Proof. Without loss of generality, we assume $\sup _{n\ge 0} \lVert T^n \rVert = 1$ if $\bigcup _n \mbox {supp}(\Delta _n) \subset \mathbb N_0$ and assume $\sup _{n\in \mathbb Z} \lVert T^n \rVert = 1$ otherwise. We have

$$ \begin{align*} \bigg\lVert \bigg(\sum_n |T_{\Delta_n}f|^s \bigg)^{1/s} \bigg\rVert & = \bigg\lVert \bigg(\sum_n \bigg|\sum_k \Delta_n(k) T^kf\bigg|^{s} \bigg)^{1/s} \bigg\rVert \\ & \lesssim \bigg\lVert \bigg(\sum_n \bigg|\sum_{k\neq 0} \int_{|t|<1/|2k| } \hat \Delta_n(t) e(kt) \, dt \, T^kf \bigg|^{s} \bigg)^{1/s} \bigg\rVert \\ &\quad + \bigg\lVert \bigg(\sum_n \bigg|\sum_{k\neq 0} \int_{1/|2k|<|t|<1/2} \hat \Delta_n(t) e(kt) \, dt \, T^kf \bigg|^{s} \bigg)^{1/s} \bigg\rVert \\ & \quad + \bigg(\sum_n | \Delta_n(0) |^{s} \bigg)^{1/s} \lVert f \rVert \\ & = \mbox{I} + \mbox{II}+ \bigg(\sum_n | \Delta_n(0) |^{s} \bigg)^{1/s} \lVert f \rVert. \end{align*} $$

Applying the Stein–Minkowski integral inequality, we obtain the following estimate for the first term:

$$ \begin{align*} \mbox{I}& = \bigg\lVert \bigg(\sum_n \bigg|\sum_{k\neq 0} \int_{|t|<1/|2k|} \hat \Delta_n(t) e(kt) \, dt \, T^kf\bigg|^{s} \bigg)^{1/s} \bigg\rVert \\ & \le \lVert f\rVert \sum_{k\neq 0} \bigg(\sum_n \bigg|\int_{|t|<1/ |2k| } \hat \Delta_n(t) e(kt) \, dt \bigg|^{s} \bigg)^{1/s}\\ & \le \lVert f\rVert \int_{|t|<1/2} \frac1{|t|} \bigg(\sum_n | \hat \Delta_n(t) |^s \bigg)^{1/s} \, dt = A \lVert f\rVert. \end{align*} $$

Also

$$ \begin{align*} \bigg(\!\sum_n | \Delta_n(0) |^{s} \bigg)^{1/s} \le \int_{|t|<1/2} \bigg(\sum_n | \hat \Delta_n(t) |^s \bigg)^{1/s} \, dt \le A. \end{align*} $$

For the second term, by the Stein–Minkowski integral inequality, we have

$$ \begin{align*} \mbox{II} & \le \bigg\lVert \bigg(\sum_n \bigg|\sum_{k\neq 0} \int_{1/|2k| <|t|<1/2} \hat \Delta_n(t) e(kt) \, dt \ T^kf \bigg|^{s} \bigg)^{1/s} \bigg\rVert \\ & \le \bigg\lVert\!\sum_{k\neq 0} \bigg(\sum_n \bigg|\int_{1/|2k|<|t|<1/2} \hat \Delta_n(t) e(kt) \, dt \bigg |^s \bigg)^{1/s} |T^kf | \bigg\rVert\\ & \le \lVert f\rVert \sum_{k\neq 0} \bigg(\sum_n \bigg|\int_{1/|2k|<|t|<1/2} \hat \Delta_n(t) e(kt) \, dt \bigg|^s \bigg)^{1/s}. \end{align*} $$

Since $\Delta _n(t), \Delta ^{\prime }_n(t)$ and $e(kt)$ are 1-periodic, $\Delta _n(-t)=\overline {\Delta _n(t)}$ , and $\Delta ^{\prime }_n(-t)=\overline {\Delta ^{\prime }_n(t)}$ , we estimate

(4) $$ \begin{align} & \bigg|\int_{1/|2k| <|t|<1/2} \hat \Delta_n(t) e(kt) \, dt \bigg| \le \bigg|\int_{1/|2k| <|t|<1/2} \hat \Delta^{\prime}_n(t) \frac{e(kt)}{2\pi k} \, dt \bigg| \nonumber \\ & \qquad + \bigg|\frac{\hat \Delta_n(1/|2k| ) e(k/|2k| )}{2\pi k} - \frac{\hat \Delta_n(-1/|2k| ) e(-k/|2k| )}{2\pi k} \bigg| \nonumber\\ & \quad \lesssim \bigg|\int_{1/|2k| <|t|<1/2} \hat \Delta^{\prime\prime}_n(t) \frac{e(kt)}{4\pi^2 k^2} \, dt \bigg| + \bigg| \frac{\hat \Delta_n(1/|2k| ) }{2\pi k} \bigg| + \bigg| \frac{\hat \Delta^{\prime}_n(1/|2k| )}{4\pi^2 k^2}\bigg|. \end{align} $$

By the Stein–Minkowski integral inequality,

$$ \begin{align*} \sum_{k\neq 0} \bigg(\sum_n \bigg|\int_{1/|2k|<|t|<1/2} \hat \Delta^{\prime\prime}_n(t) \frac{e(kt)}{4\pi^2 k^2} \, dt \bigg |^s \bigg)^{1/s} & \lesssim \sum_{k\neq 0} \frac 1{k^2} \!\int_{1/|2k|<|t|<1/2} \bigg(\sum_n \bigg | \hat \Delta^{\prime\prime}_n(t) \bigg |^s \bigg)^{1/s} \, dt \\ & \lesssim \int_{0<|t|<1/2} |t| \, \bigg(\sum_n \bigg | \hat \Delta^{\prime\prime}_n(t) \bigg |^s \bigg)^{1/s}\, dt = B. \end{align*} $$

Thus, $\mbox {II}\lesssim (B+C+D) \lVert f\rVert $ .

Proof of Theorem 2.4

Let $\Delta _n$ be the measure on the integers defined by

$$ \begin{align*}\hat\Delta_n=n^{\alpha/s}\hat\mu^n (1-\hat\mu)^r,\end{align*} $$

that is, $T_{\Delta _n}=n^{\alpha /s} T_{\mu }^n (I-T_{\mu })^r$ . Using Lemma 3.1 with $\Delta _n$ , it suffices to show that the corresponding terms defined in the lemma, $A, B, C, D$ , are bounded.

We assume $\alpha \ge -1$ because, if $\alpha <-1$ , ${\mathbf {Q}_{\alpha ,s,r}}f \le {\mathbf {Q}_{-1,s,r}}f$ .

If $\alpha>-1$ , dominating the sum by an integral and using property $\text {BA}_1$ (i), we obtain

$$ \begin{align*} \sum_n n^{\alpha} |\hat\mu(t)|^{ns} \le \sum_n n^{\alpha} (1-c h(t))^{ns} \lesssim \frac 1{h(t)^{\alpha+1}}. \end{align*} $$

If $\alpha =-1$ ,

$$ \begin{align*} \sum_n \frac 1n |\hat\mu(t)|^{ns} \le \sum_n \frac 1n (1-ch(t))^{ns}= |\!\ln(1-(1-ch(t))^s) | \lesssim \frac 1{h(t)^{\gamma}} \end{align*} $$

for any $\gamma>0$ .

For term A, using properties $\text {BA}_2$ , we estimate

(5) $$ \begin{align} A & =\int_{|t|<1/2} \frac1{|t|} \bigg(\sum_n \bigg| \hat\Delta_n(t) \bigg|^s \bigg)^{1/s} \, dt \nonumber \\ & = \int_{|t|<1/2} \bigg(\sum_n n^{\alpha} | \hat\mu(t)|^{ns} \bigg)^{1/s} \frac{|1-\hat\mu(t)|^{r}}{|t|} \, dt \nonumber \\ & \lesssim \begin{cases} \displaystyle\int_{0<t<1/2} h(t)^{r-(\alpha+1)/s-1} \, h'(t) \, dt & \mbox{ for } \alpha>-1, \\[12pt] \displaystyle \int_{0<t<1/2} h(t)^{r-\gamma/s-1} \, h'(t) \, dt & \mbox{ for } \alpha=-1. \end{cases} \end{align} $$

When $\alpha>-1$ the integral is finite for $sr>\alpha +1$ , and when $\alpha =-1$ the integral is finite for $r>0$ since we can choose $\gamma $ arbitrarily small, say $\gamma =sr/2$ .

With $B_n=T^n (I-T)^r$ ,

(6) $$ \begin{align} |\hat B^{\prime}_n(t)| = | n \hat\mu^{n-1}(t) (\hat \mu'(t)) (1-\hat\mu(t))^r - \hat\mu^{n}(t) \hat \mu'(t)(1-\hat\mu(t))^{r-1}|, \end{align} $$

and by $\text {BA}_1$ ,

$$ \begin{align*} |\hat B^{\prime\prime}_n(t)| & = | n(n-1) \hat\mu^{n-2}(t) (\hat \mu'(t))^2 (1-\hat\mu(t))^r \\ & \quad +n \hat\mu^{n-1}(t) \hat \mu"(t)(1-\hat\mu(t))^r - 2n r \hat\mu^{n-1}(t) (\hat \mu'(t))^2 (1-\hat\mu(t))^{r-1} \\ & \quad + r (r-1) \hat\mu^{n}(t) (1-\hat\mu(t))^{r-2} (\hat\mu'(t))^2 - r \hat\mu^{n}(t) (1-\hat\mu(t))^{r-1} \hat \mu"(t) | \\ & \lesssim [ n^2 |\hat\mu^{n-2}(t)| h^{2}(t) + n |\hat\mu^{n-1}(t)| h(t) + |\hat\mu^{n}(t)| ] \frac{h^{r-1}(t)|\hat\mu'(t)|}{|t|}. \end{align*} $$

Thus, for $\alpha>-1$ ,

(7) $$ \begin{align} \bigg(\!\sum_n |\hat \Delta^{\prime\prime}_n (t)|^s\bigg)^{1/s}= & \bigg(\sum_n n^{\alpha} |\hat B^{\prime\prime}_n (t)|^s\bigg)^{1/s} \lesssim h(t)^{(r-1)-(\alpha+1)/s} \frac{|\hat\mu'(t)|}{|t|}. \end{align} $$

Since $(1+\alpha )< s r $ ,

$$ \begin{align*} B= & \int_{|t|<1/2} |t| \, \bigg(\sum_n |\hat\Delta^{\prime\prime}_n(t)|^s \bigg)^{1/s} \, dt \lesssim \int_{0<t<1/2} h(t)^{(r-1)-(\alpha+1)/s } \, h'(t) \, dt <\infty. \end{align*} $$

When $\alpha =-1$ , choosing $0<\gamma <r$ , the estimate is

$$ \begin{align*} \int_{|t|<1/2} |t| \, \bigg(\sum_n |\hat\Delta^{\prime\prime}_n(t)|^s \bigg)^{1/s} \, dt \lesssim \int_{0<t<1/2} h(t)^{r-\gamma-1 } \, h'(t) \, dt <\infty. \end{align*} $$

For the remaining terms, we address the case $\alpha>-1$ since the estimates for the case $\alpha =-1$ follow similar arguments.

For $(1+\alpha )< s r $ , by $\text {BA}_2$ and dominating the kth term by an integral over intervals $(1/{k+1}, 1/k)$ , we have

$$ \begin{align*} C= & \sum_{k\neq 0} \frac 1{|k|} \bigg(\sum_n |\hat\Delta_n(1/|2k| )|^s \bigg)^{1/s} \nonumber \\ \lesssim & \sum_{k\neq 0} \frac 1{|k|} \bigg(\sum_n n^{\alpha} |\hat\mu(1/|2k|)|^{ns} |1-\hat\mu(1/|2k|)|^{s r} \bigg)^{1/s} \nonumber \\ \lesssim & \sum_{k> 0} \frac 1k h(1/2k)^{r - (\alpha+1)/s} \lesssim c+ \int_0^{1/2} \frac{h(t)^{r - (\alpha+1)/s}}t \, dt \nonumber \\ < & c+ \int_0^{1/2} h(t)^{r - (\alpha+1)/s-1} h'(t)\, dt <\infty. \end{align*} $$

From (6) and $\text {BA}_1$ (ii),

(8) $$ \begin{align} \bigg(\sum_n |\hat\Delta^{\prime}_n(t )|^s\bigg)^{1/s} = \bigg(\sum_n n^{\alpha} |\hat B^{\prime}_n(t )|^s\bigg)^{1/s} \lesssim h(t)^{r-1-(\alpha+1)/s} |\mu'(t)| \lesssim \frac{h(t)^{r-(\alpha+1)/s}}{|t|}. \end{align} $$

Thus,

$$ \begin{align*} D = & \sum_{k\neq 0} \frac 1{|k|^2} \bigg(\sum_n |\hat\Delta^{\prime}_n(1/|2k| )|^s \bigg)^{1/s} \lesssim \sum_{k\neq 0} \frac 1{|k|} h(1/2k)^{r - (\alpha+1)/s} <\infty. \\[-51pt] \end{align*} $$

Proof of Corollary 2.5

Let $\{n_k\}$ be an increasing sequence in $\mathbb N$ such that ${n_{k+1}-n_k\sim n_k}$ . Then we have $n_{k+1}=n_{k+1}-n_k+n_k\lesssim 2n_k$ .

Applying Abel’s summation,

$$ \begin{align*} \sum_{j=1}^u j^{\beta } T_{\mu}^{j-1}(1-T_{\mu})f & = \sum_{j=1}^u j^{\beta } T_{\mu}^{j-1}- \sum_{j=1}^u j^{\beta } T_{\mu}^jf = \sum_{j=0}^{u-1} (j+1)^{\beta } T_{\mu}^{j}f- \sum_{j=1}^u j^{\beta } T_{\mu}^jf\\ & = \sum_{j=0}^{u-1} ((j+1)^{\beta } - j^{\beta } )T_{\mu}^{j}f- u^{\beta } T_{\mu}^uf\\ & = \sum_{j=1}^{u} (j^{\beta } - (j-1)^{\beta } )T_{\mu}^{j-1}f- u^{\beta } T_{\mu}^uf. \end{align*} $$

Thus,

(9) $$ \begin{align} u^{\beta } |T_{\mu}^{u} f| &\le \sum_{j=1}^{u} (j^{\beta }-(j-1)^{\beta } )|T_{\mu}^{j-1} f| + \sum_{j=1}^u j^{\beta } |T_{\mu}^{j-1}(1-T_{\mu})f| \nonumber \\ &\lesssim \sum_{j=1}^{u} j^{\beta -1} |T_{\mu}^{j-1} f| + \sum_{j=1}^{u} j^{\beta } |T_{\mu}^{j-1}(1-T_{\mu})f|. \end{align} $$

Decomposing $n\in (n_k,n_{k+1}]$ as $n=n_{k-1}+l$ with $l\in (n_k-n_{k-1},n_{k+1}-n_{k-1}]$ , and applying the above to the function $T_{\mu }^{n_{k-1}}(1-T_{\mu })^rf$ with $u=l$ ,

$$ \begin{align*} n^{\beta } |T_{\mu}^{n}(1-T_{\mu})^r f| &\lesssim \frac{n^{\beta}}{l^{\beta}}\bigg[\!\sum_{j=1}^{l} j^{\beta -1} |T_{\mu}^{j+n_{k-1}-1} (1-T_{\mu})^r f| + \sum_{j=1}^{l} j^{\beta} |T_{\mu}^{j+n_{k-1}-1}(1-T_{\mu})^{r+1}f| \bigg] \\ &\lesssim \sum_{j=n_{k-1}}^{n_{k+1}-1} j^{\beta -1} |T_{\mu}^j (1-T_{\mu})^r f| + \sum_{j=n_{k-1}}^{n_{k+1}-1} j^{\beta} |T_{\mu}^{j} (1-T_{\mu})^{r+1} f | \end{align*} $$

because ${n}/{l} \le ({n_{k+1}}/{n_k-n_{k-1}})\lesssim ({2n_k}/{n_{k-1}})\lesssim 4$ . Thus,

$$ \begin{align*} &\bigg(\!\sum_k \max_{n_{k}<n\le n_{k+1}} n^{\beta s}|T_{\mu}^{n}(I-T_{\mu})^{r} f|^s \bigg)^{1/s}\le \sum_k \max_{n_{k}<n\le n_{k+1}} n^{\beta}|T_{\mu}^{n}(I-T_{\mu})^{r} f| \nonumber \\ & \quad \lesssim \sum_k \sum_{j=n_{k-1}}^{n_{k+1}-1} j^{\beta -1} |T_{\mu}^j (1-T_{\mu})^r f| + \sum_k \sum_{j=n_{k-1}}^{n_{k+1}-1} j^{\beta} |T_{\mu}^j (1-T_{\mu})^{r+1} f | \nonumber \\ & \quad \lesssim \ {\mathbf{Q}_{\beta-1,1,r}}f + {\mathbf{Q}_{\beta,1,r+1}}f. \end{align*} $$

By Theorem 2.4, both generalized square functions are bounded in Y when $0\le \beta < r$ . Therefore is also bounded in Y.

By [Reference Cohen, Cuny and Lin5], $\lim _{n\to \infty } n^{r} \lVert T_{\mu }^n(I - T_{\mu })^{r}f \rVert = 0$ .

From these results, it follows that $n^{\beta } |T_{\mu }^n(I - T_{\mu })^{r}f|\to 0$ almost everywhere as well.

Proof of Proposition 2.6

We prove only the result for the convolution measures with $\hat \Delta _{k,r}=n_k^{\beta } ({\hat \mu }^{n_k} - {\hat \mu }^{n_{k+1}})(1-\hat \mu )^r$ because the other case follows similar arguments. It suffices to verify that $\{\Delta _{k,r}\}_k$ satisfy the conditions of Lemma 3.1.

If $\delta \ge 0$ ,

$$ \begin{align*} \sum_k n_k^{s\delta} (n_{k+1}-n_k) |\hat\mu(t)|^{n_k s}\lesssim \frac1{h(t)^{s\delta+1}}. \end{align*} $$

We also have

(11) $$ \begin{align} \bigg(\!\sum_k|\hat\Delta_{k,r}(t)|^{s}\bigg)^{1/s} & \le \bigg(\sum_k n_k^{\beta s} |\hat\mu(t)|^{sn_k} \, |1-\hat\mu(t)^{(n_{k+1}-n_k)}|^s\bigg)^{1/s} |1-\hat\mu(t)|^r \nonumber \\ & \le \bigg(\sum_k n_k^{\beta s} (n_{k+1}-n_k)^s |\hat\mu(t)|^{s n_k} \bigg)^{1/s} |1-\hat\mu(t)|^{r+1} \nonumber \\ & \lesssim \bigg(\sum_k n_k^{\beta s+\gamma (s-1)} (n_{k+1}-n_k) |\hat\mu(t)|^{s n_k} \bigg)^{1/s} |h(t)|^{r+1} \nonumber \\ & \lesssim \frac{h(t)^{r+1}}{ h(t)^{\beta+\gamma+(1-\gamma)/s}}. \end{align} $$

Under the assumptions for $\gamma $ , $\beta $ and s we have $\beta +\gamma +(1-\gamma )/s <1+r$ .

Then, since $|h(t)|\lesssim |t| h'(t)$ ,

$$ \begin{align*} A = \int_{|t|<1/2} \frac 1{|t|} \bigg(\sum_k |\hat\Delta_{k,r}(t)|^{s} \bigg)^{1/s} \, dt \lesssim \int_{|t|<1/2} \frac{h(t)^{r} \, h'(t)}{ h(t)^{\beta+\gamma+(1-\gamma)/s}} \, dt =K <\infty, \end{align*} $$

and dominating the terms of the sum by an integral over intervals of length $l^{-2}$ ,

$$ \begin{align*} C & = \sum_{l\neq 0} \frac 1{|l|}\bigg(\!\sum_k |\hat\Delta_{k,r}(1/|2l| )|^s \bigg)^{1/s} \lesssim \sum_{l\neq 0} \frac {h(1/|2l|)^{1+r-(\gamma+\beta)-(1-\gamma)/s}}{|l|} \lesssim K <\infty. \end{align*} $$

Let $B_k=\hat \mu ^{n_k}(t)- \hat \mu ^{n_{k+1}}(t)$ . Then

$$ \begin{align*} | B^{\prime}_k(t) | &\le n_k | \hat \mu^{n_k-1}(t)| \, | 1-\hat\mu(t)^{n_{k+1}-n_k} | \,|\hat\mu'(t) | + (n_{k+1}-n_k) | \hat\mu(t) |^{n_{k+1}-1} \, |\hat\mu'(t)|\\ & \lesssim n_k^{1+\gamma} | \hat \mu^{n_k-1}(t)| \, |\hat\mu'(t)| |1-\hat\mu(t)| + n_k^{\gamma} | \hat \mu^{n_{k+1}-1}(t)| \, |\hat\mu'(t)| \end{align*} $$

and

$$ \begin{align*} B^{\prime\prime}_k & = n_k(n_k-1) \hat\mu^{n_k-2}(t) (\hat \mu'(t))^2 - n_{k+1}(n_{k+1}-1) \hat\mu^{n_{k+1}-2}(t)(\hat \mu'(t))^2 \nonumber \\ &\quad + n_k \hat\mu^{n_k-1}(t) \hat \mu"(t) - n_{k+1} \hat\mu^{n_{k+1}-1}(t) \hat \mu"(t) \\ & = n_k(n_k-1)\hat\mu^{n_k-2}(t) (1-\hat\mu^{n_{k+1}-n_k}(t)) (\hat \mu'(t))^2 \nonumber\\ & \quad + [n_k(n_k-1)-n_{k+1} (n_{k+1}-1)] \hat\mu^{n_{k+1}-2} (\hat \mu'(t))^2 \nonumber\\ & \quad + n_k \hat\mu^{n_k-1}(t) (1-\hat\mu^{n_{k+1}-n_k}(t)) \hat \mu"(t) \nonumber\\ & \quad +(n_k-n_{k+1}) \hat\mu^{n_{k+1}-1}(t) \hat \mu"(t).\nonumber \end{align*} $$

Since

$$ \begin{align*} \hat\Delta^{\prime}_{k,r} = n_k^{\beta} [B^{\prime}_k(t) (1-\hat\mu(t))^r + B_k(t) r (1-\hat\mu(t))^{r-1} \hat\mu'(t)], \end{align*} $$

we have

(12) $$ \begin{align} \bigg(\sum_k |\hat\Delta^{\prime}_{k,r}|^s \bigg)^{1/s} \lesssim \frac {h^r(t) |\hat\mu'(t)| }{h(t)^{(\gamma+\beta)+(1-\gamma)/s }}. \end{align} $$

Estimating

$$ \begin{align*} |n_k(n_k-1)-n_{k+1}(n_{k+1}-1)| \lesssim n_k^{\gamma}(n_{k+1}+n_k+1) \lesssim n_k^{1+\gamma}, \end{align*} $$

we have

$$ \begin{align*} & |n^{\beta} B_k"(t) (1-\hat\mu(t))^r| \lesssim [ n_k^{2+\gamma+\beta} \hat\mu^{n_k-2}(t) h(t)^2 + n_k^{1+\gamma+\beta} |\hat\mu(t)|^{n_{k+1}-1} h(t) \\ &\quad + n_{k}^{1+\gamma+\beta} |\hat\mu(t)|^{n_{k+1}-2} h(t)+ n_{k}^{\gamma+\beta} |\hat\mu(t)|^{n_{k+1}-1} ] \frac{h(t)^r |\hat\mu'(t)|}{|t|}. \end{align*} $$

Since

$$ \begin{align*} & |\hat\Delta^{\prime\prime}_{k,r}| \sim n_k^{\beta} [|B^{\prime\prime}_k(t)| |1-\hat\mu(t)|^r + |B^{\prime}_k(t)| |1-\hat\mu(t)|^{r-1} |\hat\mu(t)| \\ &\quad + |B_k(t)| |1-\hat\mu(t)|^{r-2} |\hat\mu'(t)|^2 + |B_k(t)| |1-\hat\mu(t)|^{r-1} |\hat\mu"(t)| ], \end{align*} $$

we have

(13) $$ \begin{align} \bigg(\!\sum_k |\hat\Delta^{\prime\prime}_{k,r}(t)|^s\bigg)^{1/s} \lesssim \frac{h(t)^r}{h(t)^{(\gamma+\beta)+(1-\gamma)/s } } \frac{|\hat\mu'(t)|} {|t|}, \end{align} $$

and

$$ \begin{align*} B = \int_{|t|<1/2} |t|\,\bigg(\!\sum_k |\hat\Delta^{\prime\prime}_{k,r}(t)|^s\bigg)^{1/s} \, dt \lesssim \int_{0<t<1/2} \frac {h(t)^r \ h'(t)}{h(t)^{(\gamma +\beta)+(1-\gamma)/s }}\, dt<\infty. \end{align*} $$

For the last term, from equation (12) and $|t\hat \mu '(t)|\lesssim h(t)$ , we have

$$ \begin{align*} D \lesssim \sum_{l\neq0} \frac 1{l^2} \frac {h(1/|2l|)^r |\hat \mu'(1/|2l|)|}{h(1/|2l|)^{(\gamma+\beta)+(1-\gamma)/s }} \lesssim \sum_{l\neq 0} \frac {h(1/|2l|)^{r +1 -(\gamma+\beta)-(1-\gamma)/s}}{|l|} \lesssim K <\infty. \end{align*} $$

Proof of Proposition 2.7

Let $\Delta _{n,r}f=T_{\mu }^n(I-T_{\mu })^r f$ , and $\{n_k\}$ be any increasing sequence. By Abel summation,

$$ \begin{align*} \sum_{u=n}^m u^{\beta} T_{\mu}^u(I-T_{\mu})f=\sum_{u=n+1}^{m+1} (u^{\beta} - (u-1)^{\beta})T_{\mu}^uf + n^{\beta}T_{\mu}^n -(m+1)^{\beta} T_{\mu}^{m+1}f.\end{align*} $$

Hence,

$$ \begin{align*} | n_k^{\beta}T_{\mu}^{n_k}f -n_{k+1}^{\beta} T_{\mu}^{n_{k+1}}f | \lesssim \bigg|\sum_{u=n_k}^{n_{k+1}-1} u^{\beta} T_{\mu}^u(I-T_{\mu})f \bigg| + \bigg|\sum_{u=n_k+1}^{n_{k+1}} (u^{\beta} - (u-1)^{\beta})T_{\mu}^uf \bigg|. \end{align*} $$

Applied to the function $(I-T_{\mu })^r f$ , we obtain

(14) $$ \begin{align} |D_{k,\beta}f| = & |n_k^{\beta} \Delta_{n_k,r} f - n_{k+1}^{\beta} \Delta_{n_{k+1},r}f | \lesssim \sum_{u=n_k+1}^{n_{k+1}} u^{\beta-1} | \Delta_{u,r}f| + \sum_{u=n_k}^{n_{k+1}-1} u^{\beta} |\Delta_{u,r+1}f|. \end{align} $$

Since $s\ge 1$ ,

$$ \begin{align*} \bigg(\!\sum_k |D_{k,\beta}f|^s \bigg)^{1/s} \le \sum_k |D_{k,\beta}f| \lesssim & \sum_u u^{\beta-1 } |\Delta_{u,r}f| + \sum_u u^{\beta} |\Delta_{u,r+1}f|. \end{align*} $$

Then

$$ \begin{align*} \lVert n^{\beta}\Delta_{n,r}f \rVert_{v(s)} \lesssim {\mathbf{Q}_{\beta-1 ,1,r}}f + {\mathbf{Q}_{\beta ,1,r+1}}f,\end{align*} $$

which, by Theorem 2.4, are bounded on Y because $\beta <r$ .

Since $ \lVert n^{\beta } \Delta _{n,r}f \rVert _{o(s)}\le \lVert n^{\beta } \Delta _{n,r}f \rVert _{v(s)}$ the result also holds for oscillation norms.

Proof of Theorem 1.7

By [Reference Dungey9], there exist a power-bounded operator R and $a\in (0,1)$ such that $T=I-(I-R)^{a} = R_{\nu _{a}}$ , where $\nu _{a}$ is the probability measure on $\mathbb Z$ defined in equation (2). Therefore Theorem 2.4, Corollary 2.5 and Propositions 2.6 and 2.7 apply to T.

Proof. There are several versions of the ergodic decomposition of measures [Reference Cornfeld, Fomin and Sinai6, Reference Walters28]. We use the formulation based on conditional expectation on a standard probability space [Reference Sarig26]. (The argument works for any other formulation.)

Let $\mathrm {Inv}(\tau )=\{E \in \mathcal B : E = \tau ^{-1}E\}$ . Let $m_y$ be the conditional probabilities with respect to $\mathrm {Inv}(\tau )$ . (Refer to [Reference Sarig26] to see how the $\{m_y\}$ are constructed.) Then, for m-almost every $y\in X$ , $m_y$ is $\tau $ -invariant and ergodic and $\int _X f \,dm= E(E(f|\mathrm {Inv}(\tau ))= \int _X \int _X f \,dm_y \,dm(y)$ . Then

$$ \begin{align*} m(|U_{\tau}f|>\unicode{x3bb}) & = \int_X m_y(|U_{\tau}f|>\unicode{x3bb} ) \,dm(y) \\ & \le \int_X \frac C{\unicode{x3bb}} \int_X |f| \,dm_y \, dm(y)=\frac C{\unicode{x3bb}} \int_X |f|\, dm. \\[-46pt] \end{align*} $$

Proof. By Lemma 5.1, it suffices to consider the case when $(X, \mathcal B, m,\tau )$ is an ergodic measure-preserving system.

Given $\unicode{x3bb}>0$ and $f\in L^1$ , decompose f via the ergodic Calderón–Zygmund decomposition, with the sets $B,B_i, E_i$ and the associated good and bad functions g and b defined as in Proposition 5.2.

Since Q is sublinear, we have

$$ \begin{align*} m(x:|Q f(x)|>2\unicode{x3bb})\le m(x:|Q g(x)|>\unicode{x3bb}) + m(x:|Q b(x)|>\unicode{x3bb}).\end{align*} $$

By the $L^2$ -boundedness of Q, $m(x:|Q g(x)|>\unicode{x3bb} ) \lesssim {\lVert g\rVert _2^2}/{\unicode{x3bb} ^2} \le {2\lVert g\rVert _1}/{\unicode{x3bb} } \le {2\lVert f\rVert _1}/{\unicode{x3bb} }$ .

Let $E_i^*=E_i\cup \tau ^{i}E_i \cup \tau ^{2 i}E_i \cup \tau ^{-i} E_i \cup \tau ^{-2i} E_i$ . By property (e) of the ergodic Calderón–Zygmund decomposition,

$$ \begin{align*} m(x:|Q b(x)|> \unicode{x3bb})\le & m\bigg(\bigcup_i E^*_i\bigg)+ m\bigg(x\in \bigg(\bigcup_i E^*_i\bigg)^{c}: |Q b(x)|>\unicode{x3bb}\bigg)\\ \le & 5 \sum_i m(E_i) + \frac 1{\unicode{x3bb}} \int_{(\bigcup_i E^*_i)^{*c}} |Q b|\, dm \\ \le & 10 \frac{\lVert f\rVert_1}{\unicode{x3bb}} + \frac 1{\unicode{x3bb}} \sum_i \int_{(E^*_i)^{*c}} |Q b_i|\, dm. \end{align*} $$

Noting that each $b_i$ is a $\unicode{x3bb} $ -atom, apply assumption (b) to obtain

$$ \begin{align*} \int_{(\bigcup_i E^*)^c} |Q b(x)| \,dm \le \sum_i \int_{E_i^{*c}} |Q b_i(x)| \,dm \le C \unicode{x3bb} \sum_i m(E_i) \le 2 C \|f\|_1,\end{align*} $$

completing the proof.

Proof. Let a be a $\unicode{x3bb} $ -atom and let B, d, E and $E^*$ as in Definition 5.3. Since a is supported on E, and using the cancelation property (c) of Definition 5.3,

$$ \begin{align*}\tau_{\Delta_n} a(x)=\sum_{j} \Delta_n(j) a(\tau^j x) = & \sum_{j: \tau^jx\in B} \sum_{l=1}^{d} (\Delta_n(j+l) - \Delta_n(j)) a(\tau^{j+l} x). \end{align*} $$

Then

$$ \begin{align*} (Q a(x))^s =& \sum_{k=1}^{\infty} \max_{n\in J_k} \bigg|\sum_{j: \tau^jx\in B} \sum_{l=1}^{d} (\Delta_n(j+l) - \Delta_n(j)) a(\tau^{j+l} x)\bigg|^s\\ = & \sum_{k=1}^{\infty} \max_{n\in J_k} \bigg|\sum_{j: \tau^jx\in B} \sum_{l=1}^{d}a(\tau^{j+l}x) \int_{|t|<0.5} \hat\Delta_n(t) (e((j+l)t)- e(jt) ) \, dt \bigg|^s. \end{align*} $$

Let $I(x)$ and $II(x)$ be the expressions obtained by replacing the integral over $\{|t|<0.5\}$ with integrals over $\{|t|<1/|j|\}$ and $\{1/|j|<|t|<0.5\}$ , respectively. We have

$$ \begin{align*} I(x)&= \bigg(\sum_{k=1}^{\infty} \max_{n\in J_k} \bigg|\sum_{j: \tau^jx\in B} \sum_{l=1}^{d}a(\tau^{j+l}x) \int_{|t|<1/|j|} \hat\Delta_n(t) (e((j+l)t)- e(jt) ) \, dt \bigg|^s \bigg)^{1/s}, \\II(x)&= \bigg(\sum_{k=1}^{\infty} \max_{n\in J_k} \bigg|\sum_{j: \tau^jx\in B} \sum_{l=1}^{d}a(\tau^{j+l}x) \int_{1/|j|<|t|<0.5} \hat\Delta_n(t) (e((j+l)t)- e(jt) ) \, dt \bigg|^s\bigg)^{1/s}. \end{align*} $$

By the triangular inequality of the $\ell ^s$ -norm,

$$ \begin{align*}Q a(x)\le I(x)+II(x).\end{align*} $$

From the construction of $E^*$ , if $x\in (E^*)^c$ and $\tau ^jx\in B$ , we must have $|j|\ge 2d>1$ .

We estimate $I(x)$ as follows:

$$ \begin{align*} I(x)& = \bigg(\sum_{k=1}^{\infty}\max_{n\in J_k} \bigg|\sum_{j: \tau^jx\in B}\sum_{l=1}^{d}a(\tau^{j+l}x)\int_{|t|<1/|j|} \hat\Delta_n(t) (e((j+l)t)- e(jt) ) \, dt \bigg|^s \bigg)^{1/s}\\ & \le \! \sum_{j: \tau^jx\in B} \sum_{l=1}^{d} |a(\tau^{j+l}x)| \int_{|t|<1/|j|} \bigg(\sum_{k=1}^{\infty} \max_{n\in J_k} |\hat\Delta_n(t)|^s \bigg)^{1/s} |e((j+l)t)- e(jt) | \, dt \\ & \le \! \sum_{j: \tau^jx\in B} \sum_{l=1}^{d} |l| \ |a(\tau^{j+l}x)| \int_{|t|<1/|j|} \bigg(\sum_{k=1}^{\infty} \max_{n\in J_k} |\hat\Delta_n(t)|^s \bigg)^{1/s} |t| \, dt \\ & \le 2\unicode{x3bb} d \int_{|t|<0.5} \bigg(\sum_{k=1}^{\infty} \max_{n\in J_k} |\hat\Delta_n(t)|^s \bigg)^{1/s} |t| \sum_{|j|>1} 1_{[x:x\in \tau^{-j}B]}(x) 1_{[j:|j|<1/|t|]}(j) \, dt. \end{align*} $$

Then

$$ \begin{align*} \int_{(E^*)^c} I(x) \,dm & \le 2\unicode{x3bb} d \int_{|t|<0.5} \bigg(\sum_{k=1}^{\infty} \max_{n\in J_k} |\hat\Delta_n(t)|^s\bigg)^{1/s} |t| \sum_{1<|j|<1/|t|} m((E^*)^c\cap \tau^{-j}B) \\ & \le 4\unicode{x3bb} d \ m(B) \int_{|t|<0.5} \bigg(\sum_{k=1}^{\infty} \max_{n\in J_k} |\hat\Delta_n(t)|^s\bigg)^{1/s} \, dt \le \ C \unicode{x3bb} m(E). \end{align*} $$

Handling $II(x)$ requires integration by parts. We will need the following estimates. Because $\Delta _n(t),\Delta ^{\prime }_n(t),e(kt)$ are 1-periodic, the same procedure as (4) yields, for ${|j|>2|l|}$ ,

$$ \begin{align*} & \bigg|\int_{|t|>1/|j|} \hat\Delta_n(t) (e((j+l)t)- e(jt) ) \, dt \bigg| \\ &\quad \lesssim \bigg|\int_{|t|>1/|j|} \hat\Delta^{\prime}_n(t) \bigg(\frac{e((j+l)t)}{j+l}- \frac{e(jt)}{j} \bigg) \, dt \bigg| + \bigg| \hat\Delta_n\bigg(\frac{1}{|j|}\bigg) \bigg| \frac{|l|}{j^2}. \end{align*} $$

From [Reference Bellow and Calderón3], since $2|l|<2d<|j|$ , we have

$$ \begin{align*} \bigg|\frac{e((j+l)t)-1}{(j+l)^2}- \frac{e(jt)-1}{j^2} \bigg| \lesssim \frac{|l||t|}{|j|^2}. \end{align*} $$

Thus,

$$ \begin{align*} & \int_{|t|>1/|j|} \hat\Delta^{\prime}_n(t) \bigg(\frac{e((j+l)t)}{(j+l)}- \frac{e(jt)}{j} \bigg) \, dt\\ &\quad \lesssim \bigg|\int_{|t|>1/|j|} \hat\Delta^{\prime\prime}_n(t) \bigg(\frac{e((j+l)t)-1}{(j+l)^2}- \frac{e(jt)-1}{j^2} \bigg) \, dt \bigg|\\ & \qquad+ \bigg| \hat\Delta^{\prime}_n\bigg(\frac{1}{|j|}\bigg) \bigg| \frac{|l|}{|j|^3} \\ & \quad \le \int_{|t|>1/|j|} |\hat\Delta^{\prime\prime}_n(t)| |t| \frac{|l|}{|j|^2} \, dt+ \bigg| \hat\Delta^{\prime}_n\bigg(\frac{1}{|j|}\bigg) \bigg| \frac{|l|}{|j|^3}. \end{align*} $$

Accordingly, $II(x)$ is dominated by three parts $II_1$ , $II_2$ and $III_3$ estimated below.

$$ \begin{align*} II_{1}(x) & = \bigg(\sum_{k=1}^{\infty} \max_{n\in J_k} \bigg|\sum_{j: \tau^jx\in B} \sum_{l=1}^{d} a(\tau^{j+l} x) \int_{1/|j|<|t|<0.5} |\hat\Delta^{\prime\prime}_n(t)| \frac{l |t|}{|j|^2} \, dt \bigg|^s\bigg)^{1/s}\\ & \le d \sum_{j: \tau^jx\in B} \sum_{l=1}^{d} |a(\tau^{j+l} x)| \int_{1/|j|<|t|<0.5} \bigg(\sum_{k=1}^{\infty} \max_{n\in J_k} |\hat\Delta^{\prime\prime}_n(t)|^s\bigg)^{1/s} \frac{|t|}{|j|^2} \, dt\\ & \le 2\unicode{x3bb} d \int_{|t|<0.5} \bigg(\sum_{k=1}^{\infty} \max_{n\in J_k} |\hat\Delta^{\prime\prime}_n(t)|^s\bigg)^{1/s} |t| \sum_{|j|>1/|t|} 1_{[\tau^jx\in B]}(x) \frac{1}{|j|^2} \, dt. \end{align*} $$

Integrating over $(E^*)^c$ ,

$$ \begin{align*} \int_{(E^*)^c} II_{1}(x) \,dm & \le 2\unicode{x3bb} d \int_{|t|<0.5} \bigg(\sum_{k=1}^{\infty} \max_{n\in J_k} |\hat\Delta^{\prime\prime}_n(t)|^s\bigg)^{1/s} |t| \sum_{|j|>1/|t|} \frac{1}{|j|^2} m(B)\\ & \lesssim \unicode{x3bb} m(E) \int_{|t|<0.5} \bigg(\sum_{k=1}^{\infty} \max_{n\in J_k} |\hat\Delta^{\prime\prime}_n(t)|^s\bigg)^{1/s} |t|^2 \, dt= C \unicode{x3bb} m(E). \end{align*} $$

Lastly, we need to handle two remaining pieces:

$$ \begin{align*} II_{2}(x) & = \int_{(E^*)^c} \bigg(\sum_k \max_{n\in J_k} \bigg|\sum_{j: \tau^jx\in B} \sum_{l=1}^{d} |a(\tau^{j+l} x)| \bigg| \hat\Delta_n\bigg(\frac{1}{|j|}\bigg) \bigg| \frac{|l|}{|j+l|^2} \bigg|^s\bigg)^{1/s} \,dm\\ & \lesssim \int_{(E^*)^c } d \sum_{j: \tau^jx\in B} \sum_{l=1}^{d} |a(\tau^{j+l} x)| \bigg(\sum_k \max_{n\in J_k} \bigg| \hat\Delta_n\bigg(\frac{1}{|j|}\bigg) \bigg|^s\bigg)^{1/s} \frac{1}{j^2} \,dm\\ & \lesssim \unicode{x3bb} m(E) \sum_{|j|>1} \frac 1{j^2} \bigg(\sum_k \max_{n\in J_k} \bigg| \hat\Delta_n\bigg(\frac{1}{|j|}\bigg) \bigg|^s \bigg)^{1/s} \le C \unicode{x3bb} m(E) \end{align*} $$

and

$$ \begin{align*} II_{3}(x) & = \int_{(E^*)^c } \bigg(\sum_k \max_{n\in J_k} \bigg|\sum_{j: \tau^jx\in B} \sum_{l=1}^{d} |a(\tau^{j+l} x)| \bigg| \hat\Delta^{\prime}_n\bigg(\frac{1}{|j|}\bigg) \bigg| \frac{|l|}{|j+l|^3} \bigg|^s\bigg)^{1/s} \,dm \\ & \lesssim \int_{(E^*)^c } d \sum_{j: \tau^jx\in B} \sum_{l=1}^{d} |a(\tau^{j+l} x)| \bigg(\sum_k \max_{n\in J_k} \bigg| \hat\Delta^{\prime}_n\bigg(\frac{1}{|j|}\bigg) \bigg|^s\bigg)^{1/s} \frac{1}{|j|^3} \,dm\\ & \lesssim \unicode{x3bb} m(B) d \sum_{|j|>1} \bigg(\sum_k \max_{n\in J_k} \bigg| \hat\Delta^{\prime}_n\bigg(\frac{1}{|j|}\bigg) \bigg|^s \bigg)^{1/s} \frac 1{|j|^3} \le C \unicode{x3bb} m(E). \end{align*} $$

Combining the estimates for $I(x),II_{1}(x), II_{2}(x)$ and $II_{3}(x)$ , we obtain the desired result:

$$ \begin{align*} \int_{(\bigcup_i E^*)^c } Qa(x)\, dm \le \ C \unicode{x3bb} \ m(E). \\[-47pt] \end{align*} $$

Proof of Theorem 1.9

Proof of Part A. Assuming that ${\mathbf {Q}_{{\alpha ,s,r}}}$ is bounded on $L^2$ , we simply need to check the requirements in Theorem 5.5 for $\hat \Delta _n=n^{\alpha /s} \hat \mu ^n(1-\hat \mu )^{r}$ and the sequence $n_k=k$ , $r\ge 0$ .

First note that, under the assumptions for h, if $(1\kern1pt{+}\kern1pt\alpha )\kern1pt{\le}\kern1pt sr\kern-0.4pt$ , ${\int _{|t|<0.5} h^{r-(1+\alpha )/s}(t) \, dt\kern1pt{<}\kern1pt\infty \kern-0.3pt.}$ Using estimates for $\Delta _n(t)$ in equation (5), for $(1+\alpha )\le sr$ we have

$$ \begin{align*} \int_{|t|<0.5} \bigg(\sum_{n=1}^{\infty} |\hat\Delta_n(t)|^s\bigg)^{1/s} \, dt\le \int_{|t|<0.5} h^{r-(1+\alpha)/s}(t)\, dt <\infty \end{align*} $$

and

$$ \begin{align*} \sum_{|j|>1} \frac{1}{j^2} \bigg(\sum_n \bigg| \hat\Delta_n\bigg(\frac{1}{|j|}\bigg) \bigg|^s \bigg)^{1/s} \le & \sum_{|j|>1} \frac 1{j^2} h\bigg(\frac{1}{j}\bigg)^{r-(1+\alpha)/s} \lesssim \int_0^1 h(t)^{r-(\alpha+1)/s}\, dt <\infty. \end{align*} $$

From property $\text {BA}_1$ (ii) and equation (8),

$$ \begin{align*} \sum_{j>1} \frac{1}{|j|^3} \bigg(\sum_n \bigg| \hat\Delta^{\prime}_n\bigg(\frac{1}{|j|}\bigg) \bigg|^s\bigg)^{1/s} \le & \sum_{j>1} \frac 1{j^2} h\bigg(\frac{1}{j}\bigg)^{r-(\alpha+1)/s} <\infty. \end{align*} $$

By the estimates for $|\hat \Delta ^{\prime \prime }_n(t)|$ in equation (7) and $\text {BA}_1$ (ii),

$$ \begin{align*} \int_{|t|<0.5} \bigg(\sum_{n=1}^{\infty} |\hat\Delta^{\prime\prime}_n(t)|^s\bigg)^{1/s} |t|^2 \, dt & \le \int_{|t|<0.5} h(t)^{(r-1)-(\alpha+1)/s}|\hat\mu'(t)| |t| \, dt \\ & \le \int_{|t|<0.5} h(t)^{r-(\alpha+1)/s}\, dt <\infty. \end{align*} $$

Proof of Part B. Since $\lVert \{n^{\beta }\tau _{\mu }^n(I-\tau _{\mu })^r f\} \rVert _{v(s)}$ is bounded on $L^2$ , by Proposition 5.4 it suffices to check the conditions of Theorem 5.5 for the corresponding sublinear operator Q defined by $\hat \Delta _n=n^{\beta }({\hat \mu }^n(1-\hat \mu )^r)$ .

Because $s>1$ we can choose $0<\delta <\min (1,(s-1)/(s+1))$ . Starting with the sequence $\{2^k\}$ , enlarge it by adding $N_k=2^{k(1-\delta )}$ equally spaced points in between; call these $\{r_{k,j}\}_{j=0,N_k-1}$ . These points have the properties $r_{k,0}=2^k$ and $r_{k,j+1}-r_{k,j}\sim 2^k/N_k=2^{\delta k}\sim r_{k,j}^{\delta }$ . Denote the full collection of these by $\{r_j\}$ with the indices rearranged appropriately.

Let $B_{n,r}=n^{\beta } \tau _{\mu }^{n}(I-\tau _{\mu })^r$ and $K_n=B_{r_j,r}$ for $r_j\le n<r_{j+1}$ . Also let $I_j=[r_j,r_{j+1}]$ . We have

By properties of variational norms [Reference Jones, Kaufman, Rosenblatt and Wierdl13–Reference Jones, Seeger and Wright15],

$$ \begin{align*} \lVert \{ (B_{n,r} -K_n) a\} \rVert_{v(s)} \lesssim & \bigg(\sum_j \lVert\{ (B_{n,r} -B_{r_j,r}) a: n\in I_j \} \rVert_{v(s)}^s \bigg)^{1/s}.\end{align*} $$

Using Abel summation as in equation (14), bound the last term by

$$ \begin{align*} & \lVert \{ (B_{n,r} -B_{r_j,r}) a : n\in I_j\} \rVert^s_{v(s)} \lesssim \sum_{n\in I_j} |(B_{n,r} -B_{r_j,r}) a |^s \\ & \quad\lesssim r_j^{\delta} \bigg(\sum_{n=r_j+1}^{r_{j+1}} n^{\beta-1} |\tau_{\mu}^{n} (1-\tau_{\mu})^r a| + \sum_{n=r_j}^{r_{j+1}-1} n^{\beta} |\tau_{\mu}^{n} (1-\tau_{\mu})^{r+1} a| \bigg)^s \\ & \quad\lesssim r_j^{\delta(1+s)} \sum_{n=r_j}^{r_{j+1}} n^{(\beta-1)s} |\tau_{\mu}^{n} (1-\tau_{\mu})^r a|^s + r_j^{\delta(1+s)} \sum_{n=r_j}^{r_{j+1}-1} n^{\beta s} |\tau_{\mu}^{n} (1-\tau_{\mu})^{r+1} a|^s. \end{align*} $$

Thus,

(16) $$ \begin{align} \lVert \{ (B_{n,r} -K_n) a\} \rVert_{v(s)}\lesssim {\mathbf{Q}_{(\beta-1)s+\delta(1+s),s,r}}a+ {\mathbf{Q}_{\beta s+\delta(1+s),s,r+1}}a. \end{align} $$

Since $\beta \le r$ and $\delta (1+s)+1<s$ , these are bounded on $L^1$ .

When $r>0$ ,

$$ \begin{align*} & \lVert \{K_n a\} \rVert_{v(s)} \lesssim \sum_j | (r_j^{\beta} \tau_{\mu}^{r_j}- r_{j+1}^{\beta} \tau_{\mu}^{r_{j+1}}) (1-\tau_{\mu})^r a| \\ & \quad\lesssim \sum_j r_j^{\beta} |(\tau_{\mu}^{r_j}- \tau_{\mu}^{r_{j+1}}) (1-\tau_{\mu})^r a| + \sum_j ( r_{j+1}^{\beta} -r_j^{\beta}) |\tau_{\mu}^{r_{j+1}} (1-\tau_{\mu})^r a| \\ & \quad\lesssim \sum_j r_j^{\beta} |(\tau_{\mu}^{r_j}- \tau_{\mu}^{r_{j+1}}) (1-\tau_{\mu})^r a| + \sum_j r_{j+1}^{\beta-1+\delta} |\tau_{\mu}^{r_{j+1}} (1-\tau_{\mu})^r a|. \end{align*} $$

Since $\beta \le r$ , Lemma 5.6(a) applies to the first term with $s=1$ , and Lemma 5.6(b) applies to the second term.

When $r=0$ and $\beta =0$ , we have

$$ \begin{align*} \lVert \{K_n a\} \rVert_{v(s)} \lesssim \sum_j | ( \tau_{\mu}^{r_j}- \tau_{\mu}^{r_{j+1}}) a|. \end{align*} $$

Thus, Lemma 5.6(a) applies.

In either case, we obtain

(17) $$ \begin{align} \int_{(E^*)^c} \lVert \{K_n a(x)\} \rVert_{v(s)} \, dm \lesssim \unicode{x3bb} m(E). \end{align} $$

Finally, from equations (15), (16) and (17),

$$ \begin{align*} \int_{(E^*)^c} \lVert \{ B_{n,r} a\} \rVert_{v(s)} \,dm \le & \int_{(E^*)^c} [ \lVert \{ (B_{n,r} -K_n) a\} \rVert_{v(s)} + \lVert \{K_n a(x)\} \rVert_{v(s)}] \, dm\\ \lesssim & \ \unicode{x3bb} m(E). \end{align*} $$

The result for the oscillation norm follows from noticing that $ \lVert B_{n,r}a \rVert _{o(s)}\le \lVert B_{n,r}a \rVert _{v(s)}$ .

Proof of Theorem 1.10(d)

By equation (10), if $s\ge 2$ , then

$$ \begin{align*} \bigg(\!\sum_k \max_{n_k<n\le n_{k+1}} n^{r s} |\tau_{\mu}^n(1-\tau_{\mu})^r f |^s \bigg)^{1/s} \lesssim {\mathbf{Q}_{2r-1,2,r}}f + {\mathbf{Q}_{2(r+1)-1,1,r+1}}f. \end{align*} $$

Thus, the result follows from Theorem 1.10(a) with $s=2$ .

Proof. For simplicity, we only prove case (a) because case (b) follows from similar estimates.

By Theorem 5.5, it suffices to check the assumptions of that theorem for $\hat \Delta _n=r_n^{\beta }({\hat \mu }^{r_n}-{\hat \mu }^{r_{n+1}})(1-\hat \mu )^r$ and the sequence $n_k=k$ .

From equation (11), and since $\beta +\delta +(1-\delta )/s\le r+1$ ,

$$ \begin{align*} \bigg(\sum_{n=1}^{\infty} |\hat\Delta_n(t)|^s\bigg)^{1/s} \lesssim & \frac {h(t)^{r+1}}{h(t)^{\beta+\delta +(1-\delta )/s} } \lesssim 1. \end{align*} $$

Hence,

$$ \begin{align*} \int_{|t|<0.5} \bigg(\sum_{n=1}^{\infty} |\hat\Delta_n(t)|^s\bigg)^{1/s} \, dt \lesssim 1 \quad\mbox{and}\quad \sum_{|j|>0} \frac 1{|j|^2} \bigg(\sum_{n=1}^{\infty} \bigg|\hat\Delta_n\bigg(\frac{1}{|j|}\bigg)\bigg|^s\bigg)^{1/s} <\infty. \end{align*} $$

By equation (12),

$$ \begin{align*} \bigg(\sum_{n=1}^{\infty} |\hat\Delta^{\prime}_n(t)|^s\bigg)^{1/s} \lesssim \frac{h(t)^r }{h(t)^{\beta+\delta+(1-\delta)/s}} |\hat\mu'(t)|, \end{align*} $$

and by $\text {BA}_1$ (ii), $|t \hat \mu '(t)| \lesssim h(t)$ ,

$$ \begin{align*} \sum_{|j|>0} \frac 1{|j|^3} \bigg(\sum_{n=1}^{\infty} |\hat\Delta^{\prime}_n (1 /|j|)|^s\bigg)^{1/s} & \lesssim \sum_{|j|>0} \frac 1{|j|^3} \frac{h({1}/{|j|})^r}{h({1}/{|j|})^{\,\beta+\delta+(1-\delta)/s} } |\hat\mu'\bigg(\frac{1}{|j|}\bigg)|\\ & \lesssim \sum_{|j|>0} \frac 1{|j|^2}h\bigg(\frac{1}{|j|}\bigg)^{r+1-(\beta+\delta+(1-\delta)/s)} \lesssim \sum_{|j|>0} \frac 1{|j|^2}<\infty. \end{align*} $$

By equation (13) and $\text {BA}_1$ ,

$$ \begin{align*} \int_{|t|<0.5} \bigg(\sum_{n=1}^{\infty} |\hat\Delta^{\prime\prime}_n(t)|^s\bigg)^{1/s} |t|^2 \, dt \lesssim \int_{|t|<0.5} \frac{h(t)^r}{h(t)^{\,\beta+\delta+(1-\delta)/s} }\frac{|\hat\mu'(t)|}{|t|} |t|^2 \, dt \lesssim 1. \end{align*} $$