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With $\sum_{\nu \ge 0}\phi_\nu(\xi)=1$ be a Littlewood-Paley partition of unity we find that $u=\sum_{\nu \ge 0}\phi_\nu(D)u$ and thus since $$\Vert u\Vert_{B^\alpha_{\infty, \infty}}=\sup_{\nu\in \mathbb N} 2^{\nu \alpha}\Vert\phi_\nu(D)u\Vert_{L^\infty},$$ we get $$\Vert u\Vert_{L^\infty}\le \sum_{\nu \ge 0}2^{\nu \alpha}\Vert\phi_\nu(D)u\Vert_{L^\infty}... 1 Come on, for the first property just use the fact that$$ \int_{-n}^n K(x) dx = \sum_{j = -n+1}^{n-1} \phi(j) + \tfrac12(\phi(-n)+\phi(n)) $$has a finite limit as n \to \infty, together with convergence of$$ \biggl| \int_{-a}^a K(x) dx - \int_{-\lfloor a\rfloor}^{\lfloor a\rfloor} K(x) dx \biggr| \leqslant |\phi(-\lfloor a\rfloor-1)| + |\phi(-\lfloor a\...

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$\newcommand\R{\mathbb R}\newcommand{\Z}{\mathbb{Z}}\newcommand{\ep}{\epsilon}\newcommand{\fl}{\lfloor#1\rfloor}$The answer is yes to each of your two questions. Let $a_n:=\phi(n)$. Then \begin{equation*} K(x)=\sum_{n\in\Z}a_n R(x-n). \end{equation*} Note that for all $j\in\Z$ we have $K(j)=a_j$ and $K$ linear (or, more exactly, affine) on the ...

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$\newcommand{\Z}{\mathbb{Z}}\newcommand{\ep}{\epsilon}$Let $a_n:=\phi(n)$. Then \begin{equation} K(x)=\sum_{n\in\Z}a_n 1(n-1/2\le x<n+1/2). \end{equation} So, $K(x)=a_0=0$ if $1/2\le x<1/2$. So, for $\ep\in(0,1/2)$, \begin{equation} I_\ep:=\int_{1/\ep<|x|<\ep}K(x)\,dx=\int_{|x|<\ep}K(x)\,dx =\sum_{n\in\Z}a_n J_n, \end{equation} ...

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Section 6.1 of Yufei Zhao's notes Graph theory and Additive Combinatorics discusses Roth's theorem in finite fields. The version of the Roth's theorem states that a set $X$ in the vector space $\mathbb{F}_3^n$ without three-term arithmetic progressions (3-AP) has size at most $O(3^n/n)$. The argument is as follows: "Decompose the original object into ...

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