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3

First, we need to show that $u \in L^n(\Omega)$. You can show by a change of variable that $$\int_\Omega |\ln \ln (1 + \frac{1}{|x|})|^n dx \leq C \int_0^1 |\ln \ln (1 + \frac{1}{r})|^n r^{n-1} dr$$ The function $|\ln \ln (1 + \frac{1}{r})|^n r^{n-1}$ is continuous on $]0,1]$, so the only problem may be on 0. But we have that for r smaller than 1/e $$1 ... 3 Necas' book "Direct Methods in the Theory of Elliptic Problems" is a wonderful guide for the topics that you mentioned above, although the proofs are very abstract and most steps are omitted. Evans' book is more understandable and also includes the topics above, but not in the most general settings of the theorems. If you just start studying the Sobolev ... 2 Since U is a bounded domain, its closure F=\bar U is compact. Therefore any (smooth) function F\to\mathbb R is compactly supported and so C_0^\infty(F)=C^\infty(F). Consequently W^{1,2}_0(F)=W^{1,2}(F). It is possible that someone prefers to work with a closed set instead of an open one (as in the case of manifolds with boundary) and means ... 2 Sobolev spaces are useful because are complete function spaces with a norm that reflects the differentiability of functions (unlike L^p norm) has nice geometry (unlike C^k norm) allows approximation by C^\infty functions (unlike C^k norm) "Nice geometry" means: uniformly convex norm (often, even inner-product norm). This property gives ... 2 First of all, the bible book Evans PDE contains some very nice exercise in Sobolev space chapter, and they are relatively easy, you could do them first. The next book you could try is Leoni's Sobolev space. Chapter 10, 11, 12 contains a lot of exercises, some of them are tricky, but it worth the time. After you finish those two books, you may want to try ... 2 We can identify H^1_0 with H^{-1} by Riesz Representation Theorem, because H^1_0 is Hilbert space. But, in this case, is not a "good" idea to do this identification. You can consult Functional Analysis, Sobolev Spaces and Partial Diferential Equations, by Brezis, in the page 136, he makes a comment about it, and you will understand that choice of ... 2 This is not correct, the norm of Du, D^2u, etc, takes the norm of all individual entries (partial derivatives). So$$ \|v\|_{H^2}^2 = \|v\|_{L^2}^2 + \|\frac{\partial v}{\partial x_1}\|_{L^2}^2 + \|\frac{\partial v}{\partial x_1}\|_{L^2}^2 + \|\frac{\partial^2 v}{\partial x_1^2}\|_{L^2}^2 +\|\frac{\partial^2 v}{\partial x_1\partial x_2}\|_{L^2}^2 + ...

2

I think theorem 4 (iii) on page 131 in this book is good enough to answer your question.

2

As remarked in the comment, the construction is identical to the case where one minimizes $R(u)$. First of all, observe that $$\tilde R(u) = R(u) + \frac{\int_M \alpha |u|^2 }{\|u\|^2_{2}} \ge R(u) - \|\alpha\|_\infty \ge \lambda_1 - \|\alpha\|_\infty.$$ Thus $\tilde R$ is bounded below. So we can take $u_1, u_2, \cdots \in H$, so that $\|u_i\|_2 = 1$ ...

2

Let $\Delta_M$ and $\nabla_M$ be the Laplace-Beltrami operator and the gradient on $M$ and denote the variable on $[0,\infty)$ by $t$. Then $\Delta=\Delta_M+\partial_t^2$ and $\nabla u=(\nabla_Mu,\partial_tu)$. Let me also denote the divergence of a vector field $V$ on $M$ by $\nabla_M\cdot V$. Assume that $u$ and $v$ are smooth and compactly supported. ...

2

Let me give you a hint: First, let us define $$v_\epsilon:= \epsilon^{-\frac{n}{p^*}}u\left(\frac{x}{\epsilon}\right)$$ Can you compute $\|v_\epsilon\|_{L^p}$ and $\|\nabla v_\epsilon \|_{L^p}$ in term of $u$? Try to write it down explicitly. Then you will know why it is bounded in $W^{1,p}$ Secondly, the fact that $v_\epsilon$ has no convergent ...

2

The statement can be proven using the properties of the trace mapping $\tau$, where $\tau:H^1(\Omega) \to L^2(\partial \Omega)$ is continuous, and $\tau v = v|_{\partial \Omega}$ for continuous $v\in C(\bar \Omega)$. Due to the density of $C_c^\infty(\Omega)$ in $H^1_0(\Omega)$, there are functions $v_k \in C_c^\infty(\Omega)$ converging to $u$ in ...

1

This is more of a problem relating to weak convergence in $L^{2}$. Since $f_{n}\to f$ weakly in $H^{1}\left(\Omega\right)$, we know $g_{n}:=\nabla f_{n}\to\nabla f=:g$ weakly in $L^{2}\left(\Omega\right)$. Now, we require the following fact: If $x_{n}\rightharpoonup x$ weakly and $y_{n}\to y$ in norm, then $\left(x_{n}\right)_{n}$ is bounded (lets say by ...

1

By definition we have $|D^ku|^2=\sum_{|\alpha|=k}|D^{\alpha}u|^2$ then $$\int |y|^{2k}|\hat{u}|^2dx\leq C\int |D^ku|^2dx.$$ Note that if $a,b>0$ then $a+b\leq 2\max\{a,b\}$ and thus $(a+b)^s\leq 2^s(a^s+b^s)$, $\forall s>0$. It follows that $(1+|y|)^{2k}\leq C(1+|y|^{2k})$ and this implies $$\int (1+|y|)^{2k}|\hat{u}|^2dx\leq C\int ... 1 You have trivially that$$C^{\infty}_c(\mathbb{R^+}) \subset C^{\infty}_0(\mathbb{R^+})$$But W^{1,2}_0(\mathbb{R^+}) is usually defined as the closure of C^{\infty}_c(\mathbb{R^+}) for the W^{1,2} norm$$W^{1,2}_0(\mathbb{R^+}) := \overline{C^{\infty}_c(\mathbb{R^+})} $$1 There are two theorems you need to cite. The theorem 1 states that W^{2,2}(\mathbb R)=W^{2,2}_0(\mathbb R), which can be find in page 217, remark 13 in this book Theorem 2 states that C_0^\infty(I) is dense in W_0^{2,2}(I) for any interval I, of course for I=\mathbb R. You can find this theorem in p211 theorem 8.7 for the version of W^{1,2} in ... 1 Let's first be clear about what he means by u^\pm. He's writing this for the restriction of \bar{u} (which is defined on all of B as above) to B^\pm, respectively. In other words, u^+ = u in B^+ and u^- = \bar{u} in B^-. Now he wants to consider these as functions restricted to the set B \cap \{x_n =0\} \subset \mathbb{R}^{n-1}, which IS ... 1 Fix y \in [0,\infty). Let f_N(x,y) := \sum_{k=1}^N e^{-y\sqrt{\lambda_k}} (u,\varphi_k) \varphi_k(x) be the partial sum, and let g_N := f-f_N be the tail of the series. Consider the integral\int_{\Omega } g_N^2: writing g_N^2 as a product of two sums and multiplying out the terms, then only the diagonal terms survive after integrating because the ... 1 By Lax-Milgram there exists only one \theta\in W_0^{1,2} such that $$\int_\Omega\nabla\theta\cdot\nabla\phi dx=\int_\Omega f\phi dx$$ for all \phi\in W_0^{1,2}. So in fact T is well defined, linear and continuous from W_0^{1,2} to W_0^{1,2} with norm |\theta|_{W_0^{1,2}}. Obviously it is compact, since the range of T ... 1 No it is not: The mapping u\mapsto (-\Delta u,\gamma u), where \gamma is the trace operator, is an isomorphism from H^1_\Delta(\Omega) onto L^2(\Omega)\times H^{-1/2}(\partial\Omega) (see P. Grisvard, Elliptic Problems in Nonsmooth domains, Theorem 2.5.2.1). On the other hand, the trace operator \gamma:H^1(\Omega)\rightarrow H^{1/2}(\partial\Omega) ... 1 There are some issues which must be considered. For instance, the image of g is always a set of zero measure in \mathbb{R}^n, thus, you can redefine F and \nabla F in this set to be anything. How do you approach this? Another issue is the following: Let n=3 and x=(x_1,x_2,x_3). Define F(x)=|x|^{-1/4}-1 for |x|\le 1 and F(x)=0 for ... 1 Using Green's first identity what you would find is$$\int_{\Omega} \Delta u \phi = \int_{\partial \Omega} \phi \langle\nabla u, \nu \rangle - \int_{\Omega} \langle\nabla u, \nabla \phi\rangle$$But \phi is 0 on the boundary so really you just distribute the Laplacian into two gradients, and pick up a minus sign. 1 Yes: Call A the set of Lebesgue points of u (which is dense in \Omega). Then u restricted to A is Lipschitz and therefore there is a unique Lipschitz extension v to \bar{A}=\Omega. Since |\Omega\setminus A|=0 we get that v is a representative of u, and therefore "u is Lipschitz". 1 The inclusions H_0^1 \subset L^2 holds per definition. To show the inclusion L^2 \subset H^{-1} we take a v \in L^2, then for every u \in H_0^1 we have by the Cauchy–Schwarz inequality $$(v,u)_{L^2} \leq \Vert v \Vert_{L^2} \Vert u \Vert_{L^2} \leq \Vert v \Vert_{L^2} \Vert u \Vert_{H_0^1}.$$ Therefore the Map I_v : u ... 1 You can show that the bilinear form is coercive in H^2. To do so, you have to use the inequality$$ \|u\|_{H^1} \le c \|u''\|_{L^2} \quad \forall u\in H^2_0(\Omega). $$Then by Lax-Milgram theorem you obtain existence of a unique solution for all f. 1 It does has strict inequality: Let f_1 satisfies -\Delta f_1 = \lambda f_1 and \|f_1\|_{L^2(\Omega_1)} = 1 on \Omega_1. If \Omega_1 \subset \Omega_2, then the function$$g(x) = \begin{cases} f_1(x) & \text{ if } x\in \Omega_1 \\ 0 & \text{ if not.}\end{cases}$$is a W^{1, 2}_0(\Omega) function, \|g\|_{L^2(\Omega_2)} = ... 1 First you can prove (by approximating u and v by smooth functions) that$$ D(uv) = u Dv + vDu. $$Assume first p<n and q<n. Then using Sobolev embedding theorem, we know that$$ u\in L^{\frac{pn}{n-p}}(\Omega), \quad v\in L^{\frac{qn}{n-q}}(\Omega). $$Using Hoelder inequality one can prove that f\in L^p(\Omega), g\in L^q(\Omega) implies ... 1 Suppose you want to find a number r whose square r^{2} is 2. That has no meaning for numerical analysis because all numbers on a computer are rational, and \sqrt{2} is not rational. It wasn't until the late 1800's that Mathematicians found a logically consistent way to define a real number. But once such a beast could be defined, then one can prove ... 1 Let K be the support of u. For any test function \phi with support in \mathbb{R}^n\setminus K$$ \int_{\mathbb{R}^n}D^\alpha u\,\phi\,dx=(-1)^{|\alpha|}\int_{\mathbb{R}^n}u\,D^\alpha\phi\,dx=0  because the support of $u$ and $\phi$ are disjoint. This means that $D^\alpha u$ is supported no $K$.

1

For example, for $\Omega=\mathbb{R}$, the distribution $\delta_0$ defined by $\delta_0(\varphi)=\varphi(0)$ belongs to $H^{-1}(R)$(in fact $\delta\in H^s(R)$ for $s<-1/2$) but don't belong to $L^2(\mathbb{R})$.

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