论文标题
希尔伯特转型的完全收敛
Complete convergence of the Hilbert transform
论文作者
论文摘要
假设$ \ {a_j \} \ in \ ell^1 $，并假设对于任何序列$（t_n）$的整数的任何序列$（t_n）$退出常数$ c_1> 0 $ 0 $，这样的$ \ sharp \ sharp \ left \ left \ weft \ { \ Mathcal {b} _n-t_n} \！\！\！\ rish {1.9ex} \ hbox {$ \ scriptssize \ prime $} \; \ frac {a_ {a_ {k+i}} {i} \ right |>λ\ right \} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ sharp \ left \ left \ weft \ {k \ in \ mathbb {z}：\ sup_ {n \ geq 1}} \！\！\ rish {1.9ex} \ hbox {$ \ scriptSize \ prime $} \;; \ frac {a_ {a_ {k+i}} {i} \ right |>λ\ right \}，所有$λ> 0 $，其中$ \ mathcal {b} _n = \ { - n = \ { - n = \ { - n， - （n-1）， - （n-1）， - （n-1）， - （n-2），\ dots，\ dots，\ dots，n-2，n-2，n-1，n \} $。然后，有一个常数$ C_2> 0 $，不取决于序列$ \ {a_j \} $，因此$ \ sum_ {n = 1}^\ infty \ infty \ sharp \ sharp \ left \ left \ {k \ in \ mathbb {z} \！\！\ rish {1.9ex} \ hbox {$ \ scriptSize \ prime $} \;; \ frac {a_ {k+i}} {i} \ right |>λ\ right \} \ leq \ frac {c_2}λ\ sum_ {i = - \ \ iffty}^{\ infty}^{\ infty} {\ infty} | 令$（x，x，\ mathscr {b}，μ）$为一个度量空间，$τ：x \ x $ to x $是可逆性的可逆变换，并假设l^1（x）$中的$ f \ in l^1（x）$，以便对任何序列$（t_n）$的整数中的$ c_1> $ ege se use c_1 $ gee 1} \ left | \ sum_ {i \ in \ mathcal {b} _n-t_n} \！\！\！\！ \ frac {f（τ^ix）} {i} \ right | >λ\ right \} \ leqc_1μ\ left \ {x：\ sup_ {n \ geq 1} \ left | \ sum_ {i \ in \ nathcal {b} _n _n} _n} \！\！\！\！\！\！\！\！ \ frac {f（τ^i x）} {i} \ right |>λ\ right \} $$对于所有$λ> 0 $，其中$ \ mathcal {b} _n = \ { - n = \ { - n， - n， - （n-1）， - （n-1）， - （n-1）， - （n-2），\ dots，\ dots，n-dots，n-2，n-1，n-1，n \ \} $。然后存在一个常数$ C_2> 0 $，该$不取决于$ f $，因此$$ \ sum_ {n = 1}^\inftyμ\ left \ left \ {x：\ left | \ sum_ {i = -n}^{n}^{n} \！\！\！\！\！\！\！\！\！ \; \ frac {f（τ^ix）} {i} \ right |>λ\ right \} \ leq \ frac {c_2}λ\ | f \ | f \ | _1 $ _1 $ _1 $$，用于$λ> 0 $。
Suppose that $\{a_j\}\in \ell^1$, and suppose that for any sequence $(t_n)$ of integers there exits a constant $C_1>0$ such that $$\sharp\left\{k\in\mathbb{Z}:\sup_{n\geq 1}\left|\sum_{i\in \mathcal{B}_n-t_n} \!\!\!\raise{1.9ex}\hbox{$\scriptsize\prime$}\; \frac{a_{k+i}}{i}\right|>λ\right\}\\ \leq C_1\sharp\left\{k\in\mathbb{Z}:\sup_{n\geq 1}\left|\sum_{i\in \mathcal{B}_n} \!\!\raise{1.9ex}\hbox{$\scriptsize\prime$}\; \frac{a_{k+i}}{i}\right|>λ\right\},$$ for all $λ>0$, where $\mathcal{B}_n=\{-n, -(n-1), -(n-2),\dots , n-2, n-1, n\}$. Then there is a constant $C_2>0$ which does not depend on the sequence $\{a_j\}$ such that $$\sum_{n=1}^\infty\sharp\left\{k\in\mathbb{Z}:\left|\sum_{i=-n}^{n} \!\!\raise{1.9ex}\hbox{$\scriptsize\prime$}\; \frac{a_{k+i}}{i}\right|>λ\right\}\leq\frac{C_2}λ\sum_{i=-\infty}^{\infty}|a_i|$$ for all $λ>0$. Let $(X,\mathscr{B},μ)$ be a measure space, $τ:X\to X$ an invertible measure-preserving transformation, and suppose that $f\in L^1(X)$ such that for any sequence $(t_n)$ of integers there exists a constant $C_1>0$ such that $$μ\left\{ x: \sup_{n\geq 1}\left|\sum_{i\in \mathcal{B}_n-t_n}\!\!\!\raise{1.9ex}\hbox{$\scriptsize\prime$}\; \frac{f(τ^ix)}{i}\right| >λ\right\}\leq C_1μ\left\{x: \sup_{n\geq 1}\left|\sum_{i\in \mathcal{B}_n}\!\!\raise{1.9ex}\hbox{$\scriptsize\prime$}\; \frac{f(τ^i x)}{i}\right|>λ\right\} $$ for all $λ>0$, where $\mathcal{B}_n=\{-n, -(n-1), -(n-2),\dots , n-2, n-1, n\}$. Then there exists a constant $C_2>0$ which does not depend on $f$ such that $$\sum_{n=1}^\inftyμ\left\{x:\left|\sum_{i=-n}^{n}\!\!\raise{1.9ex}\hbox{$\scriptsize\prime$} \;\frac{f(τ^ix)}{i}\right|>λ\right\}\leq\frac{C_2}λ\|f\|_1$$ for all $λ>0$.
