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No, this is not true. You only have $$(W^{1,p}_0(\Omega))^*=W^{-1,p}(\Omega)$$ However, the dual space of $W^{1,p}(\Omega)$ is not identified, although it is smaller then $W^{-1,p}(\Omega)$.

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Actually you can get, for a fixed $\epsilon>0$, that for any $u\in W^{m,p}(\Omega)$ $$\|{u}\|_{W^{m,p}}\leq \epsilon\|{D^\alpha u}\|_{L^p}+C_{\epsilon}\|{u}\|_{L^q},$$ where $|{\alpha}|=m$. Usually we just take $q=p$ but if domain is good enough and by embedding we could extend $q$ to where embedding would go. This theorem states that the extreme term ...

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We have $\|\partial^\beta u\|_{L^p(U)} \leq \|u\|_{W^{k-1,p}(U)}$ Also, by the theorem, $W^{k,p}(U) \rightarrow W^{k-1,p}(U)$ is compact. Now, suppose there is no $C_\epsilon$. Then we get a sequence $u_n \in W^{k,p}(U)$ (normalize so these are all norm 1) violating it with $n$ in place of $C_\epsilon$. That is, $\|u_n\|_{W^{k-1,p}} \geq \epsilon + n ... 0 so convergence in Lp implies there exists subsequence unk converging pointwise a.e. to u, and convergence in L∞ implies unk converges uniformly (and hence pointwise) a.e. to some v, so that v=u a.e.. 0 You can use the fact that$L^p$spaces are a Radon-Riesz spaces, for p>1. and the case$p=1$i don't have a proof but you can use the fact that each of the$L^p$-spaces,$1 ≤ p < ∞$, has the property that each sequence on the unit sphere that converges almost everywhere converges also in norm. 1 Let$C_n=q_1^n|_\Sigma-q_2^n|_\Sigma$be the the sequence of constant functions, which as you alread have noted, converge to$q_1|_\Sigma-q_2|_\Sigma$in$H^{1/2}(\Sigma)$. Therefore,$C_n \to q_1|_\Sigma-q_2|_\Sigma$in$L^2(\Sigma)$, or equivalently, $$\int_\Sigma |C_n-(q_1|_\Sigma-q_2|_\Sigma)|^2d\Sigma\to 0.\tag{1}$$ There are some ways to prove now ... 2 The function$x\mapsto \|x\|^{2s-4}x_i^2$is homogeneous of degree$2s-4+2=2s-2$. So it is integrable on the unit ball if and only if$2s-2>-n$. This is a special case of general fact about homogeneous functions. (I.e., those with$f(tx)=t^df(x)$for all$x$and all$t>0$.) Indeed, suppose$f$is homogeneous of degree$d$and is not zero a.e. ... 0 First notice that, by Hölder's inequality, we have$cu\in L^p$for$1/p=1/q+1/6$(since$u\in L^6$by the Sobolev embedding). Next your inequality gives$u\in W^{2,p}$for this same$p$. Applying Sobolev embedding again, we get$\nabla u \in L^{p*}$where as usual$1/p^* =1/p-1/n$. If we substitute we get $$\frac{1}{p^*} = \frac{1}{q} +\frac{1}{6} ... 1 First of all, the problem should be$$u(x):=\ln\left(\ln\left(1+\frac{1}{|x|}\right)\right) $$but not$$u(x)=\ln\left(\ln\left(\frac{1}{1+|x|}\right)\right)$$as you stated. Next, we have$$ \partial_i u(x) = = \frac{1}{\ln(1+\frac{1}{|x|})}\frac{x_i}{|x|^3}\frac{1}{1+\frac{1}{|x|}}$$Hence we have$$ |\nabla u| \approx ... 1 I think you probably mean$\xi > \eta$? The overall inequality is $$|u(\xi) - u(\eta)|^2 \leq (b-a)\int_a^b (u')^2~dx.$$ Write this as $$u(\xi)^2 - 2 u(\xi)u(\eta) + u(\eta)^2 \leq (b-a)\int_a^b (u')^2~dx.$$ Integrate in$\xi$from$a$to$b$.$u(\eta)^2$does not depend on$\xi$, so we obtain a factor of$(b-a)$. Similarly with the integral on the ... 2 The derivative of$u$is $$u'(x)= \begin{cases} 2x\cos\frac{1}{x} + \sin\frac{1}{x}&0<x\leq 1\\ 0 & x=0 \end{cases}.$$ In particular, note that since$\left|h\cos\frac{1}{h}\right|\leqslant |h|,$$$u'(0) = \lim_{h \rightarrow 0} \frac{h^2\cos\frac{1}{h}-0}{h}=\lim_{h \rightarrow 0} h\cos\frac{1}{h}=0.$$ So$u$has a bounded derivative and ... 1 Choose any$g \in \mathcal{C}^{\infty}[0,1]$that is identically$1$near$0$and identically$0$near$1$. Then, for any continuously differentiable$f$on$[0,1]$, you have $$f(0)=-fg|_{0}^{1} = -\int_{0}^{1}f(t)g'(t)+f'(t)g(t)\,dt.$$ That's enough to give you constants$C$and$D$such that $$|f(0)| \le ... 1 For unboundedness from above, just construct a sequence (u_n) that oscillates (with increasingly "stronger" oscillations) around 1 (i.e. \|\nabla u_n\|_{L^2} is going to infinity, and \|u_n\|_{L^2}\equiv 1). For the lower bound, note that$$\int Vu^2 dx\geq -\|V\|_{L^\infty}\|u\|_{L^2}=-\|V\|_{L^\infty}.$$Edit: for the unboundedness from above, to ... 0 Assuming that your Fourier transform is defined by \mathcal{F}(\phi) = \int_{\Bbb R^n} \phi(x) e^{-i\bf{x}\cdot \bf{\xi}}\, d\bf{\xi}, it is by definition that for non-integer s, (1 - \triangle)^s is the operator with Fourier multiplier (1 + |\xi|^2)^s (if the Fourier integral had kernel e^{-2\pi i\bf{x}\cdot\bf{\xi}} instead of ... 0 We actually use the fact that C_c^\infty(R^N) is dense in H^s(R^N) for each s\in R. (Or you could use \mathcal{S}(R^N) instead of C_c^\infty. not really matters) Anyhow, take (u_n)\subset C_c^\infty be such that u_n\to u in H^s(R^N). Now, by embedding result, we have$$ \|u_n-u\|_{L^\infty(R^N)}\leq C\|u_n-u\|_{H^s(R^N)}\to 0 $$Now fix ... 0 This is the result of Trace operator. Namely, we have there exists a linear bounded operator T: H^(\Omega)\to L^2(\partial \Omega), provided that \Omega has at least Lipschitz boundary. That is, you have for u\in H^1(\Omega),$$\|u\|_{L^1(\partial \Omega)}\leq C\|u\|_{H^1(\Omega)} $$Let define set D:=\{u\in H^1(\Omega\setminus\Omega_1):\, ... 0 For this question, you only need to use Sobolev embedding. Once we have \|u-u_\Omega\|_{L^p}\leq C\|Du\|_{L^p}, we have u-u_\Omega\in W^{1,p}(\Omega). Hence, the sobolev embedding tells you that$$ \|u-u_\Omega\|_{L^{p^*}} \leq C\| D(u-u_\Omega)\|_{L^p}=C\|Du\|_{L^p} $$Done. 1 Let's consider the following: Fix the sequence u_n\to u weakly in H_0^1(\Omega) as you suggest in your post. Define \Omega_k:=\Omega\cap B(0,k). Then each \Omega_k is compact and hence we could consider to use Rellich theorem. But before we do that, let me point out that the boundary \partial \Omega_k may not be smooth, actually it can even be ... 1$$\|(\phi_iu)_{\epsilon_i}-\phi_iu\|_{L^\infty(\Omega)}< \frac{\eta}{2^i} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(4)$$and$$\left|\|\nabla (\phi_iu)_{\epsilon_i}\|_{L^\infty(\Omega)}- \|\nabla (\phi_iu)\|_{L^\infty(\Omega)}\right|<\frac{\eta}{2^i} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(5)$$So far I am comfortable and confident ... 0 In two dimensions it is possible to take advantage of the "enstrophy miracle". This is not possible in three dimensions. 1 Not really a direct answer to your question, but Terry Tao has a splendid article on the Navier-Stokes Millenium problem, accessible via arXiv: "Localisation and compactness properties of the Navier-Stokes global regularity problem" 1 We have, with B=B(0,1), multiplying by u both sides of the equation and integrating by parts (recall u vanishes at the boundary)$$ \int_B |\nabla u|^2 +u^2 dx \leq \int_B |\nabla u |^2 dx + V(x)u^2 dx = -\int_B fudx \leq \| u\|_{L^2}\| f\|_{L^2}, $$where the first inequality is because V\geq1 and the last one is Cauchy-Schwarz. Therefore we get$$ ... 0 Work first with Schwartz functions. We calculate $$|u(x)| \leq \| \check{u} \|_{L^1} = \int_{\mathbb{R}^n} (1+|\xi|^2)^{-s/2}(1+|\xi|^2)^{s/2} |\check{u}| d \xi.$$ Now just notice that if$s>n/2$then$(1+|\xi|^2)^{s/2}\in L^2(\mathbb{R}^n)$so you can apply Hölder's inequality and conclude that $$\| u\|_{L^\infty} \leq C(n,s) \| u\|_{H^s}.$$ Now ... 1 You missed half of them. You need the normalized half-period functions$\sin(nx/2)$for$n=1,2,3,\cdots$. These are the solutions of the Sturm-Liouville eigenvalue problem $$-f'' = \lambda f,\\ f(0)=0,\;\;f(2\pi)=0.$$ The eigenvalues are$\lambda = (n/2)^{2}$for$n=1,2,3,\cdots$, and the normalized eigenfunctions ... 1 First, embed into$C^{0,1-N/p}$(Sobolev-Morrey); then use compact embedding between Hölder spaces. Yes, Hölder spaces don't get a very detailed treatment in PDE books (or any book I know, for that matter). The properties are generally bad: nonseparable, nonreflexive... not much to work with. As a reference, I suggest Chapter 3 of the book Lectures on ... 1 This is not true without some additional assumptions on the boundary. Just to make life simple, the following example uses$N=2$, but it is easy to modify for any$N \ge 2$. Fix$p > q \ge 1$. Let$\Omega$be a domain in the plane which consists of squares$(Q_k)_{k=0}^\infty$, of side length$2^{-k}$centered at points on the$x$-axis, together with ... 0 I agree with above, this inequality is false in general. But what you may look at, or instead that you could use the equivalent norm in$W^{m,p}$, in which we have $$\|u\|_{L^p(\Omega)}\leq \epsilon\|D^\alpha u\|_{L^p(\Omega)}+C_\epsilon \|u\|_{L^p(\Omega)}$$ where$|\alpha|=m$. If$\Omega$is finite and$m=1$, you could instead have ... 1 If$u\in H^1(\mathbb{R})$, then in particular it's absolutely continuous, so that the fundamental theorem of calculus gives $$|u(x+h)-u(x)|\leq \int_0^1 |u'(x+th)||h|dt$$ So that squaring and integrating over$\mathbb{R}$we get (using Jensen's inequality and Fubini's theorem) $$\left\|\frac{u(\cdot +h)-u(\cdot)}{h}\right\|_2^2 \leq \| \int_0^1 ... 0 See this notes by Terry Tao. Also, Sobolev multiplication below the borderline. 1 Isn't it the case that \partial_t \psi_t exists, and thus so does \displaystyle \int_{B_1} \partial_t |\nabla \psi_t|^p dx \ \ = \ \ \frac{d }{dt} \int_{B_1} |\nabla \psi_t|^p dx Or am I missing something? 1 You were done. Consider just$$\left| \int (f - f_n) \psi' dx \right| \leq \int |f-f_n||\psi'| dx \leq ||\psi'||_q||f-f_n||_p $$So the limit is 0 0 The estimate that I can prove, and that doesn't involve crazy exponents, is:$$ [u]_{W^{2k}_2(B_1)} \leq N \left( \| Lu \|_{L^2(B_2)} + \| u\|_{L^2(B_2)}\right),\qquad \text{(1)} $$where L=\sum (D^j)^{2k}. Notice that the inequality you write in the question has to fail (take for example k=1 and u any harmonic function), so that the term \| ... 0 I am afraid that it is rather straight-forward: We have that$$ Dw_k \cdot Dv_k -|Du|^2 =Dw_k\cdot Dv_k-Dw_k\cdot Du+Dw_k\cdot Du-Du\cdot Du= Dw_k\cdot (Dv_k- Du)+Du\cdot(Dw_k-Du), $$and hence$$ Dw_k \cdot Dv_k -|Du|^2 \le \lvert Dw_k\cdot (Dv_k- Du)\rvert+\lvert Du\cdot(Dw_k-Du)\rvert \le \lvert Dw_k\rvert\lvert Dv_k- Du\rvert+\lvert Du\rvert \lvert ... 0 By the Cauchy-Schwarz inequality, we have $$\int_U |Dw_k| |Dv_k - Du|\,dx \le \sqrt{\int_U |Dw_k|^2} \sqrt{\int_U |Dv_k - Du|^2}.$$ Since the Sobolev norm has$\|f\|_{H^1}^2 = \int_U (|f|^2 + |Df|^2) \ge \int_U |Df|^2$, we have $$\int_U |Dw_k| |Dv_k - Du|\,dx \le \|w_k\|_{H^1} \|v_k - u\|_{H^1}.$$ Since$w_k$converges in$H^1$norm,$\sup_k \|w_k\|_{H^1} ...

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