Recently, I have read Tuomas Hytonen's paper On Petermichl's Dyadic Shift And The Hilbert Transform and got into trouble in a certain part of his article.
In the first place, we should have some definitions. The scaled dyadic systems is defined as $$ r\mathcal{D}^{\beta}:=\{rI : I \in \mathcal{D}^{\beta} \}, $$ with $r \in [1,2)$. The dyadic shift associated to $r\mathcal{D}^{\beta}$ is $$ 山^{\beta,r}f := \sum_{I \in r\mathcal{D}^{\beta}} H_{I}\left< h_I,f \right> = \sum_{j\in \mathbb{Z}}\sum_{I \in r\mathcal{D}^{\beta}_{j}} H_{I}\left< h_I,f \right>. $$
Following is where I am puzzled (see page 3 of his paper) :
Note that the partial sums of the j-series are differences of two conditional expectations of $山^{\beta,r}f$, and hence dominated by $M^{\beta,r}(山^{\beta,r}f)$, where $M^{\beta,r}$ is the dyadic maximal operator related to the scaled dyadic system $r\mathcal{D}^{\beta}$.
That is, there exists a constant $C>0$ such that $$ \left| \sum^{n}_{j=m}\sum_{I \in r\mathcal{D}^{\beta}_{j}} H_{I}\left< h_I,f \right> \right|\leq C M^{\beta,r}(山^{\beta,r}f), \tag{1} $$ for every pair of numbers $(m,n)$, s.t. $-\infty <m<n<\infty$.
I have been trying to verify this inequality many days, but still think out nothing. Can anyone help me to prove inequality $(1)$ ? Thanks !
Let $(\mathscr{G}_n)_{n=1}^\infty$ be an increasing sequence of finite $\sigma$-algebras and define $\mathbb{E}(\cdot \mid \mathscr{G}_n)$ to be the conditional expectation with respect to $\mathscr{G}_n$. We prove $$ \lvert \mathbb{E}(f \mid \mathscr{G}_n) \rvert \leq C M_n f $$ pointwise, where $$ M_nf(x) = \sup \{ m(E)^{-1}\int_E \lvert f\rvert \ :\ x \in E \in \mathscr{G}_n \}. $$ where $m$ denotes measure. But this is obvious since $$ \mathbb{E}(f \mid \mathscr{G}_n) = \sum m(E)^{-1} \left( \int_E \lvert f \rvert\right) \mathbb{1}_E $$ where the sum is over the partition defined by the finite $\sigma$-algebra $\mathscr{G}_n$. Thus in the supremum defining $Mf$, the conditional expectation is itself considered.
Now it should be the case that $$ \mathbb{E}(山f \mid \mathscr{G}_n) = \sum_{j=0}^n \sum H_I \langle f, h_I \rangle $$ with $\mathscr{G}_n$ the $\sigma$ algebra generated by the scaled dyadic system at depth $n$. I might have an index-off-by-one error here: look at the minimal $n$ for which $H_I$ is $\mathscr{G}_n$-measurable. Thus the partial sums is a difference of conditional expectations. Now note that you can replace the maximal functions $M_n$ with a more refined maximal function, which finishes the proof.
Corrections
The $\sigma$-algebra only needs to be locally finite in the following sense. Around each point $x$ there is a smallest element of the $\sigma$-algebra containing $x$, which is an interval. Maybe we only need it to have nonzero measure, but it doesn't matter in this case. Furthermore as defined in the paper we should have $\mathscr{G}_n = r \mathscr{D}^\beta_n$ is a decreasing sequence of $\sigma$-algebras, as $n$ increases.
A locally finite $\sigma$-algebra defines a countable partition by $\{ U_x \}$ with $U_x = \cap_{x \in E \in \mathscr{G}_n} E$.
I think $\mathscr{G}_n = r\mathscr{D}^\beta_n$ as defined is a translate of the set of dyadic intervals $r\mathscr{D}^0_n$ of size $r2^{n}$, and so is locally finite.
Note that $H_I$ is $\mathscr{G}_{n-2}$ measurable if $I \in r \mathscr{D}^\beta_n$, because $h_I$ is $\mathscr{G}_{n-1}$ measurable for $I \in \mathscr{G}_{n}$. This is because $I_- \in \mathscr{G}_{n-1}$ and $I^+ \in \mathscr{G}_{n-1}$. Also for $I \in \mathscr{G}_{n-k}$ with $k > 0$ we have $\mathbb{E}(H_I \mid \mathscr{G}_{n-2}) = 0$. This is because $H_I$ will have average value $0$ over all sets in the partition defined by $\mathscr{G}_{n-2}$. The amended formula reads $$ \mathbb{E}(山f \mid \mathscr{G}_{n-2}) = \sum_{j=n}^\infty \sum_{I \in \mathscr{G}_j = r \mathscr{D}^\beta_j} H_I \langle f, h_I \rangle. $$