# Interlude: Atomic decomposition of L(log L)^r

This is going to be a shorter post about a technical fact that will be used in concluding the proof of the Tao-Wright lemma.
What we are going to see today is an atomic decomposition of the Orlicz spaces of $L (\log L)^r$ type. Surprisingly, I could find no classical references that explicitely state this useful little fact – some attribute it to Titchmarsh, Zygmund and Yano; indeed, something resembling the decomposition can be found for example in Zygmund’s book (Volume II, page 120). However, I could only find a proper statement together with a proof in a paper of Tao titled “A Converse Extrapolation Theorem for Translation-Invariant Operators“, where he claims it is a well-known fact and proves it in an appendix (the paper is about reversing the implication in an old extrapolation theorem of Yano [1951], a theorem that tells you that if the operator norms $\|T\|_{L^p \to L^p}$ blow up only to finite order as $p \to 1^{+}$, then you can “extrapolate” this into an endpoint inequality of the type $\|Tf\|_{L^1} \lesssim \|f\|_{L(\log L)^r}$).

Briefly stated, the result is as follows. We will consider only $L(\log L)^r ([0,1])$, that is the Orlicz space of functions on $[0,1]$ with Orlicz/Luxemburg norm

$\displaystyle \|f\|_{L(\log L)^r ([0,1])} = \inf \Big\{\mu > 0 \text{ s.t. } \int_{0}^{1} \frac{|f(x)|}{\mu} \Big(\log \Big(2 + \frac{|f(x)|}{\mu}\Big)\Big)^{r} \,dx \leq 1 \Big\}.$

Our atoms will be quite simply normalised characteristic functions: that is, for any measurable set $E \subset [0,1]$ we let $a_E$ denote the atom associated to $E$, given by

$\displaystyle a_E := \frac{\mathbf{1}_E}{\|\mathbf{1}_E\|_{L(\log L)^r}};$

obviously $\|a_E\|_{L(\log L)^r} = 1$.
The statement is then the following.

Atomic decomposition of $L(\log L)^r$:
Let $f \in L(\log L)^{r}([0,1])$. Then there exist measurable sets $(E_j)_j$ and coefficients $(\alpha_j)_j$ such that

$\displaystyle f = \sum_{j} \alpha_j a_{E_j}$

and

$\displaystyle \sum_{j} |\alpha_j| \lesssim \|f\|_{L(\log L)^r}.$

The reader should compare the above decomposition to the one we gave for Lorentz spaces in an old post. The statements and the relative proofs are quite related.
Other kinds of decompositions are also possible. See for example the paper “On an optimal decomposition in Zygmund spaces” by Fiorenza and Krbec (mathscinet reference).

Before we start proving things, we should ask ourselves a few basic questions. First of all, what is that normalisation factor $\|\mathbf{1}_E\|_{L(\log L)^r}$ in the definition of an atom? It is clear from the definition that it is the smallest $\mu > 0$ such that

$\displaystyle \frac{|E|}{\mu} \Big(\log^{+} \frac{1}{\mu}\Big)^r \leq 1, \ \ \ \ \ \ \ (1)$

where we are writing $\log^{+} t := \log(2 + t)$ for shortness. It is immediate that the optimal $\mu$ is a function of $|E|$ only, which is good. A simple calculation shows that for sure

$\displaystyle \|\mathbf{1}_E\|_{L(\log L)^r} \leq |E| \Big(\log^{+} \frac{1}{|E|}\Big)^r$

(by taking $\mu$ in (1) equal to the RHS). A more laborious calculation shows that this is actually tight, that is if we take $\mu = \gamma |E| \Big(\log^{+} \frac{1}{|E|}\Big)^r$ for some $\gamma \ll 1$ (how small depends only on $r$) then this $\mu$ violates (1) and therefore $\|\mathbf{1}_E\|_{L(\log L)^r} \gtrsim |E| \Big(\log^{+} \frac{1}{|E|}\Big)^r$ as well (this is essentially due to the fact that the logarithm grows slower than any positive power).
Secondly, having completed the step above, we should ask ourselves what the $L(\log L)^r$ norm of a simple function is. Here the answer is not very satisfactory I’m afraid. Indeed, take simple function $\sum_{j} \beta_j \mathbf{1}_{E_j}$, with $E_j$ disjoint measurable sets and let $E := \bigcup_j E_j$. In general we can only say that

$\displaystyle \sum_{j} |\beta_j| |E_j| \Big(\log^{+} \frac{1}{|E|}\Big)^r \lesssim \Big\| \sum_{j} \beta_j \mathbf{1}_{E_j} \Big\|_{L(\log L)^r} \lesssim \sum_{j} |\beta_j| |E_j| \Big(\log^{+} \frac{1}{|E_j|}\Big)^r;$

the RHS is due to the triangle inequality for Orlicz norms and the discussion above; and the LHS is due to Jensen’s inequality (the function $\Phi(t):= t (\log^{+} t)^r$ being convex) and the laborious calculations mentioned above that I have omitted. The RHS can be seen to be tight by taking all coefficients $\beta_j$ extremely small except for one (in which case we can also make the LHS quite smaller by fiddling with the $E_j$) and the LHS can be seen to be tight by taking them all equal to 1 (in which case we can also make the RHS much larger, always by fiddling with the $E_j$). In general, calculating $\big\| \sum_{j} \beta_j \mathbf{1}_{E_j} \big\|_{L(\log L)^r}$ implies solving an optimisation problem: finding the smallest $\mu$ such that

$\displaystyle \sum_{j} |E_j| \Phi\Big(\frac{\beta_j}{\mu}\Big) \leq 1,$

which is not a straightforward task.

## Proof of the atomic decomposition

That said, we are ready to start the proof. Using function $\Phi(t) = t (\log^{+} t)^r$ as defined above, we will write $\Phi(L)$ for $L(\log L)^r$ in order to save some room (as is actually more customary in the literature on Orlicz spaces).
Let $f \in \Phi(L)([0,1])$ be the given function, assumed positive for convenience (without loss of generality). We assume furthermore for convenience that the support of $f$ has measure $\ll 1$ (something that we can achieve by splitting the function into $O(1)$ pieces). As usual we let $f^\ast$ denote the non-increasing rearrangement of $f$, that is

$\displaystyle f^\ast (s) := \inf \big\{ t > 0 \text{ such that } |\{x \in [0,1] : f(x) > t\}| \leq s \big\}.$

We will split the function $f$ according to its magnitude, though in such a way that the supports will have dyadic sizes (this makes the calculations easier). For $k \in \mathbb{Z}$ define then

$\displaystyle E_k := \{x \in [0,1] \text{ s.t. } f^\ast (2^{k+1}) < f(x) \leq f^\ast (2^k)\}$

and observe that $|E_k| \sim 2^k$ (here inevitably $k < -C$ for some large positive constant $C$ because of our assumptions). We notice that the function $f \mathbf{1}_{E_k}$ has $\Phi(L)$ norm bounded by $f^\ast(2^k) \|\mathbf{1}_{E_k}\|_{\Phi(L)} \lesssim f^\ast(2^k) |E_k| \big(\log^{+} 1/|E_k|\big)^r \sim f^\ast (2^k) 2^k |k|^r$ (by the discussion above about the Orlicz norm of characteristic functions). We can therefore normalise by this factor and write

\displaystyle \begin{aligned} f = \sum_{k < -C} f \mathbf{1}_{E_k} = & \sum_{k < -C} f^\ast (2^k) 2^k |k|^r \frac{f \mathbf{1}_{E_k}}{f^\ast (2^k) 2^k |k|^r} \\ =: & \sum_{k < -C} f^\ast (2^k) 2^k |k|^r \, f_k. \end{aligned}

The function $f_k$ is supported on $E_k$ and we have the bound $f_k \lesssim a_{E_k}$ (recall the definition of atom: $a_E := \mathbf{1}_E (\|\mathbf{1}_E\|_{\Phi(L)})^{-1}$). We make two claims now, which together will allow us to conclude:

• Claim 1: we can write $f_k = \sum_{j} \alpha^{(k)}_j a^{(k)}_j$ for atoms $a^{(k)}_j$ and coefficients $\alpha^{(k)}_j$ such that $\sum_{j} |\alpha^{(k)}_j| \lesssim 1$.
• Claim 2: we have $\sum_{k < -C} f^\ast (2^k) 2^k |k|^r \lesssim \|f\|_{\Phi(L)}$.

Indeed, using Claim 1 and the decomposition above we can write ${f}$ as

$\displaystyle f = \sum_{k < -C} f^\ast (2^k) 2^k |k|^r \, f_k = \sum_{k < -C} \sum_j \, [f^\ast (2^k) 2^k |k|^r \alpha^{(k)}_j] \, a^{(k)}_j,$

a linear combination of atoms, and we have by Claims 1 & 2 that

$\displaystyle \sum_{k < -C} \sum_j \, f^\ast (2^k) 2^k |k|^r |\alpha^{(k)}_j| \lesssim \|f\|_{\Phi(L)},$

as desired. It therefore remains to verify the claims.

### Proof of Claim 1

This claim is consequence of the more general claim that if $0 \leq g \leq a_G$ then $g = \sum_{j} \alpha_j a_j$ with $a_j$ atoms and $\sum_{j} \alpha_j \lesssim 1$.

Remark: Notice that this is not an atomic decomposition of ${g}$ because $\|g\|_{\Phi(L)}$ might be much smaller than $\sum_{j} \alpha_j$.

Indeed, any bounded function can be written in a standard way as

$\displaystyle g = \sum_{j} \gamma_j \mathbf{1}_{G_j}$

with $G_j \subset G$ and coefficients such that $\sum_j \gamma_j = \|g\|_{L^\infty}$. But then

$\displaystyle g = \sum_{j} \gamma_j \mathbf{1}_{G_j} = \sum_{j} \gamma_j \|\mathbf{1}_{G_j}\|_{\Phi(L)} \, a_{G_j}$

and since $|G_j| \big(\log^{+} 1 / |G_j| \big)^r \leq |G| \big(\log^{+} 1 / |G| \big)^r$ we have

$\displaystyle \sum_{j} \gamma_j \|\mathbf{1}_{G_j}\|_{\Phi(L)} \lesssim \sum_{j} \gamma_j |G| \big(\log^{+} 1 / |G| \big)^r = \|g\|_{L^\infty} |G| \big(\log^{+} 1 / |G| \big)^r \lesssim 1;$

letting $\alpha_j := \gamma_j \|\mathbf{1}_{G_j}\|_{\Phi(L)}$ gives the claim.

### Proof of Claim 2

This claim can be proven by a standard computation. Indeed, since $f^\ast$ is non-increasing we have

\displaystyle \begin{aligned} \sum_{k < -C} f^\ast(2^k) 2^k |k|^r \leq & \sum_{k < -C} \int_{2^{k-1}}^{2^k} f^\ast(s) (\log^{+} 1/s)^r \,ds \\ = & \int_{0 \leq s \ll 1} f^\ast (s) (\log^{+} 1/s)^r \,ds. \end{aligned}

We will split this integral into two parts: one where $f^\ast(s) \leq s^{-1/2} \, 2\|f\|_{\Phi(L)}$ and its complement (nothing special about the exponent $1/2$; any exponent $< 1$ would also work). We have then

\displaystyle \begin{aligned} \int_{\{s \ll 1 : f^\ast(s) \leq s^{-1/2} \, 2\|f\|_{\Phi(L)} \}} f^\ast (s) (\log^{+} 1/s)^r \,ds \leq & 2 \|f\|_{\Phi(L)} \int_{\{s \ll 1\}} \frac{1}{s^{1/2}} \Big(\log^{+} \frac{1}{s} \Big)^r \,ds \\ \lesssim & \|f\|_{\Phi(L)}, \end{aligned}

which is an acceptable contribution. For the other part, notice that $\log^{+} 1/s \sim \log^{+} 1/s^{1/2}$ since $s \ll 1$; thus we have instead

\displaystyle \begin{aligned} \int_{\{s \ll 1 : f^\ast(s) > s^{-1/2} \, 2\|f\|_{\Phi(L)} \}} f^\ast (s) (\log^{+} 1/s)^r \,ds \lesssim & \int_{\{s \ll 1 \}} f^\ast(s) \Big(\log^{+} \frac{f^\ast(s)}{2\|f\|_{\Phi(L)}} \Big)^r \,ds \\ = & 2 \|f\|_{\Phi(L)} \int_{\{s \ll 1 \}} \frac{f^\ast(s)}{2 \|f\|_{\Phi(L)}} \Big(\log^{+} \frac{f^\ast(s)}{2 \|f\|_{\Phi(L)}} \Big)^r \,ds \\ = & 2 \|f\|_{\Phi(L)} \int_{\{x \ll 1 \}} \frac{f(x)}{2 \|f\|_{\Phi(L)}} \Big(\log^{+} \frac{f(x)}{2 \|f\|_{\Phi(L)}} \Big)^r \,dx \\ \lesssim & \|f\|_{\Phi(L)}, \end{aligned}

where the second-to-last line is due to the equidistribution of $f$ and $f^\ast$ and the last line is simply due to the definition of $\|f\|_{\Phi(L)}$ itself. The proof of the claim is concluded, and so is the proof of the atomic decomposition.