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The celebrated decomposition theorem of Fefferman and Stein shows that every function ${f \in \mathrm{BMO}({\bf R}^n)}$ of bounded mean oscillation can be decomposed in the form

$\displaystyle f = f_0 + \sum_{i=1}^n R_i f_i \ \ \ \ \ (1)$

modulo constants, for some ${f_0,f_1,\dots,f_n \in L^\infty({\bf R}^n)}$, where ${R_i := |\nabla|^{-1} \partial_i}$ 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 ${f_i}$ is an ${L^\infty({\bf R}^n)}$ function, then the Riesz transform ${R_i f_i}$ is well defined as an element of ${\mathrm{BMO}({\bf R}^n)}$, 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 ${(f_0,f_1,\dots,f_n)}$ is defined to take values on a sphere, and the iterative construction to build these functions from ${f}$ 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 ${f}$, in the sense that each of the ${f_i, i=0,\dots,n}$ depend linearly on ${f}$. Strictly speaking this is easily accomplished using the axiom of choice: take a Hamel basis of ${\mathrm{BMO}({\bf R}^n)}$, 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 ${\mathrm{BMO}({\bf R}^n)}$ by definition of Hamel basis. But these linear operations have no reason to be continuous as a map from ${\mathrm{BMO}({\bf R}^n)}$ to ${L^\infty({\bf R}^n)}$. So the correct question is whether the decomposition can be made continuously linear (or equivalently, boundedly linear) in ${f}$, that is to say whether there exist continuous linear transformations ${T_i: \mathrm{BMO}({\bf R}^n) \rightarrow L^\infty({\bf R}^n)}$ such that

$\displaystyle f = T_0 f + \sum_{i=1}^n R_i T_i f \ \ \ \ \ (2)$

modulo constants for all ${f \in \mathrm{BMO}({\bf R}^n)}$. Note from the open mapping theorem that one can choose the functions ${f_0,\dots,f_n}$ to depend in a bounded fashion on ${f}$ (thus ${\|f_i\|_{L^\infty} \leq C \|f\|_{BMO}}$ for some constant ${C}$, however the open mapping theorem does not guarantee linearity. Using a result of Bartle and Graves one can also make the ${f_i}$ depend continuously on ${f}$, 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 ${T_0,T_1,\dots,T_n}$ 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 ${\mathrm{BMO}}$ to ${L^\infty}$, 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.