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2

There is no such thing as the extension operator. The validity of norm inequality for $P$ may depend on what extension operator is used. For example, one can extend Sobolev functions on the unit ball to Sobolev functions on $\mathbb R^n$ so that the extension vanishes for $|x|>2$. In such a case, the embedding results for bounded domains still apply. ...

2

Take the your case $1 \leq q < p^{*}$ where $u \in W^{1,p}(U)$. Let $Pu := v$ where $P$ is an extension operator $P: W^{1,p}(U) \rightarrow W^{1,p}(\mathbb{R}^{n})$. Consider $v \in W^{1,p}(\mathbb{R}^{n})$ where $\text{spt}(v) \subset V$, it follows then that $v \in W^{1,p}(V)$. By Gagliardo-Nirenberg-Sobolev we have: $||v||_{L^{p^{*}}(\mathbb{R}^{n})} ... 2 Let$f:\mathbb{C}\to\mathbb{C}$be the function$f(x,y) = (x+iy) (x^2 + y^2)$. Its derivative is $$\mathrm{d}f = (x^2 + y^2) \mathrm{d}(x + i y) + 2(x + i y) (x \mathrm{d}x + y \mathrm{d}y)$$ which satisfies $$|\mathrm{d}f| \leq C(x^2 + y^2)$$ Integrating along the straight line joining$v,w\in \mathbb{C}$gives $$|f(w) - f(v) | \leq C ... 2 By Parseval, we have \hat f\in L^2 and |\cdot|^2 \hat f\in L^2, because the Laplacian becomes the multiplication by |\xi|^2 in the Fourier space. Since$$ 2|\xi_j\xi_k| \leq |\xi_j|^2+|\xi_k|^2 \leq |\xi|^2, $$and the mixed derivative \partial_j\partial_k becomes the multiplication by \xi_j\xi_k, we conclude that \partial_j\partial_k f\in L^2 ... 2 In the case n = 1 or (n \ge 2 and p> n), you have the embedding from W^{1,p} to L^\infty. In this case \overline{C_c^\infty}\subsetneq W^{1,p}(\mathbb R^n). Indeed, every limit point of \overline{C_c^\infty} is continuous and zero at 0. In the other case, you get \overline{C_c^\infty} = W^{1,p}(\mathbb R^n). This can be seen by taking ... 2 I don't know how much this will help, but: The simplest instance of a Holder norm controlled by a Sobolev norm is that of the Fundamental Theorem of Calculus. Let us suppose we have a function f:\mathbb{R}\to\mathbb{R} such that its distributional derivative f' is defined, and such that we have \int |f'|^p \mathrm{d}x < \infty. Then we can estimate ... 2 Assume that A^{-1}: Y\to L^2(\Omega) is bounded, i.e.$$\|A^{-1} f\|_2\leq c\|f\|_Y,\ \forall\ f\in Y\tag{1}$$where c>0 is a constant. We get from (1) and from your inequality that$$\|A^{-1} f\|_2\leq cM\|A^{-1}f\|_{H^{-\alpha}},\ \forall f\in Y$$or equivalently$$\|g\|_2\leq cM\|g\|_{H^{-\alpha}},\ \forall \ g\in L^2(\Omega)$$Do you know ... 2 There is a more general result that we can state: Let X,Y be Banach spaces and T:X\to Y a compact operator. Then, for each sequence x_n\in X such that x_n\to x weakly, we have that Tx_n\to Tx strongly. Proof: Because x_n converge weakly, we can assume that x_n is bounded. By definition of compactness, we have that ... 2 Consider \Omega = \{ x \in \mathbb R^2 : \frac12 < |x| < 1, x_1 > 0\}, and let u(x) = \frac x{|x|^2}. Then \nabla\cdot u = 0 and \nabla \times u = 0. The condition u \times n=0 holds on the circular parts of the boundary, and the condition u \cdot n=0 holds on the straight parts of the boundary. But clearly \|u\|_1 \ne 0. 2 We have$$|u(x',x_n)|\leqslant \int_0^{x_n}|\partial_n u(x',t)|\mathrm dt\leqslant \lVert \partial_n u(x',\cdot)\rVert_px_n^{1-1/p}.$$The integration with respect to x_n produces the constant \frac 1p. Indeed, we have$$|u(x',x_n)|^p\leqslant\lVert \partial_n u(x',\cdot)\rVert_p^p\int_0^Lx_n^{p-1}\mathrm dx_n,$$and the last integral is ... 2 No, it's not true. Let \Omega be the unit ball. Pick a function v\in H_0^1(\Omega) that is unbounded from above near 0, say v(x)=\max(0,\sqrt{-\log |x|}-1). Define u_n(x)=1-v(x)/n. The sequence converges to u\equiv 1 in the H^1 norm. The set \{u>2/3\} is all of \Omega, but the set \{u_n>1/3\} is only a spherical shell, for every ... 2 1 - First prove that for each subsequence b_k of a_k, there is a subsequence c_k of b_k such that$$|c_k-c_l|\geq \delta,\ \forall k,l$$for some \delta>0. 2 - Write$$I(1,k,l)=\int_{|(a_k-a_l)\cdot\xi|<1 } |1-e^{-2\pi i (a_k - a_l)\cdot \xi}|^2|\widehat{\phi}(\xi)|^2(1+|\xi|^2)^t d\xiI(2,k,l)=\int_{|(a_k-a_l)\cdot\xi|\geq1 } ... 2 Your reasoning is correct. If$f$cannot be redefined on a set of measure zero to become a continuous function, then$f$is not in$H^1(-1,1)$. Chances are that sooner or later you will encounter Sobolev functions of more than one variable. Those need not have a continuous representative. For example,$f(x,y)=\sqrt{-\log(x^2+y^2)}$is in$H^1(D)$where ... 1 You can get an$L^2$convergence result from basic energy methods. It seems that you're assuming that$f = f(x)$, i.e.$f$is not time-dependent. Let's go ahead and take the slightly more general$f \in H^{-1}$. I also assume that you mean$H_0^1(\Omega)$and$H^{-1} = (H_0^1)^*$. Produce a weak solution$v \in H_0^1to the elliptic problem $$... 1 For the first one: you still need to prove$$\lim\|u_m\|_{p^*} = \|u\|_{p^*}.$$That is, you need to show that u_m \to u in L^{p^*}. Hint: Use the GNS to show that u_m is a Cauchy sequence. The same comment applies to 2. In order to prove 3 from 2, you still need an extension property for U, that is, for each u \in W^{1,p}(U), there exists v \in ... 1 I think gerw has the missing piece of your proof in the hint that you should use Cauchy sequences instead of using G-N-S and ||u_{m}-u||_{L^{p^{*}}} <= C|||Du_{m}-Du||_{L^{p}}. Having said that I don't think his answer was written in a very informative way. To adapt your proof for question (1) use the following: Since you showed that \{Du_{m}\}_{m} ... 1 For the second inequality I refer you to Brezis Corollary 9.14. For the first inequality, remember that \|u\|_\infty\leq \|u\|_{C^{0,\gamma}(\overline{U})}, therefore$$\int_U |u|^p\leq\int_U\|u\|_\infty^p\leq |U|\|u\|_{C^{0,\gamma}(\overline{U})}^p$$where |U| is the Lebesgue measure of U. 1 The first inequality holds as mentioned by Tomas. The second inequality I think is a direct result of a Theorem which extends from Morrey's Inequality: If U is bounded, open subset of \mathbb{R}^{n}, and suppose \partial U is C^{1}. If n < p \leq \infty and u \in W^{1,p}(U). Then u has a version u^{*} \in C^{0,\gamma}(\bar{U}) such that ... 1 We know that f_i \to f weakly in W^{1,p}(\Omega) and f_i \to g strongly in L^{2p}, by the compact embedding. Note that g\in L^p (I assume \Omega is bounded). If we can show that g\in W^{1,p}(\Omega), and f_i \to g weakly in W^{1,p}(\Omega), then g=f since weak limit is unique. This will then answer your question. Claim 1: g\in ... 1 It may help to consider first the case n=1 and k=2. The fact that \eta_Ru\to u in L^p follows from a monotone convergence argument. Indeed, since \eta_R(x)=1 if |x|\leqslant R,$$\int_\mathbb R|\eta_R(x)u(x)-u(x)|^p\mathrm dx=\int_{\{|x|\gt R\}}|u(x)|^p\mathrm dx.$$We can use the product rule for derivatives: we get (\eta_R u)'=\eta_R ... 1 Recall, that for (x_1, x_2) \in \mathbb R^2, we have$$ \|(x_1,x_2)\|_1 = |x_1| + |x_2| \le 2^{(p-1)/p} \, \|(x_1,x_2)\|_p = 2^{(p-1)/p} \, (|x_1|^p + |x_2|^p)^{1/p}. $$Hence,$$|F(u)|^p \le \big| M \, |u| + |F(0)| \big|^p \le 2^{p-1} \, (M^p \, |u|^p + |F(0)|^p ).This shows \int |F(u)|^p dx < \infty. 1 First note that you have \begin{align*}b(u, u) &= \int_0^T((u(t), \ u(t)))dt- \int_0^T(u(t), \ u'(t))dt \\ &= \int_0^T((u(t), \ u(t)))dt- \frac12 \, \int_0^T \frac{d}{dt}(u(t), \ u(t))dt \\ &= \int_0^T((u(t), \ u(t)))dt- \frac12 \, (u(T), \ u(T)) + \frac12 \, (u(0), \ u(0)) \\ &= \int_0^T((u(t), \ u(t)))dt+ \frac12 \, (u(0), \ u(0)). ... 1 Suppose ad absurdum that there is a sequence u_n\in H^1 such thata(u_n,u_n)\to 0,\ \ \|u_n\|_{H^1}=1$$In other words$$\tag{1} \int_0^2 |u_n'|^2+\left(\int_0^1 u_n\right)^2\to 0,\ \int_0^2u_n^2+\int_0^2|u_n'|^2=1$$We conclude from (1) that$$\int_0^2 |u'_n|\to 0,\ \int_0^2u_n^2\to 1\tag{2}$$Assume without loss of generality that ... 1 To every second order elliptic PDE$$Lu = f$$on \Omega \subset \mathbb{R}^n, where L is an elliptic operator and f a measurable function, is associated a positive-definite, bounded, symmetric coefficient matrix$$Q(x) = [q_{ij}(x)], i,j = 1,\ldots n.$$That is, the quadratic form$$\mathcal{Q}(x,\xi) = \xi^\top Q(x) \xi$satisfies, for some$c,C > ...

1

Here is a sketch of a counterexample: Let $\{x_n\}$ be a sequence of distinct points in $\Omega$ converging to a point $x \in \partial \Omega$. For each $n$ let $f_n \in W_0^{1,p}(\Omega)$ be a positive function that is unbounded in a neighborhood of $x_n$ and satisfies $\|f_n\|_{W^{1,p}(\Omega)} < 2^{-n}$. (You can do this if $p < N$). Let \$f = ...

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