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4

This argument works if $U$ is bounded and has Lipschitz boundary. Since $(u_h)$ converges weak-star in $W^{1,\infty}(U)$, it is bounded in $W^{1,\infty}(U)$ and also bounded in $W^{1,n+1}(U)$ since $U$ is bounded. The space $W^{1,n+1}(U)$ is continuously embeddded into the space of H"older continuous functions $C^{0,1/n}(\bar U)$, Morrey embedding theorem. ...

3

Your phenomenon cannot be true as we have $W^{1,p}(0,1) \subseteq W^{1,q}(0,1)$ if $1 \leq q < p \leq \infty$. And indeed, for $1 \leq p < \infty$, we have \int_0^1 u'(x)^p \, dx = \frac{1}{2^p} \int_0^1 x^{-\frac{p}{2}} \, dx = \begin{cases} -\frac{1}{2^p \left( \frac{p}{2} - 1 \right)} \left(1 - \lim_{x \to 0} x^{-\frac{p}{2} + 1} \right) & ... 3 The closure of the domain in L^2 is simply L^2: Obviously it holds C_0^\infty(0,1)\subset D(A_0). The set of smooth function is dense in L^2(0,1), hence its closure is L^2(0,1). This implies that the closure of D(A_0) is L^2(0,1) as well. 2 Let us start by discussing the following Neumann problem: \begin{align} - \Delta u &= f \quad \textrm{ in } \Omega, \\ \frac{\partial u}{\partial \nu} &= 0 \quad \textrm{ on } \partial \Omega. \end{align} Here, \Omega \subset \mathbb{R}^n is a bounded Lipschitz domain. As Michal has already mentioned, a solution u of this boundary value ... 2 If you use (u, v)_{H_0^1} = (\nabla u, \nabla v)_{L^2}, you get R = -\Delta, as you have shown. And there is another scalar product on H_0^1, such that R = I - \Delta. Which one? 2 So first of all, H^1 does not really have a restriction map, even to interior points, much less boundary points (except in one dimension, where there is a continuous embedding into C^0). The better way of thinking about this is to start out with smooth test functions, then look at the equation that you get and identify the solution space and the test ... 2 It is usual to assume some regularity property for \Gamma. For example, in Evan's PDE, if \Omega is bounded and has a C^1 boundary, then there is (bounded) trace operatorT : H^1(\Omega) \to L^2(\Gamma)$$so that T u = u|_\Gamma if u \in H^1(\Omega) \cap C(\overline\Omega). It is also proved that$$\{ H^1(\Omega) : Tu = 0\} = H^1_0(\Omega),...

2

Let me handle the case $d = 1$ for simplicity of notation. Since $H$ is a Hilbert space, $F$ is represented by an element $g \in H$. Thus, for all $\phi \in \mathcal{D} \subset H$ with $\operatorname{supp} \phi = K$ we have $$|F(\phi)| = \left| \left<\phi, g \right>_H \right| \leq ||g||_H ||\phi||_H = ||g||_H \left( \int_{\Omega} \phi \cdot \phi + \... 2 The space \tilde{C}^k is not complete with respect to the norm ||.||_{k,p} -- assuming the norm ||.||_p is the L^p norm$$||f||_p^p := \int_\Omega |f|^p dx$$The reason is that convergence in this kind of norm does not preserve differentiability. Edit: provide simple example to reply to a comment: look, for example, at the functions$$f_n(x) = |x|^{...

2

Strictly speaking, the answer is no. $f \in W^{2,p}$ is really an equivalence class of functions which are equal a.e. It doesn't even exactly make sense to speak of their pointwise regularity. But even if you "open up" the equivalence class and look at the functions inside, not all of them have pointwise regularity. (For instance, you can do whatever you ...

1

Adding to @WORDS's answer, I thought up a way to prove $W^{1,1}\subseteq AC$. Here it is. Let $u\in W^{1,1}(a,b)$. Let $\rho_n$ be the standard mollifiers. We have: \begin{align*} \rho_n\ast u,\rho_n\ast u'\in{}&\mathcal{C}^\infty(a,b)\,\,\forall n; \\ \rho_n\ast u\to{}&u\,\,\text{in }L^1(a,b); \\ \rho_n\ast u'\to{}&u'\,\,\text{in }L^1(a,b); \\ \...

1

Let $\tilde V$ denote the closure of $\mathfrak D$ with respect to the $H^1$-norm. We want to show $\mathcal V = \tilde V$. Clearly, $\tilde V$ is a closed subspace of $\mathcal V$. Now let $f\in \mathcal V$ be given such that $f(v)=0$ for all $v\in \tilde V$. This holds in particular for all $v\in \mathfrak D$. Hence it follows that $f$ is a gradient, i.e....

1

To answer your additional question: Without assumptions on the boundary the statement is false. You can pretty much do the same thing as in the unbounded case. More precisely: look at the bounded part of a spiral (with some thickness) with infinite length. Parameterized along the length of the spiral from the outside in, just let $f_n=0$ on $[0,n]$ and $f_n =... 1 That's not true. For example, when$n=1$,$\sin x$is an element in$W^{1,p}_0(0,\pi)$since it is a$W^{1,p}$-limit of elements in$C^\infty_c((0,\pi))$. However,$\sin x$does not have compact support$(0,\pi)$. In general, if$\Omega$has a nice boundary (for example if it's$C^1$),$W^{k,p}_0(\Omega)$are those in$W^{k,p}(\Omega)$which vanishes at the ... 1 If$N > 1$and$\Omega = B(0,1)$then$u(x) = \log \log \left( 1 + \frac{1}{||x||} \right) \in W^{1,N}(\Omega)$, the function is continuous on$B(0,1) \setminus \{ 0 \}$and unbounded at the origin and so does not equal a.e to any continuous function defined on$B(0,1)$. If$N = 1$then any Sobolev function has a (locally) absolutely continuous ... 1 Your explanation for the choice of the function space is right. Note that$u \in H_0^2(\Omega)$if and only if$u = 0$and$\nabla u = 0$on$\partial \Omega$. This, of course, is not the right function space for your Neumann problem! 1 let's take$\Omega = [0,1]$,$\varphi,f \in C^\infty_c((0,1))$. Then $$\langle f'',\varphi \rangle = \int_0^1 f''(x) \varphi (x) dx = f'(1)\varphi (1)-f'(0)\varphi (0) - \int_0^{1} f'(x) \varphi '(x) dx$$$\varphi,f$have their support strictly inside$(0,1)$so$f'(0)\varphi (0) = f'(1)\varphi (1)= 0$and $$\langle f'',\varphi \rangle =-\langle f',\... 1 Yes, your argument is correct. You first prove the inequality on a dense subspace and both sides of the inequality depend continuously (w.r.t. H^1) on v. Hence, you can pass to the limit via density and obtain the inequality on all of H_0^1. This approach also works for many other properties of H_0^1. 1 Since the mapping$$ \iota\colon\mathcal{D}\to H $$is a continuous linear operator (which is the "generalization" of bounded operators to the setting of locally convex spaces), there is a (continuous) transpose$$ \iota^t \colon H' \to \mathcal{D}' $$satisfying$$ \langle \varphi, \iota^t(T)\rangle = \langle \iota(\varphi), T\rangle$$for all$\varphi\in\...

1

This is not possible. Every $v \in W^{2,4}(\Omega)$ satisfies $\nabla v \in C(\bar\Omega)$. Hence, if $\varphi^\varepsilon \in W^{2,4}(\Omega)$ converges in $W^{1,\infty}(\Omega)$ towards $v$, we get $\nabla v \in C(\bar\Omega)$, since the derivatives converge in $L^\infty(\Omega)$. But $\nabla \varphi \in L^\infty(\Omega)$ can be discontinuous.

1

You can find the general statement and proof in Chapter 6 of the book "Sobolev Spaces" by Robert A. Adams and John. J. F. Fournier.

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