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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 ...


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 $$(-\Delta)^s u(x) = \sum_{i=1}^\infty (\lambda_i)^s u_i \phi_i(x).$$


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


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, ...


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} ...


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

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 ...


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 ...


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

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 ...


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 ...


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 ...


1

Such an estimate can be found in Grisvard "Elliptic problems in nonsmooth domains", Theorem 1.5.1.10. It basically says $$ \delta \|u\|_{L^2(\partial\Omega)}^2 \le \|\mu\|_{C^1(\bar\Omega)} \left(\epsilon^{1/2}\|\nabla u\|_{L^2(\Omega)}^2 + (1+\epsilon^{-1/2}) \|u\|_{L^2(\Omega)}^2 \right) $$ for all $\epsilon\in(0,1)$, $u\in H^1(\Omega)$. The vector ...


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 ...


1

Yes, this statement ist true since $H^1(\Omega)$ is a subspace of $L^2(\Omega)$. Moreover, both conditions, i.e., $f=g$ a.e. in $\Omega$ and $f=g$ in $L^2(\Omega)$, are equivalent by definition, since $L^2(\Omega)$ is constructed as a quotient space by the subspace of functions with $f=0$ a.e.


1

The form $a$ is bilinear because the integral is linear. The continuity is a little more complicated: $$a(u,v)=\int_{0}^1 u'v'+\int_{0}^1 uv\leq \left(\int_{0}^1(u')^2 \right)^{1/2}\left(\int_{0}^1(v')^2 \right)^{1/2}+\left(\int_{0}^1u^2 \right)^{1/2}\left(\int_{0}^1v^2 \right)^{1/2}$$ $$\leq \left(\int_{0}^1u^2 ...


1

Yes, the Poincaré is the place to begin. Don't use up all of the good stuff, though. Stash an $\epsilon $ of it away: $$\int_\Omega |\nabla u|^2 \ge \epsilon \int_\Omega |\nabla u|^2 + (1-\epsilon)\lambda_1 \int_\Omega u^2$$ where $\epsilon $ is small enough so that $(1-\epsilon)\lambda_1\ge \lambda$. Then estimate $$1-\cos u = 2\sin^2 ...


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

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 ...


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

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 ...


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 ...


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)\ ...


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

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 ...


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

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 ...


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 ...


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 ...



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