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Consider the following statement, which does not involve any PDE: this is multivariable calculus only. If $f:\mathbb R^n\to \mathbb R$ is $C^1$ smooth and $f(0,0,0)=0$, then for every $R>0$ there exists $C>0$ such that $|f(z)|\le C|z|$ whenever $|z|\le R$. Proof: by the mean-value inequality we have $|f(z)-f(0)|\le |z|\sup_{|\xi|<R} ... 2 Physical interpretation can be very helpful, but ultimately we need$E$to be such that we can infer something about$dE/dt$from the PDE infer what we want about the solution from$E$The consideration usually begins with 1. Note that$u_tu_{tt} = \frac{1}{2}(u_t)^2$. So, if we have a hyperbolic PDE$u_{tt}=F(x,u,Du,D^2u)$, it is reasonable to ... 1 The Sobolev embedding theorem implies that in two dimensions,$H^1$(which is$W^{1,2}$) embeds into$L^q$for every$1\le q<\infty$. (By the way, it's not enough for$\Omega$to be bounded; we need some assumption on$\partial \Omega$, such as smoothness. This is assumed at the beginning of the paper.) But in (9) the authors use$H^{3/4}$(which is ... 0 Think of$L^2$as the space of bounded functionals on$L^2$. In particular, your bounded sequence is a bounded sequence of linear functionals$f_n: L^2\to \mathbb R$. If you have a space$Z$compactly embedded into$L^2$, then the restrictions of$f_n$to the unit ball of$Z$have a convergent subsequence (by the Arzelà–Ascoli theorem). Also, the ... 1 It appears that my (first) edition of the book does not contain this statement, but I think I understand it. The elements of$H^{-1} $are bounded linear functionals on$H^1_0$: $$H^{-1}=\{f:H_0^1\to\mathbb R \ ; \ |f(u)|\le C\|u\|_{H^1}\}$$ Then what do we mean by saying that$L^2\subset H^{-1}$? It means that some functionals on$H^1_0$admit a bound by ... 1 Yes, and you actually can say a bit more. Every bounded Lipschitz domain is a Sobolev extension domain (classical theorem of Calderón and Stein, referenced here). Thus we can assume that$u\in W^{1,p}(\mathbb R^n)$. If$p>n$, then$u$is Hölder continuous and the matter is trivial. So let's assume$1<p\le n$. I consulted Fine Regularity of ... 2 Yes, you must show that the tempered distribution$(1+|\xi|^2)^{s/2}\,\hat{u}$is given by integration against an$L^2$function. The multiplication by$(1+|\xi|^2)^{s/2}$does not change whether or not it is an a.e. pointwise function, but it does affect square integrability. For example, the Fourier transform of Dirac$\delta$is (up to a constant) the ... 1$D((0,T)\times \Omega)\subset L^2((0,T);H^1(\Omega))\subset L^1_{loc}((0,T);H^1(\Omega))$. Define $$f(v) = \int_0^Tdt \langle f(t,\cdot),v (t,\cdot)\rangle_{H^{-1};H^1},$$ then $$|f(v)|\le \int_0^Tdt \| f(t,\cdot)\|_{H^{-1}}\|v (t,\cdot)\|_{H^1} \le | K |\int_0^Tdt \| f(t,\cdot)\|_{H^{-1}}\sqrt{\|v (t,\cdot)\|_{L^\infty}^2+\|\nabla_x v ... 0 Proposition 3.1 of Michael Taylor's PDE book (first volume) confirms that this is so for a compact manifold (which is what I was looking for). 0 Both of the following books have good treatments of these ideas: Sobolev Spaces on Riemannian Manifolds by Emmanuel Hebey Some Nonlinear Problems in Riemannian Geometry by Thierry Aubin 1 It is true and it is dense. If f\in C^\infty([0,T]\times \Gamma) then \partial^n_t u clearly is C^\infty is t and belongs to H^1(\Gamma), for all t\in[0,T], as it is sufficiently smooth with respect to the variables which paprametrize \Gamma. Also, as \Gamma is compact and smooth, H^1(\Gamma) is the completion C^\infty(\Gamma). 2 Let w=\dfrac{\partial v}{\partial x}. Since v\in W^{1,\infty}, we have w\in L^\infty. The space L^\infty is dual to L^1. Thus, pairing a fixed L^\infty function with a convergent sequence of L^1 functions produces a convergent sequence of numbers. This is why$$\int b(u_k) w\to \int b(u)w$$For the other integral, note that a(u_k)w ... 1 In a pure study of Partial Differential Equations, somehow you have to come up with the existence of a function to solve some kind of equation. How do you establish the existence of a function? There are a few techniques. One is to use the Riesz representation theorem on L^{2} spaces (Lax-Milgram is just an extension of Riesz representation.) Another is to ... 0 Yes, that's actually true. First, note that the inequality$$ ||u||_{L^2} + ||\Delta u ||_{L^2} \leq c ||u||_{H^2} $$is always true (the constant may vary due to your definition though). Hence it remains to show that$$ ||u||_{H^2} \leq C \left( ||u||_{L^2} + ||\Delta u ||_{L^2} \right). $$This follows from the fact that the Laplacian is elliptic and can ... 1 We have, using integration by parts$$ \int_{Q}\varphi^2(\Delta v)v_t= -\int_{Q}\nabla (\varphi^2 v_t)\cdot\nabla v= -2\int_{Q}(\nabla\varphi\cdot\nabla v)\varphi v_t-\int_{Q}(\nabla v_t\cdot\nabla v)\varphi^2. $$But$$ \int_{Q}(\nabla v_t\cdot\nabla v)\varphi^2=\frac{1}{2}\int_{Q}\frac{\partial}{\partial t}|\nabla ... 1 If you want to learn the basics of Sobolev Spaces, you can consult Brezis or Evans books. If you need a more detailed knowledge of the subject then, you can consult Adams book, which as cited above by @dinosaur, or you can take a look on Leoni's book. I apologizem but with respect to FEM I can't say nothing. 1 So$\Delta u$is still only defined in a weak sense, in that there is a distribution$w$such that $$\int u \Delta \phi = \int w \phi$$ for all$\phi \in C_0^\infty$. So, if $$\int_\Omega u_t \phi + \nabla u \cdot \nabla \phi = \int_\Omega f\phi$$ for all$\phi$, then integrating by parts $$\int_\Omega u_t \phi - u \Delta \phi = \int_\Omega f\phi ... 0 You may view W^{1,\infty}(\Omega) as a subspace of L^\infty(\Omega)^{n+1} (via u \mapsto (u,\nabla u), which can be identified with the dual of L^1(\Omega)^{n+1}. Thus, for any bounded sequence u_n, there is a subsequence (without relabeling), such that u_n \stackrel{*}{\rightharpoonup} u and \partial_i u_n \stackrel{*}{\rightharpoonup} v_i in ... 2 Yes. More specifically, if$$ \limsup_{n\to\infty} \frac {\|f^{(n)}\|^{1/n}_{L^2}} {n}=s<\infty, $$then f can be extended analytically in$$ \Omega=\{x+iy: x\in(0,1),\,\lvert y\rvert<1/(s\mathrm{e})\}\subset\mathbb C. $$In the case \|f^{(n)}\|^{1/n}_{L^2}\le a, it extends analytically in (0,1)\times\mathbb R. It does not extend however to an ... 2 In fact a_nb_n\in L^1(\Omega) not L^2(\Omega), and$$ \int_\Omega a_nb_n\,dx \to \int_\Omega ab\,dx. $$Indeed,$$ a_nb_n-ab=a_n(b_n-b)+(a_n-a)b $$Then$$ \Big|\int_\Omega a_n(b_n-b)\,dx\,\Big|\le \|a_n\|_{L^2}\|b_n-b\|_{L^2}\le M\|b_n-b\|_{L^2}\to 0, $$and$$ \int_\Omega (a_n-a)b \to 0, $$as a_n-a\rightharpoonup 0. EDIT by OP: this ... 2 You can, if you wish, define a Sobolev space on a non-open set [0,1]. For example: A function f:[0,1]\to \mathbb R belongs to W^{1,p}([0,1]) if there exists g\in L^p([0,1]) and c\in \mathbb R such that f(x) = c+\int_0^x g for a.e. x. The definition is the easy part. The hard part is figuring out what to do with this definition. Has it ... 1 I wish to elaborate on the example provided by David Mitra. Assume that \Omega=(0,2\pi) and f_n(x)=\sin nx, and assume that this sequence has a subsequence which converged almost everywhere, i.e., f_{n_k}(x)\to f(x), almost everywhere in [0,2\pi]. Clearly \lvert f(x)\rvert\le 1, and f measurable, as a limit of measurable functions. Note also ... 3 \Omega is bounded, so there is a K such that \Omega \subset \mathbb{R}^n\times (-\infty,K). Then it's basically John's hint from the other question: For u \in C_0^\infty(\Omega), we have$$u(x,0) = - \int_0^K \frac{\partial u}{\partial x_{n+1}}(x,t)\,dt.$$Hölder's inequality gives$$\lvert u(x,0)\rvert \leqslant \left(\int_0^K \left\lvert ... 1 Daniel Fischer gave a good reason already, but here is a concrete example:$(a,b)=(-1,1)$, $$f(x)= \begin{cases}\frac{n}{n-1}(1-x), \quad &x>1/n \\ nx, \quad &0\le x\le 1/n \\ 0, \quad &x<-1/n\end{cases}$$ The functions$f_n$converge in$L^2$norm to $$f(x)= \begin{cases}1-x, \quad &x>0 \\ 0, \quad &x\le 0 \end{cases}$$ which ... 3 As Lion said: the integral of$fu$is nonnegative, but not because of any pointwise bound for$fu$. It's integration by parts:$\int_\Omega (-u\,\Delta u ) = \int_{\Omega}|\nabla u|^2\ge 0$. To give a simple example: let$u (x)=x(x-1)(x-4) $in one dimension, with$\Omega=(0,4)$. Then$f(x) = -u''(x) = 10-6x$. The product$fu$changes sign at$1$and at ... 1 As @Yiorgos pointed out, when$k>\frac{n}{2}$(and the boundary is good) we have that$H_0^k(\Omega)\subset C(\overline{\Omega})$and hence your statement is true. On the other hand, for$k<\frac{n}{2}$this might not be true and an counter example for your statement can be found here or here. 1 What you say is true (for sufficiently smooth boundary) for$u\in H^k_0(\Omega)$, with$\Omega$open in$\mathbb R^n$, if $$k>\frac{n}{2},$$ according to the standard Sobolev imbedding. In the case of$H^1_0(\Omega)$, the boundary trace of its elements are identified with elements of$H^{1/2}(\partial\Omega)$. But such traces are not defined pointwise, ... 1 Take a look in Brezis book page 303 (Estimates near the boundary). First, fix a point$p\in \partial\Omega\setminus\overline{K}$. Let$\theta_i,U_i$be as in the book, consider the function$v=\theta_i u$and assume without loss of generality that$\partial K$is not contained in$\overline{U_i}$. As the trace of$u$near$p$is zero, you have that$v\in ...

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Since you are not assuming integrability of mixed partials, this is not so much a Sobolev embedding situation, but more of elliptic regularity result. Luckily everything is in $L^2$, which leads to conclusion quickly. Consider the Fourier transform $\hat f(\eta,\xi)$ where $(\eta,\xi)\in\mathbb R^2$. Your assumptions imply $$|\eta|^k \hat f \in L^2, \quad ... 0 Let \delta>0 and define f_\delta:\mathbb{R}\to [0,\infty) by$$ f_\delta(x) = \left\{ \begin{array}{rl} \sqrt{x^2+\delta^2}-\delta &\mbox{ if $x\ge0$} \\ 0 &\mbox{ otherwise} \end{array} \right. $$Note that f_\delta\in C^1 and f_\delta(x)\to g(x) for all fixed x, where g(x)=x if x\ge 0 and g(x)=0 if x<0. Consider ... 0 The function f(x)=e^{-|x|} has a weak derivative f'(x)=-{\rm sgn\,}(x)\,e^{-|x|}, while f,f'\in L^2(\mathbb{R}). Hence f\in W^{1,2}(\mathbb{R}). But f' does not have a weak derivative, since its disribution derivative f''=e^{-|x|}+2\delta\, e^{-|x|}=e^{-|x|}+2\delta in \mathcal{D}'(\mathbb{R}). Hence f\notin ... 2 Most of this is repeated from my comments following the answer to this question, but the basic problem is here: I remember that we identify H with its dual, but not V with its dual. Actually, it's the other way around; it's ok to identify V = L^2 with its dual, but if we also insist on identifying H with its dual, we have exactly the problem ... 1 I'm assuming you're starting with the one-dimensional case, which is what I'll address here. The one-dimensional case is a little bit special, because some of your intuition about what a weak derivative 'should be' breaks down. For instance, a step function has a jump discontinuity and does not have a weak derivative at that point. On the other hand, ... 0 Let I_1u=\|\nabla u\|_2^2, I_2u=\|u\|_2^2 and I_3u=\|u\|_{2\sigma+2}^{2\sigma+2}. Note that$$\langle I'_1u,v\rangle=2\int\nabla u\nabla v,\ \ \langle I'_2u,v\rangle=2\int u v,\ \langle I'_3u,v\rangle=(2\sigma+2)\int |u|^{2\sigma+1}u v\tag{1}$$If u\neq 0, we apply the quotient rule to conclude that$$\langle ...

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Hint: Define $T:W^{1,p}(\Omega)\to L^p(\Omega)\times L^p(\Omega,\mathbb{R}^n)$ by $$Tu=(u,\nabla u)$$ With suitable norms on $W^{1,p}(\Omega)$ and $L^p(\Omega)\times L^p(\Omega,\mathbb{R}^n)$, you can prove that $T$ is an isometry.

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The trick is here that the norms (and scalar products) of $L^2(U)$ and $H^1(U)$ are not the same, which means that the closure of the orthogonal basis (or complete orthogonal set) will be different, depending on the process of completion. The general idea is that the inclusion of the derivative in the Sobolev norm will not only make sure that for a Cauchy ...

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So, it looks like $\langle \cdot,\cdot \rangle$ indicates the duality pairing between either two $L^2$ (or generally $L^p$) spaces or between $H^1_0$ and its dual $H^{-1}$. Now, if we have an $H^1_0$ function $f$ and an $H^{-1}$ "function" $g$, it's all well and good to write $\langle f,g \rangle$, but how do you actually calculate it? After all, $g$ could ...

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Well, let me forget $f$. Then, notice that $$\int u_t\phi(u)=-\int (\nabla \phi(u))^2,$$ where we used the boundary conditions and the definition of $\phi$ to remove the boundary term $$\int_{\partial\Omega}\phi(u)\nabla\phi(u)nds=0.$$ Now $$\int u_t\phi(u)=\int u_t u\textbf{1}_{u<0}+\int u_t (u-1)\textbf{1}_{u>1}=\int u_t ... 0 The function f(x)=\mathrm{e}^{-x} does not belong to any W^{m,2}(\mathbb R), m\ge 0, because if it did, then we would have that f\in L^2(\mathbb R), which is not true since$$ \int_{-\infty}^\infty |f(x)|^2\,dx=\int_{-\infty}^\infty \mathrm{e}^{-2x}\,dx=\infty. $$0 I think it might not even be true it's equivalent to a norm like that. My approach would be to show if (f_n,g_n,h_n) is Cauchy in this norm, then (f'_n + g_n+a h_n) , (h'_n-af_n) and (g'_n) converge to some L^2 functions, by completeness of L^2. Now you should be able to write all 6 of f_n,g_n,h_n,f'_n,g'_n,h'_n in terms of these three to get ... 1 Sweeping out the trash, reformulate what indeed is needed to be established:$$ \bigcap_{k=0}^{m-[\frac{n}{2}]-1}C^k\bigl([0,T];H^{m-k}(\Omega)\bigr)\subset C^{m-[\frac{n}{2}]-1}\Bigl(\overline{Q}_T\Bigr), $$which in fact immediately follows by an a priori inclusion$$ \bigcap_{k=0}^{l}C^k\bigl([0,T];C^{l-k}(\overline{\Omega})\bigr)\subset ...

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This is related to a problem posed by Tosio Kato in 1953 and not resolved until 2002, a few years after Kato's death. Take a look at the references on this page to Kato's Conjecture: http://www.math.missouri.edu/~hofmann/ I believe that you are essentially correct because the form domain is related to the square root of the elliptic operator, and ...

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It suffices to establish that any sequence $\,u_k=u_k(|x|)\to 0\,$ weakly convergent in $\,H_{rad}^1(\mathbb{R}^n)\,$ will be srongly convergent in $\,L^{2\sigma+2}(\mathbb{R}^n)$, which might be ruined by the unboundedness of $\,\mathbb{R}^n$ if it were not for the remarkable inequality $\,(\ast\ast)\,$ that provides cutting off the infinity.  Indeed, ...

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