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0

The proof that you are looking for can be found in the page 258 of the Evans book:


1

Because $W_0^{1,p}$ is compactly embedded in $L_p$, the embedding operator is obviously compact so it maps weakly convergent sequences in strongly convergent, i.e a weakly convergent subsequence in $W_0^{1,p}$ is strongly convergent in $L_p$. Finally, you can extract a further subsequence, which is pointwise a.e convergent: see Does convergence in $L^{p}$ ...


1

1) also can be justified by that $H^1$ is a reflexive Banach space, so from the bounded sequence you can extract a weakly convergent subsequence with a limit in $H^1(\Omega)$ and because the embedding operator is compact that means that it maps weakly convergent sequence to strongly convergent, i.e $u(.,t_{k_l})\rightarrow w$ strongly in $L_2$ (here, we ...


0

I think Nate's response answers your question; I'm just writing a bit of an explanation of what is going on in the proof. We are looking for a polynomial $q$ which approximates $v$ well, up to the higher order derivatives (which we can estimate using $|v|_{k+1,p}$). Let's think about this abstractly for a moment. We have some Banach space $B$ with a ...


1

In the proof, we have $f_1, \dots, f_N$ chosen to be a basis for the dual $P_k^*$, where $N = \dim P_k$. This means that the map $T : P_k \to \mathbb{R}^N$ defined by $T(p) = (f_1(p), \dots, f_N(p))$ is an isomorphism. In particular, it is surjective. So given $v \in W^{k+1, p}(\Omega)$ there exists $q \in P_k$ such that $T(q) = (f_1(v), \dots, f_N(v))$. ...


2

If you note $\|\cdot \|_\infty$ the sup ess norm, you have : $$\| u - \tilde{u} \|_{\infty} \leq \| u - u_k\|_{\infty}+ \| u_k - \tilde{u} \|_{\infty}$$ Now let $\epsilon > 0$, As $u_k \to \tilde{u}$ uniformly, there exist $k_1$ such that $\forall k > k_1, \ \| \tilde{u} - u_k\|_{\infty} < \frac{\epsilon}{2}$ As $u_k \to u$ in $W^{1,2}_0$, it ...


1

This is a classical result. You could find the answer on this book,page 176, section 5.2.3, theorem 4


2

It seems that what you're looking for is an interpolation space : https://en.wikipedia.org/wiki/Interpolation_space I don't remember if these interpolation spaces between $BV$ and $L^1$ are "usual" spaces or not


1

This is always true, as $\Omega \setminus (\Omega_1\cup \Omega_2)$ can only contain the boundaries of $\Omega_1$ and $\Omega_2$, which have zero measure due to the regularity assumptions on the domains.


0

Weak convergence in $H=H^1_0$ implies strong convergence in $L^2$ because $H^1_0$ is compactly embedded in $L^2$, see: https://en.wikipedia.org/wiki/Sobolev_inequality#Sobolev_embedding_theorem


2

One approach is via the Newtonian potential. You'll need the following ingredients. Harmonic functions are $C^\infty$ smooth. Let $v = \Gamma*(\Delta u)$; this is a function such that $\Delta v=\Delta u$ Since $\nabla \Gamma$ is locally integrable and $\Delta u$ is bounded, we can differentiate under the integral sign: $\nabla (\Gamma*(\Delta u)) = ...


2

Do you know the regularity of Elliptic equation? The answer of your question is not just a Sobolev space problem, it is an Elliptic PDE problem. The answer is by the regularity of Laplace equation, you can boost up the regularity of solution based on the smoothness of your boundary. Check this book, Section 6.3.1 Theorem 1 and Theorem 2. In those theorems, ...


1

Yes, it is true. This is actually an general idea for space involving several order of derivatives: "the extreme terms in a sum often already suffice to control the intermediate terms". Notice that by extreme we mean both highest order and the lowest order. For example, $W^{3,p}$ norm of $u$ can be controlled by using only $L^p$ norm of $u$ and the $L^p$ ...


0

Please do not use the same symbol for your function and test functions. You are confusing yourself. We assume $u\in H$ and $(u_n)\subset C_c^1$ such that $u_n\to u$ in $H$, which in particular means that $u_n\to u$ in $L^2$ and $\partial_i u_n\to v_i$ in $L^2$ for each $i=1,\ldots, N$. Then we observe that, for any $\varphi\in C_c^1$, $$ \int_\Omega ...


1

Yes, a very simple scaling argument shows $$TV(u,\alpha) = \alpha \, TV(u,1).$$ Hence, $BV(\Omega, \alpha) = BV(\Omega)$ for all $\alpha > 0$.


1

Shouldn't also the Sobolev norm exist (be finite)? Yes, but this condition is automatically satisfied because of the way that the space was defined. In fact, by definition $$L_2(\Omega)=\left \{f:\Omega\to\mathbb{R};\;f\text{ is measurable and }\int_\Omega |f(x)|^2 \ dx<\infty\right\}.$$ So, "$u \in L_2(\Omega)$" implies $$\int_\Omega |u(x)|^2 \ ...


-1

You're looking at the square of the norm. In the last line of example 4.17, note the square root of the sum of squares. It's natural to see the squares of the eigenvalues in the square of the norm. Both the norm and the square are going to be finite, and can be used to define $H^s$, but the square isn't going to be homogeneous (like you're expecting). ...


1

Seems to me that the definition given for $H^s$ can't be right for $s<0$; it "must" be that $H^s$ is actually the space of tempered distributions $f$ such that $\hat f\in L^2(\mu)$, where $d\mu(\xi)=(1+|\xi|^2)^{s/2}\,d\xi$. Assuming so, this is easy: Say $(f_n)$ is Cauchy in $H^s$. Then $(\hat f_n)$ is Cauchy in $L^2(\mu)$. So $\hat f_n\to g$ in ...


0

Just look at the definition: since $\eta(x)=0$ for all $x \in \mathbb{R}^N$ with $|x| \geq 1$, it follows that $\operatorname{supp}\eta \subset \overline{B(0,1)}$. Hence $\eta \in C^\infty_c(\mathbb{R}^N)$.


0

@Yongyong's answer is very good indeed. Here I just want to point out the idea behind this remark is that $W^{1,1}$ is not a reflexive space. (Note $W^{1,p}$ is reflexive for any $1<p<\infty$). Hence, when we try to use weak compactness to extract a weak subsequence, the limiting function may not lie in $W^1{1,1}$, and hence the theorem fails. If you ...


0

Consider the function \begin{align} &f(x)=-1,x\in[-1,0];\\ &f(x)=1,x\in(0,1]. \end{align} Now for sufficiently small $h>0$, and some compact set which is a subset of $[-\alpha,\alpha]$ for some $\alpha\in(0,1)$, we have: for $x\in[-\alpha,-h]\cup(0,\alpha]$, \begin{equation} \frac{1}{h}|u(x+h)-u(x)|=0; \end{equation} for $x\in(-h,0)$, ...


1

By density it is enough to consider smooth functions. The arguments work for general Sobolev functions as well, but there is less to worry about with smooth functions. Consider first $i_{k,l}$ for $0\leq l\leq k$. Take a function $u\in C^\infty$. Its squared norm is $$ \|u\|_{H^k}^2=\sum_{|\alpha|\leq k}\|\partial^\alpha u\|_{L^2}. $$ Now $$ ...


2

The question first is whether $u_n^+$ is in $W_0^{1,p}$ for all $n$ if $u_n\in W_0^{1,p}$. Since $u\in W_0^{1,p}$, there exists a sequence $\phi_m^n\in C_0^\infty$ converges to $u_n$ in $W_0^{1,p}$. Then $\phi_m^{n,+}$ converges to $u_n^+$ in $L^p$. It is also not difficult to prove that $\phi_m^{n,+}$ is Lipschitz continuous, then it has a.e. derivative. ...


2

As PhoemueX already said, you need a function, which has zero boundary values, but the derivative is non-zero at the boundary. In particular, you can take $$ u(x) = \prod_{i=1}^d x_i \, (1-x_i). $$ Then, it is easy to see that $u(x) = 0$ for $x \in \partial[(0,1)^d]$, but $\nabla u(x) \ne 0$ for $x \in \partial[(0,1)^d]$. Since $u$ is smooth on $[0,1]^d$, ...


1

Let $u$ be a vector field in $L^2(\Omega)^n$. Then a function $q\in L^1_{loc}(\Omega)$ is the divergence (in the distributional sense) of $u$ if it satisfies $$ \int_\Omega u\cdot \nabla \phi = -\int_\Omega q \phi $$ for all functions $\phi\in C_0^\infty(\Omega)$. This is written $q = \textrm{div} u$. Thus, the definition is similar to the definition of a ...


0

In (3) and (4), the absolute value sign is disregarded. Since the function $(1-|x_2|)\Phi(\cdot,x_2)$ is an even function w.r.t. $x_2$, hence we have $$-\int_{-1}^1(1-|x_2|)\Phi(x_2,x_2)dx_2$$ and $$\int_{-1}^1(1-|x_2|)\Phi(-x_2,x_2)dx_2.$$ rather than $$-\int_{-1}^1(1-|x_2|)\Phi(|x_2|,x_2)dx_2$$ and $$\int_{-1}^1(1-|x_2|)\Phi(-|x_2|,x_2)dx_2.$$


0

It appears that the answer to the question is indeed negative. Here is a counterexample. Let $\Omega = I = (0,1)$. Define $V:=H_0^1(\Omega)$, $H:=L^2(\Omega)$ with the scalar products $(u,v)_V = \int_\Omega \nabla u\cdot \nabla v $ and $(u,v)_H=\int_\Omega u v $. The example exploits the fact that the mapping $u\mapsto u^+$ is unbounded from $V^*$ to ...


2

Yes, and here's why: A $W^{1,p}_0$ functions extends (by $0$), to a $W^{1,p}$ function on a larger ball. The Morrey oscillation inequality gives Hölder continuity with a bound depending only on the $W^{1,p}$ norm. By the Arzelà-Ascoli theorem, the unit ball of $W^{1,p}_0$ is precompact in the $L^\infty$ norm.


2

The claim you wish to conclude is false. Pick $u\in C_0^\infty(\mathbb R^n)$ so that $u(0)=0$ but $\nabla u(0)\neq0$. If $p\geq1$ is not a natural number, then there is $s>0$ so that $u^p\notin H^s(\mathbb R^n)$ (because classical derivatives fail to exist to orders above $p$) although $u\in H^s(\mathbb R^n)$ for all $s$. This works in any dimension, ...


1

On Evans and Gariepy's book, page 130, Theorem 4 (iv), is the result you want. Remember "a.e." is the key element here.


0

I'm just going to put this up here for my own reference. If someone thinks it's not correct please explain. Thanks. I think he means div $u$ as a derivative in the distributional sense. That is we define $\left\langle \operatorname{div} \nabla p, \phi\right\rangle := -\left \langle\nabla p,\nabla \phi\right\rangle$ for all smooth compactly supported ...


0

The result is: If $u \in W^{1,p}(\Omega)$ on an open set $\Omega$, and if $A \subset \mathbb{R}$ is a null set, then $\nabla u = 0$ a.e. on $u^{-1}(A)$. The result is due to Stampacchia. If anyone can refer me to an English proof I'd be grateful. My source is Nonlinear Analysis by Gasinski and Papageorgiou (Remark 2.4.26).


0

In the first part, you haven't done anything wrong. Your answer, $$(p-2)|Du|^{p-3}\frac{Du}{|Du|}\frac{\partial}{\partial x_i}(Du)$$ is the same as $$(p-2)|Du|^{p-4}\sum_j\frac{\partial^2 u}{\partial x_i\partial x_j}\frac{\partial u}{\partial x_j}$$ (I'll remark that often people differentiate $|F|^p$ by writing it as $(|F|^2)^{p/2}$, so that the derivative ...


0

The bilinear form \begin{align} B:H^1_0(a,b)\times H_0^1(a,b)&\longrightarrow \mathbb{R}\\ (w,v)&\longmapsto\int_a^b w_x(x)v_x(x)\ dx \end{align} is continuous and coercive. Furthermore, the linear functional \begin{align} \Lambda: H_0^1(a,b)&\longrightarrow \mathbb{R}\\ v &\longmapsto-\int_a^b f(x)v(x)\ dx \end{align} is continuous. Thus, by ...


0

No, $|D^2u|$ is not the same as $\frac{\nabla u}{|\nabla u|}\, D^2 u$. One of them is a number and the other is a vector. One involves first order derivative and the other doesn't. The argument you read is flawed: the gradient is not the sum of derivatives, nor can it be estimated by this sum. Instead, use the fact that the square of the gradient is the ...


1

It appears that you are using the notation $\langle \cdot, \cdot \rangle_{\mathbb{H}}$ for the inner product on you Hilbert space $W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ and $\langle \cdot,\cdot\rangle_{L^2}$ for the dual pairing with the space $W^{-2,2}_0(\Omega)$, which is the co-domain of the operator $L$. The term $\langle Lu,u\rangle_{\mathbb{H}}$ then ...


0

Notation: For an arbitrary linear functional $g:V\to\mathbb{R}$, $\langle g, v\rangle$ is the value of $g$ at $v\in V$. The meaning of $$u_n\rightharpoonup u\quad\text{in}\quad L^2(0,T;H^{-1}(\Omega))\tag{A}$$ is $$\int_0^T\langle u_n(s),w(s)\rangle\ ds\to\int_0^T\langle u(s),w(s)\rangle\ ds\quad\text{in}\quad \mathbb{R}, \quad \forall\ w\in ...


0

There is a disproof of this statement in general. Take $ L^2([-1,1],\mathbb{C})$, and $a_n(x)= \exp(2\pi i n x)$ and $b_n (x) = \exp(-2\pi i n x)$. In this case, both $a_n \rightharpoonup a = 0$ and $b_n\rightharpoonup b=0$, but $a_n (x)\cdot b_n(x) =1$ for all $x$. Clearly $c_n=c =1$. Then $$\lim\limits_{n \to \infty} \int_{-1}^1 a_n(x)\cdot b_n(x)\,dx=2$$ ...


0

You first have to check that $u$ has a weak derivative. You have an obvious candidate for the weak gradient (just take the gradient separately in the four pieces), and you just have to check that it works. You might also have a theorem that saves you from calculations at this point, having a weak derivative should still be checked somehow. Then you have ...


1

For any $s>1/2$ the norm of $H^s$ controls the $L^\infty$ norm. Indeed, when the series of Fourier coefficients is absolutely convergent, their sum bounds $\sup|u|$. By the Cauchy-Schwarz inequality, $$ \|u\|_{L^\infty}\le \sum_{n \in \mathbb{Z}} |\hat u_n| \le C(s,P) \left( \sum_{n \in \mathbb{Z}} \bigg(1 + \frac{4 \pi^2 n^2}{P^2}\bigg)^{s} ...


0

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 ...


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 $L^2(\mathbb{R})$.


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 $C^{\infty}$ ...


1

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. ...


0

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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