In the previous two quarters, we have been focusing largely on the “soft” side of real analysis, which is primarily concerned with “qualitative” properties such as convergence, compactness, measurability, and so forth. In contrast, we will begin this quarter with more of an emphasis on the “hard” side of real analysis, in which we study estimates and upper and lower bounds of various quantities, such as norms of functions or operators. (Of course, the two sides of analysis are closely connected to each other; an understanding of both sides and their interrelationships, are needed in order to get the broadest and most complete perspective for this subject.)

One basic tool in hard analysis is that of *interpolation*, which allows one to start with a hypothesis of two (or more) “upper bound” estimates, e.g. and , and conclude a family of intermediate estimates (or maybe , where is a constant) for any choice of parameter . Of course, interpolation is not a magic wand; one needs various hypotheses (e.g. linearity, sublinearity, convexity, or complexifiability) on in order for interpolation methods to be applicable. Nevertheless, these techniques are available for many important classes of problems, most notably that of establishing boundedness estimates such as for linear (or “linear-like”) operators from one Lebesgue space to another . (Interpolation can also be performed for many other normed vector spaces than the Lebesgue spaces, but we will just focus on Lebesgue spaces in these notes to focus the discussion.) Using interpolation, it is possible to reduce the task of proving such estimates to that of proving various “endpoint” versions of these estimates. In some cases, each endpoint only faces a portion of the difficulty that the interpolated estimate did, and so by using interpolation one has split the task of proving the original estimate into two or more simpler subtasks. In other cases, one of the endpoint estimates is very easy, and the other one is significantly more difficult than the original estimate; thus interpolation does not really simplify the task of proving estimates in this case, but at least clarifies the relative difficulty between various estimates in a given family.

As is the case with many other tools in analysis, interpolation is not captured by a single “interpolation theorem”; instead, there are a family of such theorems, which can be broadly divided into two major categories, reflecting the two basic methods that underlie the principle of interpolation. The *real interpolation method* is based on a divide and conquer strategy: to understand how to obtain control on some expression such as for some operator and some function , one would divide into two or more components, e.g. into components where is large and where is small, or where is oscillating with high frequency or only varying with low frequency. Each component would be estimated using a carefully chosen combination of the extreme estimates available; optimising over these choices and summing up (using whatever linearity-type properties on are available), one would hope to get a good estimate on the original expression. The strengths of the real interpolation method are that the linearity hypotheses on can be relaxed to weaker hypotheses, such as sublinearity or quasilinearity; also, the endpoint estimates are allowed to be of a weaker “type” than the interpolated estimates. On the other hand, the real interpolation often concedes a multiplicative constant in the final estimates obtained, and one is usually obligated to keep the operator fixed throughout the interpolation process. The proofs of real interpolation theorems are also a little bit messy, though in many cases one can simply invoke a standard instance of such theorems (e.g. the Marcinkiewicz interpolation theorem) as a black box in applications.

The *complex interpolation method* instead proceeds by exploiting the powerful tools of complex analysis, in particular the maximum modulus principle and its relatives (such as the Phragmén-Lindelöf principle). The idea is to rewrite the estimate to be proven (e.g. ) in such a way that it can be embedded into a family of such estimates which depend holomorphically on a complex parameter in some domain (e.g. the strip . One then exploits things like the maximum modulus principle to bound an estimate corresponding to an interior point of this domain by the estimates on the boundary of this domain. The strengths of the complex interpolation method are that it typically gives cleaner constants than the real interpolation method, and also allows the underlying operator to vary holomorphically with respect to the parameter , which can significantly increase the flexibility of the interpolation technique. The proofs of these methods are also very short (if one takes the maximum modulus principle and its relatives as a black box), which make the method particularly amenable for generalisation to more intricate settings (e.g. multilinear operators, mixed Lebesgue norms, etc.). On the other hand, the somewhat rigid requirement of holomorphicity makes it much more difficult to apply this method to non-linear operators, such as sublinear or quasilinear operators; also, the interpolated estimate tends to be of the same “type” as the extreme ones, so that one does not enjoy the upgrading of weak type estimates to strong type estimates that the real interpolation method typically produces. Also, the complex method runs into some minor technical problems when target space ceases to be a Banach space (i.e. when ) as this makes it more difficult to exploit duality.

Despite these differences, the real and complex methods tend to give broadly similar results in practice, especially if one is willing to ignore constant losses in the estimates or epsilon losses in the exponents.

The theory of both real and complex interpolation can be studied abstractly, in general normed or quasi-normed spaces; see e.g. this book for a detailed treatment. However in these notes we shall focus exclusively on interpolation for Lebesgue spaces (and their cousins, such as the weak Lebesgue spaces and the Lorentz spaces ).

** — 1. Interpolation of scalars — **

As discussed in the introduction, most of the interesting applications of interpolation occur when the technique is applied to operators . However, in order to gain some intuition as to why interpolation works in the first place, let us first consider the significantly simpler (though rather trivial) case of interpolation in the case of scalars or functions.

We begin first with scalars. Suppose that are non-negative real numbers such that

indeed one simply raises (1) to the power , (2) to the power , and multiplies the two inequalities together. Thus for instance, when one obtains the geometric mean of (1) and (2):

One can view and as the unique log-linear functions of (i.e. , are (affine-)linear functions of ) which equal their boundary values and respectively as .

Example 1If and for some and , then the log-linear interpolant is given by , where is the quantity such that .

The deduction of (3) from (1), (2) is utterly trivial, but there are still some useful lessons to be drawn from it. For instance, let us take for simplicity, so we are interpolating two upper bounds , on the same quantity to give a new bound . But actually we have a refinement available to this bound, namely

for any sufficiently small (indeed one can take any less than or equal to ). Indeed one sees this simply by applying (3) with with and and taking minima. Thus we see that (3) is only sharp when the two original bounds are comparable; if instead we have for some integer , then (6) tells us that we can improve (3) by an exponentially decaying factor of . The geometric series formula tells us that such factors are absolutely summable, and so in practice it is often a useful heuristic to pretend that the cases dominate so strongly that the other cases can be viewed as negligible by comparison.

Also, one can trivially extend the deduction of (3) from (1), (2) as follows: if is a function from to which is log-convex (thus is a convex function of , and (1), (2) hold for some , then (3) holds for all intermediate also, where is of course defined by (5). Thus one can interpolate upper bounds on log-convex functions. However, one certainly cannot interpolate lower bounds: lower bounds on a log-convex function at and yield no information about the value of, say, . Similarly, one cannot extrapolate upper bounds on log-convex functions: an upper bound on, say, and does not give any information about . (However, an upper bound on coupled with a *lower* bound on gives a lower bound on ; this is the contrapositive of an interpolation statement.)

Exercise 2Show that the sum , product , or pointwise maximum of two log-convex functions is log-convex.

Remark 3Every non-negative log-convex function is convex, thus in particular for all (note that this generalises the arithmetic mean-geometric mean inequality). Of course, the converse statement is not true.

Now we turn to the complex version of the interpolation of log-convex functions, a result known as Lindelöf’s theorem:

Theorem 4 (Lindelöf’s theorem)Let be a holomorphic function on the strip , which obeys the boundfor all and some constants . Suppose also that and for all . Then we have for all and , where is of course defined by (5).

Remark 5The hypothesis (7) is a qualitative hypothesis rather than a quantitative one, since the exact values of do not show up in the conclusion. It is quite a mild condition; any function of exponential growth in , or even with such super-exponential growth as or , will obey (7). The principle however fails without this hypothesis, as one can see for instance by considering the holomorphic function .

*Proof:* Observe that the function is holomorphic and non-zero on , and has magnitude exactly on the line for each . Thus, by dividing by this function (which worsens the qualitative bound (7) slightly) we may reduce to the case when for all .

Suppose we temporarily assume that as . Then by the maximum modulus principle (applied to a sufficiently large rectangular portion of the strip), it must then attain a maximum on one of the two sides of the strip. But on these two sides, and so on the interior as well.

To remove the assumption that goes to zero at infinity, we use the trick of giving ourselves an epsilon of room. Namely, we multiply by the holomorphic function for some . A little complex arithmetic shows that the function goes to zero at infinity in (the factor decays fast enough to damp out the growth of as , while the damps out the growth as ), and is bounded in magnitude by on both sides of the strip . Applying the previous case to this function, then taking limits as , we obtain the claim.

Exercise 6With the notation and hypotheses of Theorem 4, show that the function is log-convex on .

Exercise 7 (Hadamard three-circles theorem)Let be a holomorphic function on an annulus . Show that the function is log-convex on .

Exercise 8 (Phragmén-Lindelöf principle)Let be as in Theorem 4, but suppose that we have the bounds and for all and some exponents and a constant . Show that one has for all and some constant (which is allowed to depend on the constants in (7)). (Hint: it is convenient to work first in a half-strip such as for some large . Then multiply by something like for some suitable branch of the logarithm and apply a variant of Theorem 4 for the half-strip. A more refined estimate in this regard is due to Rademacher.) This particular version of the principle gives theconvexity boundfor Dirichlet series such as the Riemann zeta function. Bounds which exploit the deeper properties of these functions to improve upon the convexity bound are known assubconvexity boundsand are of major importance in analytic number theory, which is of course well outside the scope of this course.

** — 2. Interpolation of functions — **

We now turn to the interpolation in function spaces, focusing particularly on the Lebesgue spaces and the weak Lebesgue spaces . Here, is a fixed measure space. It will not matter much whether we deal with real or complex spaces; for sake of concretness we work with complex spaces. Then for , recall (see 245B Notes 3) that is the space of all functions whose norm

is finite, modulo almost everywhere equivalence. The space is defined similarly, but where is the essential supremum of on .

A simple test case in which to understand the norms better is that of a *step function* , where is a non-negative number and a set of finite measure. Then one has for . Observe that this is a log-convex function of . This is a general phenomenon:

Lemma 9 (Log-convexity of norms)Let that and . Then for all , and furthermore we havefor all , where the exponent is defined by .

In particular, we see that the function is log-convex whenever the right-hand side is finite (and is in fact log-convex for all , if one extends the definition of log-convexity to functions that can take the value ). In other words, we can interpolate any two bounds and to obtain for all .

Let us give several proofs of this lemma. We will focus on the case ; the endpoint case can be proven directly, or by modifying the arguments below, or by using an appropriate limiting argument, and we leave the details to the reader.

The first proof is to use Hölder’s inequality

when is finite (with some minor modifications in the case ).

Another (closely related) proof proceeds by using the log-convexity inequality

for all , where is the quantity such that . If one integrates this inequality in , one already obtains the claim in the normalised case when . To obtain the general case, one can multiply the function and the measure by appropriately chosen constants to obtain the above normalisation; we leave the details as an exercise to the reader. (The case when or vanishes is of course easy to handle separately.)

A third approach is more in the spirit of the real interpolation method, avoiding the use of convexity arguments. As in the second proof, we can reduce to the normalised case . We then split , where is the indicator function to the set , and similarly for . Observe that

and similarly

and so by the quasi-triangle inequality (or triangle inequality, when )

for some constant depending on . Note, by the way, that this argument gives the inclusions

This is off by a constant factor by what we want. But one can eliminate this constant by using the tensor power trick. Indeed, if one replaces with a Cartesian power (with the product -algebra and product measure ), and replace by the tensor power , we see from many applications of the Fubini-Tonelli theorem that

for all . In particular, obeys the same normalisation hypotheses as , and thus by applying the previous inequality to , we obtain

for every , where it is key to note that the constant on the right is independent of . Taking roots and then sending , we obtain the claim.

Finally, we give a fourth proof in the spirit of the complex interpolation method. By replacing by we may assume is non-negative. By expressing non-negative measurable functions as the monotone limit of simple functions and using the monotone convergence theorem, we may assume that is a simple function, which is then necessarily of finite measure support from the finiteness hypotheses. Now consider the function . Expanding out in terms of step functions we see that this is an analytic function of which grows at most exponentially in ; also, by the triangle inequality this function has magnitude at most when and magnitude when . Applying Theorem 4 and specialising to the value of for which we obtain the claim.

Exercise 10If , show that equality holds in Lemma 9 if and only if is a step function.

Now we consider variants of interpolation in which the “strong” spaces are replaced by their “weak” counterparts . Given a measurable function , we define the *distribution function* by the formula

This distribution function is closely connected to the norms. Indeed, from the calculus identity

and the Fubini-Tonelli theorem, we obtain the formula

for all , thus the norms are essentially moments of the distribution function. The norm is of course related to the distribution function by the formula

Exercise 11Show that we have the relationshipfor any measurable and , where we use to denote a pair of inequalities of the form for some constants depending only on . (Hint: is non-increasing in .) Thus we can relate the norms of to the dyadic values of the distribution function; indeed, for any , is comparable (up to constant factors depending on ) to the norm of the sequence .

Another relationship between the norms and the distribution function is given by observing that

for any , leading to Chebyshev’s inequality

(The version of this inequality is also known as Markov’s inequality. In probability theory, Chebyshev’s inequality is often specialised to the case , and with replaced by a normalised function . Note that, as with many other Cyrillic names, there are also a large number of alternative spellings of Chebyshev in the Roman alphabet.)

Chebyshev’s inequality motivates one to define the *weak norm* of a measurable function for by the formula

thus Chebyshev’s inequality can be expressed succinctly as

It is also natural to adopt the convention that . If are two functions, we have the inclusion

and hence

this easily leads to the quasi-triangle inequality

where we use as shorthand for the inequality for some constant depending only on (it can be a different constant at each use of the notation). [Note: in analytic number theory, it is more customary to use instead of , following Vinogradov. However, in analysis is sometimes used instead to denote “much smaller than”, e.g. denotes the assertion for some sufficiently small constant .]

Let be the space of all which have finite , modulo almost everywhere equivalence; this space is also known as weak . The quasi-triangle inequality soon implies that is a quasi-normed vector space with the quasi-norm, and Chebyshev’s inequality asserts that contains as a subspace (though the norm is not a restriction of the norm).

Example 12If with the usual measure, and , then the function is in weak , but not strong . It is also not in strong or weak for any other . But the “local” component of is in strong and weak for all , and the “global” component of is in strong and weak for all .

Exercise 13For any and , define the(dyadic) Lorentz normto be norm of the sequence , and define theLorentz spacebe the space of functions with finite, modulo almost everywhere equivalence. Show that is a quasi-normed space, which is equivalent to when and to when . Lorentz spaces arise naturally in more refined applications of the real interpolation method, and are useful in certain “endpoint” estimates that fail for Lebesgue spaces, but which can be rescued by using Lorentz spaces instead. However, we will not pursue these applications in detail here.

Exercise 14Let be a finite set with counting measure, and let be a function. For any , show that(Hint: to prove the second inequality, normalise , and then manually dispose of the regions of where is too large or too small.) Thus, in some sense, weak and strong are equivalent “up to logarithmic factors”.

One can interpolate weak bounds just as one can strong bounds: if and , then

for all . Indeed, from the hypotheses we have

and

for all , and hence by scalar interpolation (using an interpolation parameter defined by , and after doing some algebra) we have

As remarked in the previous section, we can improve upon (11); indeed, if we define to be the unique value of where and are equal, then we have

for some depending on . Inserting this improved bound into (9) we see that we can improve the weak-type bound (10) to a strong-type bound

for some constant . Note that one cannot use the tensor power trick this time to eliminate the constant as the weak norms do not behave well with respect to tensor product. Indeed, the constant must diverge to infinity in the limit if , otherwise it would imply that the norm is controlled by the norm, which is false by Example 12; similarly one must have a divergence as if .

Exercise 15Let and . Refine the inclusions in (8) to

Define the *strong type diagram* of a function to be the set of all for which lies in strong , and the *weak type diagram* to be the set of all for which lies in weak . Then both the strong and weak type diagrams are connected subsets of , and the strong type diagram is contained in the weak type diagram, and contains in turn the interior of the weak type diagram. By experimenting with linear combinations of the examples in Example 12 we see that this is basically everything one can say about the strong and weak type diagrams, without further information on or .

Exercise 16Let be a measurable function which is finite almost everywhere. Show that there exists a unique non-increasing left-continuous function such that for all , and in particular for all , and . (Hint: first look for the formula that describes for some in terms of .) The function is known as thenon-increasing rearrangementof , and the spaces and are examples ofrearrangement-invariant spaces. There are a class of useful rearrangement inequalities that relate to its rearrangements, and which can be used to clarify the structure of rearrangement-invariant spaces, but we will not pursue this topic here.

Exercise 17Let be a -finite measure space, let , and be a measurable function. Show that the following are equivalent:

- lies in , thus for some finite .
- There exists a constant such that for all sets of finite measure.
Furthermore show that the best constants in the above statements are equivalent up to multiplicative constants depending on , thus . Conclude that the modified weak norm , where ranges over all sets of positive finite measure, is a genuine norm on which is equivalent to the quasinorm.

Exercise 18Let be an integer. Find a probability space and functions with for such that for some absolute constant . (Hint: exploit the logarithmic divergence of the harmonic series .) Conclude that there exists a probability space such that the quasi-norm is not equivalent to an actual norm.

Exercise 19Let be a -finite measure space, let , and be a measurable function. Show that the following are equivalent:

- lies in .
- There exists a constant such that for every set of finite measure, there exists a subset with such that .

Exercise 20Let be a measure space of finite measure, and be a measurable function. Show that the following two statements are equivalent:

- There exists a constant such that for all .
- There exists a constant such that .

** — 3. Interpolation of operators — **

We turn at last to the central topic of these notes, which is interpolation of operators between functions on two fixed measure spaces and . To avoid some (very minor) technicalities we will make the mild assumption throughout that and are both -finite, although much of the theory here extends to the non--finite setting.

A typical situation is that of a linear operator which maps one space to another , and also maps to for some exponents ; thus (by linearity) will map the larger vector space to , and one has some estimates of the form

for all respectively, and some . We would like to then interpolate to say something about how maps to .

The complex interpolation method gives a satisfactory result as long as the exponents allow one to use duality methods, a result known as the Riesz-Thorin theorem:

Theorem 21 (Riesz-Thorin theorem)Let and . Let be a linear operator obeying the bounds (13), (14) for all respectively, and some . Then we havefor all and , where , , and .

Remark 22When is a point, this theorem essentially collapses to Lemma 9 (and when is a point, this is a dual formulation of that lemma); and when and are both points; this collapses to interpolation of scalars.

*Proof:* If then the claim follows from Lemma 9, so we may assume , which in particular forces to be finite. By symmetry we can take . By multiplying the measures and (or the operator ) by various constants, we can normalise (the case when or is trivial). Thus we have also.

By Hölder’s inequality, the bound (13) implies that

for all and , where is the dual exponent of . Similarly we have

for all , that are simple functions with finite measure support. To see this, we first normalise . Observe that we can write , for some functions of magnitude at most . If we then introduce the quantity

(with the conventions that in the endpoint case ) we see that is a holomorphic function of of at most exponential growth which equals when . When instead , an application of (15) shows that ; a similar claim obtains when using (16). The claim now follows from Theorem 4.

The estimate (17) has currently been established for simple functions with finite measure support. But one can extend the claim to any (keeping simple with finite measure support) by decomposing into a bounded function and a function of finite measure support, approximating the former in by simple functions of finite measure support, and approximating the latter in by simple functions of finite measure support, and taking limits using (15), (16) to justify the passage to the limit. One can then also allow arbitrary by using the monotone convergence theorem. The claim now follows from the duality between and .

Suppose one has a linear operator that maps simple functions of finite measure support on to measurable functions on (modulo almost everywhere equivalence). We say that such an operator is of *strong type * if it can be extended in a continuous fashion to an operator on to an operator on ; this is equivalent to having an estimate of the form for all simple functions of finite measure support. (The extension is unique if is finite or if has finite measure, due to the density of simple functions of finite measure support in those cases. Annoyingly, uniqueness fails for of an infinite measure space, though this turns out not to cause much difficulty in practice, as the conclusions of interpolation methods are usually for finite exponents .) Define the *strong type diagram* to be the set of all such that is of strong type . The Riesz-Thorin theorem tells us that if is of strong type and with and , then is also of strong type for all ; thus the strong type diagram contains the closed line segment connecting with . Thus the strong type diagram of is convex in at least. (As we shall see later, it is in fact convex in all of .) Furthermore, on the intersection of the strong type diagram with , the operator norm is a log-convex function of .

Exercise 23If with the usual measure, show that the strong type diagram of the identity operator is the triangle . If instead with the usual counting measure, show that the strong type diagram of the identity operator is the triangle . What is the strong type diagram of the identity when with the usual measure?

Exercise 24Let (resp. ) be a linear operator from simple functions of finite measure support on (resp. ) to measurable functions on (resp. ) modulo a.e. equivalence that are absolutely integrable on finite measure sets. We say areformally adjointif we have for all simple functions of finite measure support on respectively. If , show that is of strong type if and only if is of strong type . Thus, taking formal adjoints reflects the strong type diagram around the line of duality , at least inside the Banach space region .

Remark 25There is a powerful extension of the Riesz-Thorin theorem known as theStein interpolation theorem, in which the single operator is replaced by a family of operators for that vary holomorphically in in the sense that is a holomorphic function of for any sets of finite measure. Roughly speaking, the Stein interpolation theorem asserts that if is of strong type for with a bound growing at most exponentially in , and itself grows at most exponentially in in some sense, then will be of strong type . A precise statement of the theorem and some applications can be found in Stein’s book on harmonic analysis.

Now we turn to the real interpolation method. Instead of linear operators, it is now convenient to consider *sublinear operators* mapping simple functions of finite measure support in to -valued measurable functions on (modulo almost everywhere equivalence, as usual), obeying the homogeneity relationship

and the pointwise bounds

and

for all , and all simple functions of finite measure support.

Every linear operator is sublinear; also, the absolute value of a linear (or sublinear) operator is also sublinear. More generally, any *maximal operator* of the form , where is a family of sub-linear operators, is also a non-negative sublinear operator; note that one can also replace the supremum here by any other norm in , e.g. one could take an norm for any . (After and , a particularly common case is when , in which case is known as a *square function*.)

The basic theory of sublinear operators is similar to that of linear operators in some respects. For instance, continuity is still equivalent to boundedness:

Exercise 26Let be a sublinear operator, and let . Assume that either is finite, or has finite measure. Then the following are equivalent:

- can be extended to a continuous operator from to .
- There exists a constant such that for all simple functions of finite measure support.
- can be extended to a operator from to such that for all and some .
Show that the extension mentioned above is unique. Finally, show that the same equivalences hold if is replaced by throughout.

We say that is of *strong type * if any of the above equivalent statements (for ) hold, and of *weak type * if any of the above equivalent statements (for ) hold. We say that a linear operator is of strong or weak type if its non-negative counterpart is; note that this is compatible with our previous definition of strong type for such operators. Also, Chebyshev’s inequality tells us that strong type implies weak type .

We now give the real interpolation counterpart of the Riesz-Thorin theorem, namely the Marcinkeiwicz interpolation theorem:

Theorem 27 (Marcinkiewicz interpolation theorem)Let and be such that , and for . Let be a sublinear operator which is of weak type and of weak type . Then is of strong type .

Remark 28Of course, the same claim applies to linear operators by setting . One can also extend the argument toquasilinearoperators, in which the pointwise bound is replaced by for some constant , but this generalisation only appears occasionally in applications. The conditions can be replaced by the variant condition (see Exercise 31, Exercise 33), but cannot be eliminated entirely: see Exercise 32. The precise hypotheses required on are rather technical and I recommend that they be ignored on a first reading.

*Proof:* For notational reasons it is convenient to take finite; however the arguments below can be modified without much difficulty to deal with the infinite case (or one can use a suitable limiting argument); we leave this to the interested reader.

By hypothesis, there exist constants such that

for all simple functions of finite measure support, and all . Let us write to denote for some constant depending on the indicated parameters. By (9), it will suffice to show that

By homogeneity we can normalise .

Actually, it will be more slightly convenient to work with the dyadic version of the above estimate, namely

see Exercise 11. The hypothesis similarly implies that

The basic idea is then to get enough control on the numbers in terms of the numbers that one can deduce (20) from (21).

When , the claim follows from direct substitution of (18), (19) (see also the discussion in the previous section about interpolating strong bounds from weak ones), so let us assume ; by symmetry we may take , and thus . In this case we cannot directly apply (18), (19) because we only control in , not or . To get around this, we use the basic real interpolation trick of *decomposing* into pieces. There are two basic choices for what decomposition to pick. On one hand, one could adopt a “minimalistic” approach and just decompose into two pieces

where and , and the threshold is a parameter (depending on ) to be optimised later. Or we could adopt a “maximalistic” approach and perform the dyadic decomposition

where . (Note that only finitely many of the are non-zero, as we are assuming to be a simple function.) We will adopt the latter approach, in order to illustrate the dyadic decomposition method; the former approach also works, but we leave it as an exercise to the interested reader.

From sublinearity we have the pointwise estimate

which implies that

whenever are positive constants such that , but for which we are otherwise at liberty to choose. We will set aside the problem of deciding what the optimal choice of is for now, and continue with the proof.

From (18), (19), we have two bounds for the quantity , namely

and

From construction of we can bound

and similarly for , and thus we have

for . To prove (20), it thus suffices to show that

It is convenient to introduce the quantities appearing in (21), thus

and our task is to show that

Since , we have , and so we are reduced to the purely numerical task of locating constants with for all such that

We can simplify this expression a bit by collecting terms and making some substitutions. The points are collinear, and we can capture this by writing

for some and some . We can then simplify the left-hand side of (22) to

Note that is positive and is negative. If we then pick to be a sufficiently small multiple of where (say), we obtain the claim by summing geometric series.

Remark 29A closer inspection of the proof (or a rescaling argument to reduce to the normalised case , as in preceding sections) reveals that one establishes the estimatefor all simple functions of finite measure support (or for all , if one works with the continuous extension of to such functions), and some constant . Thus the conclusion here is weaker by a multiplicative constant from that in the Riesz-Thorin theorem, but the hypotheses are weaker too (weak-type instead of strong-type). Indeed, we see that the constant must blow up as or .

The power of the Marcinkiewicz interpolation theorem, as compared to the Riesz-Thorin theorem, is that it allows one to weaken the hypotheses on from strong type to weak type. Actually, it can be weakened further. We say that a non-negative sublinear operator is *restricted weak-type * for some if there is a constant such that

for all sets of finite measure and all simple functions with . Clearly restricted weak-type is implied by weak-type , and thus by strong-type . (One can also define the notion of *restricted strong-type * by replacing with ; this is between strong-type and restricted weak-type , but is incomparable to weak-type .)

Exercise 30Show that the Marcinkiewicz interpolation theorem continues to hold if the weak-type hypotheses are replaced by restricted weak-type hypothesis. (Hint: where were the weak-type hypotheses used in the proof?)

We thus see that the strong-type diagram of contains the interior of the restricted weak-type or weak-type diagrams of , at least in the triangular region .

Exercise 31Suppose that is a sublinear operator of restricted weak-type and for some . Show that is of restricted weak-type for any , or in other words the restricted type diagram is convex in . (This is an easy result requiring only interpolation of scalars.) Conclude that the hypotheses in the Marcinkiewicz interpolation theorem can be replaced by the variant .

Exercise 32For any , let be the natural numbers with the weighted counting measure , thus each point has mass . Show that if , then the identity operator from to is of weak-type but not strong-type when and . Conclude that the hypotheses cannot be dropped entirely.

Exercise 33Suppose we are in the situation of the Marcinkiewicz interpolation theorem, with the hypotheses replaced by . Show that for all and there exists a such thatfor all simple functions of finite measure support, where the Lorentz norms were defined in Exercise 13. (Hint: repeat the proof of the Marcinkiewicz interpolation theorem, but partition the sum into regions of the form for integer . Obtain a bound for each summand which decreases geometrically as .) Conclude that the hypotheses in the Marcinkiewicz interpolation theorem can be replaced by . This Lorentz space version of the interpolation theorem is in some sense the “right” version of the theorem, but the Lorentz spaces are slightly more technical to deal with than the Lebesgue spaces, and the Lebesgue space version of Marcinkiewicz interpolation is largely sufficient for most applications.

Exercise 34For , let be -finite measure spaces, and let be a linear operator from simple functions of finite measure support on to measurable functions on (modulo almost everywhere equivalence, as always). Let , be the product spaces (with product -algebra and product measure). Show that there exists a unique (modulo a.e. equivalence) linear operator defined on linear combinations of indicator functions of product sets of sets , of finite measure, such thatfor a.e. ; we refer to as the

tensor productof and and write . Show that if are of strong-type for some with operator norms respectively, then can be extended to a bounded linear operator on to with operator norm exactly equal to , thus(Hint: for the lower bound, show that for all simple functions . For the upper bound, express as the composition of two other operators and for some identity operators , and establish operator norm bounds on these two operators separately.) Use this and the tensor power trick to deduce the Riesz-Thorin theorem (in the special case when for , and ) from the Marcinkiewicz interpolation theorem. Thus one can (with some effort) avoid the use of complex variable methods to prove the Riesz-Thorin theorem, at least in some cases.

Exercise 35 (Hölder’s inequality for Lorentz spaces)Let and for some . Show that , where and , with the estimatefor some constant . (This estimate is due to O’Neil.)

Remark 36Just as interpolation of functions can be clarified by using step functions as a test case, it is instructive to use rank one operators such aswhere are finite measure sets, as test cases for the real and complex interpolation methods. (After understanding the rank one case, I then recommend looking at the rank two case, e.g. , where could be very different in size from .)

** — 4. Some examples of interpolation — **

Now we apply the interpolation theorems to some classes of operators. An important such class is given by the *integral operators*

from functions to functions , where is a fixed measurable function, known as the *kernel* of the integral operator . Of course, this integral is not necessarily convergent, so we will also need to study the sublinear analogue

which is well-defined (though it may be infinite).

The following useful lemma gives us strong-type bounds on and hence , assuming certain type bounds on the rows and columns of .

Lemma 37 (Schur’s test)Let be a measurable function obeying the boundsfor almost every , and

for almost every , where and . Then for every , and are of strong-type , with well-defined for all and almost every , and furthermore

Here we adopt the convention that and , thus and .

*Proof:* The hypothesis , combined with Minkowski’s integral inequality, shows us that

for all ; in particular, for such , is well-defined almost everywhere, and

Similarly, Hölder’s inequality tells us that for , is well-defined everywhere, and

Applying the Riesz-Thorin theorem we conclude that

for all simple functions with finite measure support; replacing with we also see that

for all simple functions with finite measure support, and thus (by monotone convergence) for all . The claim then follows.

Example 38Let be a matrix such that the sum of the magnitudes of the entries in every row and column is at most , i.e. for all and for all . Then one has the boundfor all vectors and all . Note the extreme cases , can be seen directly; the remaining cases then follow from interpolation.

A useful special case arises when is an

-sparsematrix, which means that at most entries in any row or column are non-zero (e.g. permutation matrices are -sparse). We then conclude that the operator norm of is at most .

Exercise 39Establish Schur’s test by more direct means, taking advantage of the duality relationshipfor , as well as Young’s inequality for . (You may wish to first work out Example 38, say with , to figure out the logic.)

A useful corollary of Schur’s test is Young’s convolution inequality for the convolution of two functions , , defined as

provided of course that the integrand is absolutely convergent.

Exercise 40 (Young’s inequality)Let be such that . Show that if and , then is well-defined almost everywhere and lies in , and furthermore that(Hint: Apply Schur’s test to the kernel .)

Remark 41There is nothing special about here; one could in fact use any locally compact group with a bi-invariant Haar measure. On the other hand, if one specialises to , then it is possible to improve Young’s inequality slightly, towhere , a result of Beckner; the constant here is best possible, as can be seen by testing the inequality in the case when are Gaussians.

Exercise 42Let , and let , . Young’s inequality tells us that . Refine this further by showing that , i.e. is continuous and goes to zero at infinity. (Hint:first show this when , then use a limiting argument.)

We now give a variant of Schur’s test that allows for weak estimates.

Lemma 43 (Weak-type Schur’s test)Let be a measurable function obeying the boundsfor almost every , and

for almost every , where and (note the endpoint exponents are now excluded). Then for every , and are of strong-type , with well-defined for all and almost every , and furthermore

Here we again adopt the convention that and .

*Proof:* From Exercise 17 we see that

for any measurable , where we use to denote for some depending on the indicated parameters. By the Fubini-Tonelli theorem, we conclude that

for any ; by Exercise 17 again we conclude that

thus is of weak-type . In a similar vein, from yet another application of Exercise 17 we see that

whenever and has finite measure; thus is of restricted type . Applying Exercise 30 we conclude that is of strong type (with operator norm ), and the claim follows.

This leads to a weak-type version of Young’s inequality:

Exercise 44 (Weak-type Young’s inequality)Let be such that . Show that if and , then is well-defined almost everywhere and lies in , and furthermore thatfor some constant .

Exercise 45Refine the previous exercise by replacing with the Lorentz space throughout.

Recall that the function will lie in for . We conclude

Corollary 46 (Hardy-Littlewood-Sobolev fractional integration inequality)Let and be such that . If , then the function , defined asis well-defined almost everywhere and lies in , and furthermore that

for some constant .

This inequality is of importance in the theory of Sobolev spaces, which we will discuss in a subsequent lecture.

Exercise 47Show that Corollary 46 can fail at the endpoints , , or .

*Update*, Apr 6: another exercise added; note renumbering.

*Update*, Apr 8: some formatting errors fixed.

*Update*, Sep 14: definition of sublinearity fixed.

## 121 comments

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30 March, 2009 at 10:15 am

AnonFollowing up on an old comment: https://terrytao.wordpress.com/2007/05/10/open-question-boundedness-of-the-trilinear-hilbert-transform/#comment-33400

I have often wondered if there is a counterexample to a multilinear interpolation statement of the following form. If T(f,g) is a multilinear operator given by convolution with an appropriate kernel (where we have the bilinear Hilbert Transform in mind as an example) such that T(f,1) and T(1,g) are bounded linear operators (say from L^p to L^p, for a range of p’s), then T(f,g) satisfies some Holder-type multilinear bounds (after ruling out degenerate cases such as $\int f(x+t)g(x+t)/t$). At a high level my question is, should all of these multilinear results people have been working on lately really follow from some multilinear interpolation theorem that hasn’t been found yet?

30 March, 2009 at 12:25 pm

Anonfyi, example 2, 10 are displaying errors.

[Fixed – T]30 March, 2009 at 6:04 pm

The discussion of Real Analysis:1 « Liu Xiaochuan’s Weblog[…] title of this post is:’Interpolation of L^p spaces‘, of which I have no idea at all before […]

31 March, 2009 at 5:21 am

Terence TaoDear Anon,

Well, it’s easy to cook up bilinear operators T(f,g) which actually vanish when f=1 or g=1, but are non-trivial otherwise (e.g. let’s work on a compact domain such as the circle, and consider the product of the mean zero version of f with the mean zero version of g), so it is quite unlikely that one can get much control on a general T(f,g) out of knowledge of T(f,1) and T(1,g) for various f, g.

This is in contrast to the linear setting, where we of course have the T(1) theorem. The difference there is that if one looks at the bilinear form for a Calderon-Zygmund operator T and starts decomposing it into various components, one soon sees that the dominant components either come from the terms where g is localised and f is slowly varying (and can thus be modeled to some extent by a multiple of 1), or when f is localised and g is slowly varying, or when f and g are both localised and bounded to the same interval. These three cases can be handled by an appropriate hypothesis on T(1), an appropriate hypothesis on , and a weak boundedness hypothesis respectively.

For the bilinear Hilbert transform, we have the frequency modulation invariance which dethrones the privileged nature of the function 1. Instead, a more reasonable hypothesis would be to assume control of things like and for arbitrary frequencies (and one also needs to consider the adjoint expression ). In some ways, one can view the existing proofs of the boundedness of the bilinear Hilbert transform as a reduction to these model cases; the general trilinear form is decomposed into many components, which are then reassembled into “trees” in phase space, each of which can be estimated by a paraproduct analysis resembling that needed to understand (variants) of expressions such as . So the “divide and conquer” philosophy underlying real interpolation is still very much present in the proofs, even if they don’t show up on the level of formal interpolation theorems. (This tends to be typical with any advanced form of harmonic analysis; the general philosophies are common to many problems, but are difficult to formalise as a universal lemma or other abstract tool; instead, each individual operator or expression needs to be manged “by hand”.)

On the other hand, I could well imagine that the abstract combinatorics of collecting tiles into trees, trees into forests, and then summing up over forests could be formalised into some sort of interpolation result concerning L^p-like norms of sequences of numbers indexed by tiles. This could perhaps provide a unifying framework for many of the bilinear Hilbert transform-like results out there in the literature.

31 March, 2009 at 9:34 pm

anon2I believe the statement of LindelĂ¶fâ€™s theorem (theorem 1) in equation (7) has some typos.

Also, I think the sentence beginning “Thus we see that (3) is only sharp when…” has a few typos that make it a bit confusing.

31 March, 2009 at 10:22 pm

Anonymousin the definition of distribution function the image may be infinity. Why do we take ?

31 March, 2009 at 10:31 pm

Anonymousin lemma 5, in the second condition is it or ?

1 April, 2009 at 8:20 am

Terence TaoThanks for the corrections! (But I believe the exponent in Lemma 5 is indeed as stated.)

1 April, 2009 at 9:27 am

AnonymousThanks I was not sure I just wanted to check it.

1 April, 2009 at 10:11 am

AnonymousIn second paragraph of the proof of Theorem 1, it seems the limit is .

[Corrected, thanks – T.]2 April, 2009 at 5:07 am

liuxiaochuanDear Professor Tao:

I got stuck in Exercise 4. Are you sure the mutiplier in the hint is correct?

2 April, 2009 at 7:07 am

Terence TaoDear Liuxiaochuan: Ah, yes. For that multiplier to work, one has to apply a Lindelof-type theorem to a half-strip rather than a strip. (It is possible to work on the strip directly, but one has to use a more sophisticated multiplier.) I’ve adjusted the hint accordingly.

2 April, 2009 at 10:07 pm

liuxiaochuanDear Professor Tao:

I still have soubts about exercise 4. I am wondering if the multiplier should be for upper half strip, instead of

3 April, 2009 at 7:05 am

Terence TaoDear Liuxiaochuan,

I think the multiplier is correct as it stands. One needs the imaginary part of the logarithmic factor to decay like O(1/t) so as not to create an exponential term in the magnitude of the multiplier. With , the imaginary part is close to (or , etc. depending on what branch of the logarithm one takes), and this will lead the multiplier growing or decaying exponentially in t (depending on what and are).

One can take instead of if one is worried about the pole at the origin, but if one is working in a sufficiently high half-strip, this is pole is not a problem.

3 April, 2009 at 4:45 pm

liuxiaochuanDear Professor:

Thank you so much, I finally got this exercise! ha!

4 April, 2009 at 10:54 am

PDEbeginnerDear Prof. Tao,

There seems some problem for the choices of above excercise 5. I guess the first is . I was wondering if you could help to explain a little or give a hint about the motivation of using Lorentz space. Is it related to the Lorentz transform in the special relativity? The Lorentz space seems related to Besov space (I tried to find some explanation on wiki, unfortunately it only gives a very brief definition.)

Thanks a lot in avdance.

4 April, 2009 at 11:08 am

Terence TaoThanks for the correction! Lorentz spaces were created by the mathematician George Lorentz, who I believe is unrelated to the physicist Hendrik Lorentz of Lorentz transformation fame.

Lorentz spaces are a “logarithmic” refinement of the Lebesgue spaces that can help discern finer integrability properties of a function. For instance, consider a function f on [0,1] of the form . Knowing which spaces f belongs to tells you what a is, but not what b is; but knowing what spaces f lies in reveals both indices. (The same, incidentally, is true for the Besov space norms, although the additional exponent in the Besov space definition measures distribution in dyadic frequency ranges , rather than distribution in magnitudes .)

6 April, 2009 at 2:58 pm

254C, Notes 2: The Fourier transform « What’s new[…] Young’s inequality (Exercise 25 of Notes 1) we know that is defined a.e., and lies in ; indeed, we […]

8 April, 2009 at 11:11 am

SamirDear Professor Tao,

Do you think you could provide a little advice for exercise 2? I’m surprisingly having more trouble with that than some of the later ones! Thanks!

Samir

8 April, 2009 at 12:34 pm

SamirHi Professor Tao,

I think I’ve proved the assertion in ex. 2, so feel free to disregard that last comment unless it serves the many other readers of your blog, or current students in 245C.

11 April, 2009 at 11:47 pm

liuxiaochuanDear Professor Tao:

In the last sentence before Lemma 2, I believe what you really meant is ‘ a log-convex function of , if I’m right.

12 April, 2009 at 2:07 am

liuxiaochuanDear Professor Tao:

In Exercise 5, should as a step function also be asked to only take two values(one be 0)?

12 April, 2009 at 10:06 am

Terence TaoDear Liuxiaochuan,

Actually I do mean 1/p, i.e. the function is a (non-strictly) log-convex function of t. In interpolation, 1/p is the more important parameter than p.

The property of being a step function (i.e. being equal to for some A, E) is equivalent to that of taking two values, one of them zero.

14 April, 2009 at 6:59 pm

liuxiaochuanDear Professor Tao:

In exercise 6, I don’t underdtand the first line after the formular. Are there any typos?

14 April, 2009 at 8:18 pm

Terence TaoAh, wordpress ate the < and > signs again. Fixed now.

16 April, 2009 at 2:10 pm

SamirDear Professor Tao,

In lemma 2, you state: is a log-convex function of . Shouldn’t this read: is a log-convex function…? Alternatively, if we leave the log function as written, then it is a convex function, correct?

Samir

[Corrected, thanks – T]20 April, 2009 at 10:37 pm

liuxiaochuanDear Professor Tao:

I failed to figure out exercise 8, even with the hint. Could your please help me out? How to dispose of the regions of X where f is too large or too small?

21 April, 2009 at 5:53 am

PDEbeginnerDear Prof. Tao and Xiaochuan,

I also tried this exercise, and did it in the following way: W.l.o.g. we assume , by , we have . Hence, If I am wrong, please help to delete it.

21 April, 2009 at 8:27 am

liuxiaochuanDear PDEbeginner:

I guess you are right. Thank you so much.

But in this way, the hint is not used, maybe Professor Tao’s hint points out another method of this exercise. Let’s just wait for his summarization. :)

21 April, 2009 at 10:55 am

Terence TaoThere is more than one way to solve the problem. The way the hint suggests is to discard all the points x for which |f(x)| exceeds 1 (this is the empty set) or is less than (this is not the empty set, but gives an acceptable contribution to the norm of ). The remainder can be dealt with by the distributional formula for the L^p norm.

I’ve put the solution on the tricki article

http://www.tricki.org/article/Control_level_sets

as it is a good example of the technique.

23 April, 2009 at 4:43 am

liuxiaochuanDear Professor Tao:

In(10), I guess is not defined by , but rather , am I right?

23 April, 2009 at 9:38 am

Terence TaoActually, I want to keep as the harmonic average , in order to keep the final constant as . When performing the scalar interpolation, one would need to use a different parameter, say , defined by , but if one does the algebra one can check that the constant in (11) is indeed at the end of the day.

25 April, 2009 at 7:04 am

liuxiaochuanDear Professor Tao:

In exercise 12, why can the inequality ensure that the quasi-norm is not equivalent to an actual norm? I think it should be at least larger than n.

27 April, 2009 at 10:23 am

Terence TaoDear Liuxiaochuan,

There was a misprint in that exercise; the should be .

30 April, 2009 at 9:34 pm

245C, Notes 4: Sobolev spaces « What’s new[…] integration . By the Hardy-Littlewood-Sobolev theorem on fractional integration (Corollary 7 of Notes 1) we conclude […]

30 June, 2009 at 5:34 pm

liuxiaochuanDear Professor Tao:

In exercise 13, what is p’ when ?

30 June, 2009 at 7:00 pm

Terence TaoDear Liuxiaochuan,

will be a negative number in this case.

1 July, 2009 at 8:19 am

AnonymousDear Prof. Tao.

It looks like that by the trick of collapsing to a point in Marcinkiewicz’s interpolation theorem we can expect the following inequality to be true:

$latex$

Is this right? Thank you very much

[I am guessing you are encountering the bug that the < and > signs inside a latex environment in WordPress are sometimes interpreted as HTML. Try < and > instead. -T]1 July, 2009 at 11:35 am

AnonymousIÂ´ll try again…

It looks like that by the trick of collapsing to a point in Marcinkiewiczâ€™s interpolation theorem we can expect the following inequality to be true:

Thanks in advance.

1 July, 2009 at 9:35 pm

Terence TaoYes, this is correct (assuming the usual hypotheses on , of course). One can also establish this bound more directly by using the distribution function of f.

16 July, 2009 at 9:58 am

SpencerDear All,

Perhaps someone reading can help me out here…

I think understand the strengthening of (3) to get (6) but am getting confused about the strengthening of the bound on the distribution function in (11) to get the next line.

Is this performed in a similar way? – If so perhaps I am just lost in the algebra of the exponents. And what is the relevance of t_0? Or just notationally convenient?

Many Thanks to anyone who can help,

Spence

16 July, 2009 at 10:10 am

Terence TaoDear Spence,

I would recommend drawing a graph of the quantities , , , and as functions of t, to see what is going on. (The picture is clearest if one uses a log-log plot, so that all functions become linear.)

16 July, 2009 at 3:22 pm

SpencerDear Professor Tao,

Much clearer now. Thank you very much.

Best,

Spencer

17 July, 2009 at 6:03 am

SpencerDear Prof. Tao,

In the last paragraph of the Riesz-Thorin proof you have the sentence beginnning: “But one can extend the claim to any (keeping simple with finite measure support)…..”

You describe a decomposition of and then how to approximate the two parts separately. However, as it stands at the moment, is seems to me that the two approximations are exactly the same and work would as one. I may be wrong but perhaps the second occurence of is in fact ?

I’m also fairly sure that the phantom equation (18) is in fact line (16), though the link does snap back to (16).

Best Regards,

Spencer

17 July, 2009 at 7:55 am

Terence TaoSpencer: Thanks for the corrections!

11 August, 2009 at 1:37 pm

PDEbeginnerDear Prof. Tao,

There seems some problem for (6). The following case seems a counterexample:

,

Take .

I think the RHS should be .

11 August, 2009 at 1:47 pm

Terence Tao(6) is specific to the special case , as mentioned a few lines earlier.

15 August, 2009 at 12:07 pm

PDEbeginnerDear Prof. Tao,

In the proof of Theorem 1, the term of seems to be .

15 August, 2009 at 1:41 pm

PDEbeginnerOops, I made a mistake in the above comments, because I printed out all the notes several months ago and read the old version. I will be more careful before making a comments.

8 September, 2009 at 12:17 am

PDEbeginnerDear Prof. Tao,

I think should be $[0,\infty] \times [0,1]$ just above exercise 15.

There seems some problems in the definition of sublinear operator just above exercise 17: and . If we take , there seems some problems in exercise 17. On the other hand, the definition of sublinear function in wiki also has no absolute value.

[Corrected, thanks – T.]7 October, 2009 at 10:24 am

PDEbeginnerDear Prof. Tao,

I just finished reading this note, there are two small corrections (I think):

1. In the proof of Theorem 4, the expression below ‘To prove (20)’ seems to forget a sum over .

2. In the last exercise, ‘(7)’ should be ‘Corollary 7’

I also have a question: Can we establish the weak-type Schur’s test by a more direct approach as in exercise 24? I tried to do it, but failed.

Thanks a lot in advance!

7 October, 2009 at 3:06 pm

Terence TaoThanks for the corrections!

One can prove Schur’s test directly if one knows about the duality between and , but this requires a small amount of knowledge of Lorentz spaces (actually much of the theory required is already buried in the above notes in various places, particularly in the exercises). Ultimately the proof is basically the same as the one given here, though, after one unpacks all the definitions and theory.

3 January, 2010 at 10:21 pm

254A, Notes 1: Concentration of measure « What’s new[…] The idea is then to estimate the tail of and by two different means. With , the point is that the variables have been made bounded by fiat, and so the more powerful large deviation inequalities can now be put into play. With , in contrast, the underlying variables are certainly not bounded, but they tend to have small zeroth and first moments, and so the bounds based on those moment methods tend to be powerful here. (Readers who are familiar with harmonic analysis may recognise this type of divide and conquer argument as an interpolation argument.) […]

5 January, 2010 at 4:20 pm

254A, Notes 0: A review of probability theory « What’s new[…] The idea is then to estimate the tail of and by two different means. With , the point is that the variables have been made bounded by fiat, and so the more powerful large deviation inequalities can now be put into play. With , in contrast, the underlying variables are certainly not bounded, but they tend to have small zeroth and first moments, and so the bounds based on those moment methods tend to be powerful here. (Readers who are familiar with harmonic analysis may recognise this type of divide and conquer argument as an interpolation argument.) […]

22 March, 2010 at 11:58 am

JacquesDear Pr. Tao, it seems to me that there is a typo in the statement of the Riesz-Thorin theorem. In the definition of $p_theta$ and $q_theta$, there should be brackets around 1-\theta.

[Corrected, thanks – T.]4 April, 2010 at 8:22 am

pavel zorinDear Prof. Tao,

some errata for your proof of the Marcinkiewicz interpolation theorem:

In equation (22), there should be no summation over m.

In the last equation (“simplify the left-hand side of (22) toâ€¦”), the summation goes over n instead of m, and the exponent of is , not .

This affects the choice of the c’s, i.e. with would do. The normalization sequence satisfies $d(n) = O(1)$ automatically given .

I also wonder whether, at the end of the proof of the Riesz-Thorin theorem, the decomposition of f into a bounded and a finitely supported function could be omitted in case (when the sum of the norms of the components is greater then the norm of f itself). It does provide continuity of (with norm 2), so that the conclusion follows because it actually has norm 1 on a dense subspace (of simple functions), but can one do without it?

best regards, pavel

4 April, 2010 at 9:21 am

Terence TaoThanks for the correction!

One can try to extend the main argument directly to general functions rather than just to simple functions with finite measure support; from a conceptual viewpoint this is more natural, but from a technical viewpoint it basically requires one to establish the continuity of to beforehand to justify various formal operations, which is inconvenient as this is part of what one is trying to prove in the first place.

7 April, 2010 at 2:13 pm

pavel zorinDear Prof. Tao,

I am experiencing slight difficulties with Exercise 21 (Marcinkiewicz theorem, Lorentz space version).

Doing the same calculations as in the given proof of the special case, I obtain (this time with and ), as a sufficient condition the existence of summable (over m) c’s such that

.

Now, assuming the easiest case , the exponent of is less than 1 for one of the i's (unless ). Even then, I could not think of an approach to finding c's that uses all the information: whatever I came up with looked like it would work without any assumptions on a's and was therefore false. Could you give some further indications?

kind regards, pavel

7 April, 2010 at 3:02 pm

Terence TaoAh, yes, this is a little delicate, in part because the decomposition of f used here is actually not the optimal one for this problem – it decomposes the height of the function dyadically (vertical dyadic decomposition), when in fact it is the width that ought to be dyadically decomposed (horizontal dyadic decomposition). One can still recover from this point by decomposing the sequence into dyadic pieces, but this is quite messy.

Another approach is to use frequency envelopes. Pick a small and replace by the slightly larger envelope . The point of doing so is that the obey a Lipschitz property but are still summable in . One then chooses to be adapted to the crossover point between the two quantities in the (modifying the term by some multiple of ); as long as the Lipschitz parameter is small enough, one will still be able to close the argument.

I may elaborate on the frequency envelope approach in a subsequent blog post.

14 April, 2011 at 4:28 pm

Wenying GanDear Prof. Tao:

I think there is a typo in $Ex 8$. There should be a index 1/p for log(1+|X|). Or the result we need to prove is incorrect. Thanks in advance.

Best

Wenying Gan

[Corrected, thanks – T.]4 November, 2012 at 6:13 pm

Gandhi ViswanathanDear Professor Tao,

The “Control Level Sets” tricki page you cite earlier also lacks the (1/p) index mentioned in this comment about Ex 8. Can we assume the same correction is needed in the tricki ?

4 November, 2012 at 6:21 pm

Terence TaoYes; I edited the tricki page accordingly.

4 November, 2012 at 6:49 pm

Gandhi ViswanathanThanks!

3 May, 2011 at 6:22 pm

Stein’s interpolation theorem « What’s new[…] where are functions depending on in a suitably analytic manner, for instance taking for some test function , and similarly for . If are chosen properly, will depend analytically on as well, and the two hypotheses (1), (2) give bounds on and for respectively. The Lindelöf theorem then gives bounds on intermediate values of , such as ; and the Riesz-Thorin theorem can then be deduced by a duality argument. (This is covered in many graduate real analysis texts; I myself covered it here.) […]

4 April, 2012 at 12:50 am

XDear Prof. Tao,

There is one thing unclear for me In your statement of Lindelof’s Theorem. Suppose is very small, then if , then the bound goes to . So I think the theorem doesn’t cover directly the case where the bound is, for example, .

Could you please help explain a bit?

Thank you!

X

4 April, 2012 at 7:14 am

Terence TaoThe factor will damp out the growth as .

31 October, 2012 at 9:10 am

Gandhi ViswanathanDear Professor Tao, I noticed that the quantity defined via in the complex analysis (4th) proof of Lemma 2 above is identical to the quantity defined below Lemma 2 in the 2nd proof (via Jensen’s inequality).

4 March, 2013 at 3:13 pm

Lior SilbermanIn Remark 2, should be .

[Corrected, thanks – T.]23 September, 2013 at 10:54 am

AnonymousDear Professor Tao,

I was wandering if for $p1$ such that $\frac{1}{p}+\frac{1}{q}=2$ one has a young type inequality for the convolution. That is $||f*g||_{1}\leq C||f||_{p} ||g||_{q}$

Thanks!

27 April, 2014 at 12:34 pm

FanFor remark 8, is it necessary that the Haar measure is “bi”-invariant? Is the commutativity of the convolution implicitly used anywhere?

27 April, 2014 at 2:02 pm

Terence TaoThe bi-invariance of Haar measure is needed in order to ensure the map is measure-preserving, otherwise one has to be very careful how to define convolution, as various definitions of convolution that are equivalent in the bi-invariant setting may now cease to be equivalent.

27 April, 2014 at 2:20 pm

FanThanks.

5 May, 2014 at 6:49 am

urbanoDear Professor Tao,

is it also possible to show that ?

4 February, 2015 at 3:40 am

Felix V.Dear Prof. Tao, thanks for the nice post.

One small comment: In equation (7) in the statement of LindelĂ¶f’s theorem, I believe you want to put |t| instead of t in the exponent of the right hand side. Otherwise, the theorem still holds true, but the statement is needlessly weak.

Best regards, Felix V.

[Corrected, thanks – T.]19 February, 2015 at 4:36 am

Felix V.Dear Prof. Tao,

I think I found a problem in Exercise 17: For sublinear operators,

it seems that “boundedness” in the sense of

for all step functions does not(!) imply that T can be extended

continuously to all of .

My counterexample is the following: Define for simple functions

Here, can be regarded as an -space on a singleton

set. For simple functions, Tf is indeed well-defined with ,

which implies

and

so that T is indeed sublinear and “bounded”.

But T itself is not continuous (and thus admits no continuous extension)

with respect to the -norm: If we take

and ,

then , but $Tf=e^{i\pi}\left\Vert f\right\Vert _{2}=-1$ and

for .

Best regards, Felix V.

19 February, 2015 at 4:40 am

Felix V.Oh, I just noted that you require to have values in . But you also claim that every linear operator is sublinear, which certainly does not hold if we require values in .

19 February, 2015 at 9:56 am

Terence TaoSorry, there was an additional hypothesis in the definition of sublinearity (namely that ) that was not corrected in the post (though it was noted in the errata for the published version of these notes).

25 November, 2015 at 10:07 am

AnonTypo? In Riesz-Thorin Theorem missing parantheses around 1-\theta?

[Corrected, thanks – T.]2 December, 2015 at 12:06 am

AnonymousDear Prof. Tao,

I guess that in Lemma 9, $p$ and $L^p(X)$ should be replaced by $p_\theta$ and $L^{p_\theta}(X)$, resp.

[Actually, I prefer to leave the qualitative portion of the lemma (which I presume is the part you are referring to) in terms of a parameter instead of a parameter, as the latter is only useful for the quantitative estimate. -T.]11 December, 2015 at 5:19 pm

AnonymousAbout Exercise 16,

I think Exercise 16 should be corrected a little bit. If substitute t with 0 and f*(0) respectively, we might have the measure of X is infinite and ||f||_{inf}<= f*(0)

http://math.stackexchange.com/questions/257365/questions-related-to-distribution-function-and-its-inverse

In this site, altering some of the definitions and conditions makes things going well.

It would be appreciated if you check.

[The case should not have been present and is now deleted – T.]12 December, 2015 at 3:55 am

AnonymousDear Prof. Tao,

In Exercise 17, is p` is the dual exponent of p?

[Corrected, thanks – T.]26 December, 2015 at 4:52 pm

Jiwoong JangDear Prof. Tao,

In the exercise about the marcinkiewicz theorem on the Lorentz spaces, I see your point that adopting the frequency envelope b_m of a_m works, but only in the particular case of $r<=q_\theta$. If r is large, I still do not know how to eliminate the r portion of the exponent of b_m.

Could you give me a direction to go? Or, Is it enough to prove all the things? I`m now considering the quasi-metric structure of $l^{q_\theta/r}(\mathbb{Z})$.

Best

Jang

1 June, 2016 at 6:58 pm

Anonymous… from the calculus identity

and the Fubini-Tonelli theorem, we obtain the formula…Some typo here? The left hand side is a function in while the right hand side is a number?

[The on the RHS should be , now fixed – T.]27 June, 2016 at 9:08 am

BensonDear Prof,

Is there any work done on application of any of the interpolation theorems to signal and image processing . I will like to know.

11 October, 2016 at 6:58 am

Lecture 9. Diffusion semigroups in L^p | Research and Lecture notes[…] proof of the theorem can be found in thisÂ postÂ by […]

3 December, 2016 at 5:10 am

VMTHi Professor,

I have some little corrections to your instructive proof of Marcinkiewicz interpolation:

in (22) there should be only the sum over n.

In the last display of the proof, m should be replaced with n and with .

I guess one should then take a sufficiently small multiple of , where .

Finally, there are three references to (16), (19) which should be replaced with (18), (19).

[Corrected, thanks – T.]23 January, 2017 at 3:41 pm

Anonymous(1) In Section 3, what is the definition of ? Is it the same or different from the “direct sum”?

[No: see https://en.wikipedia.org/wiki/Minkowski_addition – T.](2) In Stein and Weiss’s

Introduction to Fourier Analysis on Enclidean Spaces, the linear operator in the Riesz-Thorin Theorem is defined on a set of simple functions of sets of finite measures. Are that version equivalent to the one in this note? Is there any historical reason or theoretical convenience in Stein-Weiss’s version?[It makes little difference in practice, since a bounded linear operator on a dense subclass of a normed vector space can always be uniquely extended to a bounded linear operator on the full space. -T.]25 January, 2017 at 4:37 am

AnonymousIn remark 29 (lines 4 and 7), it seems that the constant (with several subscripts) should not contain the subscript .

Is the best possible constant (as a function of its subscripts) known up to a multiplication by an absolute constant?

[Yes – I believe this is discussed for instance in Bergh and Lofstrom. -T.]21 February, 2017 at 11:13 am

Anonymous– Right after (6), would you elaborate what the following remark means?

The geometric series formula tells us that such factors are absolutely summable, and so in practice it is often a useful heuristic to pretend that the cases dominate so strongly that the other cases can be viewed as negligible by comparison.

Do you have an example when one would like to sum the factors ?

– “However, one certainly cannot interpolate lower bounds: (i)lower bounds on a log-convex function {\theta \rightarrow A_\theta} at {\theta=0} and {\theta=1} yield no information about the value of, say, {A_{1/2}}. (ii)Similarly, one cannot extrapolate upper bounds on log-convex functions: an upper bound on, say, {A_0} and {A_{1/2}} does not give any information about {A_1}. (However, an upper bound on {A_0} coupled with a lower bound on {A_{1/2}} gives a lower bound on {A_1}; this is the contrapositive of an interpolation statement.)”

I’m a little confused with this remark. For (i), if one looks at the log-convex function (5), then (3) means one can interpolate lower bounds?

25 February, 2017 at 10:30 am

Terence TaoSee for instance the argument deriving (12) from (11) (here we use a continuous variable instead of a discrete one , but the principle is the same. See also the proof of Theorem 27.)

The function in (5) is log-convex as well as log-concave, which is why one can interpolate lower bounds for that function. However, most log-concave functions will not be log-convex.

21 February, 2017 at 5:20 pm

AnonymousIn the second proof Lemma 9, one can divide by its or norm to achieve one of the normalizations. How can one do it simultaneously for both? Why is the measure also allowed to change?

25 February, 2017 at 10:33 am

Terence TaoIf is a measure space, then so is for any . It is then a simple matter of algebra to find, given any measure space and any in both and of that space, constants such that . If we know how to prove Lemma 9 in this case, a little further algebra then recovers Lemma 9 in the general case.

23 February, 2017 at 3:03 pm

AnonymousIn the third proof of Lemma 9, typos in the tensor power:

The component x_m should be x_M and the underlying measure space for the tensor power should be X^M

[Corrected, thanks – T.]25 February, 2017 at 1:37 pm

AnonymousOne more in

we see from many applications of the Fubini-Tonelli theorem that

the left hand side should be

[Corrected, thanks – T.]5 March, 2017 at 2:52 pm

AnonymousIn the set of your notes (http://www.math.ucla.edu/~tao/247a.1.06f/notes1.pdf) on harmonic analysis, the first sentence of the introduction to the Vinogradov notation is very confusing (page 5):

In (modified) Vinogradov notation, the notation (read: is less than or

comparable to ) or is used synonymously with or .

I think you mean

(resp. ) is used synonymously with

or

just as indicated in this post?

[Correction added to 247a page, thanks -T.]10 March, 2017 at 3:38 pm

AnonymousIn the first proof of Lemma 9, HĂ¶lder is used. It is said in your 247a notes 1 that Lemma 9 actually implies HĂ¶lder also.

The hint there says that one should consider the convexity of with respect to a measure for some constants .

I don’t understand what this really means. Would you elaborate a little bit more?

[Apply Lemma 9 with replaced by and replaced by , and for a carefully selected choice of . -T]11 March, 2017 at 5:15 pm

Anonymous1. Suppose one wants to prove the Holder with . Suppose in Lemma 9, with . One can then choose any with , then let

and

.

The constants do not need to be unique, right?

2. It is said in the hint that one needs to first reduce to the case of finite measure space and and everywhere non-vanishing simple functions. Why is that so? It seems that one can get the proof by directly manipulate the constants so that one can match the exponents of the two inequalities (Holder and the log-convexity).

[One needs various norms to be finite and non-zero in order to justify some algebraic manipulations, such as raising these norms to a negative power. -T]17 November, 2017 at 1:11 pm

AnonymousWhy can one replace by in Lemma 9? Isn’t Lemma 9 only true when one replace by where is a (nonnegative) constant?

26 November, 2017 at 10:19 am

Terence TaoLemma 9 is true for arbitrary non-negative measures , which gives one the (powerful) freedom to multiply by arbitrary non-negative weights, not just constants.

17 November, 2017 at 1:38 pm

AnonymousI attempted what was suggested directly:

which seems not quite true…

26 November, 2017 at 10:30 am

Terence TaoThis is a correct inequality. One now has to select the parameters carefully to obtain the desired conclusion.

20 November, 2017 at 6:45 pm

AnonymousSuppose one wants to prove HĂ¶lder for . Assume the log-convexity for . Apply Lemma 9 with replaced by and replaced by . Comparing the exponents, we want

So if we let such that and

then we have the desired result.

Where one needs to first reduce to the case of finite measure space and f and g everywhere non-vanishing simple functions?

26 November, 2017 at 10:31 am

Terence TaoIf some of the exponents are negative, you will have a division by zero problem if one did not prepare in advance by reducing to the case when were everywhere non-vanishing.

10 March, 2017 at 4:39 pm

AnonymousIn 247a notes 1 Problem 5.9, it is said that one can differentiate twice with respect to to get the log convexity of the norm. Why should one use instead of ?

[The function is convex, but the function is usually not convex. -T.]10 March, 2017 at 4:43 pm

AnonymousAs it is mentioned in the notes, this approach is messy. Has anyone performed the calculation ever to show that this is indeed doable? (Lots of terms would appear when calculating the second derivative. Still possible to see that the second derivative is positive?)

13 March, 2017 at 7:32 am

AnonymousAnother version of (7) in Theorem is given in the 247a notes 1:

Are these two versions equivalent? (Since it could be that and a positive constant is missing in front of the "inner" exponential function in (7), I'm not sure about this.)

[Yes, they are equivalent. Note that one has the freedom to reduce if desired, in particular one can assume that is positive. -T]13 March, 2017 at 10:45 am

AnonymousAccording to this set of notes

https://terrytao.wordpress.com/2016/09/22/246a-notes-1-complex-differentiation/

holomorphic functions are defined on open subsets of . What does the function being holomorphic on the strip (which is closed) mean? (What does one mean that is holomorphic on the boundary?)

[It means that the function extends to be holomorphic on some open neighbourhood of the strip. -T]13 March, 2017 at 11:43 am

AnonymousIn the proof of Theorem 4, would you elaborate why is the function holomorphic and non-zero on ?

Isn’t it true that one needs to define using the complex logarithm, which is not holomorphic at ? Moreover, what branch one needs to use for the function ?

[One only needs to take logarithms of the positive real constants ; at no point does one need to take logarithms of the complex variable . -T]17 March, 2017 at 7:37 am

AnonymousIn the last step of passage to the limit of Theorem 21 (Riesz-Thorin theorem), one decomposes into a bounded function and a function of finite measure support and then approximates each piece by simple functions. How would one put the two approximations together to get the approximation for ?

18 March, 2017 at 6:37 am

AnonymousIn Lemma 9 and Theorem 21; the exponents are in while the are in . Can we change the range for to be also in ? (Or is there a typo so that the should also in ?)

18 March, 2017 at 6:43 am

AnonymousIt is said at the beginning of section 3 that “much of the theory here extends to the non--finite setting.”

Is the -finite assumption implicitly used somewhere in the proof of Theorem 21? (Or is the theorem true in the non--finite setting?)

19 March, 2017 at 6:09 am

AnonymousIn Lemma 9, what norm should one use for the space ?

In Theorem 21, what norm should one use for the Minkowski sums and ?

22 March, 2017 at 6:05 am

Terence TaoI don’t believe I using a norm for these spaces anywhere in my text, but the standard norm on the intersection of two normed vector spaces is defined as (one can also use the equivalent norm ), and the standard norm on the sum is .

19 March, 2017 at 10:16 am

AnonymousIn the definition of the distribution function, several other books (e.g. Folland) use instead of in the formula

In practice of doing analysis (in the case that ), how much difference would this make? For instance, would it affect the proof of Mincinkiewicz interpolation theorem?

6 December, 2017 at 7:10 am

AnonymousInstead of being right-continuous, one gets being left-continuous here. That’s the only difference I think. In practice, may be not a big deal.

6 December, 2017 at 7:08 am

jackIt is said in Wikipedia (https://en.wikipedia.org/wiki/Lorentz_space) that the Lorentz space norm with is the area of the largest rectangle with sides parallel to the coordinate axes that can be inscribed in the graph.

However, direct calculation shows that the area of any inscribed rectangle is

What goes wrong? Can it be explained by Exercise 13?

6 December, 2017 at 7:15 am

AnonymousThe Lorentz norm of an indicator function can be calculated easily. But when it goes to general functions, even just simple functions, the calculation becomes much more difficult. What is this norm really about?

Maybe its property such as is more important than what it really is?

6 December, 2017 at 12:33 pm

Terence TaoYes, the Lorentz space norm of this function is equal to . As it turns out the dyadic Lorentz space norm is also equal to . (In general, the two expressions will be comparable, but not identical.) I find the dyadic norm to be a bit more tractable in computations (for instance it is easy to compute for linear combinations of a few indicator functions, particularly if the sets involved are disjoint and the coefficients are lacunary).

16 December, 2017 at 12:32 pm

AnonymousIn Stein-Weiss, the following inequality is used for the proof of ( and ):

How is this inequality done?

[Upper bound the integral, and then use the monotonicity of norms. -T]16 December, 2017 at 7:54 pm

AnonymousUsing the monotonicity of norms, one has

It seems that one cannot get the exponent in right.

Rewriting the RHS as

what should one use?

17 December, 2017 at 11:54 am

AnonymousIf in Stein-Weiss, is a typo which should be , then the first inequality here has already given the conclusion.