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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 ...


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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..


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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.


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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 ...


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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. ...


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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} ...


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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 ...


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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 ...


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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 ...


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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 ...


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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 ...


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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 ...


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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 ...


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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):\, ...


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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.


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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 ...


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$$\|(\phi_iu)_{\epsilon_i}-\phi_iu\|_{L^\infty(\Omega)}< \frac{\eta}{2^i} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(4)$$ 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 ...


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In two dimensions it is possible to take advantage of the "enstrophy miracle". This is not possible in three dimensions.


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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"


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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 $$ ...


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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 ...


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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 ...


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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 ...


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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 ...


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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 ...


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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 ...


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See this notes by Terry Tao. Also, Sobolev multiplication below the borderline.


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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?


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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 $0$


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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 $\| ...


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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 ...


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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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