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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 all $r\in(0,1/2)$ 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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