89 lines
2.8 KiB
TeX
89 lines
2.8 KiB
TeX
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\documentclass{article}
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\input{../note-setup-leftsidebox}
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\title{\LaTeX{} Note Template}
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\author{}
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\date{}
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\begin{document}
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\section{Demo}
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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|
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\le{} C \sum_{|\alpha|=m} \int_{\Omega} |x-z|^{-n+m} |D^\alpha u(z)|\,dz
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\le{} C' d^{m-\frac{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}[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{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{lemma}[xxx]
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Let $f \in L^p(\Omega)$ for $p \ge 1$ and $m \ge 1$ and let $g(x) = \int_\Omega |x-z|^{-n+m} |f(z)|\,dz$.
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Then $\Vert g \Vert_{L^p(\Omega)} \le C_{m,n} d^m \Vert f\Vert_{L^p(\Omega)}$.
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\end{lemma}
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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{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{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{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{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{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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\end{document}
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