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Yes, $H^s$ embeds continuously to $L^2=H^0$ for any $s\geq0$. (And it fails for all $s<0$. In fact, $H^s$ embeds continuously to $H^r$ whenever $r\leq s$.) The norm on $H^s(\mathbb R^n)$ is (up to a constant that you are free to choose) $$\|f\|_{H^s} = \|w(\cdot)^s\hat f(\cdot)\|_{L^2},$$ where $\hat f$ is the Fourier transform of $f$ and ...

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 $$\int (1+|y|)^{2k}|\hat{u}|^2dx\leq C\int ... 0 For example, for \Omega=\mathbb{R}, the distribution \delta_0 defined by \delta_0(\varphi)=\varphi(0) belongs to H^{-1}(R)(in fact \delta\in H^s(R) for s<-1/2) but don't belong to L^2(\mathbb{R}). 1 Note that in step one, it is given that for any multi-index \alpha with \|\alpha\| \le k, D^\alpha u_n converges uniformly to some function u_\alpha. In particular, u_n \to u  uniformly. You do not know, a priori, how all these u_\alpha's are related. In particular, you do not know u_\alpha = D^\alpha u. (Actually, from step 1 alone you do ... -1 Another approach without convolutions: Since Du is an L^p(0,1) function, hence L^1(0,1) in particular, for each x \in (0,1) the integral w(x) = \int_0^x Du(t)\,dt makes sense. By the fundamental theorem of calculus for Lebesgue integrals, w is absolutely continuous and w' = Du almost everywhere. Intuitively, we should expect that w differs ... 0 A graph will really help. I've sketched a graph of a function F_{a,b,h,k} that cannot be used directly as a test function, but it can after mollification. The derivative F' of F is shown below the graph of F. The exact assumptions are that 0 < a-h < a \le b < b+k < 1. That way, after you mollify with a compactly supported C^{\infty} ... 0 This is only possible if you know that (u_n') is uniformly bounded in L^2(0,T;V^*). Then write$$ \left|\int_0^T \langle u_n'-u',w\rangle \right| \le \left|\int_0^T \langle u_n'-u',w-w_j\rangle \right|+\left|\int_0^T \langle u_n'-u',w_j\rangle \right|\\ \le \|u_n'-u'\|_{L^2(0,T;V^*)}\|w-w_j\|_{L^2(0,T;V)}+\left|\int_0^T \langle u_n'-u',w_j\rangle ...

0

That is correct. Here are some examples where a higher order weak derivative exists but a lower order weak derivative does not exist: (p.15 remark 2.19(ii)) http://bolzano.iam.uni-bonn.de/~beck/seltop/topics_pde.pdf http://math.7starsea.com/post/308 Crostul mentioned that if ALL weak derivatives of order $n$ exist, then lower order weak derivatives exist. ...

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I think here you probability need $\Omega$ to be bounded as well. (however since $u\in W_0^{1,2}(\Omega)$, you do not need smooth boundary condition) And I will assume by $d\lambda^n$ you just mean $x\in \mathbb R^n$, the standard integration notation. For your first question, since $u\in W_0^{1,2}(\Omega)$, by Sobolev embedding you have $u\in L^{p^*}$ ...

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 $\gamma:H^1(\Omega)\rightarrow H^{1/2}(\partial\Omega)$ ...

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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 ...

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Yes, this is one way to do it. I would suggest you to read this book, chapter 8 for a complete introduction on 1 dimensional Sobolev space. Also, this note collects more results, but without proof, of Sobolev function properties in one dimensions.

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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 $$1 ... 0 The 1 in parentheses is useless; it doesn't scale if you multiply u by a constant. So your inequality can hold for all u\in H^1 if and only if$$\|u\|_{2^*}\leq \epsilon (\|\nabla u\|_2+\|u\|_p^{\frac{p}{2}})+C_\epsilon \|u\|_2$$holds. And here the scaling of the variable u_t = u^{n/2^*}u(tx) becomes an issue: namely, \|u_t\|_{2^*} and \|\nabla ... 2 As remarked in the comment, the construction is identical to the case where one minimizes R(u). First of all, observe that$$\tilde R(u) = R(u) + \frac{\int_M \alpha |u|^2 }{\|u\|^2_{2}} \ge R(u) - \|\alpha\|_\infty \ge \lambda_1 - \|\alpha\|_\infty.$$Thus \tilde R is bounded below. So we can take u_1, u_2, \cdots \in H, so that \|u_i\|_2 = 1 ... 0 For completeness, here is the counterexample for p=2^*. Let u be any nice compactly supported function. For t>0, define u_t(x) = t^{n/p} u(tx). This function has the same L^2 norm as u. Also, \nabla u_t(x) = t^{1+n/p} \nabla u(tx) which implies$$\|\nabla u\|_2^2 = t^{2(1+n/p) - n}\|\nabla u\|^2=\|\nabla u\|^2$$because n/p = n(n-2)/(2n) = ... 2 Since U is a bounded domain, its closure F=\bar U is compact. Therefore any (smooth) function F\to\mathbb R is compactly supported and so C_0^\infty(F)=C^\infty(F). Consequently W^{1,2}_0(F)=W^{1,2}(F). It is possible that someone prefers to work with a closed set instead of an open one (as in the case of manifolds with boundary) and means ... 0 That's not possible. The first eigenspace is always one dimensional: Let \phi_1 be the first eigenfunction. Then as \nabla |\phi_1| = \pm \nabla \phi_1 almost everywhere, R(\phi_1) = R(|\phi_1|) and so |\phi_1| is also an eigenfunction with eigenvalue \lambda_1. By maximum principle, |\phi_1|>0 on \Omega_1, which implies that \phi_1 is ... 1 It does has strict inequality: Let f_1 satisfies -\Delta f_1 = \lambda f_1 and \|f_1\|_{L^2(\Omega_1)} = 1 on \Omega_1. If \Omega_1 \subset \Omega_2, then the function$$g(x) = \begin{cases} f_1(x) & \text{ if } x\in \Omega_1 \\ 0 & \text{ if not.}\end{cases}$$is a W^{1, 2}_0(\Omega) function, \|g\|_{L^2(\Omega_2)} = ... 1 You have trivially that$$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^+})} $$3 Necas' book "Direct Methods in the Theory of Elliptic Problems" is a wonderful guide for the topics that you mentioned above, although the proofs are very abstract and most steps are omitted. Evans' book is more understandable and also includes the topics above, but not in the most general settings of the theorems. If you just start studying the Sobolev ... 1 By Lax-Milgram there exists only one \theta\in W_0^{1,2} such that $$\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 ... 0 The function u_\delta is defined so that it is continuous and has compact support. You can make a C^\infty function with compact support out of it by mollification, since the support of u_\delta is at positive distance from \partial \Omega. (This is helpful if H_0^1 is defined as the closure of C^\infty functions with compact support.) It ... 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 ... 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 ... 0 This theorem basically tells you that among "almost every" lines parallel to the axis, the function behaves like an absolutely continuous function. In another words, this theorem tells you that the Sobolev function behaves very good at the most of points, and the bad points have measure 0 and hence in some case is ignorable. Maybe you should compare ... 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 ... 2 I think theorem 4 (iii) on page 131 in this book is good enough to answer your question. 0 What you have is is that the average value of u on \partial B (which is \frac{\int_{\partial B(x,r/2) } u}{\sigma_{n-1}}) is greater than or equal to \frac{5r}{2}. Since the average value of the function u is at least \frac{5r}{2}, there must exist a point on the domain of integration such that the value is at least \frac{5r}{2}. 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 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$$But \phi is 0 on the boundary so really you just distribute the Laplacian into two gradients, and pick up a minus sign. 0 I know two methods to answer you question if the \partial \Omega is regular enough. (1) You could obtain this result by Fourier transform. http://www.math.ucla.edu/~tao/254a.1.01w/notes2.dvi page10 (2) By the theory of regularity of the equation -\Delta u = f, \|u\|_{H^2}\le \|f\|_{L^2}. 1 Suppose you want to find a number r whose square r^{2} is 2. That has no meaning for numerical analysis because all numbers on a computer are rational, and \sqrt{2} is not rational. It wasn't until the late 1800's that Mathematicians found a logically consistent way to define a real number. But once such a beast could be defined, then one can prove ... 1 The inclusions H_0^1 \subset L^2 holds per definition. To show the inclusion L^2 \subset H^{-1} we take a v \in L^2, then for every u \in H_0^1 we have by the Cauchy–Schwarz inequality $$(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}.$$ Therefore the Map I_v : u ... 0 First notice that v\in W^{1,2}(B) and then consider u as a continuous extension of v in \Omega. 2 Sobolev spaces are useful because are complete function spaces with a norm that reflects the differentiability of functions (unlike L^p norm) has nice geometry (unlike C^k norm) allows approximation by C^\infty functions (unlike C^k norm) "Nice geometry" means: uniformly convex norm (often, even inner-product norm). This property gives ... 0 You can write$$ v -u^+= \max(u,u^1)-u^+ = \max(u^+-u^1+u^-,0) + u^1-u^+. $$Now by assumption u^1-u^+\in H^1_0(B_R). Moreover, u^+-u^1=0 on the boundary, and \max(u^-,0)=0 on the boundary. Hence the trace of v-u^+ is zero. And v-u^+\in H^1_0(B_R). 1 There are two theorems you need to cite. The theorem 1 states that W^{2,2}(\mathbb R)=W^{2,2}_0(\mathbb R), which can be find in page 217, remark 13 in this book Theorem 2 states that C_0^\infty(I) is dense in W_0^{2,2}(I) for any interval I, of course for I=\mathbb R. You can find this theorem in p211 theorem 8.7 for the version of W^{1,2} in ... 1 There are some issues which must be considered. For instance, the image of g is always a set of zero measure in \mathbb{R}^n, thus, you can redefine F and \nabla F in this set to be anything. How do you approach this? Another issue is the following: Let n=3 and x=(x_1,x_2,x_3). Define F(x)=|x|^{-1/4}-1 for |x|\le 1 and F(x)=0 for ... 2 Let me give you a hint: First, let us define$$v_\epsilon:= \epsilon^{-\frac{n}{p^*}}u\left(\frac{x}{\epsilon}\right) $$Can you compute \|v_\epsilon\|_{L^p} and \|\nabla v_\epsilon \|_{L^p} in term of u? Try to write it down explicitly. Then you will know why it is bounded in W^{1,p} Secondly, the fact that v_\epsilon has no convergent ... 1 Yes: Call A the set of Lebesgue points of u (which is dense in \Omega). Then u restricted to A is Lipschitz and therefore there is a unique Lipschitz extension v to \bar{A}=\Omega. Since |\Omega\setminus A|=0 we get that v is a representative of u, and therefore "u is Lipschitz". 0 Once you have:$$ \left\| \psi' \right\|_{L^2}^2 = -\int_\mathbb{R}\psi\psi''\,dx \tag{1}$$the inequality:$$ \left\| \psi' \right\|_{L^2}^2 \leq \left\|\psi\right\|_{L^2}\cdot\left\|\psi''\right\|_{L^2}\tag{2}$$just follows from the Cauchy-Schwarz inequality. Nothing strange in the minus sign in the RHS of (1): just think to any non-negative test ... 1 You can show that the bilinear form is coercive in H^2. To do so, you have to use the inequality$$ \|u\|_{H^1} \le c \|u''\|_{L^2} \quad \forall u\in H^2_0(\Omega). $$Then by Lax-Milgram theorem you obtain existence of a unique solution for all f. 2 This is not correct, the norm of Du, D^2u, etc, takes the norm of all individual entries (partial derivatives). So$$ \|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 + ...

0

The Hölder maps are very straightforward to define in the usual way if you have a metric on both the domain and range. If $N$ is just a differentiable manifold, I guess the local version of Hölder continuity is invariant under local diffeomorphisms. So that would still be meaningful, although not the global version. Concerning the Sobolev spaces, these are ...

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

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 ...

0

Note that the domain has to be smaller than the unit ball, otherwise the function has a singularity at $|x|=\frac1{e-1}<1$. I choose $B(0,1/e)$. Take a smooth test function $\phi\in C_0^\infty(B(0,1/e))$. You need to verify $$\int_{B_1} u \nabla \phi = - \int_{B_1} \nabla u \phi$$ with $\nabla u$ given by your formula. The trick is to integrate ...

1

The relations are as follows: $$F_{\operatorname{div} v}(\vec \xi) = i \vec\xi\cdot F_v,\qquad F_{\operatorname{curl} v}(\vec \xi) = i \vec\xi\times F_v \tag{1}$$ They are easier to remember if one uses the notation $\nabla \cdot v$ and $\nabla \times v$, and notes that on the Fourier side, differentiation (i.e., $\nabla$) corresponds to multiplication by ...

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 ...

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