# Tag Info

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The $W^{1,p}$ norm of $|r-r_k|^\alpha$ is bounded: $$\int_{|x|\le1}|x-r_k|^{-\alpha p}\,dx=\int_{|x-r_k|\le1}|x|^{-\alpha p}\,dx\le\int_{|x|\le2}|x|^{-\alpha p}\,dx.$$ Similarly for the gradient. Let $C$ be a bound. Then $$\bigl\|2^{-k}\,|x-r_k|^{-\alpha}\bigr\|_{1,p}=2^{-k}\,\bigl\||x-r_k|^{-\alpha}\bigr\|_{1,p}\le C\,2^{-k}.$$ Since $\sum ... 2 You know already that the weak derivative (call it$v$for the moment) is continuous. By definition,$v$is called a weak derivative if for all$\phi \in C^\infty_0(a, b)$, we have $$\int_a^b \phi' (x) u(x) dx = - \int_a^b \phi(x) v(x) dx.$$ By approximations, the same formula holds when$\phi$is of compact support and Lipschitz. Now fix any$c, d\in (a, ...

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Suppose $v \in H^1(T)$ is not essentially constant and $\int_T v \neq 0$. Define $a=-\frac{\int_T v^2}{\int_T v}$. Then $v$ and $a+v$ are orthogonal in $L^2$ (check it) but not in $H^1$ (the derivatives are equal everywhere, so the inner product is the integral of a nonnegative function which is nonzero on a set of positive measure). The first assumption is ...

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I would argue as follows. Assume that $n\ge 2$ and let $$T_i f (x)=\int_{\mathbb{R}^n} f(y)\frac{x_i-y_i}{\lvert x-y\rvert^{n}}\chi_\Omega(y)\, dy.$$ Here $\chi_\Omega(y)$ is $1$ if $y\in \Omega$ and $0$ otherwise. The kernel of this operator can be estimated in absolute value by the kernel of the Riesz potential: $$\left\lvert \frac{x_i-y_i}{\lvert ... 1 Since u\in C^1(\overline U), |u| is Lipschitz, hence absolutely continuous. The fundamental theorem of calculus applies to such functions: the value at x_n=0 is related to the integral of x_n-derivative in the usual way (the other boundary term is zero since \zeta is compactly supported). (From a comment by John Ma) 0 Answering my own question: (Could you check my proof please? That would help me a lot. I need to know if I am wrong or not.) Since f and f_n are continuous and bounded they are Riemann integrable and by uniform convergence f_n\rightrightarrows f it holds that \sum_{i=1,...,N_n}f_n\left(x_{i,n}\right)\chi_{Q_{i,n}}\stackrel{L^1}{\rightarrow}f. (also ... 0 On a bounded domain, Sobolev spaces are nested in the same way and for the same reason that Lebesgue spaces are nested: W^{k,p}\subset W^{k,q} if p\ge q. (More precisely, the identity map is a bounded operator from W^{k,p} to W^{k,q}.) So, you can obtain the claimed inequality by considering the composition of inclusion W^{1,2} \to W^{1,q} with ... 1 A connection on E is a map \nabla: \Gamma(E) \to \Gamma(E \otimes T^*M) satisfying certain conditions. By having it act as the Levi-Civita connection on T^*M you inductively also have connections \nabla: \Gamma(E \otimes (T^*M)^{\otimes_k}) \to \Gamma(E \otimes (T^*M)^{\otimes_{k+1}}). \nabla^k means a composition of a bunch of these to get a map ... 0 After not receiving an answer here, I got it from MathOverflow. The answer is "yes" there is the characterization. 1 I read V(\nabla u)-V(\nabla u) as the function defined by V((\nabla u)(\xi)). So if 1<p<2, then we have from the second inequality in$$c^{-1}(\left|\xi\right|^{2}+\left|\eta\right|^{2})^{\frac{p-2}{2}}\left|\xi-\eta\right|^{2}\leq\left|V(\xi)-V(\eta)\right|^{2}\leq ...

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Using (3) in the form $$|V(\xi) - V(\eta)|^2 \leq K(|\xi|^2 + |\eta|^2)^{\frac{p-2}{2}} |\xi - \eta|^2 \le K (|\xi| + |\eta|)^{p-2} |\xi - \eta|^2$$ we get from (1) $$\int_B |V(\nabla u) - V(\nabla v)|^2 \le K \int_B (|\nabla u| + |\nabla v|)^{p-2} |\nabla u- \nabla v|^2 \leq \int_B |\nabla u|^p - |\nabla v|^p$$ and an application of (2) yields the ...

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Maybe this can help someone one day. Since I saw this argument in a paper, maybe it is important. Since $u - \varphi \in W^{1,p}_{0}(\Omega)$, we have $|u - \varphi| \in W^{1,p}_{0}(\Omega).$ Note that $||u(x)| - |\varphi (x)|| \leq |u(x) - \varphi (x)|,$ for all $x$ then $||u| - |\varphi|| \in W^{1,p}_{0}(\Omega)$. For $j>j_0$ we have for all x that ...

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The case $p=2$ is special, because the mean minimizes $\int |u-c|^2$ among all $c\in\mathbb{R}$. It does not have such a property for $p\ne 2$, and there is no reason for the stated inequality to hold. As PhoemueX noted, the Sobolev space is not really relevant: any $L^p$ function can be approximated by smooth functions in $L^p$ norm, and the quantities in ...

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In a sense, yes. Precisely, we have the following result: Let $I$ be an interval (possibly unbounded) and $f\in W^{1,p}(I)$ with $1 \leq p \leq \infty$. Then there exists a function $\tilde{f}\in C(\overline{I})$ such that $f = \tilde{f}$ a.e. on $I$ and $$\int_y^x f'(t)\,dt=\tilde{f}(x)- \tilde{f}(y),\qquad \forall\ x,y\in \overline{I}.$$ ...

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First, I am not sure what the author means by Banach's theorem, but it seems like the relevant result from functional analysis which he's using is the bounded inverse theorem: if $T:(X,\left\|\right\|_{X})\rightarrow (Y,\left\|\right\|_{Y})$ is a continuous (i.e. bounded) bijective linear map between two Banach spaces $X$ and $Y$, then $T^{-1}: ... 1 You don't need$a_n'$to converge uniformly to the sign function; it's enough to have them uniformly bounded and converging pointwise. (If your chosen$a_n$functions don't do this, then pick other ones that do.) Then you can use the dominated convergence theorem to show that$a_n'(u) u_{x_i} \to \operatorname{sgn}(u) u_{x_i}$in$L^p$. 3 Step 1: Assume$u \in W^{1,1}(\mathbb{R}) \cap H^1(\mathbb{R}) \cap C^\infty(\mathbb{R}), then $$u(x) = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{ix\xi}\hat{u}(\xi)\, d\xi,$$ so that by Holder's inequality, Minkowski's inequality and Plancherel's identity we have \begin{align} |u(x)| \le &\ \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}|\hat{u}(\xi)|\, d\xi \\ = ... 1 You should use the divergence theorem: Divergence Theorem:(see "Finite element methods for Maxwell's equations" by Peter Monk, Theorem 3.24) Let\Omega\subset \mathbb R^3$be a bounded Lipschitz domain with a unit outward normal$\nu$. Then 1) the mapping $$\gamma_\nu: \left[ C^\infty(\overline\Omega)\right]^3\rightarrow H^{-1/2}(\partial \Omega)$$ ... 0 Another example: continuous functions with unbounded derivative at either$0$or at$1$such that$f'\notin L^p(I)$. The function$f(x)=x^\alpha\in L^p\quad\forall \alpha>0\quad$, because it is bounded.$f'=\alpha \frac{1}{x^{1-\alpha}} \Rightarrow \quad$if$\quad p(1-\alpha)\ge 1\quad\Leftrightarrow \alpha\leq\frac{p-1}{p}$then$\quad (f')^p\notin ...

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Given a general $H^1$ function, it would be at best a $H^{\frac{1}{2}}$ function, consider the case where $T[u]\in H^{\frac{1}{2}}(\partial\Omega)\setminus H^1(\partial\Omega)$, and $N=2$. The function you have defined can be represented in polar coordinates by a function $w(r,\theta)=v(x)$, where $r=|x|$ and $\theta=\arctan\frac{x_2}{x_1}$. Notice that it ...

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