# Tag Info

4

The "real" definition for weak convergence is: $u_n\rightharpoonup u$ in $H^1(\Omega)$ if for each bounded linear functional $f$ on $H^1(\Omega): \langle f,u_n\rangle\rightarrow \langle f,u\rangle$ as $n\to\infty$. We want to show that both deffinitons are equivalent. 1) Note that for arbitrary functions $g_0,g_1,...,g_n\in L^2(\Omega)$ the functional ...

4

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 $$\int_y^x f'(t)\,dt=\tilde{f}(x)- \tilde{f}(y),\qquad \forall\ x,y\in \overline{I}.$$ ...

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 \\ = ...

3

Assume that there is such an operator $T$. Take $w\equiv 1$ in $U$. Then approximate $u$ by a sequence of functions $w_k \in C_c^\infty(U)$, $w_k\to w$ in $L^p(U)$, which works due to density. Then $u:=1-w$ is zero, and $u_k:=1-w_k$ converges to $u$ in $L^p(U)$. Then $Tu=0$ and $Tu_k=1$, hence $Tu_k$ cannot converge to $Tu$. Contradiction.

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 $c, d\in (a, ... 2 The$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

Yes, your idea is correct. Every function $u$ with weak gradient $\nabla u\in L^\infty(\mathbb{R}^n)$ has a Lipschitz representative (e.g., by the argument in the second paragraph here). Being Lipschitz and in $L^2$, it has to be essentially bounded. Indeed, if $|u(x_0)|>M$, then $|u|>M/2$ on the ball of radius $M/(2L)$ centered at $x_0$, where $L$ is ...

2

Expanding on John Ma's comment: consider the bilinear form on $W^{2,2}$ defined by $$B(u,v)=\int_\Omega \Delta u\Delta v$$ Only two derivatives of each function are involved, so this form is bounded on $W^{2,2}$: $$|B(u,v)|\le C\|u\|_{W^{2,2}}\|v\|_{W^{2,2}}\tag{1}$$ Hence, for every $u\in W^{2,2}$ the map $v\mapsto B(u,v)$ is a bounded linear functional ...

1

Informally, $H^s_p$ should consist of functions whose derivatives of orders up to $s$ are in $L^p$. Let's consider the case $s=1$ and $p=2$ for simplicity. We need functions $f$ such that $f,\nabla f\in L_2$. On the Fourier side, this means we want $\hat f\in L_2$ and $\xi\hat f\in L^2$. How to combine these conditions into one? Requiring ...

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 ...

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The #3 of that Wikipedia talk page requires $1<p<N$ and $$1 \ge \frac1q > \frac1p - \frac{1}{N-1}\frac{p-1}{p}$$ So, as written, it does not apply to $p=2=N$. But on a bounded domain, $W^{1,2}$ continuously embeds into $W^{1,2-\epsilon}$, and since the inequality $$\frac12 > \frac1{2-\epsilon} - \frac{1-\epsilon}{2-\epsilon}$$ holds for small ...

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Yes, because $H_0^1(\Omega)$ is a convex set which is norm-closed. -- Svetoslav Reference: Convex set weakly closed if and only if strongly closed as well

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For Sobolev spaces, I like to think about the case $n=1$ first, since there, the Sobolev functions can be easily characterized. Using this, we will show that the claim is false for $s \geq 2$. To this end, take $f \in C_c^\infty(\Bbb{R})$ with $f(x) = x$ for $|x| < 1$. Then $|f(x)| = |x|$ for $|x| < 1$ and hence the weak derivative of $|f|$ is given ...

1

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 ...

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 $$c^{-1}(\left|\xi\right|^{2}+\left|\eta\right|^{2})^{\frac{p-2}{2}}\left|\xi-\eta\right|^{2}\leq\left|V(\xi)-V(\eta)\right|^{2}\leq ... 1 You don't need a_n' to converge uniformly to the sign function; it's enough to have them uniformly bounded and converging pointwise. (If your chosen a_n functions don't do this, then pick other ones that do.) Then you can use the dominated convergence theorem to show that a_n'(u) u_{x_i} \to \operatorname{sgn}(u) u_{x_i} in L^p. 1 You should use the divergence theorem: Divergence Theorem:(see "Finite element methods for Maxwell's equations" by Peter Monk, Theorem 3.24) Let \Omega\subset \mathbb R^3 be a bounded Lipschitz domain with a unit outward normal \nu. Then 1) the mapping$$\gamma_\nu: \left[ C^\infty(\overline\Omega)\right]^3\rightarrow H^{-1/2}(\partial \Omega)$$... 1 In case of \Omega=(-1,1), the distance function is d_\Omega(x)=\min(1-x,1+x), so this Jacobi weight is not of one of the types you mention. However, close to the boundary, this weight behaves like a power-type weight. For instance, if x is close to 1, there is the lower bound$$ (1-x)^\alpha(1+x)^\beta \ge d_\Omega(x)^{\alpha} \ x\ge 1, $$and this ... 1 You need a constant in there:$$\| (|u| - j)^{+}\|_{L^{p}(A_j)} \leq C|A_j|^{\frac{1}{n}} \| \nabla u \|_{L^{p}(A_j)}$$Let q =(1/p-1/n)^{-1}. By the Sobolev-Poincaré inequality (using the fact that the function w:=(|u| - j)^{+} has zero trace), we have$$\| w\|_{L^{q}(A_j)} \leq C \| \nabla w \|_{L^{p}(A_j)} \le C \| \nabla u \|_{L^{p}(A_j)}$$By ... 1 Notice that u \to |u| is a Lipschitz map, hence we can apply the chain rule for Sobolev functions to find the weak derivative of |u|. (recall that Lipschitz functions are differentiable a.e.). A proof of this fact can be found in any introductory book on Sobolev spaces. Alternatively, we can prove it directly. Consider$$f_{\epsilon}(t) = ...

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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)

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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 $T^{-1}: ... 1 I like your approach to the problem, we can slightly modify your idea to get the result. I want to use the following fact about sequences of real numbers: Let$\{x_n\} \subset \mathbb{R}$. If there is$x$such that every subsequence$\{x_{n_k}\}$admits a further subsequence that converges to$x$, then$\{x_n\}$converges to$x$. Let$\{f_{n_k}\} ...

1

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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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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