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I think the best answer is to look at the paper Di Fazio, G., Hakim, D.I. & Sawano, Y. Elliptic equations with discontinuous coefficients in generalized Morrey spaces, European Journal of Mathematics (2017) 3: 728. DOI: 10.1007/s40879-017-0168-y and the references therein. As a quick summary the paper says in the introduction $a_{ij} \in W^{1,n}(\Omega)... 0 I finished by point 3. I promized. The paper is disponible here for more information. And I recal that$u^\varepsilon$are regular engouh (continuous) so that measurability in any subsequent space is valid. The case$p=1$. The authors say the sequence$u^\varepsilon$uniformely bounded in$L^\infty(0,T;L^1(\mathbb R^n))$is weakly relatively compact in$L^\...

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You have already solved it with $(u,v)=\int \hat{u}(\xi)\tilde{v}(\xi)d\xi=\int\hat{u}(\xi)<\xi>^s\tilde{v}(\xi)<\xi>^{-s}d\xi$ is a isomorphism of $H^{-s}$ and the dual of $H^{s}$

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For $\Rightarrow$, let $\alpha$ be such that $|\alpha| \leq k$, let $f=D^{\alpha}u$, let $\hat{u}$ be the $L^2$ (and distributional) Fourier transform of $u$. $f$ is both a $\mathcal{S}’$ distribution and a $L^2$ function, hence its (distributional) Fourier transform is (up to a scalar function) $\xi^{\alpha}\hat{u}$, and it must also be a $L^2$ function. ...

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It is easy to see that $u^+$ or $u^-$ have as much weak derivatives as $u$, just set $$\Omega^+:=\{x\in\Omega: u(x)\ge 0\},\quad\Omega^-:=\{x\in\Omega: u(x)\le 0\}$$ Hence $$\int_{\Omega} u^+(x)\partial_j\varphi(x)\, dx=\int_{\Omega} u(x)\partial_j\varphi(x)\chi_{\Omega^+}(x)\, dx=-\int_{\Omega}\partial_j u(x)\chi_{\Omega^+}(x)\varphi(x)\, dx$$ where $\... 1 The second definition is bogus. In fact, if you only assure that$u(t) \in W^{1,\infty}$for (almost) all$t \in (0,T)$, the function $$t \mapsto \|u(t)\|_{W^{1,\infty}}$$ might fail to be measurable, hence the left-hand side of (1) is not even well defined. Moreover, the separability condition is crucial. There are weakly measurable$u$, for which (1) ... 1 For the poisson equation the natural space of the weak formulation is$H^{-1}$. However it is wrong or at least bad style to write it as an integral. It is advised to use the bracket notation$(f,\phi)_{H^{-1}}$. Notice that for all functions$f\in L^2$there exists an associated element$f^*\in H^{-1}$given by $$(f^*,\phi)_{H^{-1}}=\int f \phi dx$$ To ... 1 Here is an idea. If you would have equality, then you get $$\| \frac{\mathrm d}{\mathrm d t} \Pi_C\gamma (t) \| = \|\dot\gamma(t)\|$$ for a.a.$t$. Moreover, the projection is firmly non-expansive. Hence, $$\| \gamma(t_2) - \gamma(t_1) \|^2 \ge \| \Pi_C \gamma(t_2) - \Pi_C \gamma(t_1) \|^2 + \| \Pi_C \gamma(t_2) - \gamma(t_2) - \Pi_C \gamma(t_1) + \gamma(... 2 Why not take a concrete example L^2(0,1) is the inner product space \langle f,g\rangle= \int_0^1 f(x)\overline{g(x)}dx, H^1(0,1) is the inner product space ( f,g) =\langle f,g\rangle+\langle f',g'\rangle e_n = \frac{e^{2i \pi nx}}{\sqrt{1+4\pi^2 n^2}} is an orthonormal basis of H^1(0,1). The map f \mapsto (g \mapsto (g,\overline{f})) is an ... 4 There exists a compact embedding H^1_0(\Omega)\to H^{-1}(\Omega). There also exists an isometric isomorphism H^1_0(\Omega)\to H^{-1}(\Omega). There is nothing contradictory about this, because these are two different maps. (If you view a map from a normed space to its dual as a bilinear form on the space, the first map corresponds to the L^2 inner ... 2 Then answer is yes. Here is a proof: For each j=1,\ldots,n put f_j:=\partial_jg_1 \chi_{\Omega_1}+\partial_j g_2 \chi_{\Omega_2}. Clearly, f_j\in L^2(\Omega'). If we can show that the weak (distributional) derivative of f coincides with f_j, we obtain f\in H^1(\Omega'). To this end, we simply verify for any \phi\in C^\infty_0(\Omega') that: \... 1 Sure. If M is a compact manifold, and k is a non-negative integer, then H^k(M) is the space of function u\in L^2(M) with the property that for any \ell smooth vector fields X_1,\cdots, X_\ell on M, with \ell\leq k, we have X_1\cdots X_\ell u\in L^2(M). For a reference, you can see Michael Taylor's PDE I text. 0 This does not seem to be true. Consider the analogous problem in one dimension, where B reduces to (-1, 1) and B_+=(0,1). Let f=-2. Then the unique solution to$$ \begin{cases} u''=-2, & (0, 1), \\ u(0)=u(1)=0, \end{cases} $$is u(x)=x(1-x), and its even extension is$$ v(x)=|x|(1-|x|), $$which is NOT the solution to$$ \begin{cases} w''=-... 0 It's not true that$\mathcal C_c^\infty (I)$is dense in$W^{1,p}(I)$when$I\neq \mathbb R$. What is true is$\mathcal C_c^\infty (\mathbb R)$is dense in$W^{1,p}(\mathbb R)$. And indeed, if$I\neq \mathbb R$is an interval, we define$W_0^{1,p}(I)$as the closure of$\mathcal C_c^\infty (I)$in$W^{1,p}(I)$. 2 By Poincaré, $$\int{|u|^2} \leq C\int{|\nabla u|^2} \leq C\int{|fu|}.$$ By Cauchy-Schwarz, $$\|u\|^2_{L^2} \leq C\|f\|_{L^2}\|u\|_{L^2}.$$ Thus$\|u\|_{L^2} \leq C\|f\|_{L^2}$. Now, with the same estimation, $$\|\nabla u\|_{L^2}^2 \leq \int{|fu|} \leq \|f\|_{L^2}\|u\|_{L^2} \leq C\|f\|^2_{L^2}.$$ Therefore, $$\|u\|^2_{W^{1,2}_0} = \|u\|^2_{L^2}+\|\... 2 You have to use the Poincaré inequality again. This gives$$\lVert u\rVert_{L^2}^2\le C_1\lVert \nabla u\rVert^2_{L^2}\le C_2\lVert f\rVert_{L^2}\lVert \nabla u\rVert_{L^2}, $$so \lVert u\rVert_{L^2}^2 + \lVert \nabla u\rVert^2_{L^2}\le C_3\lVert f\rVert_{L^2}\lVert \nabla u\rVert_{L^2}, for some big enough constant C_3>0. This is the inequality you ... 7 To motivate Sobolev spaces, let me pose a motivating problem. Let \Omega be a smooth, bounded domain in {\Bbb R}^n and let f be a C^\infty function on \Omega. Prove that there exists a C^2 function u satisfying -\Delta u = f in \Omega and u = 0 on the boundary of \Omega. As far as PDE's go, this is the tamest of the tame: it's a ... 1 No. Consider f \in \cap_{s < 0}H^s(\mathbb{R}) defined by \hat f(t) = (1+t^2)^{-1/4}. This function is not in L^2. 0 I can't tell if you've miscopied or misread, but your statement about w' is wrong. What Brezis actually writes (p. 213, section 8.2) is$$ w' = G'(v)v' = p|v|^{p-1}v' $$Now, w' = G'(v)v' is just the chain rule like you say, and for G'(v) we calculate:$$ G'(v) = (p-1)|v|^{p-2}\cdot \mathop{sgn} v\cdot v + |v|^{p-1}$$which simplifies to (p-1)|v|^{p-... 0 We know that u\in W^{2,2}(\Omega') because \|D^2 u\|_{L^2(\Omega')}<\infty and u\in W^{1,2}(\Omega'). Now, take a cut-off function \eta\in C_c^{\infty}(\Omega) such that 0\leq \eta \leq 1, \eta=1 in \Omega', \|\nabla \eta\|_{L^{\infty}(\Omega)}\leq C_1(\Omega',\Omega), \|D^2 \eta\|_{L^{\infty}(\Omega)}\leq C_2(\Omega',\Omega). Then ... 1 Thank you so much for your answer! I actually managed to resolve my confusion. When calculating the given example in the book, one comes to a point where the decisive question is: Is$$ \int_0^a \frac{r}{(r \log(r))^p} \, \mathrm{d} r < \infty$$for$a < 1$. But this integral is only finite if$p \leq 2\$, so actually the counterexample in the ...

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