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| 1 | +\chapter{Almost-Periodicity} |
| 2 | +\label{chap:ap} |
| 3 | + |
| 4 | + |
| 5 | +\begin{lemma}[Marcinkiewicz-Zygmund inequality] |
| 6 | +\label{mzi} |
| 7 | +\lean{other_marcinkiewicz_zygmund'} |
| 8 | +\leanok |
| 9 | +Let $m\geq 1$. If $f:G\to\bbr$ is such that $\bbe_x f(x)=0$ and $\abs{f(x)}\leq 2$ for all $x$ then |
| 10 | +\[\bbe_{x_1,\ldots,x_n} \abs{\sum_{i=1}^n f(x_i)}^{2m} \leq (4mn)^{m}.\] |
| 11 | +\end{lemma} |
| 12 | + |
| 13 | +\begin{proof} |
| 14 | +\leanok |
| 15 | +Let $S$ be the left-hand side. Since $0=\bbe_y f(y)$ we have, by the triangle inequality, and H\"{o}lder's inequality, |
| 16 | +\[S=\bbe_{x_1,\ldots,x_n}\left\lvert \sum_i f(x_i) - \bbe_{y_i} f(y_i)\right\rvert^{2m} = \bbe_{x_1,\ldots,x_n} \left\lvert\bbe_{y_i}\brac{\sum_i f(x_i)-f(y_i)}\right\rvert^{2m}\leq \bbe_{x_1,\dots,y_n} \left\lvert \sum_i f(x_i)-f(y_i)\right\rvert^{2m}.\] |
| 17 | +Changing the role of $x_i$ and $y_i$ makes no difference here, but multiplies the $i$ summand by $\{-1,+1\}$, and therefore for any $\epsilon_i\in\{-1,+1\}$, |
| 18 | +\[ S \leq\bbe_{x_1,\ldots,y_n}\left\lvert \sum_i \epsilon_i(f(x_i)-f(y_i))\right\rvert^{2m}.\] |
| 19 | +In particular, if we sample $\epsilon_i\in\{-1,+1\}$ uniformly at random, then |
| 20 | +\[ S \leq\bbe_{\epsilon_i} \bbe_{x_1,\ldots,y_n}\left\lvert \sum_i \epsilon_i(f(x_i)-f(y_i))\right\rvert^{2m}.\] |
| 21 | +We now change the order of expectation and consider the expectation over just $\epsilon_i$, viewing the $f(x_i)-f(y_i)=z_i$, say, as fixed quantities. For any $z_i$ we can expand $\bbe_{\epsilon_i} \lvert \sum_i \epsilon_iz_i\rvert^{2m}$ and then bound it from above, using the triangle inequality and $\abs{z_i}\leq 4$, by |
| 22 | +\[4^{2m}\sum_{k_1+\cdots+k_n=2m}\binom{2m}{k_1,\ldots,k_n}\abs{\bbe\epsilon_1^{k_1}\cdots \epsilon_n^{k_n}}.\] |
| 23 | +The inner expectation vanishes unless each $k_i$ is even, when it is trivially one. Therefore the above quantity is exactly |
| 24 | +\[\sum_{l_1+\cdots+l_n=m}\binom{2m}{2l_1,\ldots,2l_n}\leq m^mn^m,\] |
| 25 | +since for any $l_1+\cdots+l_n=m$, |
| 26 | +\[\binom{2m}{2l_1,\ldots,2l_n}\leq m^m\binom{m}{l_1,\ldots,l_n}.\] |
| 27 | +This can be seen, for example, by writing both sides out using factorials, yielding |
| 28 | + |
| 29 | +\[\frac{(2m)!}{(2l_1)!\cdots (2l_n)!}\leq \frac{(2m)!}{2^mm!}\frac{m!}{l_1!\cdots l_n!}\leq m^m\frac{m!}{l_1!\cdots l_n!}.\] |
| 30 | +\end{proof} |
| 31 | + |
| 32 | + |
| 33 | +\begin{lemma}[Complex-valued Marcinkiewicz-Zygmund inequality] |
| 34 | +\label{mzi_complex} |
| 35 | +\leanok |
| 36 | +\lean{complex_marcinkiewicz_zygmund} |
| 37 | +Let $m\geq 1$. If $f:G\to\bbc$ is such that $\bbe_x f(x)=0$ and $\abs{f(x)}\leq 2$ for all $x$ then |
| 38 | +\[\bbe_{x_1,\ldots,x_n} \abs{\sum_{i=1}^n f(x_i)}^{2m} \leq (16mn)^{m}.\] |
| 39 | +\end{lemma} |
| 40 | + |
| 41 | +\begin{proof} |
| 42 | +\uses{mzi} |
| 43 | +\leanok |
| 44 | +Test. |
| 45 | +\end{proof} |
| 46 | + |
| 47 | + |
| 48 | +\begin{lemma} |
| 49 | +\label{random_approx_expect} |
| 50 | +\lean{lemma28} |
| 51 | +\leanok |
| 52 | +Let $\epsilon>0$ and $m\geq 1$. Let $A\subseteq G$ and $f:G\to \bbc$. If $k\geq 64m\epsilon^{-2}$ then the set |
| 53 | +\[L=\{ \tup{a}\in A^k : \|\tfrac{1}{k}\sum_{i=1}^kf(x-a_i)-\mu_A\ast f\|_{2m}\leq \epsilon \| f\|_{2m}\}.\] |
| 54 | +has size at least $\lvert A \rvert^k/2$. |
| 55 | +\end{lemma} |
| 56 | + |
| 57 | +\begin{proof} |
| 58 | +\uses{mzi_complex} |
| 59 | +\leanok |
| 60 | +Note that if $a\in A$ is chosen uniformly at random then, for any fixed $x\in G$, |
| 61 | +\[\bbe f(x-a_i)= \frac{1}{\abs{A}}\sum_{a\in A}f(x-a)=\frac{1}{\abs{A}}\ind{A}\ast f(x)=\mu_A\ast f(x).\] |
| 62 | +Therefore, if we choose $a_1,\ldots,a_k\in A$ independently uniformly at random, for any fixed $x\in G$ and $1\leq i\leq k$, the random variable $f(x-a_i)-f\ast \mu_A(x)$ has mean zero. By the Marcinkiewicz-Zygmund inequality Lemma~\ref{mzi}, therefore, |
| 63 | +\begin{multline*} |
| 64 | +\bbe\abs{ \frac{1}{k}\sum_i f(x-a_i)-f\ast \mu_A(x)}^{2m} \leq \\(16m/k)^mk^{-1} \bbe \sum_i \abs{f(x-a_i)-f\ast \mu_A(x)}^{2m}. |
| 65 | +\end{multline*} |
| 66 | +We now sum both sides over all $x\in G$. By the triangle inequality, for any fixed $1\leq i\leq k$ and $a_i\in A$, |
| 67 | +\begin{align*} |
| 68 | +\sum_{x\in G} \abs{f(x-a_i)-f\ast \mu_A(x)}^{2m} |
| 69 | +&\leq 2^{2m-1}\sum_{x\in G}\abs{f(x-a_i)}^{2m}+\sum_{x\in G}\abs{f\ast \mu_A(x)}^{2m}\\ |
| 70 | +&\leq 2^{2m-1}\brac{\norm{f}_{2m}^{2m}+\norm{f\ast \mu_A}_{2m}^{2m}}. |
| 71 | +\end{align*} |
| 72 | +We note that $\norm{\mu_A}_1=\frac{1}{\abs{A}}\sum_{x\in A}\ind{A}(x)=\abs{A}/\abs{A}=1$, and hence by Young's inequality, $\norm{f\ast \mu_A}_{2m}\leq \norm{f}_{2m}$, and so |
| 73 | +\[\sum_{x\in G} \abs{f(x-a_i)-f\ast \mu_A(x)}^{2m}\leq 2^{2m}\norm{f}_{2m}^{2m}.\] |
| 74 | +It follows that |
| 75 | +\[\bbe_{a_1,\ldots,a_k\in A}\norm{\frac{1}{k}\sum_i\tau_{a_i}f-f\ast \mu_A}_{2m}^{2m}\leq |
| 76 | +(64m/k)^m\norm{f}_{2m}^{2m}.\] |
| 77 | +In particular, if $k\geq 64\epsilon^{-2}m$ then the right-hand side is at most $(\frac{\epsilon}{2}\norm{f}_{2m})^{2m}$ as required. |
| 78 | +\end{proof} |
| 79 | + |
| 80 | + |
| 81 | +\begin{lemma} |
| 82 | +\label{aps_in_translates} |
| 83 | +\lean{just_the_triangle_inequality} |
| 84 | +\leanok |
| 85 | +Let $A\subseteq G$ and $f:G\to \bbc$. Let $\epsilon>0$ and $m\geq 1$ and $k\geq 1$. Let |
| 86 | +\[L=\{ \tup{a}\in A^k : \|\tfrac{1}{k}\sum_{i=1}^kf(x-a_i)-\mu_A\ast f\|_{2m}\leq \epsilon \| f\|_{2m}\}.\] |
| 87 | +If $t\in G$ is such that $\tup{a}\in L$ and $\tup{a}+(t,\ldots,t)\in L$ then |
| 88 | +\[\| \tau_t(\mu_A\ast f)-\mu_A\ast f\|_{2m}\leq 2\epsilon \|f\|_{2m}.\] |
| 89 | +\end{lemma} |
| 90 | + |
| 91 | +\begin{proof} |
| 92 | +\leanok |
| 93 | +Test. |
| 94 | +\end{proof} |
| 95 | + |
| 96 | + |
| 97 | +\begin{lemma} |
| 98 | +\label{lots_of_diagonals} |
| 99 | +\lean{big_shifts} |
| 100 | +\leanok |
| 101 | +Let $A\subseteq G$ and $k\geq 1$ and $L\subseteq A^k$. Then there exists some $\tup{a}\in L$ such that |
| 102 | +\[\#\{ t\in G : \tup{a}+(t,\ldots,t)\in L\}\geq \frac{\lvert L\rvert}{\lvert A+S\rvert^k}\lvert S\rvert.\] |
| 103 | +\end{lemma} |
| 104 | + |
| 105 | +\begin{proof} |
| 106 | +\leanok |
| 107 | +Test. |
| 108 | +\end{proof} |
| 109 | + |
| 110 | + |
| 111 | +\begin{theorem}[$L_p$ almost periodicity] |
| 112 | +\label{lp_ap} |
| 113 | +\lean{AlmostPeriodicity.almost_periodicity} |
| 114 | +\leanok |
| 115 | +Let $\epsilon\in (0,1]$ and $m\geq 1$. Let $K\geq 2$ and $A,S\subseteq G$ with $\lvert A+S\rvert\leq K\lvert A\rvert$. |
| 116 | +Let $f:G\to \bbc$. There exists $T\subseteq G$ such that |
| 117 | +\[\lvert T\rvert \geq K^{-512m\epsilon^{-2}}\lvert S\rvert\] |
| 118 | +such that for any $t\in T$ we have |
| 119 | +\[\| \tau_t(\mu_A\ast f)-\mu_A\ast f\|_{2m}\leq \epsilon \| f\|_{2m}.\] |
| 120 | +\end{theorem} |
| 121 | + |
| 122 | +\begin{proof} |
| 123 | +\uses{random_approx_expect, lots_of_diagonals, aps_in_translates} |
| 124 | +\leanok |
| 125 | +Test. |
| 126 | +\end{proof} |
| 127 | + |
| 128 | + |
| 129 | +\begin{theorem}[$L_\infty$ almost periodicity] |
| 130 | +\label{linfty_ap} |
| 131 | +\lean{AlmostPeriodicity.linfty_almost_periodicity} |
| 132 | +\leanok |
| 133 | +Let $\epsilon\in (0,1]$. Let $K\geq 2$ and $A,S\subseteq G$ with $\lvert A+S\rvert\leq K\lvert A\rvert$. |
| 134 | +Let $B,C\subseteq G$. Let $\eta=\min(1,\lvert C\rvert/\lvert B\rvert)$. There exists $T\subseteq G$ such that |
| 135 | +\[\lvert T\rvert \geq K^{-4096\lceil \lo{\eta}\rceil\epsilon^{-2}}\lvert S\rvert\] |
| 136 | +such that for any $t\in T$ we have |
| 137 | +\[\| \tau_t(\mu_A\ast 1_B\ast \mu_C)-\mu_A\ast 1_B\ast \mu_C\|_{\infty}\leq \epsilon.\] |
| 138 | +\end{theorem} |
| 139 | + |
| 140 | +\begin{proof} |
| 141 | +\leanok |
| 142 | +\uses{lp_ap} Let $T$ be as given in \ref{lp_ap} |
| 143 | +with $f=1_B$ and $m=\lceil \lo{\eta}\rceil$ and $\epsilon=\epsilon/e$. (The size bound on $T$ follows since $e^2\leq 8$.) Fix $t\in T$ and let $F=\tau_t(\mu_A\ast 1_B)-\mu_A\ast 1_B$. We have, for any $x\in G$, |
| 144 | +\[(\tau_t(\mu_A\ast 1_B\ast \mu_C)-\mu_A\ast 1_B\ast \mu_C)(x)=F\ast \mu_C(x)=\sum_y F(y)\mu_{C}(x-y)=\sum_yF(y)\mu_{x-C}(y).\] |
| 145 | +By Hölder's inequality, this is (in absolute value), for any $m\geq 1$, |
| 146 | +\[\norm{F}_{2m}\norm{\mu_{x-C}}_{1+\frac{1}{2m-1}}.\] |
| 147 | +By the construction of $T$ the first factor is at most |
| 148 | +$\frac{\epsilon}{e}\| 1_B\|_{2m}=\frac{\epsilon}{e}\lvert B\rvert^{1/2m}$. |
| 149 | +We have by calculation |
| 150 | +\[\norm{\mu_{x-C}}_{1+\frac{1}{2m-1}}=\lvert x-C\rvert^{-1/2m}=\lvert C\rvert^{-1/2m}.\] |
| 151 | +Therefore we have shown that |
| 152 | + |
| 153 | +\[\| \tau_t(\mu_A\ast 1_B\ast \mu_C)-\mu_A\ast 1_B\ast \mu_C\|_{\infty}\leq \frac{\epsilon}{e}(\lvert C\rvert/\lvert B\rvert)^{-1/2m}.\] |
| 154 | +The claim now follows since, by choice of $m$, |
| 155 | +\[(\lvert C\rvert/\lvert B\rvert)^{-1/2m}\leq e\] |
| 156 | +(dividing into cases as to whether $\eta=1$ or not). |
| 157 | +\end{proof} |
| 158 | + |
| 159 | + |
| 160 | +\begin{theorem} |
| 161 | +\label{linfty_ap_boosted} |
| 162 | +\lean{AlmostPeriodicity.linfty_almost_periodicity_boosted} |
| 163 | +\leanok |
| 164 | +Let $\epsilon\in (0,1]$ and $k\geq 1$. Let $K\geq 2$ and $A,S\subseteq G$ with $\lvert A+S\rvert\leq K\lvert A\rvert$. |
| 165 | +Let $B,C\subseteq G$. Let $\eta=\min(1,\lvert C\rvert/\lvert B\rvert)$. There exists $T\subseteq G$ such that |
| 166 | +\[\lvert T\rvert \geq K^{-4096\lceil \lo{\eta}\rceil k^2\epsilon^{-2}}\lvert S\rvert\] |
| 167 | +such that |
| 168 | +\[\| \mu_T^{(k)}\ast \mu_A\ast 1_B\ast \mu_C-\mu_A\ast 1_B\ast \mu_C\|_{\infty}\leq \epsilon.\] |
| 169 | +\end{theorem} |
| 170 | + |
| 171 | +\begin{proof} |
| 172 | +\uses{linfty_ap} |
| 173 | +\leanok |
| 174 | +Let $T$ be as in Theorem~\ref{linfty_ap} with $\epsilon$ replaced by $\epsilon/k$. Note that, for any $x\in G$, |
| 175 | +\[\mu_T^{(k)}\ast \mu_A\ast 1_B\ast \mu_C(x)=\frac{1}{\lvert T\rvert^k}\sum_{t_1,\ldots,t_k\in T}\tau_{t_1+\cdots+t_k}\mu_A\ast 1_B\ast \mu_C(x).\] |
| 176 | +It therefore suffices (by the triangle inequality) to show, for any fixed $x\in G$ and $t_1,\ldots,t_k\in T$, that with $F=\mu_A\ast 1_B\ast \mu_C$, we have |
| 177 | +\[\lvert \tau_{t_1+\cdots+t_k}F(x)-F(x)\rvert \leq \epsilon.\] |
| 178 | +This follows by the triangle inequality applied $k$ times if we knew that, for $1\leq l\leq k$, |
| 179 | +\[\lvert \tau_{t_1+\cdots+t_l}F(x)-\tau_{t_1+\cdots+t_{l-1}}F(x)\rvert \leq \epsilon/k.\] |
| 180 | +We can write the left-hand side as |
| 181 | +\[\lvert \tau_{t_1+\cdots+t_l}F(x)-\tau_{t_1+\cdots+t_{l-1}}F(x)\rvert=\lvert \tau_{t_l}F(x-t_1-\cdots-t-{l-1})-F(x-t_1-\cdots-t-{l-1})\rvert.\] |
| 182 | +The right-hand side is at most |
| 183 | +\[\| \tau_{t_l}F-F\|_\infty\] |
| 184 | +and we are done by choice of $T$. |
| 185 | +\end{proof} |
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