# Tag Info

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 $$C_1^s \| f \|_{H^s}^2 \leq \| f \|_{s,2}^2 \leq C_2^s \| f \|_{H^s}^2 \quad \text{for s > 0}, ... 0 Your idea is right. But let me tell you what you are actually doing here. This is actually an minimizing problem in calculous of variation. (of course I believe you are already very well realized about this) Let me re-write you problem in a more standard way in calculus of variation: Finding the minimizer of functional$$ E[u]:= \|u'\|_{L^p(0,1)}$$among ... 0 Well, I think I found something. We want to show that the infimum of the norm of the derivative is attained on the intersection of our space with the unit circumference of L^p. All we need is the Relich-Kondrachov theorem, which allows us to extract a weakly convergent minimizing sequence, which is strongly convergent in L_p. Therefore if u is the ... 1 I don't know how to make a link in comment so I write it here, you don't need to take this as an answer. @Fundamental's answer is good enough, the answer is no. If you are looking for good reference for Sobolev embedding, especially focus on Trace operator, I would recommend you read Leoni's book, it explains embedding and trace in a very details way. 3 The first Laplacian you mention (sometimes called the Laplace-Beltrami operator) acts on scalar functions, that is, functions S^2 \to \mathbb{R}. The de Rham (a.k.a. Hodge) Laplacian acts on differential forms. In particular, the de Rham Laplacian acts on zero-forms, which are precisely scalar functions S^2 \to \mathbb{R}, on which it agrees with the ... 2 Let's prove your assertion. Let f be a bounded linear functional on Y. Then f \circ T is a bounded linear functional on X and as x_n converges to x weakly, we have$$ \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}$$since the right hand side involves all of the terms that the left hand side involves (details vary depending on your precise ... 0 The integral is finite for all \alpha\in\mathbb{R}. Observe that although \log r is unbounded near r=0, we always have \lim_{r\to0^+}|\log r|^{2\alpha}r=0. 2 Assume first that 1\leq p <\infty If u\in C^\infty(\bar{U}) then clearly v=F(u)\in C^1(\bar{U}) and \nabla v=F'(u)\nabla u. Now if u is a general W^{1,p} function then take a sequence u_k \to u in W^{1,p} with u_k\in C^\infty(\bar{U}) and such that u_k\to u and \nabla u_k \to \nabla u pointwise a.e. in U. Then$$ |F(u)-F(u_k)| ...

2

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

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 ...

0

$$f'(x)=-H(-x)+H(x)$$ where $H(x)=0$ when $-\infty\lt x\leq 0$ and $H(x)=1$ when $x\gt 0$

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 ...

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 ...

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 $$... 0 Suppose u\in W_0^{1,2}(\Omega)\cap W^{2,2}(\Omega) and \partial\Omega\in C^2, then the inequality will be hold. Indeed, we have the PDE \begin{cases} -\Delta u=-\Delta u &x\in\Omega\\ u=0&x\in\partial \Omega \end{cases} has solution u\in W_0^{1,2}(\Omega) and hence by outer regularity we have$$ \|u\|_{W^{2,2}(\Omega)}\leq C\|\Delta ...

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 $$\left\{ \begin{array}{ccc} -\Delta u-\alpha |u|^{\beta-1}u=0 &\mbox{ if x\in\Omega}, \\ u=0 &\mbox{if x\in\partial\Omega}. \end{array} ... 1 For n\geq 3 take f_k(x)=\min\{ |x|^{-1}, k\}. Then it's easy to see that \| f_k\|_{H^1_0(\Omega)} is uniformly bounded in k, but f_k(0)=k is unbounded. For n=2 a similar argument applies to f_k(x)=\min\{ \ln \ln(1+1/|x|), k\}. 2 First of all, we only use weak star convergence in the case of p=\infty, i.e., weak star convergence stands for convergence test against the pre-dual, not dual which is the case of weak convergence used for p<\infty. Next, for your question, you can think W^{1,p} space as N+1 fold of L^p space, i.e. you have ... 2 Notice that rewriting what you have on the second to last line of teh first chain of inequalities (you applied Holder wrong in the last line):$$ \| 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

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 ...

0

I'm not sure your question is well posed. When you say "if $|y|_Y < \infty$" then it seems you are implicitly assuming that $y$ is an element of $Y$. Indeed if $y$ were not an element of $Y$, then the expression $|y|_Y < \infty$ would not be defined.

0

Note that since $v_{m}$ is Lipshitz continuous, it is also almost everywhere differentiable (by Rademachar's Theorem). It follows then that $$\lvert \partial_k v_m(x)\rvert = \lim_{h\to 0} \left\lvert\frac{v_m(x+h\cdot e_k) - v_m(x)}{h}\right\rvert \leqslant 2\sigma$$ for $1 \leqslant k \leqslant n$, it follows then that $\lVert \nabla ... 1 Take$\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. 2 This is indeed a consequence of Hölder's inequality: $$|u(x)-u(y)| \leq \int_y^x |u'(t)|dt \leq \left( \int_y^x dt\right)^{(p-1)/p} \left( \int_y^x |u'|^p dt \right)^{1/p}$$ $$\leq|x-y|^{1-1/p} \left( \int_0^1 |u'|^p dt \right)^{1/p}.$$ 0 I don't know how your prove can work without prove the trace estimation... Here is a more standard idea. (You can find it on Evans PDE book or Leoni's Sobolev space. I would say Evans, it is easier) Suppose$\Omega$is an bounded extension domain so that the extension theorem, which you quote in your post, will work. Next, for$u\in W^{1,p}(\Omega)$you ... 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 ... 0 I would say it doesn't hold, See if this is true : Take as$\Omega:= B(0,3) - \{0\}$and take p=1.$f:= \eta log(|x|)sin(|x|)$where$\eta$is a smooth function that is$1$on$B(0;1)$and$0$outside$B(0,2)$. First$f$is in$W^{1,1}(\Omega)$:$\int_{B(0,1)} |f(x)|dx = \int_0^{2\pi}\int_0^1 |log(r)sin(r)r|dr \leq C\int_0^1 |log(r)r|dr$and an ... 1 There are multiple definitions of$H^{1/2}(\partial Ω)$which are equivalent if the boundary is regular enough. The most intuitive is probably as the range of the trace operator$tr\colon H^1(Ω) \to L^2(\partial Ω): $$H^{1/2}(\partial Ω) = \{ u ∈ L^2(\partial Ω) \;|\; ∃ \tilde u ∈ H^1(Ω)\colon u = tr(\tilde u) \}, \quad \| u \|_{H^{1/2}(\partial Ω)} = \inf ... 0 Integrating by part, we can get$$\int_\Omega u\frac{\partial^2\phi}{\partial x_i^2}\,dx=-\int_\Omega \frac{\partial u}{\partial x_i}\frac{\partial\phi}{\partial x_i}\,dx+\int_{\partial \Omega}u \frac{\partial\phi}{\partial x_i}\nu_i\,dS$$and$$\int_\Omega \frac{\partial u}{\partial x_i}\frac{\partial\phi}{\partial x_i}\,dx=-\int_\Omega \frac{\partial^2 ... 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 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].$$As you can see, ... 2 Take any u\in C_0^1(\mathbb{R}^2_+) and define$$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 ... 0 Suppose the claim is not true. Then for allr\in(0,1/2)$you have $$\int_{\partial B_r}|u-\bar{u}|^2\, \mathrm{d}x > 3\int_{B_1} |u-\bar{u}|^2\, \mathrm{d}x.$$ Integrate this inequality with respect to$r$from zero to$\frac12$; integrating the surface integrals with respect to radius gives a volume integral. You get $$\int_{B_{1/2}}|u-\bar{u}|^2\, ... 1 Notice that \text{supp}(\alpha) \subset W, so that both \alpha u_n and \alpha u vanish outside W, and then so do their gradients. Therefore$$\| \alpha u_n -\alpha u\|_{W^{1,p}(W)}=\| \alpha u_n -\alpha u\|_{W^{1,p}(\Omega)}.$$Be careful, trace zero only works when the domain is nice enough. In a general domain you have to show that u is ... 2 Ok, here it goes. We assume F is an linear bounded operator over H^1(\Omega) Let E denote the space of N+1 fold L^2(\Omega), i.e., E(\Omega):=(L^2(\Omega))^{N+1}. Then the operator T, from H^1(\Omega)\to E(\Omega) is defined by T[u]=(u,\partial_1 u,\partial_2u,\ldots,\partial_Nu) and we have T[u]\in E(\Omega). Take G:=T(H^1(\Omega)) ... 2 The first inequality is the triangle (Minkowski) inequality for the L^p norm, the second inequality is Minkowski inequality for the counting measure, http://en.wikipedia.org/wiki/Minkowski_inequality. 1 From my experience, there are three types of Laplace equation PDEs, namely, (D) for Dirichlet problem \begin{cases}\tag D -\Delta u=f\\ u=0 \end{cases} (N) for Neumann problem \begin{cases}\tag N -\Delta u=f\\ \partial_\nu u=0 \end{cases} and (R) for Robin problem \begin{cases}\tag R -\Delta u=f\\ \partial_\nu u+\alpha u=0 \end{cases} We use test function ... 2 Let H be a Hilbert space and define M by$$M=\{u\in H:\ F(u)=0\},$$where F:H\to\mathbb{R} is a C^1(H) function. Theorem: Suppose that for all u\in M, F'(u)\neq 0. Then, M is a C^1 Hilbert Manifold of H. To prove it, fix u\in H. Remember that F':H\to H^\star, so F'(u)\neq 0 means that the linear function F'(u) has non trivial ... 2 First we look for a distributional solution. Remember that, as an distribution, \Delta u is defined by$$\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 ... 0 Multiplying by a test function$v$and using the divergence theorem we obtain: $$\int_{\mathbb R^d}\nabla u\cdot\nabla v\,dx=\int_{\mathbb R^d}(f-c(u))v\,dx$$ let$v=-\Delta^{-h}_j\Delta_j^h u$for some direction$j\in\{1,...,d\}$, where$\Delta^h_ju=\frac{u(x+he_j)-u(x)}{h}$then $$\int_{\mathbb R^d}\nabla u\cdot\nabla v\,dx=\int_{\mathbb R^d}\nabla ... 0 in one dimension the Sobolev embedding theorems imply that H^{1,2}(a,b) is continuously embedded into C^0 (even C^{0,\alpha} for appropriately chosen \alpha). This implies in particular that approximation in the H^{1,2} norm implies uniform convergence hence -- of course -- pointwise convergence. Edit (responding to questions in comments): it is ... 0 The answer to your final question depends on what inner product you give L^2(I). We have \langle f, v \rangle_{V', V} = (f,v)_{L^2(I)}. To understand this properly, you should read up on Gelfand triples. This is the situation where you have V \subset H \subset V^*, where V is a Banach space continuously and densely embedded in a Hilbert space H. ... 0 Example from here works. In n\ge 2 dimensions, the function$$ u(x) = \begin{cases} \log \log (1+1/|x|),\quad &|x|<1/(e-1) \\ 0,\quad &|x|\ge 1/(e-1) \end{cases} $$is unbounded, belongs to the Sobolev class W^{1,n} (hence W^{1,2}), and vanishes in a neighborhood of the boundary of the unit ball B (hence is in W_0^{1,2}(B)). If you are ... 3 Actually you can build an example out of almost any function: Just notice that if u: B(0,1)\to \mathbb{R}, then u_r(x)=u(x/r) is defined in the ball of radius r and you get$$ \| 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)}. $$Translating appropriately you get an example in ... 1 Find functions u compactly supported inside B(x,r) so that \int_{B(x,r)} |u|^p \, dx / \int_{B(x,r)} |\nabla u|^p \, dx goes to infinity as r \to \infty. Probably something like$$ 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} $$2 We have that (see Evans in regularity part) for each f\in L^2(\Omega), the problem$$ \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 ...

0

The key idea is that $U=u$ on the boundary where $u\in H^{1/2}(S^n)$ is given. Hence, the Poincare inequality works in some sense and $\|\nabla U\|_{L^2}$ will be an equivalent norm of $\|U\|_{H^1}$. To see how, firstly, we trivially have $\|\nabla U\|_{L^2}\leq \| U\|_{H^1}$. Now for the converse. Notice that $U-u\in H_0^1$ and hence by poincare we have ...

0

See Aubin's book Some Nonlinear Problems in Riemannian Geometry.

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