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4

Ok, here is it. We will exactly follow the proof of Morrey's theorem. I will replace $$\|\nabla u\|_{L^p(B)}\leq Mr^{N/p}$$ by $$\|\nabla u\|_{L^p(Q)}\leq Mr^{N/p} \tag 1$$ where $Q$ is a cube with length $r$. I do this only intend to match the proof of Morry. Now, for any $x\in Q$ where $Q$ is a cube with length $r$ and centered at origin, we have $$... 3 Actually you can build an example out of almost any function: Just notice that if u: B(0,1)\to \mathbb{R}, then u_r(x)=u(x/r) is defined in the ball of radius r and you get$$ \| u_r\|_{p,B(0,r)} = r^{n/p}\| u\|_{p,B(0,1)}, \quad \| \nabla u_r\|_{p,B(0,r)} =r^{-1} r^{n/p} \| \nabla u\|_{p, B(0,1)}. $$Translating appropriately you get an example in ... 3 This is not true. Let's prove it for a more general setting: assume that \beta>0 is such that \beta+1 < 2, then for any \alpha>0 there exists nontrivial variational solution u of the problem$$ \left\{ \begin{array}{ccc} -\Delta u-\alpha |u|^{\beta-1}u=0 &\mbox{ if $x\in\Omega$}, \\ u=0 &\mbox{if $x\in\partial\Omega$}. \end{array} ...

3

The first Laplacian you mention (sometimes called the Laplace-Beltrami operator) acts on scalar functions, that is, functions $S^2 \to \mathbb{R}$. The de Rham (a.k.a. Hodge) Laplacian acts on differential forms. In particular, the de Rham Laplacian acts on zero-forms, which are precisely scalar functions $S^2 \to \mathbb{R}$, on which it agrees with the ...

2

The first inequality is the triangle (Minkowski) inequality for the $L^p$ norm, the second inequality is Minkowski inequality for the counting measure, http://en.wikipedia.org/wiki/Minkowski_inequality.

2

The direct answer is that since the weak derivative of a function is only defined almost everywhere, you can put whatever value of $f'(0)$ you want. Always, remember that we say $g$ is the weak derivative of $f(x)=|x|$ if for any $\phi\in C_c^\infty(\mathbb R)$, we have $$\int_{\mathbb R} f\,\phi'\,dx=-\int_{\mathbb R}g\,\phi\,dx \tag 1$$ Hence, define ...

2

First of all, we only use weak star convergence in the case of $p=\infty$, i.e., weak star convergence stands for convergence test against the pre-dual, not dual which is the case of weak convergence used for $p<\infty$. Next, for your question, you can think $W^{1,p}$ space as $N+1$ fold of $L^p$ space, i.e. you have ...

2

This is indeed a consequence of Hölder's inequality: $$|u(x)-u(y)| \leq \int_y^x |u'(t)|dt \leq \left( \int_y^x dt\right)^{(p-1)/p} \left( \int_y^x |u'|^p dt \right)^{1/p}$$ $$\leq|x-y|^{1-1/p} \left( \int_0^1 |u'|^p dt \right)^{1/p}.$$

2

We have that (see Evans in regularity part) for each $f\in L^2(\Omega)$, the problem $$\left\{ \begin{array}{cc} -\Delta u =f&\mbox{ in \Omega,} \\ u=0 &\mbox{ on \partial\Omega,} \end{array} \right.$$ has an unique solution $u\in H$ satisfying $\|u\|_{H^2}\le C\|f\|_2$. Fix $u\in H$ and write $-\Delta u=f$. Let f_n\in ... 2 Your inequality (12) is missing some squares, it should be $$\| \mathbf{u}(t) \|_{L^2(U)}^2 = \| \mathbf{u}(s) \|_{L^2(U)}^2 + 2 \int_s^t \langle \mathbf{u'}(\tau),\mathbf{u}(\tau)\rangle d\tau \tag{12}$$ With this small correction you get \begin{align} T \| \mathbf{u}(t) \|_{L^2(U)}^2 &\le \int_0^T \| \mathbf{u}(s) \|_{L^2(U)}^2 \, ds + 2 ... 2 If T: X \to Y is bounded and surjective, the Open Mapping Theorem says there is an isomorphism S:\; X/\ker(T) \mapsto Y such that T = S \circ \pi, where \pi: X/\ker(T) \to Y is the quotient map. The trouble is, a quotient of X might not be isomorphic to a closed subspace of X, so there might not be a bounded right inverse. For example, every ... 2 First we look for a distributional solution. Remember that, as an distribution, \Delta u is defined by\langle\Delta u,v \rangle=-\langle\nabla u,\nabla v\rangle,\ \forall\ v\in C_0^\infty(\Omega). \tag{1}$$From (1), we can say that a solution in the distributional sense, is a function u\in H^1(\Omega) with Tu=g satisfying$$\int_\Omega \nabla ... 2 LetH$be a Hilbert space and define$M$by $$M=\{u\in H:\ F(u)=0\},$$ where$F:H\to\mathbb{R}$is a$C^1(H)$function. Theorem: Suppose that for all$u\in M$,$F'(u)\neq 0$. Then,$M$is a$C^1$Hilbert Manifold of$H$. To prove it, fix$u\in H$. Remember that$F':H\to H^\star$, so$F'(u)\neq 0$means that the linear function$F'(u)$has non trivial ... 2 Take any$u\in C_0^1(\mathbb{R}^2_+)$and define $$u_n(x)=u\left(\frac{x}{n}\right),\ \forall\ x\in \mathbb{R}^2_+.$$ Note that $$\int_{\mathbb{R}^2_+}|u_n(x)|^2dx=n^2\int_{\mathbb{R}^2_+}|u(x)|^2dx,$$ while $$\int_{\mathbb{R}^2_+}|\nabla u_n(x)|^2dx=n\int_{\mathbb{R}^2_+}|\nabla u(x)|^2dx.$$ Therefore, $$\|u_n\|_2^2=n^2\|u\|_2^2,\ \|\nabla ... 2 Another way of looking at this is through the Lebesgue Differentiation theorem, that says: If v\in L^p_{loc}(\Omega) then for a.e. x\in \Omega we have$$ g_v(x,r):= \frac{1}{|B(x,r)|}\int_{B(x,r)} |v|^p dy \to |v(x)|^p \qquad \text{ as } r\to 0. $$Your hypothesis implies that g_{\nabla u} (x,r)\leq M and so |\nabla u(x)|^p \leq M for a.e. x. ... 2 Notice that rewriting what you have on the second to last line of teh first chain of inequalities (you applied Holder wrong in the last line):$$ \| Du\|_{2p}^{2p} \leq C\int |u||D^2 u||Du|^{2p-2} \leq C\| u\|_{\infty} \| D^2 u\|_{p}\| Du\|_{2p}^{2p-2}, $$since 2p/(2p-2)=p/(p-1) is the conjugate exponent of p. In other words,$$ \| Du\|_{2p}^2 \leq C\| ... 2 First of all, chatacterize each$\lambda_k$as $$\lambda_k=\min_{u\in \langle \varphi_1,\cdots,\varphi_{k-1}\rangle ^\bot,\ \|u\|_2=1}B[u,u],$$ where$\varphi_k$are the eigenfunctions associated with$\lambda_k$. They are orthonormal and normalized. Let $$\overline{\lambda_k}=\max_{S^{k-1}}\min_{u\in (S^{k-1})^\bot,\ \|u\|_2=1}B[u,u].$$ As you can see, ... 2 Assume first that$1\leq p <\infty$If$u\in C^\infty(\bar{U})$then clearly$v=F(u)\in C^1(\bar{U})$and$\nabla v=F'(u)\nabla u$. Now if$u$is a general$W^{1,p}$function then take a sequence$u_k \to u$in$W^{1,p}$with$u_k\in C^\infty(\bar{U})$and such that$u_k\to u$and$\nabla u_k \to \nabla u$pointwise a.e. in$U$. Then $$|F(u)-F(u_k)| ... 2 Shown u^\epsilon\to 0 weakly in L^2 is the easy part, I think you already proved it. (I looked at comments, it has a good hint) The trick part is to show \nabla u^\epsilon\to 0 weakly in L^2 as well. Applying chain rule will lead you nowhere. You need following result in Functional analysis Let X be a Banach space, S be a total subset of ... 2 Ok, here it goes. We assume F is an linear bounded operator over H^1(\Omega) Let E denote the space of N+1 fold L^2(\Omega), i.e., E(\Omega):=(L^2(\Omega))^{N+1}. Then the operator T, from H^1(\Omega)\to E(\Omega) is defined by T[u]=(u,\partial_1 u,\partial_2u,\ldots,\partial_Nu) and we have T[u]\in E(\Omega). Take G:=T(H^1(\Omega)) ... 2 Let's prove your assertion. Let f be a bounded linear functional on Y. Then f \circ T is a bounded linear functional on X and as x_n converges to x weakly, we have$$ \lim _{n \to \infty} f\circ T(x_n) = f\circ T(x), $$that is,$$\lim_{n\to \infty} f(T(x_n)) = f(T(x))$$for all bounded linear functional f. Thus T(x_n) converges weakly ... 1 The differential operators like T are not bounded from H^n to H^{n}. But T is bounded from H^n to H^{n-1}, since the latter space requires one fewer derivative than the domain. Indeed,$$\|f'\|_{H^{n-1}}\le \|f\|_{H^n}$$since the right hand side involves all of the terms that the left hand side involves (details vary depending on your precise ... 1 Find functions u compactly supported inside B(x,r) so that \int_{B(x,r)} |u|^p \, dx / \int_{B(x,r)} |\nabla u|^p \, dx goes to infinity as r \to \infty. Probably something like$$ u(y) = \prod_{k=1}^N \cos\bigl((y_k-x_k)\sqrt N \pi/(2r)\bigr) I_{|x_k-y_k| < r/\sqrt N} $$1 This exercise is the result of Chain rule: Recall that if f\in C^1(\mathbb R) with f'\in L^\infty(\mathbb R) then we have f(u)\in W^{1,p}(U) if U is bounded, and we have \partial_i f(u)=f'(u)\partial_i u in weak sense. Hence, for part (a), the function f(x)=|x| has derivative -1 or 1 and is (pise-wise) C^1, and hence the chain rule ... 1 Take \Omega=(0,1), define u_t(x) = x-t. Set$$ f(t) := \int_\Omega |\nabla u_t^+|^2 dx. $$Then it holds f(t) =1 for all t\le 0, but f(t) = 1-t for t>0. If the functional would have been differentiable, so would have been f. Thus, the functional is not differentiable. Edit: To obtain a counter-example in W-^{1,2}, my proposal would be: ... 1 For your first question, the answer can be found in paper [1] and references therein. The answer for your second question can be found here. [1]: Morrey, C. B., Jr.; Nirenberg, L. On the analyticity of the solutions of linear elliptic systems of partial differential equations. Comm. Pure Appl. Math. 10 (1957), 271–290. 1 For n\geq 3 take f_k(x)=\min\{ |x|^{-1}, k\}. Then it's easy to see that \| f_k\|_{H^1_0(\Omega)} is uniformly bounded in k, but f_k(0)=k is unbounded. For n=2 a similar argument applies to f_k(x)=\min\{ \ln \ln(1+1/|x|), k\}. 1 Yes, the norms are equivalent. All you have to show is that there exist C_1, C_2 > 0 such that$$ C_1 (1+|x|)^2 \leq 1+|x|^2 \leq C_2 (1+|x|)^2 \quad \forall x \in \mathbb R. $$This relation holds for e.g. C_1 = \frac{1}{2} and C_2 = 1 (*). Then$$C_1^s \| f \|_{H^s}^2 \leq \| f \|_{s,2}^2 \leq C_2^s \| f \|_{H^s}^2 \quad \text{for$s > 0$}, ... 1 From my experience, there are three types of Laplace equation PDEs, namely, (D) for Dirichlet problem \begin{cases}\tag D -\Delta u=f\\ u=0 \end{cases} (N) for Neumann problem \begin{cases}\tag N -\Delta u=f\\ \partial_\nu u=0 \end{cases} and (R) for Robin problem \begin{cases}\tag R -\Delta u=f\\ \partial_\nu u+\alpha u=0 \end{cases} We use test function ... 1 You're definitely on the right track, but you at some point have to use the structure of$(W^{1,p}(Ω))^*$. I would suggest a more elementary proof in the beginning: You want to prove convergence in$(W^{1,p}(Ω))^*$, i.e.$K(u, u_t) - K(u, u)\$ must go to zero in that norm. Write down the definition of the dual norm, then apply the triangle inequality and ...

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