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1

Definitely, it is!  Let $\Omega\subset\mathbb{R}^n$ be a bounded domain, $n\geqslant 1$, and $u\in W^{1,1}(\Omega)$. Function $\,u\,$ is additionally assumed to be essentially bounded on $\Omega$, so let $\,M=\|u\|_{L^{\infty}(\Omega)}\overset{\rm def}{=}\underset{x\in \Omega}{\rm ess\,sup\,}|u(x)|$.   Choose some cut-off function $\,\eta\in ...


0

It depends on your norm on $W^{m,p}(\Omega)$. You could choose \begin{equation*} \| u \|_{m,p}^2 = \sum_\alpha \|D^\alpha u \|_p^2, \end{equation*} or \begin{equation*} \| u \|_{m,p}^p = \sum_\alpha \|D^\alpha u \|_p^p. \end{equation*} Then, you similarly choose \begin{equation*} \| u \|_{p}^2 = \sum_{i=1}^N \|u_i \|_p^2, \end{equation*} or ...


1

Note that $$f_x = \frac{(1 + x^2) \frac 1 2 x^{-1/2} - x^{1/2} \cdot 2x}{(1 + x^2)^2} = \frac{\frac 1 2 x^{-1/2} - \frac 3 2 x^{3/2}}{(1 + x^2)^2}$$ For large values of $x$, we have the estimate (where $\lesssim$ means "up to some constants") $$|f_x(x)| \lesssim \frac{x^{3/2}}{x^4} = x^{-5/2}$$ For small values of $x$, we have the similar estimate ...


0

First solution: Let $S=\{u\in H_0^1(\Omega):\ J(u)=1\}$ where $J:H_0^1(\Omega)\to\mathbb{R}$ is defined by $$J(u)=\int_\Omega |u|^2=\|u\|_2^2$$ Define $I:S\to\mathbb{R}$ by $$I(u)=\int_\Omega |\nabla u|^2=\|u\|_{1,2}^2$$ i - $\lambda_1=\min_{u\in S} I(u)$ ii - If $u\in H_0^1(\Omega)$ satisfies $I(u)=\lambda _1$ then, $u$ is a eigenfunction associated with ...


1

It's closed for domains in any dimensions, in particular one does not need an embedding into $C^0$. Here is a stronger statement. For any domain $\Omega\subset \mathbb R^n$, any $p\in [1,\infty]$, and any $a,b\in \mathbb R$, the set $$F=\{u\in L^p(\Omega):a\le u \le b \text{ a.e.}\}$$ is closed in the norm topology of $L^p$. Proof. We have $u\in F$ if ...


2

The answer uses a argument of compacity, see this answer here and the comments therein. First note that by Holder inequality $$\left(\int_0^1 u\right)^2\leq \int_0^1 u^2,\ \forall u\in L^2,$$ hence, $a(u,u)\ge \|u\|_{1,2}^2-\int_0^1 u^2$ for all $u\in V$. We conclude that it is sufficient to find a positive constant $C$ such that $$\tag{1} ...


3

Let $\Omega$ be the disk of radius $r<1$ centered at the origin. Define $u(x,y) = (-\log(x^2+y^2))^{1/3}$. Clearly, this is not a continuous function. But on every line not passing through the origin, this function is Lipschitz, therefore absolutely continuous. It remains to check that its partial derivatives are in $L^2$. By the chain rule, ...


1

This is true for $p\geqslant 1$. The case $p=1$, being a limiting one, requires more efforts. It will be sufficient to establish the rule $$ \partial_j(uv)=u\partial_j v+v\partial_j u,\quad j=,\dots,n,\tag{1} $$ for the weak derivatives $\partial_j$ just in case $p=1$, with the equality $(1)$ understood in a weak sense, i.e., as an integral identity $$ ...


1

This is false; the functional is bounded from below. Begin with $h\equiv 0$ for simplicity: the functional $$ \varphi(u) = \int_{0}^{1} \left(\frac{(u')^2}{2} - \displaystyle\frac{{u }^2}{4}\right)\,dt $$ is never negative. Indeed, the 2nd form of Wirtinger's inequality says that $$\int_0^1 u^2 \le \frac{1}{\pi^2}\int_0^1 (u')^2$$ hence $$ \varphi(u) \ge ...


0

I think you made a mistake in the last integral. Substituting $u = -\ln(r)$, we get \begin{equation*} \int_0^{1/2} \frac1r \, (-\ln(r))^{2\,\alpha - 2} \, \mathrm{d}r = \int_\infty^{-\ln(1/2)} u^{2\,\alpha - 2} \, \mathrm{d}u = c \, [ u^{2 \, \alpha - 1} ]_\infty^{-\ln(1/2)}. \end{equation*} Now, for $0 < \alpha < 1/2$, this is finite, since ...


1

For the unit circle $U=\{x\in\mathbb{R}^2\,\colon\, |x|<1\}$, a routine example is something like that $$ f(x)= \begin{cases} \ln{\ln{\frac{1}{|x|}}},\quad |x|<e^{-2},\\ \ln{2}, \quad |x|\geqslant e^{-2}. \end{cases} $$ To show that $f\in L^2(U)$, take a remarkable limit $$ \lim_{|x|\to 0}|x|^{\mu}\!\!\cdot\!\ln{\frac{1}{|x|}}=0\quad ...


0

As @John pointed out in the comment, your apporach may not work, because $u'_n$ does not need to be dounded in $C^0$. One way to approach this problem is the following. We will use Theorem 4.26. from Brezis book. To this end, extend all functions $u_n$ to $\mathbb{R}$ by using Thereom 8.6. of the same book and note that the extended sequence still bounded. ...


3

I post an answer since it needs 50rep to add a comment. See Intrinsic weak derivatives and Sobolev spaces between manifolds by Alexandra Convent and Jean Van Schaftingen Gromov's compactness theorem for pseudo holomorphic curves by Rugang Ye


1

For bounded domains $\Omega\in\mathbb{R}^n$, $n\geqslant 2$, satisfying the cone condition, the spaces coincide, i.e., $D^{k,p}(\Omega)=W^{k,p}(\Omega)$.  Examples of nonsmooth bounded domains $\Omega$ for which $D^{1,p}(\Omega)\neq W^{1,p}(\Omega)$, i.e., $D^{1,p}(\Omega)\not\subset L^p(\Omega)$, can be found in "Sobolev spaces" ...


1

When $p\neq 2$ the Fourier analysis on the Schwartz space $\mathcal{S}'(\mathbb{R}^n)$ works quite well despite the fact that the Fourier transform on $L^p(\mathbb{R}^n)$ is not an isomorphism. Read about Bessel potentials, e.g., section 7.63 in the textbook Sobolev Spaces http://bookza.org/book/492535/11bd42   by R. Adams and J. Fournier. The ...


0

What your professor meant to say takes just one line which looks like that $$ \|g\|_{L^p}^p=\int\limits_{\Omega}|h(z)-\overline{h}|\,dz\leqslant C\!\!\int\limits_{\Omega}|\nabla h(z)|\,dz\leqslant Cp\!\!\int\limits_{\Omega}|\nabla g(z)|^{p-1}|\nabla g(z)|\,dz\leqslant \|g\|_{L^p}^{p-1}\|\nabla g\|_{L^p} $$ where the Hölder's inequality is applied with ...


0

The dual space $(H_0^1)^*$ is per definition the space of all linear and continuous functionals $(H_0^1)^*\to \mathbb R$. There are different ways in generating this dual space: (1) Riesz representation lets you represent each functional $f\in (H_0^1)^*$ by scalar product with an element $u_f\in (H_0^1)$ $$ f(v) = \langle u_f,v\rangle \quad\forall v\in ...


0

As you have said, the dual space of $H_0^1$ is by definition $H^{-1}$. It can be identified with $H_0^1$ via isomorphism, but whether the identification is useful or not, it depends on the problem you are studying. If, for example, you are studying abstract problems about Hilbert spaces and it's duals then, the identification is recommended and you gain a ...


0

By convolution with a smooth kernel. I sketch the construction in Theorem 15.21 in A first course in Sobolev spaces by Leoni. Choose a smooth function $\varphi$ on $\mathbb R$ so that the support of $\varphi$ is contained in $(-1,1)$ and the integral of $\varphi$ is $1$. For $y>0$, define $$ g(x,y) =e^{-y/p} \frac{1}{y} \int_{\mathbb R} ...


0

Fix a positive $\eta$ and consider the corresponding $\delta$ in the definition of equi-integrability. By Egoroff's theorem, there exists some $A$ such that $$\lim_{n\to\infty}\sup_{x\in A}|f_n(x)-f(x)|=0\quad\mbox{and}\quad \lambda(\Omega\setminus A)\lt \delta.$$ We thus obtain, using integrability of $f$, $$\int_\Omega|f_n-f|dx\leqslant ...


2

$H^{-s}$ is by definition the dual of $H^s$. I suppose that $\int f\,g\,dx$ must be understood in the sense of duality. Elements of $H^{-s}$ are not functions, but distributions, so that $f\,g$ is not defined in general. If for instance both $f$ and $g$ are in the Schwartz class, then $$ \Bigl|\int f\,g\,dx\Bigr|=\Bigl|\int\hat f\,\hat ...


1

Consider the case $\kappa = 0$, $1 < s < \lambda \leqslant 1$, the general one can be easily deduced from this one. Firstly, because $h$ is bounded in $\overline{\Omega}$ we have trivially: $\| hu \|_p \leqslant \| h \|_{\infty} \| u \|_p$. Secondly, for the Gagliardo seminorm $$ [ hu ]_{s, p}^p := \underset{\Omega \times \Omega}{\int \int} \frac{| ...


1

EDIT: I created a document in English with second-order elements: http://homepage.cem.itesm.mx/jose.luis.gomez/fem/MEF00450secondOrder.pdf (END EDIT) "...the motivation behind the ansatz equation..." is that you are looking for a linear combination of some base functions that will be the best approximation to the unknwon solution. Those base functions are ...


0

The statement is true and the bound on $u$ is not needed. The following argument are standard in deriving these "Poincare type" inequality. First of all, we assume that $\Omega\setminus D$ is of positive measure (If it is measure zero, the inequality is obviously false by considering constant function). Also $c_1 >0$ or the inequality is trivial. So we ...


0

We have that $u_n \rightharpoonup u$ in $L^2(0,T;H^{-1}(\Omega))$ if and only if $$\int_0^T \langle u_n(t), w(t) \rangle_{H^{-1}(\Omega), H^1(\Omega)} \to \int_0^T \langle u(t), w(t) \rangle_{H^{-1}(\Omega), H^1(\Omega)}$$ for all $w \in L^2(0,T;H^1(\Omega))$. Notice that the dual space of $L^2(0,T;H^1(\Omega))$ is $L^2(0,T;H^{-1}(\Omega)).$


1

Yes, if you know that $f$ and $u'$ belong to $L^2\big([0,T]; L^2(\Omega)\big)$, and $Au = f-u'$, then $ Au\in L^2\big([0,T]; L^2(\Omega)\big)$. After all, $ L^2\big([0,T]; L^2(\Omega)\big) $ is a linear space. If you are unsure because the ODE holds in some weak sense (e.g., as distributions), consider that the difference of two distributions that are ...


2

Due to the embeddings $W^{2,2}(\Omega)\hookrightarrow W^{1,p}(\Omega)$ and $W^{1,2}(\Omega)\hookrightarrow L^p(\Omega)$ with any $p\in (2,\infty)$, by Hölder's inequality it readily follows that the product $f\cdot g\cdot h\in W^{1,q}(\Omega)$ with any $q\in [1,2)$. In terms of the Sobolev space $W^{1,q}(\Omega)$, the regularity of the product cannot be ...


0

The product is not in $H^1$: (1) $fgh\in L^p(\Omega)$ for all $p<\infty$ due to the Sobolev embeddings into the $L^q$ spaces (2) the limiting terms are the products of type $fg\nabla h$, $f\nabla gh$, which are in $L^p(\Omega)$ for all $p<2$ ($f\in L^\infty(\Omega)$, $\nabla g\in L^2(\Omega)$, $h\in L^q(\Omega)$ $\forall q<\infty$ implies that the ...


1

$$ |f_n - f_{n+p}|_{L^\infty} \le |f_n - f_{n+p}|_{W^{1,1}} $$because of the inequality you have shown, hence the sequence $(f_n) $ is a Cauchy sequence in the space $L^\infty$. The uniqueness of the limit for both norms (maybe because of the convergence in distribution, which is weaker than both convergences, but I am not sure of this point) ensures that ...


1

It means that the weak derivative of $u$, $u'$ is an element of $L^2(0,T;H^{-1}(\Omega))$, and by definition it satisfies for all scalar test functions $\phi$ the identity $$\int_0^T \phi'(t) u(t) = -\int_0^T u'(t)\phi(t)$$ where the right hand side is an element of $H^{-1}(\Omega)$ (we have integrated out the time). The equality makes sense since the left ...


0

(1) You should prove the identity for smooth functions. (2) Prove that $\frac{\partial}{\partial x}:H^1_0(\Omega)\to L^2(\Omega)$ and $D_h:L^2(\Omega)\to L^2(\Omega)$ are continuous. (3) Approximate $v\in H^1_0(\Omega)$ by $v_\epsilon\in C_0^\infty(\Omega)$ with $\|v-v_\epsilon\|_{H^1(\Omega)}\le \epsilon$. (4) Conclusion: Use (1)-(3) to prove the claim ...


1

You've shown that $D_h$ commutes with $\dfrac{\partial}{\partial x}$ when applied to smooth functions. Now use the definition of weak derivative. If $v \in H^1(\Omega)$ then \begin{align*} \int_\Omega D_h v \frac{\partial \psi}{\partial x} \, dx dy &= - \int_\Omega v D_{-h} \frac{\partial \psi}{\partial x} \, dxdy \\ &= - \int v \frac{\partial ...


1

Define $T:C_0^2(\overline{\mathbb{R}^2_+})\to C_0^1(\mathbb{R})$ by $$(Tu)(x)=u(x,0),\ \forall\ x\in \mathbb{R}$$ By using the calculations in Theorem 4.1. here, we have that $$\|Tu\|_{H^1}\le C\|u\|_{H_2},\ \forall\ u\in C_0^2(\overline{\mathbb{R}^2_+}) \tag{1}$$ Moreover, note that $$\frac{d}{dx}(Tu)(x)=T\left(\frac{\partial }{\partial x}u\right)(x)\ ...


0

I think that I should use $f_n$ such that $$f_n (x)= \left\{\begin{array}\ n, \ f(x)>n \\ f, \ |f(x)|\le n \\ -n, \ f(x)<-n \end{array}\right.$$ So we have that \begin{equation} \nabla f_n (x)= \left\{\begin{array}\ 0, \ f(x)>n \\ \nabla f, \ |f(x)|\le n \\ 0, \ f(x)<-n \end{array}\right. \end{equation} And $f_n\in W^{1,1}(B), \ f_n ...


1

In the (most interesting) case $p < n$, you can argue as follows. For arbitrary $x \in \Omega$, you have $\mathrm{cap}_p(U_r(x)) \to 0$ as $r \to 0$, where $U_r(x)$ is the open ball with center $x$ and radius $r$. Fix a closed set $K \subset \Omega$ and set $k = \mathrm{cap}_p(K)$. Now, fix a countable set $M = \{x_1, x_2, \ldots\} \subset \Omega$ with ...


3

No, that formula won't work. But you can use a spectral definition. There is an orthonormal basis $\{\phi_i\}_{i=1}^\infty$ for $L^2(M)$ consisting of eigenfunctions of $-\Delta$, with corresponding eigenvalues $\{\lambda_i\}$. Then for $u = \sum_i u_i \phi_i$, put $$(-\Delta)^s u(x) = \sum_{i=1}^\infty (\lambda_i)^s u_i \phi_i(x).$$


0

Suppose that the inequality doesn't hold, so for all $n\in\mathbb{N}$ there exists a $u_n\in H$ such that $$ \|u_n\|_{H^1}>n\sum_{|\alpha|=2}\|D^\alpha u_n\|_{L^2}. $$ Further, normalizing, we can assume that $\|u_n\|_{H^1}=1$ for all $n\in\mathbb{N}$. So, $$ \frac{1}{n}>\sum_{|\alpha|=2}\|D^\alpha u_n\|_{L^2} $$ for all $n\in\mathbb{N}$. You should be ...


2

It's a lemma of J.-L. Lions, see An application of J.-L. Lion's Lemma where a book reference is given (Brezis, Functional Analysis...). You don't get quantitative control of $C$ from the compactness argument by which the lemma is usually proved. Instead, we can use the Sobolev embedding with the Peter-Paul trick as here. Below I write $\|u\|_p$ ...


1

Combine the following facts: If $u$ is harmonic, then the partial derivatives of $u$ are harmonic. If $u$ is harmonic and $\phi:\mathbb R\to\mathbb R$ is convex, then $\phi\circ u$ is subharmonic. The function $\phi(t)=t^2$ is convex. $|\nabla u|^2$ is the sum of squares of partial derivatives of $u$. Any book that deals with subharmonic functions ...


0

If $d=1$, then $X=H^1(\mathbb R)$ is complete. Below I assume $d\ge 2$. Under the $H^1$ norm, $X$ is not complete. Indeed, let $v$ be an $H^1$ function with $v(0) =1$, vanishing outside the unit ball. Define $$u(x) = \sum_{n=1}^\infty 2^{-n}v(x-3^n\xi)$$ where $\xi$ is a fixed unit vector. The series converges in $H^1$, and every partial sum is in $X$, ...


0

Yes. For any bounded function $f$ the multiplication operator $T(g)=e^fg$ is an isomorphism of $L^p$ onto $L^p$, with its inverse being $T^{-1}(g)=e^{-f}g$. Since $L_2$ is the composition of two isomorphisms, it is an isomorphism.


3

J. Nečas was not the first to establish this inequality in the special case $k=1$.  Five years earlier, Lamberto Cattabriga, assisted by Giovanni Prodi, proved an equivalent inequality $$ \|u\|_{0,p}\leqslant C\Bigl(\Bigl|\int\limits_{\Omega}u\,dx\Bigr| +\sum\limits_{j=1}^d\|\partial_j u\|_{-1,p}\Bigr) \tag{$\ast$} $$ for a bounded smooth ...


2

A distribution $\mu$ is said to be in $L_2$ if there exists a function $f \in L_2$ so that for all test functions $\phi$ we have $\mu(\phi) = \int_{\mathbb R} f(x) \phi(x) \, dx$.


2

The presence of $\chi_E$ is confusing. Here's a simpler version: If $f_n$ converge to something in $L^2$ and $f_n\to f$ in the sense of distributions, then $f_n\to f$ in $L^2$. This sort of argument (with different kinds of convergence) is pretty standard. The proof is what you wrote: let $\tilde f$ be the $L^2$ limit; then it's also a distributional ...


2

Both historically and statistically, the one and only correct name for the inequality in question $$ \int\limits_{\Omega}\!|u(x)|^2dx\leqslant C\!\int\limits_{\Omega}\!|\nabla u(x)|^2dx \quad \forall\,u\in H_0^1(\Omega)\tag{$\ast$} $$ is to be the Friedrichs inequality. Whenever the Sobolev space $H_0^1(\Omega)$ is defined as a closure of the subspace ...


2

Let $g:\mathbb{R}\to \mathbb{R}$ be a $L^\infty$ function and define $G(x)=\int_0^x g(t)dt$. Consider the functional $$I(u)=\frac{1}{2}\|\nabla u\|_2^2+\lambda\int_\Omega G(u)-\int_\Omega fu,\ \forall u\in H_0^1(\Omega)$$ I will assume here that you know, that the first term in he above sum is weakly sequentially lower semi continuous (w.s.l.s.c.). Let's ...



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