The celebrated decomposition theorem of Fefferman and Stein shows that every function of bounded mean oscillation can be decomposed in the form

modulo constants, for some , where are the Riesz transforms. A technical note here a function in BMO is defined only up to constants (as well as up to the usual almost everywhere equivalence); related to this, if is an function, then the Riesz transform is well defined as an element of , but is also only defined up to constants and almost everywhere equivalence.

The original proof of Fefferman and Stein was indirect (relying for instance on the Hahn-Banach theorem). A constructive proof was later given by Uchiyama, and was in fact the topic of the second post on this blog. A notable feature of Uchiyama’s argument is that the construction is quite nonlinear; the vector-valued function is defined to take values on a sphere, and the iterative construction to build these functions from involves repeatedly projecting a potential approximant to this function to the sphere (also, the high-frequency components of this approximant are constructed in a manner that depends nonlinearly on the low-frequency components, which is a type of technique that has become increasingly common in analysis and PDE in recent years).

It is natural to ask whether the Fefferman-Stein decomposition (1) can be made linear in , in the sense that each of the depend linearly on . Strictly speaking this is easily accomplished using the axiom of choice: take a Hamel basis of , choose a decomposition (1) for each element of this basis, and then extend linearly to all finite linear combinations of these basis functions, which then cover by definition of Hamel basis. But these linear operations have no reason to be continuous as a map from to . So the correct question is whether the decomposition can be made *continuously linear* (or equivalently, boundedly linear) in , that is to say whether there exist continuous linear transformations such that

modulo constants for all . Note from the open mapping theorem that one can choose the functions to depend in a bounded fashion on (thus for some constant , however the open mapping theorem does not guarantee linearity. Using a result of Bartle and Graves one can also make the depend continuously on , but again the dependence is not guaranteed to be linear.

It is generally accepted folklore that continuous linear dependence is known to be impossible, but I had difficulty recently tracking down an explicit proof of this assertion in the literature (if anyone knows of a reference, I would be glad to know of it). The closest I found was a proof of a similar statement in this paper of Bourgain and Brezis, which I was able to adapt to establish the current claim. The basic idea is to average over the symmetries of the decomposition, which in the case of (1) are translation invariance, rotation invariance, and dilation invariance. This effectively makes the operators invariant under all these symmetries, which forces them to themselves be linear combinations of the identity and Riesz transform operators; however, no such non-trivial linear combination maps to , and the claim follows. Formal details of this argument (which we phrase in a dual form in order to avoid some technicalities) appear below the fold.

** — 1. Formal argument — **

Suppose for contradiction that we have bounded linear maps such that (2) holds for all . We convert this to an assertion about bilinear forms. Let be the space of Schwartz functions, and let be codimension one subspace of Schwarz functions of integral zero. We define the bilinear forms

and

by the formulae

and

From (2) and integration by parts we have the identity

for all with vanishing near the origin (so that is also Schwartz), where is the vector Riesz transform. Meanwhile, from the boundedness of and Hölder’s inequality we have the bounds

for any and , where is the covector Riesz transform, thus . (Note that is indeed in due to the hypothesis that has integral zero.

We will show that there are no bilinear forms obeying the identity (3) (for the indicated range of ) and the bounds (4), (5) (for the indicated range of ), which will give the claim.

Now we exploit symmetries of the hypotheses (3), (4), (5). For any , let be the translation operator

This operator also acts on vector-valued functions by acting on each component separately. The translation operators are uniformly bounded on and , and commute with Riesz transforms and preserve Lebesgue measure . As a consequence, if the forms obey the hypotheses (3), (4), (5), then so do the translated forms

and

with implied constants in (4), (5) uniform in .

On the other hand, , being abelian, is an amenable group. Thus, if , obey the hypotheses (3), (4), (5), then so do the means

and

where is an invariant mean of . (Note from (4), (5) that the quantities , depend continuously on .) By construction, these new bilinear forms are translation-invariant in the sense that

for all , , and . Thus, to show the non-existence of bilinear forms obeying (3), (4), (5), we may assume without loss of generality that the forms also obey the translation invariance (6).

We can argue similarly to assume rotational invariance, provided we apply the rotation correctly to the vector field . More precisely, if is any orthogonal matrix (a rotation or reflection), and obey (3), (4), (5), (6), then one can check that the rotated forms

also obey these hypotheses (with implied constants uniform in ), a key point being the identity

which follows from the chain rule. Being compact, is also amenable (indeed one can simply integrate against the probability Haar measure on ), and so by arguing as before we may assume without loss of generality that the bilinear forms obey the rotational invariance

Finally, we impose dilation invariance. For any in the positive real line , we define the dilation operator by

The operator preserves , while preserves . As a consequence, if obey (3), (4), (5), (6), (7), then so do the dilated forms

with constants uniform in . As is also amenable, we may thus assume without loss of generality that the bilinear forms obey the dilation invariance

Remark 1One could have combined the translation invariance, rotational invariance, and dilation invariance steps into a single step by considering the action of the group of similarity transformations generated by translations, rotations, and dilations, which is also amenable, but I found it conceptually easier to consider the three invariances separately.

Now we use all these invariances to almost completely determine the operators . We begin with the analysis of , which is slightly simpler. First observe from the Schwartz kernel theorem that there is a (tempered) distribution on such that

for all , where by abuse of notation the right-hand side denotes the action of the distribution on the tensor product of and . The translation invariance (6) then asserts that

for all and . Since Schwartz functions in can be approximated (in Schwartz seminorms) by linear combinations of Schwartz functions in (as can be done for instance using dyadic partitions of unity and Fourier series), we conclude that

for any Schwartz .

Morally, this means that , that is to say is really just a function of . To justify this at the level of distributions, we argue as follows. Now suppose that is such that for all . If we let be a bump function supported on of total mass one, and is sufficiently large, then we have

(from translation invariance and averaging),

(because vanishes unless , and because ),

(by translating by ),

(by translating by ),

(because ). On the other hand, the integrand can be seen to be non-vanishing only when , and has uniform compact support in , with all norms . From this and the triangle inequality we have

and thus on sending we conclude that

whenever is such that , or equivalently whenever . Thus the expression only depends on the function , so there exists a distribution on such that

We then have the convolution representation

for all , where is the reflection of . By a limiting argument one can allow to be in the Schwartz space rather than . The rotation and dilation invariances of then imply that

The rotation invariance and an averaging argument implies that

is the spherically symmetric component of , and denotes probability Haar measure on . Next, if vanishes near the origin and is such that for all , and is a bump function supported on of total mass one, then (similarly to the translation invariance analysis) we have

for large enough. But the integrand is non-vanishing only when is comaparable to , and has uniform compact support in with all norms . From this and the triangle inequality we conclude that

and hence

whenever vanishes near the origin with for all . In the case when is radial, , the latter condition simplifies to . We conclude that there is a constant such that

whenever is spherically symmetric vanishes near the origin. On the other hand, if is a spherically symmetric bump function equal to one near the origin and is decreasing, then is strictly positive, while from (9) we have

We conclude that , thus vanishes whenever is spherically symmetric and vanishes near . From (10) the same is true without the spherical symmetry hypothesis; thus the distribution is supported at the origin, and hence is a linear combination of the Dirac delta function and its derivatives. To be consistent with the dilation invariance (9), it can in fact only be a constant multiple of the Dirac delta; thus

for some . But this is only consistent with (4) if (one can take for instance to be a smoothed out version of for some large , which has a bounded BMO norm, and to be a bump function supported near the origin, to obtain a contradiction for non-zero ). Thus must vanish identically.

We now apply a similar (but slightly more complicated) argument for . The Schwartz kernel theorem (and the Hahn-Banach theorem) shows that there exists a vector-valued distribution on such that

whenever and . Translation invariance then gives

whenever is such that for all . The translation averaging argument then shows that

whenever is such that for all and for all ; this implies the existence of a vector-valued distribution on such that

whenever is such that for all . In particular we have

whenever and . The rotation and dilation invariances of then imply that

The rotation invariance implies that

is the equivariant version of (in particular one has for all and . From dilation invariance as before we have

whenever vanishes near the origin with for all . For equivariant , this latter condition is equivalent to . In particular, there is a constant such that

whenever is equivariant and vanishes near the origin Combining this with (12) we conclude that

whenever vanishes near the origin. Using polar coordinates, the right-hand side can also be expressed as

for another constant ( divided by the surface area of the unit sphere). Thus, is equal to the distribution away from the origin, and thus differs from that distribution by a (vector-valued) linear combination of Dirac delta functions and its derivatives. By dilation invariance there cannot actually be any derivative terms, and from rotation invariance there cannot be any delta function terms either. Hence we have

and hence

for some absolute constant ; comparing this with (3) we see that . We conclude that for , we have for ; but one can easily show that fails to be bounded from to (even when restricted to Schwartz functions), and the claim follows.

## 11 comments

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22 November, 2018 at 12:25 am

Anonymousformatting issue: \eqerf{decomp} does not show equation

[Corrected, thanks – T.]22 November, 2018 at 1:54 am

AnonymousIs it possible to characterize the class of functions for which the decomposition (1) can be made linear? (thereby identifying the precise “obstruction for linearity” in (1))

26 November, 2018 at 9:32 am

Terence TaoI’m not sure any particular class of functions can be “blamed” for lack of bounded linear decomposition. After all, since by the axiom of choice a linear decomposition is always possible, for any finite-dimensional class of functions one can obtain a bounded linear decomposition. One could try to investigate things more quantitatively though, for instance by restricting attention to the span of some finite number of elements of some standard basis of functions (e.g. a wavelet basis) and see how the optimal constant of a linear decomposition grows with the number of elements. The averaging argument in the post is not written in such a way as to easily extract a lower bound on the growth (as it uses infinitary concepts such as invariant means on amenable groups) but it should be possible to rework the argument more finitarily (e.g. using averages over large Folner sets as a substitute for invariant means) to obtain some (rather weak) lower bound of this type.

22 November, 2018 at 6:04 pm

HoThank you Dr. Tao

25 November, 2018 at 12:31 pm

LAGRIDAHello Terrence, i like to know your opinion about this proposition :

Let the i-th prime number : .

Prouve that if and () then

26 November, 2018 at 9:49 am

Terence TaoThis claim would imply that the Jacobsthal function cannot exceed . This is likely to be false, but this has not yet been proven; the best known bounds currently are

(the right-hand bound is due to Iwaniec; the left-hand side to Ford, Green, Konyagin, Maynard, and myself). The conjectures of Maier and Pomerance suggest should in fact grow like , which would imply that the proposition is false. It may be possible to investigate things numerically for small values of (for instance using the values in OEIS A048670), and perhaps even locate a counterexample. (Also, since you are only using shifts of the form , as opposed to all shifts between and , it may be possible to adapt the methods in my paper with Ford et al., or earlier and simpler arguments using the Erdos-Rankin method, to obtain an analytic disproof.)

Incidentally, it would be preferable if one could post mathematical questions as comments to blog posts whose topic is at least somewhat related to the content of the question (one can use the various search and categorisation options on the left sidebar to locate a suitable post, e.g., this one).

28 November, 2018 at 1:17 pm

LAGRIDAThanks a lot Mr Terence for your valuable ansere

25 November, 2018 at 5:00 pm

Amir SagivHas anyone ever attempted to carry out Uchiyama’s proof into an actual algorithm?

26 November, 2018 at 9:27 am

Embedding the Heisenberg group into a bounded dimensional Euclidean space with optimal distortion | What's new[…] The isometric embedding problem also has the key obstacle that naive attempts to solve the equation iteratively can lead to an undesirable “loss of derivatives” that prevents one from iterating indefinitely. This obstacle was famously resolved by the Nash-Moser iteration scheme in which one alternates between perturbatively adjusting an approximate solution to improve the residual error term, and mollifying the resulting perturbation to counteract the loss of derivatives. The current equation (1) differs in some key respects from the isometric embedding equation , in particular being linear in the unknown field rather than quadratic; nevertheless the key obstacle is the same, namely that naive attempts to solve either equation lose derivatives. Our approach to solving (1) was inspired by the Nash-Moser scheme; in retrospect, I also found similarities with Uchiyama’s constructive proof of the Fefferman-Stein decomposition theorem, discussed in this previous post (and in this recent one). […]

27 November, 2018 at 11:32 am

almazIt seems that on the torus it was established by Bonami. That is how it is cited in the paper of E. Amar “Représentation des fonctions de BMO et solutions de l’équation…” 1979. (see Theorem 3.2 there)

I hope it is relevant.

27 November, 2018 at 7:26 pm

Terence TaoThanks for this! It seems that the argument starts similarly in that it reduces to the translation invariant case (but does not try to impose rotation or dilation invariance), so that are Fourier multipliers . Then by inspection of the BMO and norms of lacunary type Fourier series they show that the multipliers associated to cannot have an infinite number of elements of large magnitude (e.g. larger than ), and this is incompatible with (2).