# Is there a physical/geometric proof for L^2 boundedness of Bourgain's maximal function along the squares?

One problem that has bugged me for some time (though I only seriously thought about it for a month several years ago) is to give a physical proof of the L^2 boundedness of Bourgain maximal function for averages along the squares.

To be precise, define the averages: for a function $f: \mathbb{Z} \to \mathbb{C}$, let $$A_Nf(x) := N^{-1} \sum_{n=1}^N f(x-n^2)$$ and the maximal function $$Mf(x) := \sup_{N \in \mathbb{N}} |A_Nf(x)|$$ for $x \in \mathbb{Z}$.

Question: Without resorting to the circle method, can one prove that $M: \ell^2(\mathbb{Z}) \to \ell^2(\mathbb{Z})$?

In particular, I suspect that there a physical proof analogous to the proofs for $L^2$ boundedness of Stein's spherical maximal function theorem as sketched out by Laba here: https://ilaba.wordpress.com/2009/05/23/bourgains-circular-maximal-theorem-an-exposition/ or as in Schlag's work: https://math.uchicago.edu/~schlag/papers/dukecircles.pdf In particular, I am happy with a physical proof of restricted weak-type $L^2$ boundedness.

My suspicion is that Bourgain's proof, which uses the circle method, suggests that we should consider understand what happens on arithmetic progressions, decompose the characteristic function of a set into how arithmetic progressions and combine them in some way. However I could not see how to successfully deploy this strategy...

• You mean that $A_{N}f(x)=N^{-1}\sum f(x-n^{2})$ right? – Asaf Jun 12 '18 at 1:41
• @Asaf: Yes I meant the averages along the squares. My mistake. Thanks for pointing this out. – K Hughes Jun 12 '18 at 16:23

We restrict ourself on a bounded domain $[-L,L]\cap \mathbb{Z}$ and we will extend $L\rightarrow \infty$. For any $x$ let $N_x$ the maximum integer such that $$Mf(x)=\frac{1}{N_x}\sum_{n=0}^{N_x} f(x-n)$$
Proposition : the segments $[x-N_x,x]$, $x\in [-L,L]$ are interlocking ie: if $[y-N_y,y]\cap [x-N_x,x]$ then $[y-N_y,y]\subset [x-N_x,x]$ or $[x-N_x,x]\subset [y-N_y,y]$.
Proof : Let us assume that $y-N_y<x-N_x\leq y<x$. We denote $A=[y+1,x]$, $B=[x-N_x,y]$ and $C=[y-N_y,x-N_x-1]$. Let $$\bar{Af}=\frac{1}{|A|}\sum_{i\in A}f(i)$$ $$\bar{Bf}=\frac{1}{|B|}\sum_{i\in B}f(i)$$ $$\bar{Cf}=\frac{1}{|C|}\sum_{i\in C}f(i)$$ We have that $$\bar{Af}\leq Mf(x)=\frac{1}{|A|+|B|}(|A|\bar{Af}+|B|\bar{Bf})$$ and $$\bar{Bf}\leq Mf(y)=\frac{1}{|C|+|B|}(|C|\bar{Cf}+|B|\bar{Bf})$$ So $\bar{Af} \leq \bar{Bf}$ and $\bar{Bf} \leq \bar{Cf}$. Therefore $$Mf(x)\leq \frac{1}{|A|+|B|+|C|}(|A|\bar{Af}+|B|\bar{Bf}+|C|\bar{Cf})$$ and we can chose $N_x$ such that $x-N_x=y-N_y$.
Using the proposition, we have a set of interlocking segments $D_i$, there is then an natural order $D_i<D_j$ if $D_j\subset D_i$ and we can define a rank $r$ as $$r(D_j)=\max [l:\exists D_{i_1}<D_{i_2}<...<D_{i_l}<D_j ]$$ We are now ready to construct our martingales : let $$f_k(x)=\cases{\bar{D_if} \text{ if } x\in D_i \text{ and } r(D_i)=k \\ f(x) \text{otherwise}}$$ We can now easy check that $f_k$ defined an Martingale for the filtration $\mathcal{F}_k=[D_i:r(D_i)\leq k]$. And we can conclude with Doob inequality : $$\|Mf\|_{\ell^2}^2\leq (2L+1) \mathbb{E}(|\sup f_k|^2) \leq 4(2L+1) \mathbb{E}(|f|^2)=4\|f\|_{\ell^2}^2$$ (Here $\mathbb{E}(.)$ mean $\frac{1}{2L+1}\sum_{x\in [-L,L]}.$)