%! TEX root = AC.tex % vim: tw=50 % 14/11/2024 10AM By \gls{chern} (\cref{eg_2_5}), for all $\theta > 0$ we have \[ \Pbb(|X| \ge \theta) \le 4\exp \left( -\frac{\theta^2}{4} \right) ,\] and so using the fact that $\Pbb(|X| \le t) = \int_0^t \rho_X(s) \dd s$ we have \begin{align*} \|X\|_{L^{2k}(\Pbb)}^{2k} &= \int_0^\infty t^{2k} \rho_X(t) \dd t \\ &= \int_0^\infty 2k t^{2k - 1} \Pbb(|X| \ge t) \dd t &&\text{integration by parts} \\ &\le \ub{\int_0^\infty 8k t^{2k - 1} \exp \left( -\frac{t^2}{4} \right) \dd t}_{\eqdef I(K)} \end{align*} We shall show by induction on $k$ that $I(K) \le 2^{2k} \frac{(2k)^k}{4k}$. Indeed, when $k = 1$, \[ \int_0^\infty t \exp \left( -\frac{t^2}{4} \right) \dd t = \left[ -2\exp \left( -\frac{t^2}{4} \right) \right]_0^\infty = 2 \le 2 .\] For $k > 1$, integrate by parts to find that \begin{align*} I(K) &= \int_0^\infty \ub{t^{2k - 2}}_{u} \cdot \ub{t \exp \left( -\frac{t^2}{4} \right)}_{v} \dd t \\ &= \left[ t^{2k - 2} \cdot \left( -2\exp \left( -\frac{t^2}{4} \right) \right) \right]_0^\infty - \int_0^\infty (2k - 2) t^{2k - 3} \left( -2 \exp \left( -\frac{t^2}{4} \right) \right) \dd t \\ &= 4(k - 1) \int_0^\infty t^{2(k - 1) - 1} \exp \left( -\frac{t^2}{4} \right) \dd t \\ &= 4(k - 1) I(K - 1) \\ &\le 4(k - 1) 2^{2(k - 1)} \frac{(2(k - 1))^{k - 1}}{4(k - 1)} \\ &\le 2^{2k} \frac{(2k)^k}{4k} \qedhere \end{align*} \end{proof} \begin{corollary}[Rudin's Inequality] \label{coro_3_2} % Corollary 3.2 Let $\Gamma \subseteq \charG[\Fbb_2^n]$ be a linearly independent set and let $p \in [2, \infty)$. Then for any $\ft{f} \in l^2(\Gamma)$, \[ \left\| \sum_{\gamma \in \Gamma} \ft{f}(\gamma) \gamma \right\|_{L^P(\Fbb_2^n)} = O ( \sqrt{p} \|\ft{f}\|_{l^2(\Gamma)} ) .\] \end{corollary} \begin{corollary} \label{coro_3_3} % Corollary 3.3 Let $\Gamma \subseteq \charG[\Fbb_2^n]$ be a linearly independent set and let $p \in (1, 2]$. Then for all $f \in L^p(\Fbb_2^n)$, \[ \|\ft{f}\|_{l^2(\Gamma)} = O \left( \sqrt{\frac{p}{p - 1}} \|f\|_{L^p(\Fbb_2^n)} \right) .\] \end{corollary} \begin{proof} Let $f \in L^p(\Fbb_2^n)$ and write $g = \sum_{\gamma \in \Gamma} \ft{f}(\gamma) \gamma$. Then \begin{align*} \|\ft{f}\|_{l^2(\Gamma)}^2 &= \sum_{\gamma \in \Gamma} |\ft{f}(\gamma)|^2 \\ &= \langle \ft{f}, \ft{g} \rangle_{l^2(\charG[\Fbb_2^n])} \\ &= \langle f, g \rangle_{L^2(\Fbb_2^n)} &&\text{by \gls{planche}} \end{align*} which is bounded above by $\|f\|_{L^p(\Fbb_2^n)} \|g\|_{L^{p'}(\Fbb_2^n)}$ where $\frac{1}{p} + \frac{1}{p'} = 1$, using H\"older's inequality. By \nameref{coro_3_2}, \[ \|g\|_{L^{p'}(\Fbb_2^n)} = O \left( \sqrt{p'} \|\ft{g}\|_{l^2(\Gamma)} \right) = O \left( \sqrt{\frac{p}{p - 1}} \|\ft{f}\|_{l^2(\Gamma)} \right) . \qedhere \] \end{proof} Recall that given $A \subseteq \Fbb_2^n$ of density $\alpha > 0$, we had $|\Spec_\rho(\indicator{A}) \le \rho^{-2} \alpha^{-1}$. This is best possible as the example of a subspace shows. However, in this case the large spectrum is highly structured. \begin{fcthm}[Special case of Chang's Theorem] \label{thm_3_4} % Theorem 3.4 Assuming: - $A \subseteq \Fbb_2^n$ of density $\alpha > 0$ - $\rho > 0$ Then: there exists $H \le \charG[\Fbb_2^n]$ of dimension $O(\rho^{-2} \log \alpha^{-1})$ such that $H \supseteq \Spec_\rho(\indicator{A})$. \end{fcthm} \begin{proof} Let $\Gamma \subseteq \Spec_\rho(\indicator{A})$ be a maximal linearly independent set. Let $H = \langle \Spec_\rho(\indicator{A}) \rangle$. Clearly $\dim(H) = |\Gamma|$. By \cref{coro_3_3}, for all $p \in (1, 2]$, \[ (\rho\alpha)^2 |\Gamma| \le \sum_{\gamma \in \Gamma} |\ft{\indicator{A}}(\gamma)|^2 = \|\ft{\indicator{A}}\|_{l^2(\Gamma)}^2 = O \left( \frac{p}{p - 1} \|\indicator{A}\|_{L^p(\Fbb_2^n)}^2 \right) ,\] so \[ |\Gamma| = O \left( \rho^{-2} \alpha^{-2} \alpha^{2 / p} \frac{p}{p - 1} \right) .\] Set $p = 1 + (\log \alpha^{-1})^{-1}$ to get $|\Gamma| = O(\rho^{-2} \alpha^{-2} (\alpha^2 \cdot e^2) (\log \alpha^{-1} + 1))$. \end{proof} \begin{fcdefn}[Dissociated] \glsadjdefn{diss}{dissociated}{subset}% \label{defn_3_6} % Definition 3.6 Let $G$ be a finite abelian group. We say $S \subseteq G$ is \emph{dissociated} if $\sum_{s \in S} \eps_s s = 0$ for $\eps \in \{-1, 0, 1\}^{|S|}$, then $\eps \equiv 0$. \end{fcdefn} Clearly, if $G = \Fbb_2^n$, then $S \subseteq G$ is \gls{diss} if and only if it is linearly independent.