman calculus problem, namely to find a function of three variables obeying the bounds (6) and (7). The difficulty of course, is to find the function; verifying the properties is then routine. This “hunt for a Bellman function” is surprisingly subtle, and requires one to choose a surprisingly non-trivial choice of
.
The condition (7) resembles a concavity condition, and so it is natural to try to find choices which are concave in some of the variables. An initial candidate is the function
, which certainly obeys (7) and the upper bound of (6) (and which, in fact, ultimately corresponds to the original estimate (4) before we threw in the other parameters
); unfortunately it does not obey the lower bound in (6), in the case where
is large compared with
and
. So we need to tweak this function somewhat. The first step is to improve the concavity by exploiting the fact that the function only needs to be non-trivial on the region
. One can exploit this by using the candidate function
(say). This still obeys (7), but now with a bit of a gain when
is large due to the strict concavity of
. (One cannot play similar games with the
parameter as the upper and lower bounds in (6) force linear-type behaviour in
.) But we still have not fixed the problem that
is not as large as
when
is large. The solution is to use the Bellman function

where

is a concave function which is positive for small

and is equal to

for

(say). Thus we have

when

, but

becomes as large as

or so when

exceeds

. This lets us verify both sides of
(6), so one only needs to verify
(7). If

then the claim follows from the concavity of

(which does not depend on

); when

the claim follows from the concavity of

in the

variable and in the

variable.
Remark 19 Bellman function methods are in principle the sharpest and most powerful technique to prove estimates. However they are restricted to dyadic settings (or very geometric continuous settings), and are very delicate due to the need to establish very subtle concavity properties.
— 5. Fourth proof —
Finally, we present the “
approach” to the Hardy-Littlewood maximal inequality. This approach works for both the dyadic and non-dyadic maximal functions, but is restricted to establishing
boundedness (this is a fundamental limitation of the
method, that at least one of the spaces involved has to be a Hilbert space). The basic idea is rather than prove a boundedness result directly on the maximal operator
(i.e. an estimate of the form
), we prove an estimate on the square of this operator (roughly speaking, we prove an estimate of the form
). This is an example of a powerful strategy to understand an operator by raising it to a higher power, and hoping to exploit some self-cancellation. Note that we do not have to cancel the entire power (which would roughly speaking correspond to proving a bound of the form
); any nontrivial cancellation at all is exploitable.
Let us work with the non-dyadic maximal function, thus we wish to show that

for all non-negative

. As before we may restrict

to range in a finite set

, provided our bounds are independent of the choice of

. Note that because each

is already bounded on

, the maximal operator is now already bounded with
some finite operator norm. Let

denote the optimal such norm, thus

is the best constant for which

We know

is finite; our objective is to obtain the bound

. We will do this by controlling

in a nontrivial way in terms of itself, and in particular by controlling the “square” of the maximal operator by the maximal operator.
Observe that (8) is equivalent to the uniform linearised estimate

for all measurable functions

.
Let us fix the function
. Then we can define the linear operator
by

We can thus interpret

as the largest

operator norm of the

:

To compute this operator norm we make the following observation.
Lemma 20 (
identity) Let
be a continuous map from a Hilbert space to a normed vector space, and let
be its adjoint. Then 
Proof: The first identity is just duality. Then we have

which gives the lower bound in the second identity. For the upper bound, observe that for any

that

taking square roots gives the upper bound as desired.
In light of this identity we know that

Now let us take a look at what

is. Observe that

is an integral operator with kernel

Thus the adjoint is given by

and then

is given by

Note that the

integral can be computed fairly easily. First we observe that the

integral vanishes unless

, and in the latter case it enjoys a bound of

. Also,

and

. Putting this together we see that

It is natural to split this integral into the regions

and

, leading to the bound

Comparing this with the formulae for

and

, we obtain the interesting pointwise inequality

where

is the function

, of course. On the other hand, a scaling argument gives

and hence we conclude from the triangle inequality that

taking suprema in

we conclude that

, and hence (since

is known to be finite)

, and the claim follows.
The Hardy-Littlewood maximal function directly bounds averages on balls and cubes, but it also controls several other types of averages as well. For instance, we have the pointwise inequality

for any locally integrable

, any

, and any

; this can be achieved by dividing the

integral into dyadic shells

for

, as well as the ball

, and we leave the computation to the reader.
The proof of the Hardy-Littlewood maximal inequality extends to the more general setting of spaces of homogeneous type (which is similar, but different from, the notion of a homogeneous space). These are measure spaces
with a metric
, such that the open balls are measurable with positive finite measure, and that one has the doubling property

for any

. Then the Vitali-type covering lemma extends without difficulty to this setting and yields the maximal inequality

and hence

for any

.
In particular we obtain the discrete inequality

(which we need in our applications to ergodic theory below), while on the torus

with the usual Lebesgue measure we have

for

(which we will need for our applications to complex analysis).
Remark 21 One could try playing with
instead of
here, but that turns out to not work very well. The problem is that the linearised maximal operator is only well-behaved in one variable, and the
method manages to play the two well-behaved variables against each other; going the other way one achieves no obvious cancellation.
— 6. Some consequences of the maximal inequality —
The Hardy-Littlewood maximal inequality is the underlying quantitative estimate which powers many qualitative pointwise convergence results. A basic example is
Theorem 22 (Lebesgue differentiation theorem) Let
be locally integrable. Then we have the pointwise convergence
for almost every
. In fact we have the stronger estimate 
for almost every
.
Before we prove the theorem, let us make some remarks. Firstly, from the identity

and the triangle inequality it is clear that the second estimate implies the first. Secondly, observe that

is continuous at

if and only if the worst-case local fluctuation goes to zero

Thus the differentiation theorem is asserting that locally integrable functions are
almost everywhere continuous on the average, in that the average-case local fluctuation goes to zero. This is a manifestation of one of Littlewood’s three principles, namely that
measurable functions are almost continuous. If

is such that
(12) holds, we say that

is a
Lebesgue point of

, thus for locally integrable

, almost every point

is a Lebesgue point.
Proof: It suffices to establish the result for
constrained to a large ball
, where
is arbitrary. The claim is obvious if
vanishes on
, so by linearity we may assume that
is supported on
; in particular
now lies in
. Thus it suffices to establish the claim when
.
By the preceding discussion we already know that the claim is true for the continuous compactly supported functions
, which is a dense subclass of
. To pass from the dense subclass to the full class we use the Hardy-Littlewood maximal inequality now lets us reduce matters to verifying the claim on a dense subset of
. To see this, suppose that we already have proven the claim on a dense class. Then, given any
we can write
as the limit in
of a sequence
in this dense class; by refining these sequence to make the
convergence sufficiently fast (e.g.
) and using Markov’s inequality and the Borel–Cantelli lemma, we can also ensure that
converges to
pointwise almost everywhere. Now from the Hardy-Littlewood maximal inequality, we know that
converges to zero, thus
converges to zero in measure. By the triangle inequality this implies that
converges in measure also. Thus (again passing to a rapidly converging subsequence as necessary) we see that
converges to zero for almost every
. This uniform-in-
convergence lets one deduce the convergence of
from the convergence of
.
Finally, by taking a dense class such as the Schwartz class (or even the continuous functions in
) we easily verify the convergence.
Remark 23 Because the proof used a density argument, it offers no quantitative rate for the speed of convergence. Indeed the convergence can be arbitrarily slow; this is ultimately due to the implicit hypothesis that
is measurable, which is itself a qualitative assumption that offers no explicit bounds. More quantitative versions of measurability, e.g. quantifying the extent to which a measurable set can be approximated by elementary sets, can lead to more explicit bounds.
Consider for instance the function
defined by
when
and the integer part of
is even, and
otherwise, where
is a large integer. Then we see that for
,
will stay close to
for quite a while (basically for all
), and only at scales
or less will it begin to “decide” to converge to either
or
. Modifying this type of example (e.g. by a “Weierstrass example” formed by summing together a geometrically decaying series of such oscillating functions, with
going to infinity), one can concoct functions whose convergence of
to
is arbitrarily slow. (One can obtain quantitative rates by enforcing regularity conditions on the function, which essentially compactifies the space of functions that one is working with; we will see some examples of this later.)
Exercise 24 Let
be a locally integrable function on
. Call a point
a Lebesgue point if there exists a number
such that
. Show that almost every point is a Lebesgue point, and that
is equal to
almost everywhere.
Exercise 25 (Heat kernels) For any
and any
for some
, define the heat kernel
by 
Show that if
and
, then
converges both pointwise and in norm to
as
.
Now let us explore applications to ergodic theory. In particular we wish to investigate limits of the form

for various functions

and various “shift operators”

.
Let us first look at an abstract setting, in which
is a unitary operator on a Hilbert space.
Theorem 26 (Von Neumann ergodic theorem) Let
be a unitary operator on a Hilbert space. Then for any
, the limit
exists in the strong
topology.
Proof: Let us first argue formally. We have

Formally, we have the geometric series formula

which looks like it converges to zero, unless

fails to be invertible; but on the other hand when

is just

, which of course converges to the identity. So we seem to have covered two extreme cases.
Now let us make the above argument rigorous. If
is
-invariant, thus
, then
and it is clear that
converges to
. If on the other hand
is a
-difference,
for some
, then we verify the telescoping identity
. Since unitary operators preserve the norm, we thus see from the triangle inequality that
converges to zero. Also, since the averaging operators
are uniformly bounded in
(by the triangle inequality), we see that any strong limit of
-differences also has the property that
exists and equals zero.
To summarise so far, we have two closed subspaces of
for which we know convergence. The first is
, the invariant space of
; here the limit converges to the identity. The other is
, the closure of the
-differences; here the limit conveges to zero. These two spaces turn out to be orthogonal complements. To see this, first observe that they are orthogonal: if
, then
by unitarity, and hence
is orthogonal to every
-difference
and hence by continuity of inner product is orthogonal to
. To show orthogonal complement, it then suffices to show that any vector
orthogonal to
is invariant. But then
is orthogonal to
:

We rewrite the left-hand side as

and on conjugating and subtracting we conclude that

and thus

as claimed.
Note that the above argument in fact shows the stronger claim that
converges to the orthogonal projection of
to
.
Now we move to a more specialised setting, that of a measure-preserving system.
Definition 27 (Measure-preserving system) A measure-preserving system
is a probability space
(thus
) together with a bi-measurable bijection
(thus
and
are both measurable) such that
is measure-preserving, thus
for all
.
Example 28 (Circle shift) Let
be the standard circle with the usual Borel
-algebra and Lebesgue measure, and let
for some
, which may be either rational or irrational.
Remark 29 Many of the results here hold under more relaxed assumptions on
, but we will not attempt to optimise the hypotheses here.
The shift
on the base space
,
, induces a shift on sets
,
, and then also induces a map on measurable functions
by
. The use of
is natural since it ensures that
and
.
Proposition 30 (Mean ergodic theorem) Let
be a measure-preserving system, and let
for some
. Then the sequence
converges strongly in
.
Proof: The operator
is an isometry on
, and so by the triangle inequality the averaging operators
are uniformly bounded in
. Thus it suffices to prove the claim for a dense subclass of
; we shall pick
. In this class, which is embedded into the Hilbert space
, we already know from the von Neumann ergodic theorem (and the fact that
is a unitary operator on
) that the averages are convergent in
norm. But they are also uniformly bounded in
. So the claim follows from the log-convexity of
norms (for
) or by Hölder’s inequality and the finite measure of
(for
).
Exercise 31 Show by example that the mean ergodic theorem fails for
and for
.
Now we study the pointwise convergence problem. The key quantitative estimate needed is the following analogue of the Hardy-Littlewood maximal inequality.
Theorem 32 (Hardy-Littlewood maximal inequality for measure-preserving systems) Let
be a measure-preserving system. Then we have 
and 
for all
.
Proof: The claim is trivial when
, so once again the task is to prove the pointwise estimate. By monotone convergence it suffices to show that

uniformly in

.
The idea is to lift up to a space where
can be modeled by the integer shift
, at which point we can apply (10). Fix
, and let
be the finite set
, endowed with the discrete
-algebra and uniform measure, so that
is a probability space. Inside this space we have
, where
. We define the function
by

From
(10) we have
![\displaystyle \| \sup_{1 \leq N < N_0} \frac{1}{N} \sum_{n=1}^N F(x,n+m) \|_{l^{1,\infty}_m([N_0])} \lesssim \| F(x,m) \|_{l^1_m([2N_0])}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5C%7C+%5Csup_%7B1+%5Cleq+N+%3C+N_0%7D+%5Cfrac%7B1%7D%7BN%7D+%5Csum_%7Bn%3D1%7D%5EN+F%28x%2Cn%2Bm%29+%5C%7C_%7Bl%5E%7B1%2C%5Cinfty%7D_m%28%5BN_0%5D%29%7D+%5Clesssim+%5C%7C+F%28x%2Cm%29+%5C%7C_%7Bl%5E1_m%28%5B2N_0%5D%29%7D&bg=ffffff&fg=000000&s=0&c=20201002)
for all

; writing out what the

norm means, and integrating in

using Fubini’s theorem, we conclude
![\displaystyle \| \sup_{1 \leq N < N_0} \frac{1}{N} \sum_{n=1}^N F(x,n+m) \|_{L^{1,\infty}_{x,m}(X \times [N_0])} \lesssim \| F \|_{L^1(X \times [2N_0])}.](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5C%7C+%5Csup_%7B1+%5Cleq+N+%3C+N_0%7D+%5Cfrac%7B1%7D%7BN%7D+%5Csum_%7Bn%3D1%7D%5EN+F%28x%2Cn%2Bm%29+%5C%7C_%7BL%5E%7B1%2C%5Cinfty%7D_%7Bx%2Cm%7D%28X+%5Ctimes+%5BN_0%5D%29%7D+%5Clesssim+%5C%7C+F+%5C%7C_%7BL%5E1%28X+%5Ctimes+%5B2N_0%5D%29%7D.&bg=ffffff&fg=000000&s=0&c=20201002)
The right-hand side is just

. As for the left-hand side, observe that

and so on writing out the

norm we see that the left-hand side is comparable to

and the claim follows.
Remark 33 Note how a maximal inequality for the integers was “transferred” to a maximal inequality on arbitrary measure preserving systems. There are several abstract transference principles which generalise this type of phenomenon. In particular, the above argument is in fact a special case of the Calderón transference principle.
Now we can present the analogue of the Lebesgue differentiation theorem for measure-preserving systems.
Proposition 34 (Pointwise ergodic theorem) Let
be a measure-preserving system, and let
. Then the sequence
converges pointwise almost everywhere.
Proof: By repeating the argument in the Lebesgue differentiation theorem more or less verbatim (using the above maximal inequality in place of the Hardy-Littlewood inequality) it suffices to verify the claim for a dense subclass of
, such as
. Since the
norm controls the
norm, it suffices to do so for a dense subclass of
.
Now we repeat the proof of the von Neumann ergodic theorem. For the invariant part
of
, the pointwise convergence is obvious. Also, for functions of the form
with
, the convergence is also obvious. But these functions are clearly dense in
and hence in
. Since this space and
are orthogonal complements of
, we thus have demonstrated convergence of a dense subclass of
and thus of
, as desired.
Remark 35 Notice how the strategy of establishing convergence splits into two independent parts – obtaining convergence on a dense subclass, and then establishing some harmonic analysis estimate to pass to the general case. This is not the only way to achieve convergence results. Later on we shall see a variation-norm approach which relies purely on harmonic analysis estimates to obtain convergence (foregoing the dense subclass). In the opposite direction, there are more dynamical approaches (which we do not cover here) which forego the harmonic analysis component of the argument, relying instead on analysing the dynamics of the measure-preserving system by other means (such as measure-theoretic or topological means). It is not fully understood to what extent these different techniques complement each other.
Exercise 36 Let
be a measure-preserving system. Let
be the elements of
which are
-invariant (up to sets of measure zero, of course). Show that for any
and
, the averages
converge in
norm and pointwise almost everywhere to
. In particular, when
is ergodic (thus the only invariant sets have zero measure or full measure), conclude that
converges pointwise and in
norm to
.
Exercise 37 (Poincare recurrence theorem) Let
be a measure-preserving system, and let
be non-negative with
. Show that
for infinitely many
.
By combining the Lebesgue differentiation theorem with the Radon-Nikodym theorem, one easily obtains
Theorem 38 (One-dimensional Radamacher differentiation theorem) If
is Lipschitz, then it is differentiable almost everywhere, its derivative lies in
, and we have the fundamental theorem of calculus
for all
.
Proof: We may reduce to the case when
is real. The Riemann-Stietjes measure
is easily seen to be absolutely continuous with respect to Lebesgue measure, and is thus of the form
for some locally integrable
, thus
by the Radon-Nikodym theorem. The Lebesgue differentiation theorem then shows that
exists and is equal to
at every Lebesgue point of
(see Q7), and the claim follows. (The boundedness of
is clear from the Lipschitz nature of
.)
Remark 39 There are other ways to prove this theorem that do not require Radon-Nikodym differentiation, which we shall encounter later.
This one-dimensional theorem implies a multi-dimensional analogue:
Theorem 40 (Radamacher differentiation theorem) If
is Lipschitz, then
is differentiable almost everywhere, thus for almost every
there exists a vector
such that 
Proof: For simplicity of notation we take
, although the general case is similar. Theorem 38 (and Fubini’s theorem) shows that the partial derivatives
,
exist almost everywhere and are bounded. This gives us a candidate gradient
defined almost everywhere. The remaining challenge is to show total differentiability. In other words, for any
, we need to show that for almost every
, we have

whenever

is sufficiently small depending on

. Note that the control of the partial derivatives only achieve this when

is a multiple of

or

.
Fix
. We will need to make things slightly more quantitative. For any
and
, let
denote the set


Then the almost everywhere existence of

implies that the

are measurable and increase to

as

(neglecting sets of measure zero). In particular, almost every point lies in infinitely many of the

. Because of this, we know that almost every point is a horizontal and vertical Lebesgue point of

and of

for infinitely many

. (We say that

is a horizontal Lebesgue point of

if

as

, and similarly define a vertical Lebesgue point of

.) In particular, for almost every point there is an

such that one lies in

, and is a horizontal and vertical Lebesgue point of

and of

for

. From this it is not hard to show
(13) for a large density set of

near

(say of density

), basically by decomposing
![{f(x+h_1e_1+h_2e_2) - f(x) = [f(x+h_1e_1) - f(x)] + [f(x+h_1e_1+h_2e_2) - f(x+h_1 e_1)]}](https://s0.wp.com/latex.php?latex=%7Bf%28x%2Bh_1e_1%2Bh_2e_2%29+-+f%28x%29+%3D+%5Bf%28x%2Bh_1e_1%29+-+f%28x%29%5D+%2B+%5Bf%28x%2Bh_1e_1%2Bh_2e_2%29+-+f%28x%2Bh_1+e_1%29%5D%7D&bg=ffffff&fg=000000&s=0&c=20201002)
and using the Lebesgue point properties to ensure that

is usually close to

; we omit the details. The remaining exceptional values of

can then be dealt with by locating a nearby non-exceptional value of

and using the Lipschitz property.
Exercise 41 (Fundamental theorem of calculus) Let
be locally integrable, and let
(with the usual convention that
). Show that
is differentiable at every Lebesgue point of
, and that
almost everywhere.
— 7.
functions —
Another classical application of the Hardy-Littlewood maximal function lies in obtaining a satisfactory theory of (complex)
functions on the complex disk
for
. (The theory for
is more subtle and will not be dealt with here.)
Let
be a holomorphic function. From residue calculus we have

and

when

, and so on averaging, and noting that

is the complex conjugate of

when

, we obtain

or in other words, if we set

to be the function

, we have the
reproducing formula 
whenever

and

is the
Poisson kernel 
The kernel

is clearly non-negative; applying the above identity to

(or computing directly) we see also that

has mass

:

From Young’s inequality we thus obtain that for

, the

norm of

is increasing on circles:

(Note that the

case is just the maximum principle.) In particular, the quantity

always exists. We say that

is an
function if

is analytic on

and the norm

is finite. It is not hard to see that

becomes a normed vector space (and that

whenever

); a little more work shows that it is in fact complete.
Exercise 42 For any
and
, define the Fourier coefficient
of
by the formula 
Show that 
for every
and
, and that 
for every
,
and
. Also show that
for all
, and that 
for all
. (Note that these identities can be proven either by complex analysis methods or by Fourier analysis methods; it is instructive to prove them both ways and compare results.)
The kernel
can be easily verified to obey the bounds

This is enough to obtain the pointwise estimate

where

is the Hardy-Littlewood maximal function of

. From last week’s notes we conclude that for any

and

, that

converges to

in

norm, and by modifying the proof of the Lebesgue differentiation theorem we also see that it converges pointwise almost everywhere.
The definition of
strongly suggests (but does not immediately prove) that if
lies in
, then
should converge to some sort of limit
as
. This is indeed the case:
Theorem 43 Let
for
. Then
converges both in
norm and pointwise almost everywhere to a limit
.
Remark 44 The theorem fails for
, as can be seen by explicitly computing using the function
. When
, the pointwise almost everywhere claim is still true, but the convergence is not.
Proof: Let us first demonstrate weak convergence. For any
and
, we observe that

where

is the inner product on

(here we use the fact that

is real and even). Since

converges in

norm to

as

, we thus see that

converges to a limit as

. Since

converges in

norm to

as

, we conclude (from Hölder’s inequality and the uniform

bound on

) that

also converges to a limit as

, thus

converges weakly. But as

is reflexive, the closed unit ball is also weakly closed, and we conclude that

converges weakly to

for some

, thus

for all

. Replacing

by

and using
(14) we conclude

and hence (since

converges to

in

norm)

This is for all

, thus we have

Thus

converges pointwise almost everywhere and in

norm to

.
Exercise 45 (Fatou’s theorem) Let
for some
, and let
be as in Theorem 43. Show that for almost every
, we have 
whenever
is a sequence of points converging to
non-tangentially in the sense that
is uniformly bounded away from
. (Hint: first reduce to a fixed angle of non-tangentiality (e.g. all angles less than pi/2 – 1/n), and then build an appropriate “non-tangential maximal function”, formed by taking suprema over all points in a sector with apex
and this fixed angle of non-tangentiality. Use the kernel bounds to bound this maximal function by the Hardy-Littlewood maximal operator.)
— 8. The Calderón-Zygmund decomposition —
For a function
on an abstract measure space, and any threshold
one has the basic decomposition

into a “good”

piece (bounded by

), and a “bad” piece

(larger than

, but at least of small support). If for instance

, then we have the bounds

for the good piece and

for the bad piece (the last inequality being Markov’s inequality). Thus both pieces inherit the

bound of the whole, but enjoy an additional useful property. This type of decomposition underlies tools such as the layer cake decompositions and the real interpolation method.
For arbitrary measure spaces
with no additional structure, this type of decomposition is about the best one can do to split
into “good” and “bad” parts. But if
has more structure – in particular some sort of metric or dyadic structure compatible with the measure – then one can do better, in particular one can select the bad piece
to be “locally oscillating” in a certain sense. This principle, which uses the same ideas that underlie the Hardy-Littlewood maximal inequality, is formalised in the Calderón-Zygmund decomposition, a fundamental tool in the Calderón-Zygmund theory of singular integrals.
To motivate the lemma let us first give a simple one-dimensional version.
Lemma 46 (Rising sun lemma) Let
be a bounded interval, and let
be integrable, and let
be such that
. Then there exist an at most countable family
of disjoint open intervals in
such that
-
for almost every
; -
for every
.
Proof: By subtracting
from
we may normalise
. We may take
to be half-open
. Define
by
. Then
is a continuous function which starts at zero and ends up positive. Define a “maximal” version
of
by
, then
is continuous non-decreasing which starts at zero and ends up positive. Let
, then
is an open subset of
, thus
for some at most countable set of intervals
. At the endpoints of each interval we have
, while on the interior of these intervals
is necessarily constant (why?). This gives the second desired property. For the first, one observes from Q7 that at almost every point where
, we have
differentiable with
, which implies that
, and the claim follows.
As a corollary we have
Corollary 47 (One-dimensional rising sun lemma) Let the hypotheses be as in the above lemma. Then we have a decomposition
, where
is bounded above by
with
, and for each
,
is supported in
and has mean zero. Furthermore, if
and
is non-negative, then
and
.
Proof: We set
, and set
equal to
outside of
and equal to
inside
. To prove the last two claims, we have

so on summing and using the non-negativity of

from which the former claim follows. Finally, from the definition of

and the triangle inequality

which is the latter claim.
The proof of the rising sun lemma relies crucially on the ordered nature of the real line. For more general spaces, such as Euclidean spaces, it is preferable to use arguments that rely instead on metric or dyadic structure. We begin with a dyadic version.
Proposition 48 (Dyadic Calderón-Zygmund decomposition) Let
and
. Then there exists a decomposition
, where the “good” function
obeys the bounds 
ranges over a disjoint family
of dyadic cubes, and each
is supported on
, has mean zero, and has the
bound 
Furthermore, we have the inclusions 
In particular from (3) we have
.
Proof: The arguments here will closely resemble that used to prove (3). Let us say that a cube
is bad if
, and good otherwise. Call a bad cube
maximal if
is bad, but no cube strictly containing
is bad, and let
be the collection of all maximal bad cubes, thus
is a collection of disjoint dyadic cubes, and we also clearly have
on each bad cube. From the hypothesis
and monotone convergence we see that all cubes with sufficiently large side-length are automatically good. Thus every bad cube is contained in a maximal bad cube.
Now let
be a maximal bad cube. The parent
of
(i.e. the unique dyadic cube of twice the side-length containing
) is good, hence
. Since
, we conclude that
. Now set

and

All the desired properties are then easily verified, except perhaps for the claim

. This claim is clear on each separate cube

; the only difficulty is to show that

is bounded by

outside of

. But by construction we see that if

and

, then

is good, and

. On the other hand, the (dyadic version of the) Lebesgue differentiation theorem shows that

converges to

for almost every

as

. Thus

for almost every such

, and we are done.
Exercise 49 Let
be the
-algebra generated by the cubes in
, and the Borel subsets of
. Show that
.
Note that the total bad set
is given by the level set
of the dyadic maximal function. This suggests an alternate approach to the Calderón-Zygmund decomposition, in which one starts by identifying the total bad set (in this case,
), and then decomposes it into cubes or balls. A prototype decomposition of this type is
Proposition 50 (Dyadic Whitney decomposition) Let
be an open set. Then there exists a decomposition
, where
ranges over a family
of disjoint dyadic cubes, and for each
in this family
, the parent
of
is not contained in
.
Proof: Define
to be the set of all dyadic cubes
in
which are maximal with respect to set inclusion; the claim follows from the nesting property. (The condition
is needed to ensure that every cube is contained in a maximal cube; the open-ness is to get every point of
contained in at least one cube.)
The property that
is not contained in
implies the bounds

In many applications we need to complement the upper bound with a non-trivial lower bound. This can be done with a little more effort:
Proposition 51 (Whitney decomposition) Let
be an open set and let
. Then there exists a decomposition
, where
ranges over a family
of disjoint dyadic cubes, and for each
in this family
, we have
.
Proof: Let
denote those dyadic cubes
in
such that

It is not hard to show that these cubes cover

; indeed, given any

in

, all one needs to do is to locate a cube

containing

with diameter between

and

. The cubes are not disjoint; however if one lets

be those cubes in

which are maximal with respect to set inclusion, then the claim follows from the nesting property.
Note that if
is large, the cubes in this above decomposition have the property that nearby cubes
(in the sense that
) have comparable diameter, thanks to the triangle inequality


Since every cube is contained in a ball of comparable radius (with constants depending on

), we then conclude
Proposition 52 (Whitney decomposition for balls) Let
be an open set and let
. Then one can cover
by balls
such that
, and such that each point in
is contained in at most
balls.
These decompositions can be used to prove some minor variants of the Calderón-Zygmund decomposition, which we will not describe in detail here.