**0. About these notes
**

The notes that will follow are meant to be a companion to the Harmonic Analysis course that I’m giving this semester at IST. These notes are inspired, influenced and sometimes shamelessly copied from books, lecture notes of other people, research papers and online material. The whole idea and structure of the course and, in particular, the use of the blog as a general platform of communication and interaction in relevance to the course is highly influenced by similar efforts of the other people and, especially, from Terence Tao’s blog as well as his lecture notes. Be sure to check the originals!

**1. Introduction and notations **

As mentioned in the overview of the course, we will be mainly concerned with operators acting on certain function spaces, or even spaces of more rough objects such as measures or distributions. Typically we will want to study the mapping properties of such an operator, that is whether it maps one function space to another and so on. A typical estimate in this context is of the form

where are certain, usually Banach, spaces of functions or measure or distributions and are norm, or semi-norms or, in general, norm-like quantities. Thus such an estimate states that the operator takes functions (or `objects’) from the space to the space in a continuous way. Already such an estimate can reveal quite a lot for the nature and the properties of the operator .

When studying the mapping properties of an operator, it is oftentimes useful to restrict attention to a `nice’ subclass inside . If for example consists of integrable functions, a good idea is to first consider the action of on the class of smooth functions with compact support, or on the class of simple functions. These subclasses are nice or explicit enough to be able to overcome many technical difficulties in trying to define for a general object . Furthermore, when these classes are dense in the original space, there is a very natural candidate for the extension of to the whole class. It turns out that this happens whenever is bounded on the dense subclass. Another useful technique is to decompose a general function into different pieces. Since is usually linear, we can then examine the effect of on each piece and sum the pieces together. Likewise, we can decompose the operator to different components, each component being easier to control than the `whole’ operator . Finally, we combine these two ideas and decompose both the function and the operator into different pieces. Usually good control on the different pieces is expected to imply a good control on the original operator and/or function. There are however technical difficulties in putting the pieces together, justifying how each individual estimate sums up to a `global’ estimate.

Overall this course is all about estimates: Estimating the norm of a function, the norm of an operator, the norms of the different pieces of a decomposition of a function and so on. It is very useful to introduce some notation:

**Hardy notation; a constant that has an unspecified value.** Such a constant , or and so on, usually represents a numerical constant that does not depend on any of the parameters of the problem. Using this notation, we will many times using a letter, for example, to denote a generic numerical constant. Different appearances of the letter will not necessarily denote the same numerical constant, even in the same line of text. For example a very useful estimate is the following

We will use write estimates likes this in the form

which is just the statement the fact that the function behaves linearly close to . The precise values of the constants, that is, the precise slopes of the linear functions appearing in the estimate, are rarely of any importance and the do not depend on anything interesting. Taking this one step further we would write for example

when is close to and

when .

A variation of this notation is useful when a constant actually depends on one of the parameters of the problem. Thus we could write

which means that the constants may depend on and but *not* on the function . One should be careful with estimates like this. For example, the notation

is correct though not very useful as the notation `hides’ the dependence of the constant on (for example whether it is bounded in , whether it grows to infinity in and so on). On the other hand, the notation

is wrong though the estimate is actually true for *fixed* . Such a notation would imply that the sequence is uniformly bounded in which is of course not true. Such a notation is true for example in the case

**The Vinogradov notation.** Suppose that we have an estimate of the form where could be norms of functions, or operators and so on. We will write this estimate in the form

Similarly we write whenever . If we have that and then we will use the notation . This latter notation states that the quantities , are equivalent up to constants.

For example, we could write and also for close to .

If we want to state a dependence on a parameter we use a subscript. For example we write

to denote the dependence of the implied constant on and .

A lot of attention should be given when iterating this notation. While this is legitimate for a finite number of steps, an infinite number of steps can create many problems. Beware of this situation especially in inductive arguments. Never hide the dependence on the induction parameter in the Vinogradov notation!

**The Landau – big – notation.** In this notation, writing means that there exists a numerical constant such that . The big notation however is mostly useful when we want to denote a main term and an error term, and keep track of everything in a nice way. Imagine for example that we want to study the function for close to zero, say . The Taylor expansion of around zero is of the form

While it is correct that as , what happens if we want to keep track of lower order terms? Well, we could use the big- notation to write

Note that this is correct since the higher order terms and so on, are always controlled by when . This is a very useful device if we want to `carry’ the lower order terms in our calculations. For example, since

we can write

If we want to state the dependence on some parameter we use subscripts again. Thus we could write meaning that . Also note that the bound can be written in the form .

**2. Recalling notions from measure theory **

We begin this introductory lecture by reminding some basic facts from Measure theory. As mentioned in the description of the course, our first task will be to recall all the basic notions and tools from integration theory and spaces, thus defining our main setup. Our basic environment is a *measure space* , that is a set together with a -algebra of sets in and a non-negative measure on . The measure will always assumed to be -finite ( can be decomposed as a countable union of sets of finite measure). Recall that our subject is *Euclidean* harmonic analysis so, in most cases, the underlying space will be the -dimensional Euclidean space, will be the Lebesgue measure on and will be either the -algebra of *Lebesgue*-measurable sets, or the -algebra of *Borel*-measurable sets in .

Typically we will consider measurable functions ; recall here that measurable means that the pre-image of every measurable set (thus of every set in ) is a measurable set (that it is a set in ). However, we will mostly consider functions , where it is understood that is equipped with the Borel -algebra. Again, the special case of Lebesgue-measurable complex valued functions on is of particular importance. Thus the main example to keep in mind is a *Lebesgue-measurable*, complex valued function

where is equipped with the Lebesgue -algebra and is equipped with the Borel -algebra of sets in . Note these definitions and conventions here since we won’t repeat them every time we consider measurable functions.

Let us go back to the case of a general measure space . If not otherwise stated, a *set* in will mean a *measurable* set in and a function will mean a *measurable* complex valued function. For a set in , the indicator function of will be denoted by or :

A *simple* function is then a *finite linear combination of indicator functions*, that is a function defined as

where and are (measurable) sets. A standard way to identify a set in with a measurable function on is via the map .

Two functions (or sets) in a measure space will be considered one and the same object if they agree *almost everywhere*. For example, consider a set in and a subset with . For the purposes of this course, the functions and are one and the same function. If you want to be more rigorous, you have to think of a measurable function as an *equivalence class* of functions, where two measurable functions are equivalent if and only if , *-almost everywhere* (-a.e.). That is, everywhere on except maybe on a set of measure zero. We will however abuse language a bit and just refer to as a function arbitrarily choosing a representative from every equivalence class. Moreover, we can choose the member of the class that is more convenient for our purposes. To give an example of the usefulness of this principle, think of the equivalence class of functions , say on , that agree with almost everywhere. One can think of functions that behave very erratically and are equal to almost everywhere. However, the function that is identically equal to *everywhere* still belongs to the same equivalence class and is continuous, thus it qualifies as a `nice’ representative of this equivalence class.

For continuous functions however, there is no ambiguity.

Exercise 1Let be two topological spaces and suppose that is Hausdorff. Assume that is a Borel measure on such that for every open set . Prove that if are continuous and -a.e. on , then on .

Hint:Since the space is Hausdorff, `open sets separate points': for every with there exist disjoint open neighborhoods of , respectively.

**3. spaces **

Let us begin by fixing a measure space . We assume as usual that is a non-negative -finite measure on . The most important spaces of functions in this course will be the spaces , , defined as the spaces of measurable functions such that

For we define the space of essentially bounded functions , that is the space of measurable functions such that

Recall here that the essential supremum of a function is the smallest positive number such that , -almost everywhere:

We will alternatively use the notations or even whenever the underlying measure space is clear from context or not relevant for a statement.

Exercise 2Let be a simple function of finite measure support, that is, a finite linear combination of indicator functions of sets of finite measure. Show that

and that

where

As we shall shortly see, for , the quantities are norms. In order to show this, the only difficulty is the triangle (or Minkowski, in this case) inequality. For these quantities are not norms any more but we have a quasi-triangle inequality, that is a triangle inequality with a constant different than , and the spaces are still complete vector spaces.

Lemma 1Let be a measure space. For , the quantity is a norm. In particular we have the following, for all functions :

(i) (Point Separation)

(ii) (Positive Homogeneity) For all we have

(iii) (Triangle inequality)

For , (i) and (iii) still hold true. Triangle inequality is replaced by

(iii’)(Quasi-triangle inequality)

*Proof:* The statements *(i)* and *(ii)* are trivial, given the fact that we identify functions that agree -a.e. For *(iii)* we can assume that are non-zero because of *(i)*, otherwise there is nothing to prove. The case of *(iii)* is trivial so we assume that . Because of the homogeneity property *(ii)*, it is enough to prove that whenever . Since are non-zero this means that there exists with such that and . Setting and the problem reduces to showing that

whenever

We will show (1) by using a basic convexity estimate. For we consider the function given by the formula where . Then the function is convex. This means in particular that for and we have . Using the complex triangle inequality and the convexity of we can thus write

because of the normalization .

The quasi-triangle inequality is any easy consequence of the basic estimate for and , and is left as an exercise.

Exercise 3Show that the triangle inequality is an equality if and only if or for some .

Hint:Check carefully when the inequalities in the previous proof become equalities. Use the fact that for we have a.e.

For , the spaces are *Banach* spaces, that is they are normed vector spaces which are complete with respect to the corresponding norm . For we don’t have a triangle inequality. However, the quasi-triangle inequality allows us to show that the spaces are (quasi-normed) complete vector spaces.

Proposition 2For the space is a Banach space. For the space is a complete quasi-normed vector space. Furthermore, for the preceding spaces areseparable. Separability fails however for .

A very useful variation of Minkowski’s inequality is one where we `replace’ the sum by an integral (which, in a way, is also a sum!). Roughly speaking this inequality states that the norm of an integral is always smaller or equal to the integral of the norm.

Proposition 3 (Minkowski’s integral inequality)Let and be two measure spaces where the measures are -finite non-negative measures. Let be a -measurable function on the product space .

(i) If and , then

(ii) If , for -a.e. , and the function is in for -a.e. , then for -a.e. , the function is in and

Writing *(ii)* of Minkowski’s inequality also highlights the similarity to the classical triangle inequality, where one just has to think of the integral as a `generalized sum’. This is also a good trick to help you memorize the inequality. Observe that the triangle inequality is just a special case of the integral version of Minkowski’s inequality where the measure is the counting measure. You can find the proof of this inequality in most textbooks of real analysis. See for example [F].

After the triangle inequality, the next most important inequality in the spaces , is HĆ¶lder’s inequality.

Lemma 4Let and for some . Define the exponent by means of the `HĆ¶lder relationship’

Then the function and we have the norm estimate

Exercise 4Prove Lemma 4 above.

Hint:Note that the case is trivial. Assuming that homogeneity allows us to normalize , the case or being trivial. Normalizing and setting , it is enough to prove that , whenever , for suitable . Complete the proof using the fact that the function is convex, where are positive real numbers. To show this you can use the convexity of the function .

Remark 1Observe that HĆ¶lder’s inequality is invariant under the transformation and for any constants . Note also that this inequality refers to a general measure space . Replacing the measure by the measure for some constant observe that . Using this invariance and applying HĆ¶lder’s inequality with , we get

for all . We conclude that we must have the HĆ¶lder relation between the exponents ,

whenever HĆ¶lder’s inequality holds true.

** 3.1. Log-convexity of of -norms **

We will now discuss a characteristic of norms that is implicit in many parts of the discussion on spaces, especially interpolation which we will discuss at the end of this first set of notes. It is already hidden in the proof of HĆ¶lder’s inequality above.

Let us start with a function . The function is called * convex* if for every and any we have that

The same definition makes perfect sense whenever the function is defined on some interval of the real line or, in fact, on any convex subset of a vector space. Observe that the definition states that the line connecting the points and of the graph of always lies `above` the graph of the function itself. Now if a function is positive, we will say that is *log-convex* if the function is convex. In this case we must have

Proposition 5 (Log-convexity of -norms)Let and define , as

where . Thus is aconvex combinationof and . Then we have that if and

Note that this means that the function is log-convex.

*Proof:* Observing that , we apply HĆ¶lder’s inequality to to get

which proves the desired estimate.

We will give another proof that employs a notion of convexity in complex analysis and, in particular, the maximum principle. We state the following lemma which will also be useful in the rest of the notes.

Lemma 6 (Three lines lemma)Suppose that is a bounded continuous complex-valued function on the closed strip , that is analytic in the interior of . Suppose that obeys the bounds and for all . Then we have that for all .

*Proof:* First of all we can assume that otherwise there is nothing to prove. Now, consider the function for . Thus it suffices to show that for all , whenever and . First consider the case that uniformly in . Then the result follows from the maximum principle. Indeed, there is some such that for all . Now is bounded by on the boundary of the rectangle and the maximum principle implies that is also bounded by in the interior of the rectangle as well. Thus, in this case, is bounded by throughout the strip . To get rid of the condition consider the sequence of functions , for . Since is bounded, say , we have that

as , uniformly in . Observe that we still have the bounds and for , uniformly in . Thus we also conclude that for all . Letting we get that .

Remark 2Observe that if we define the function , then under the hypothesis of the three lines lemma, we get that is log-convex. Another point to observe here is that the hypothesis we have stated here is not quite optimal. Indeed, we can actually relax the condition that is bounded with the growth condition for some when . The idea of the proof is exactly the same. One first proves the result in the case that . Then we apply this for the sequence of functions .

*Proof:* We begin by making some reductions. Observe that the inequality we want to prove is invariant under the transformations and for any constants . Using these invariances it is enough to show that if then we have that , for all with . To do this, consider the entire function

Assuming that is a simple function it is easy to see that the map is bounded throughout the strip . Observe also that we have the bounds and . Using the three lines lemma we conclude that

for all and . Applying this bound for gives

for all simple functions of finite measure support and all . But this means that

for all and for simple functions of finite measure support, as we wanted to show. A limiting argument gives the log convexity for general functions.

Remark 3In fact, one can go the other direction and prove HĆ¶lder’s inequality by means of the log-convexity of the norm. Also, as in the case of HĆ¶lder’s inequality, it is not hard to verify that whenever such an estimate is true, the indices must be related as

To see this, apply the inequality replacing the measure by , where .

Exercise 5Use the three lines lemma to give a different proof of HĆ¶lder’s inequality.

Hint:Show HĆ¶lder’s inequality initially for simple functions with finite measure support. For this, apply the three lines lemma to the function

for . Use the fact that simple functions with finite measure support are dense in , . Fill in the details of the limiting argument (omitted in the previous proof).

** 3.2. Heuristic discussion and examples of spaces **

Let us now see a couple of specific examples of spaces which will come up often in this course.

Example 1The most common setting for this course will be the Euclidean setting, that is the measure space . A typical point in will be denoted by and denotes the -dimensional Lebesgue measure. For a set in we will many times write for its Lebesgue measure. Here, is the Lebesgue -algebra of subsets of and we will oftentimes omit it from the notation.

Example 2Consider the measure space , where is the -algebra of all subsets of . Here is the counting measure. Recall that for , the counting measure of is the cardinality of , typically denoted by , if is finite, and is defined to be if is infinite. Every subset of is clearly measurable with respect to . With these definitions taken as understood observe that, for example, the space is just the space of sequences on whose -th powers are summable, that is, the space of all sequences such that

These spaces come up so often in analysis that they deserve to have a special notation; we usually denote them by . Maybe this seems like an unnecessary complication to state a very simple definition. Observe, however, that once we put things in this language we automatically have all the tools from measure theory at our disposal.

Exercise 6Let be a sequence of elements in , that is, a sequence of sequences. For each positive integer we write . Assume that for each fixed , there is a complex number such that , that is, the sequence convergespointwiseto some sequence . State Lebesgue’s dominated convergence theorem in this setup. When can we interchange the limit with summation?

Example 3We denote by the torus, that is the quotient space where is the group of integral multiples of . Thus two points of are identified if the differ by an integral multiple of . There is a natural identification of functions on and -periodic functions on . The Lebesgue measure on can also be identified with the restriction of the Lebesgue measure of on the interval , or in fact, any interval in of length . Remember that the Lebesgue measure on is translation invariant. We equip with the Lebesgue -algebra. The integral of a function can thus be written aswhere is considered as a -periodic function on . The preceding definitions imply that the measure on is translation invariant. The Lebesgue spaces , , are defined in the obvious way. Since the total measure of is finite, an important feature of the spaces is that they are nested; for we have that , being the `smaller’ space. Furthermore this embedding is continuous. See also exercise 8.

We now briefly discuss why a function may fail to belong to for . For simplicity, let us focus on the real line and consider the obstructions for membershipĀ to the spaces . Very similar conclusions hold in the -dimensional Euclidean space. Roughly speaking, there are two main obstructions:

**The decay of the function at infinity.** Simply put, the function might not decay fast enough as for the integral of to be finite. The most naive example one can think of is a constant function, e.g. for some complex number . Obviously this function raised to any power cannot be integrable close to infinity. A slightly more subtle example is the function which agrees with for , i.e. . This function fails logarithmically to be in but belongs to for any . Of course we can similarly construct functions that decay even slower at infinity so that they fail to be in for as well. Thus, whenever a function belongs to some space for some , this imposes a control on the decay of at infinity. Increasing will only make things better at infinity, provided that the function already has some decay. Observe that this obstruction does not exist on a finite measure space. This is the case for the spaces for example.

**Blow up at local singularities.** Here it is enough to consider any compact set and study the behavior of the function locally. If the function is bounded on compact sets, i.e. if it is locally bounded, then the local behavior will not be an obstruction for the function to belong to some space. Things become more interesting when there is a local singularity around a point. Here we can consider again the function , close to zero this time. This function has a logarithmic singularity at zero, and thus it does not belong to . Observe here that we have forced the function to be zero away from the origin in order to isolate the obstruction. As increases to values , this function fails more and more dramatically to belong to since we raise this singularity to higher powers, thus presents a more severe singularity. The `solution’ here would be to consider the spaces for . Thus local singularities may also prevent a function from belonging to some space. Unlike the behavior at infinity, the local behavior of improves as we decrease . For example, the function fails to be in but clearly belongs to all spaces for .

Remark 4A function is in some space if and only if the function belongs to the space. Thus, there is no cancellation involved in the integrability of a function. This is an essential difference between the Lebesgue integral and the Riemann integral. The typical example here is to consider the function

Since is the harmonic series, is not Lebesgue integrable. However, is Riemann integrable since which converges. Thus, whenever a function oscillates, we expect some cancellation in its integral thatwill notbe reflected in the Lebesgue integrability of the function.

Exercise 7Based on the previous discussion, answer the following questions (it’s a simple calculation):

(i) Let be a given number. Based on the previous discussion, construct a function that belongs to for all but does not belong to . For example, for , a possible answer is the function . Also, construct a function that belongs to for all but does not belong to .

(ii) For and , consider the function . Characterize the values of as a function of so that the function belongs to . Consider all the range and calculate the norm of the function, whenever it is finite. (iii) For and , consider the function . Characterize the values of , as a function of , so that the function belongs to . Consider all the range and calculate the norm of the function, whenever it is finite.

Remark 5An interesting notion that is implicit in the previous discussion is that oflocal integrabilityof a function. A function is calledlocally integrableif for every compact set we have that

al We then write . Local integrability ignores the behavior of a function at infinity. We are thus left with only one obstruction: the possibility that has local singularities. Observe that if for any then will be locally integrable. Similarly we can define the space .

Exercise 8Give a heuristic explanation of the fact that if for any then (hint:what is the only obstruction for a function to be locally integrable?). Give a rigorous proof by means of HĆ¶lder’s inequality. Show also (which is the same) that on a finite measure space , we have that is continuously embedded in whenever . For this it is enough to show that

What is the best value of the implied constant in the previous inequality?

Exercise 9For consider the spaces of all complex sequences such that

Show that if we have that and the embedding is continuous

A space is calledgranularif there is constant such that for all measurable sets of positive measure. Show thatĀ for a granular space with constant we have that whenever with

whenever and . What is the best value of the implied constant?

Remark 6Note that the opposite embedding is true for spaces with . The explanation for this is quite simple. Sequences on (or ) cannot have local singularities so the only deciding factor for candidature to some space is decay at infinity. This also explains the embedding in this exercise. If a sequence belongs to some space, this means there is already sufficient decay at infinity for the series to be summable. Raising the exponent only improves the decay of as . A similar phenomenon occurs in general in granular spaces.

Exercise 10(i) Let and suppose that . Show that as .

(ii) If show that as .

**4. The dual space of **

Remember that for a Banach space over , its dual is the space of all bounded linear functionals . Let and define be the duality relation . For any we define the functional

by means of the formula

It is obvious that is linear and HĆ¶lder’s inequality shows that is continuous since

for all . Thus . Actually, in most cases the opposite is true, that is, *every* functional in is uniquely defined by a function in , whenever and the measure is -finite.

Theorem 7Let and . There exists a unique such that . The same is true when and the measure is -finite.

Remark 7Theorem 7 fails (in most cases) when . In fact the dual of can be characterized as a space of measures but we will not pursue that here. We have however that for all and , with -finite, we have that

Observe however that for this we need to knowa priorithat . A way to bypass this problem is to work with a dense subclass of functions. This is essentially a duality relation but the small point just mentioned doesn’t allow one to show that the dual of is (luckily since it’s not true!). It is however a very useful device since it allows very often to `linearize’ norms. Furthermore this duality relationship shows that the norm of the functional is . Thus isisometrically isomorphicto , being the dual exponent of , for and also for whenever the measure is -finite.

Exercise 11Show the duality relation (2) in the previous remark. This is essentially a consequence of HĆ¶lder’s inequality. Using this duality relation give an alternative proof of the triangle inequality.

Remark 8Density arguments allow us to restrict in (2) to belong to any dense subclass of .

**5. Weak spaces **

Going back to the example of the function , , recall that this function does not belong to . For the following estimate is obvious

On the other hand observe that for every function we have that

That is, for all functions the measure of the set behaves like .

In general, for any measure space we define for the space weak- or to be the space of all functions such that

for some constant . We define the weak- or the norm of a function to be the smaller constant such that (3) is true. Equivalently

Note that is not a norm since the triangle inequality fails. It is however a quasi-norm (the triangle inequality holds with a constant).

Exercise 12Show that for and we have the quasi-triangle inequality

Proposition 8Let . The space is continuously embedded in :

*Proof:* We just use Chebyshev’s inequality to write

for every .

Let us also recall how we can write the norm of a function in terms of the distribution function of :

Proposition 9For we have that

Exercise 13Prove Proposition 9 above.Hint:It is elementary to see that

Use Fubini’s theorem to complete the proof.

Exercise 14For and show that

*[Update 18 Feb, 2011: Error corrected in the proof of the log-convexity of the norm via the three lines lemma.]*

*[Update 28 Feb, 2011: Error corrected in the hypothesis of Exercise 2. Also typo in the Hint of Exercise 5 corrected]*

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