351 lines
11 KiB
TeX
351 lines
11 KiB
TeX
\documentclass{article}
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\input{../note-setup}
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\title{\LaTeX{} Note Template}
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\author{Author}
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\date{\today}
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\extrainfo{Github: \href{https://github.com/fenglielie/latexzero}{https://github.com/fenglielie/latexzero}}
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\begin{document}
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% \maketitle
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\makecover{../cover/cover.png}
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\section{Theorem, Proposition, Proof}
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\begin{theorem}
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If $1<p<\infty$ and $m > n/p$, or $p=1$ and $m \ge n$, there exist a constant $C = C(m,n,\gamma,p)$, such that
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\begin{equation*}
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\Vert R^m u \Vert_{L^\infty(\Omega)} \le C d^{m-n/p} |u|_{W^m_p(\Omega)}
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\end{equation*}
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for all $u \in W^m_p(\Omega)$.
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\end{theorem}
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\begin{proof}
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First, we assume that $u \in C^m(\Omega) \cap W^m_p(\Omega)$. We can use the pointwise representation of $R^mu(x)$.
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\begin{align*}
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|R^mu(x)| ={} & m \left| \sum_{|\alpha| = m} \int_{C_x} k_{\alpha}(x,z) D^\alpha u(z)\,dz \right| \notag \\
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\le{} & C \sum_{|\alpha|=m} \int_{\Omega} |x-z|^{-n+m} |D^\alpha u(z)|\,dz \notag \\
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\le{} & C' d^{m-n/p} |u|_{W^m_p(\Omega)}.
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\end{align*}
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The proof can be completed via a density argument.
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\end{proof}
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\begin{theorem}[xxx]
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If $1<p<\infty$ and $m > n/p$, or $p=1$ and $m \ge n$, there exist a constant $C = C(m,n,\gamma,p)$, such that
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\begin{equation*}
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\Vert R^m u \Vert_{L^\infty(\Omega)} \le C d^{m-n/p} |u|_{W^m_p(\Omega)}
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\end{equation*}
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for all $u \in W^m_p(\Omega)$.
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\end{theorem}
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\begin{proof}[\upshape\bfseries Proof of xxx]
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First, we assume that $u \in C^m(\Omega) \cap W^m_p(\Omega)$. We can use the pointwise representation of $R^mu(x)$.
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\begin{align*}
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|R^mu(x)| ={} & m \left| \sum_{|\alpha| = m} \int_{C_x} k_{\alpha}(x,z) D^\alpha u(z)\,dz \right| \notag \\
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\le{} & C \sum_{|\alpha|=m} \int_{\Omega} |x-z|^{-n+m} |D^\alpha u(z)|\,dz \notag \\
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\le{} & C' d^{m-n/p} |u|_{W^m_p(\Omega)}.
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\end{align*}
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The proof can be completed via a density argument.
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\end{proof}
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\begin{theorem*}
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If $1<p<\infty$ and $m > n/p$, or $p=1$ and $m \ge n$, there exist a constant $C = C(m,n,\gamma,p)$, such that
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\begin{equation*}
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\Vert R^m u \Vert_{L^\infty(\Omega)} \le C d^{m-n/p} |u|_{W^m_p(\Omega)}
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\end{equation*}
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for all $u \in W^m_p(\Omega)$.
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\end{theorem*}
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\begin{proof}
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First, we assume that $u \in C^m(\Omega) \cap W^m_p(\Omega)$. We can use the pointwise representation of $R^mu(x)$.
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\begin{align*}
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|R^mu(x)| ={} & m \left| \sum_{|\alpha| = m} \int_{C_x} k_{\alpha}(x,z) D^\alpha u(z)\,dz \right| \notag \\
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\le{} & C \sum_{|\alpha|=m} \int_{\Omega} |x-z|^{-n+m} |D^\alpha u(z)|\,dz \notag \\
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\le{} & C' d^{m-n/p} |u|_{W^m_p(\Omega)}.
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\end{align*}
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The proof can be completed via a density argument.
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\end{proof}
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\begin{proposition}
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\begin{equation*}
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Q^m u(x) = \sum_{|\lambda| < m} \left( \int_B \psi_\lambda(y) u(y)\,dy \right) x^\lambda
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\end{equation*}
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where $\psi_\lambda \in C_0^\infty(\mathbb{R}^n)$ and $\mathrm{supp}(\phi_\lambda) \in \overline{B}$.
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\end{proposition}
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\begin{proof}
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This follows from xxx if we define
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\begin{equation*}
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\psi_\lambda(y) = \sum_{\alpha \ge \lambda,|\alpha|<m}
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\frac{(-1)^{|\alpha|}}{\alpha !} a_{[\lambda,\alpha-\lambda]} D^\alpha(y^{\alpha-\lambda} \phi(y)).
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\end{equation*}
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\end{proof}
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\begin{proposition}[xxx]
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\begin{equation*}
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Q^m u(x) = \sum_{|\lambda| < m} \left( \int_B \psi_\lambda(y) u(y)\,dy \right) x^\lambda
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\end{equation*}
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where $\psi_\lambda \in C_0^\infty(\mathbb{R}^n)$ and $\mathrm{supp}(\phi_\lambda) \in \overline{B}$.
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\end{proposition}
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\begin{proof}[\upshape\bfseries Proof of xxx]
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This follows from xxx if we define
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\begin{equation*}
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\psi_\lambda(y) = \sum_{\alpha \ge \lambda,|\alpha|<m}
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\frac{(-1)^{|\alpha|}}{\alpha !} a_{[\lambda,\alpha-\lambda]} D^\alpha(y^{\alpha-\lambda} \phi(y)).
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\end{equation*}
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\end{proof}
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\begin{proposition*}
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\begin{equation*}
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Q^m u(x) = \sum_{|\lambda| < m} \left( \int_B \psi_\lambda(y) u(y)\,dy \right) x^\lambda
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\end{equation*}
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where $\psi_\lambda \in C_0^\infty(\mathbb{R}^n)$ and $\mathrm{supp}(\phi_\lambda) \in \overline{B}$.
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\end{proposition*}
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\begin{proof}
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This follows from xxx if we define
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\begin{equation*}
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\psi_\lambda(y) = \sum_{\alpha \ge \lambda,|\alpha|<m}
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\frac{(-1)^{|\alpha|}}{\alpha !} a_{[\lambda,\alpha-\lambda]} D^\alpha(y^{\alpha-\lambda} \phi(y)).
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\end{equation*}
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\end{proof}
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\section{Corollary, Lemma, Claim}
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\begin{corollary}
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Under the assumption of xxx, the following inequality holds
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\begin{equation*}
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\inf_{v \in P^{m-1}} \Vert u - v \Vert_{W^k_p(\Omega)} \le C_{m,n,\gamma} d^{m-k} |u|_{W^k_p(\Omega)}, \,\, k = 0,1,\dots,m,
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\end{equation*}
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\end{corollary}
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\begin{corollary}[xxx]
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Under the assumption of xxx, the following inequality holds
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\begin{equation*}
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\inf_{v \in P^{m-1}} \Vert u - v \Vert_{W^k_p(\Omega)} \le C_{m,n,\gamma} d^{m-k} |u|_{W^k_p(\Omega)}, \,\, k = 0,1,\dots,m,
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\end{equation*}
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\end{corollary}
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\begin{corollary*}
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Under the assumption of xxx, the following inequality holds
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\begin{equation*}
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\inf_{v \in P^{m-1}} \Vert u - v \Vert_{W^k_p(\Omega)} \le C_{m,n,\gamma} d^{m-k} |u|_{W^k_p(\Omega)}, \,\, k = 0,1,\dots,m,
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\end{equation*}
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\end{corollary*}
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\begin{lemma}
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Let $f \in L^p(\Omega)$ for $p \ge 1$ and $m \ge 1$ and let
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\[
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g(x) = \int_\Omega |x-z|^{-n+m} |f(z)|\,dz
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\]
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Then
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\begin{equation*}
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\Vert g \Vert_{L^p(\Omega)} \le C_{m,n} d^m \Vert f\Vert_{L^p(\Omega)}.
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\end{equation*}
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\end{lemma}
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\begin{lemma}[xxx]
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Let $f \in L^p(\Omega)$ for $p \ge 1$ and $m \ge 1$ and let
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\[
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g(x) = \int_\Omega |x-z|^{-n+m} |f(z)|\,dz
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\]
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Then
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\begin{equation*}
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\Vert g \Vert_{L^p(\Omega)} \le C_{m,n} d^m \Vert f\Vert_{L^p(\Omega)}.
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\end{equation*}
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\end{lemma}
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\begin{lemma*}
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Let $f \in L^p(\Omega)$ for $p \ge 1$ and $m \ge 1$ and let
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\[
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g(x) = \int_\Omega |x-z|^{-n+m} |f(z)|\,dz
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\]
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Then
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\begin{equation*}
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\Vert g \Vert_{L^p(\Omega)} \le C_{m,n} d^m \Vert f\Vert_{L^p(\Omega)}.
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\end{equation*}
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\end{lemma*}
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\begin{claim}
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$Q^m u$ is a polynomial of degree less than $m$ in $x$.
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\end{claim}
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\begin{claim}[xxx]
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$Q^m u$ is a polynomial of degree less than $m$ in $x$.
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\end{claim}
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\begin{claim*}
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$Q^m u$ is a polynomial of degree less than $m$ in $x$.
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\end{claim*}
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\section{Definition}
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\begin{definition}
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$\Omega$ is star-shaped with respect to the ball $B$ if , for all $x \in \Omega$, the closed convex hull of $\{x\} \cup B$ is a subset of $\Omega$.
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\end{definition}
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\begin{definition}[xxx]
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$\Omega$ is star-shaped with respect to the ball $B$ if , for all $x \in \Omega$, the closed convex hull of $\{x\} \cup B$ is a subset of $\Omega$.
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\end{definition}
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\begin{definition*}
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$\Omega$ is star-shaped with respect to the ball $B$ if , for all $x \in \Omega$, the closed convex hull of $\{x\} \cup B$ is a subset of $\Omega$.
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\end{definition*}
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\section{Example}
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\begin{example}
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The integral form of the Taylor remainder for $f \in C^m([0,1])$ is given by
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\begin{equation*}
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f(s) = \sum_{k=0}^{m-1}\frac{1}{k!} f^{(k)}(0) + \int_0^s \frac{1}{(m-1)!} f^{(m)}(t)(s-t)^{m-1}\,dt.
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\end{equation*}
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\end{example}
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\begin{example}[xxx]
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The integral form of the Taylor remainder for $f \in C^m([0,1])$ is given by
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\begin{equation*}
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f(s) = \sum_{k=0}^{m-1}\frac{1}{k!} f^{(k)}(0) + \int_0^s \frac{1}{(m-1)!} f^{(m)}(t)(s-t)^{m-1}\,dt.
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\end{equation*}
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\end{example}
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\begin{example*}
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The integral form of the Taylor remainder for $f \in C^m([0,1])$ is given by
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\begin{equation*}
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f(s) = \sum_{k=0}^{m-1}\frac{1}{k!} f^{(k)}(0) + \int_0^s \frac{1}{(m-1)!} f^{(m)}(t)(s-t)^{m-1}\,dt.
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\end{equation*}
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\end{example*}
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\section{Problem, Solution}
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\begin{problem}
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Calculate the integral of the function $g(x) = 3x^2$ with respect to $x$.
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\end{problem}
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\begin{solution}
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To calculate the integral of $g(x) = 3x^2$, we use the power rule for integration:
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\[
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\int 3x^2 \, dx = x^3 + C
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\]
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where $C$ is the constant of integration.
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\end{solution}
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\begin{problem}[xxx]
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Calculate the integral of the function $g(x) = 3x^2$ with respect to $x$.
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\end{problem}
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\begin{solution}[xxx]
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To calculate the integral of $g(x) = 3x^2$, we use the power rule for integration:
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\[
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\int 3x^2 \, dx = x^3 + C
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\]
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where $C$ is the constant of integration.
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\end{solution}
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\begin{problem*}
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Calculate the integral of the function $g(x) = 3x^2$ with respect to $x$.
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\end{problem*}
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\begin{solution*}
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To calculate the integral of $g(x) = 3x^2$, we use the power rule for integration:
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\[
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\int 3x^2 \, dx = x^3 + C
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\]
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where $C$ is the constant of integration.
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\end{solution*}
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\section{Remark}
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\begin{remark}
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Such a polynomial is not unique, due to the choice od cut-off function $\phi$.
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\end{remark}
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\begin{remark}[xxx]
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Such a polynomial is not unique, due to the choice od cut-off function $\phi$.
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\end{remark}
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\begin{remark*}
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Such a polynomial is not unique, due to the choice od cut-off function $\phi$.
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\end{remark*}
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\section{Note}
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\begin{note}
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The degree of $Q^m u$ is at most $m-1$.
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\end{note}
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\begin{note}[xxx]
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The degree of $Q^m u$ is at most $m-1$.
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\end{note}
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\begin{note*}
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The degree of $Q^m u$ is at most $m-1$.
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\end{note*}
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\section{lstlisting}
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\begin{lstlisting}[language=Python,caption={hello world}]
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def hello():
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print("Hello, world!")
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hello()
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\end{lstlisting}
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\begin{lstlisting}[language=Python,caption={hanoi.py}]
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step = 1
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def hanoi(n, a, b, c, depth=0):
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def move(n, a, c):
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global step
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print(" " * depth, end="")
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print(f"{step=}: move [{n}] from {a} to {c}")
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step += 1
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if n == 1:
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move(n, a, c)
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else:
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hanoi(n - 1, a, c, b, depth=depth + 1)
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move(n, a, c)
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hanoi(n - 1, b, a, c, depth=depth + 1)
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if __name__ == "__main__":
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n = int(input("Hanoi Problem, N = "))
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hanoi(n, "A", "B", "C")
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\end{lstlisting}
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\section{algorithm}
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\begin{algorithm}[H]
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\KwIn{This is some input}
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\KwOut{This is some output}
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\SetAlgoLined
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\SetNoFillComment
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\tcc{This is a comment}
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\vspace{3mm}
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some code here\;
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$x \leftarrow 0$\;
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$y \leftarrow 0$\;
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\uIf{$ x > 5$} {
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x is greater than 5 \tcp*{This is also a comment}
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}
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\Else {
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x is less than or equal to 5\;
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}
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\ForEach{y in 0..5} {
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$y \leftarrow y + 1$\;
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}
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\For{$y$ in $0..5$} {
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$y \leftarrow y - 1$\;
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}
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\While{$x > 5$} {
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$x \leftarrow x - 1$\;
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}
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\Return Return something here\;
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\caption{what}
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\end{algorithm}
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\section{cbox}
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\begin{cbox}
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Whenever you feel like criticizing any one, just remember that all the people in this world haven't had the advantages that you've had. (The Great Gatsby)
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\end{cbox}
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\end{document}
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