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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 ...


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

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


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 $[0,1]^d$, ...


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, ...


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 ...


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

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

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.


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

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

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

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 ...


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 ...


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 ...


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} ...


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

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

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


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 $L^p$ ...


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 $T(q) = (f_1(v), \dots, f_N(v))$. ...


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.


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

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 ...


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 $$ ...


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

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 ...



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