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For $x\in\mathbb{R}^d$, define $\displaystyle G(x)=\int_0^\infty t^{s-1-d/2}e^{-t}e^{-\frac{|x|^2}{4t}}dt$. Since $t^{s-1-d/2}e^{-t}e^{-\frac{|x|^2}{4t}}>0$, $0<G(x)\le\infty$ is well-defined. By Tonelli's theorem $\displaystyle||G||_1=\int_{\mathbb{R}^d}dx\int_0^\infty t^{s-1-d/2}e^{-t}e^{-\frac{|x|^2}{4t}}dt=\int_0^\infty t^{s-1-d/2}e^{-t}dt\int_{\... 3 A counterexample for$d=2$: let$\Omega$be the disk$\{x:\|x\|<\exp(-\exp(\pi))\}$, and $$f(x) = \sin \log \log \frac{1}{\|x\|}$$ This function is in$W_0^{1,2}(\Omega)\cap L^\infty(\Omega)$(relevant calculations here) but has a discontinuity at$0$, and moreover cannot be made continuous by redefining it on a set of measure zero. On the other ... 1 A function$u \in W^{1,2}(\Omega)$might be unbounded for$n \ge 2$. Hence, it is not possible by continuous functions. In the case$p > 2$, I think one can develop an argument along the following lines: There is a function$u \in W^{1,2}(\Omega)$, which has a singularity "which is stronger as the space$W^{1,p}(\Omega)$allows". Hence, the one-sided ... 0 See in this book (page 7-8): https://books.google.it/books?id=9YuDAwAAQBAJ&pg=PP1&dq=linear+functional+analysis+Cerda&hl=it&sa=X&ved=0ahUKEwjiufnZ777NAhWLVRoKHeOCCTYQ6AEIJTAA#v=onepage&q=linear%20functional%20analysis%20Cerda&f=false and for regular partition of unity, you can consider this lemma: Proof of regular version of ... 1 Why not$L^1$: In studying elliptic equation, it is most convenient to consider$L^p$space for$1<p<\infty$. The reason is that one does not have nice$L^p$-estimates $$\|u\|_{W^{2,p}(\Omega)}\le C (\|f\|_{L^p(\Omega)} + \|u\|_{L^p(\Omega)}$$ for$p=1$(Here we assume that$p(x)$is nice). Note that the above estimates is crucial in establishing ... 1 I have to admit that I don't completely understand your reformulation, but it certainly is true that$C_0^{\infty}$is dense in $$H^{-1} = \{ f\in\mathcal S' : (1+t^2)^{-1/2}\widehat{f}(t)\in L^2 \} , \quad \|f\|_{H^{-1}} = \| (1+t^2)^{-1/2}\widehat{f}\|_2 .$$ Clearly, compactly supported continuous functions are dense in$L^2((1+t^2)^{-1}\, dt)$, and if ... 0 For your interest, see the definition of the weak derivative. For your comment, by Lesbegue differentiation theorem, a measurable function has Lesbegue derivative almost everywhere (thus it is totally possible for$f(p)$not equal its Lesbegue derivative) The problem is related to Poincare inequality, which is proved in Evans (Chapter 4?). To see the ... 0 Yes, the tensor product is dense. To approximate a function$f$in$C_c^{\infty}((a_1,b_1) \times (a_2,b_2))$you can use functions of the form$p(x,y)\chi_1(x)\chi_2(y)$where$p$is any polynomial and$\chi_1,\chi_2$are cut-off functions that are equal to$1$on most of the interval, so that$\chi_1(x)\chi_2(y)\equiv 1$on the support of$f$. Then the ... 0 First of all,$\chi_{(0,2)}$isn't optimal; the minimum (on$BV$) is assumed for$u=\chi_{(0,\infty)}$, and$F(u)=1$. Now let me show that$F(u)\ge 2$for all$u\in W^{1,1}$(so the assertion about the infimum is incorrect). Consider$F_-(u)=\int_{-2}^0 |u(x)|\, dx + |Du|(-\infty,0)$. Since the derivative exists and is in$L^1$, the total variation equals$\...

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The Gauss-Green theorem (the basis of the Green identities) states that if $u,v$ are sufficiently smooth on a nice domain $\Omega$ then $$\int_\Omega \frac{\partial u}{\partial x_j} v \, dx = - \int_\Omega u \frac{\partial v}{\partial x_j} \, dx + \int_{\partial \Omega} uv \nu_j dS$$ where $\nu_j$ is the $j$th component of the external normal unit vector and ...

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This is clear if you look at the Fourier transform. If $f,f''\in L^2$ then $$\int|\hat f(\xi)|^2<\infty$$ and $$\int|\xi|^4|\hat f(\xi)|^2<\infty,$$and hence $$\int|\xi|^2|\hat f(\xi)|^2<\infty,$$because $|\xi|^2\le\max(1,|\xi|^4)$.

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Define $f'(x) = \int_0^x f''(t)d t$, then $|f'(y)|\leq (\int _0^y |f''|^2 )^{1/2}\sqrt{y}\leq ||f''||_2 \sqrt{y}$. Therefore $$\int\limits_{\mathbb{R}} f''(x) \phi (x) d x = -\int\limits_{\mathbb{R}} f'(x) \phi '(x) d x$$ And so $$\int\limits_{\mathbb{R}} f(x) \phi''(x) d x=-\int\limits_{\mathbb{R}} f'(x) \phi '(x) d x$$

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Definition: a distribution $T$ is of order $r$ if $r$ is the smallest integer such that $$|T(\phi)| \le C\sum_{j=0}^{r} \sup_\Omega |\phi^{(j)}|$$ holds with $C$ independent of $\phi$. Example: if $f\in W^{1,1}(\Omega)$, then both $f$ and $f'$ are of order $0$. Property 1: if $T$ is of order $r$, then $T'$ is of order $\le r+1$. This follows directly ...

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The same approach as in Decay of Fourier Coefficients implies Holder Continuity? works. The starting point is the Cauchy-Schwarz inequality, $$|u(x)-u(y)|^2 =\left(\sum_{k\in\mathbb{Z}} (1+k^2)^{s/2}\hat u_k \ \frac{|e^{ikx}-e^{iky}|}{(1+k^2)^{s/2}}\right)^2 \\ \le \sum_{k\in\mathbb{Z}} (1+k^2)^s|\hat u_k|^2 \ \sum_{k\in\mathbb{Z}} \frac{|e^{ikx}-e^{iky}|... 2 This should hold. Lemma Define$$ P_j f (x) = \sum_{\sigma = \pm 1, k=2^j}^{2^{j+1}} \hat{f}(\sigma k) e^{2\pi i \sigma k x}. $$If \lvert f \rvert \leq 1, then f \in C^\alpha(\mathbb{T}) for 0 < \alpha < 1 iff$$ \sup_{j \in \mathbb{Z}} 2^{j\alpha} \| P_j f \|_\infty \leq A $$for some A, and the smallest such A is comparable to the \... 0 You should know that for each f \in L^{p}_{loc}(\Omega) we can define a Distribution by: T_{f}: \mathcal{D}(\Omega) \rightarrow\mathbb{R}, where T_{f}(\phi) = \displaystyle\int_{\Omega}f\phi. In the case, T_{f} is called regular distribution. The Du Bouis Reymond Lemma states that the map f\mapsto T_{f} is one-to-one. However, there are ... 0 As clarified in the comments, we have$$\{ x \in \Omega \mid u(x) \ne 0 \} \subset V \subset \Omega,$$where V is a compact set. Hence, V has a positive distance \delta > 0 to the boundary \partial\Omega. Mollifying u with a mollifier with radius smaller than \delta produces a sequence of smooth functions in C_0^\infty(\Omega) converging ... 1 One doesn't use translations for a general domain. One uses a diffeomorphism between a part of the domain, and half-space, and then translates in the half-space. This is what Brezis does later in the text (part C); you quoted some of it in Concerning the proof of regularity of the weak solution for the laplacian problem given in Brezis. 0 Suppose \partial\Omega is C^1. Note that W^{1,2}(\Omega)=H^1(\Omega). The following two theorems (see Partial Differential Equations (chapter 5) by Evans) can answer your question: 1 Any bounded sequence in a separable Hilbert space (which is reflexive) has a weakly convergent subsequence. Added on edit: See Theorem 3.18 of the same book. 0 Since H^q(D) is a Hilbert space, you get a subsequence of \{f_k\} (without relabeling), such that f_k^{-1} \rightharpoonup v in H^q(D) for some v \in H^q(D). Using the compact embedding from Rellich, we have f_k^{-1} \to v in H^{q-1}(D). Next, check v = f^{-1}. Again there are subsequences (without relabeling) such that f_k \to f pointwise ... -1 This is (tightly related to) formulae (6.11.2) in Maz'ya's book on Sobolev spaces, 2011 edition; or Corollary 1 in §4.11.1 in the 1985 edition. 1 The better behavior of elliptic PDE with measurable coefficients in two dimensions is explained by their relation with quasiconformal maps. I know two book sources that develop this relation. Elliptic Partial Differential Equations and Quasiconformal Mappings in the Plane by Kari Astala, Tadeusz Iwaniec & Gaven Martin. Chapter 12 of Elliptic Partial ... 1 TLDR: degenerate is short for degenerate elliptic. Consider a second-order differential operator$$\mathcal{L}u\equiv \sum_{i,j=1}^n a_{ij} u_{ij}+\sum_{i=1}^n b_i u_i + cu$$where I am using subscripts on u to denote derivatives. In your case,$$\mathcal{L}V\equiv rSV_{S}+\frac{1}{2}\sigma^{2}S^{2}V_{SS}-rV.$$\mathcal{L} is said to be elliptic at a ... 1 This is always true. In fact you may take \tilde{\Omega}=\mathbb{R}^n . The reason is that if u_k\in C^\infty_c (\Omega)  is an approximation of u , in H^1 (\Omega) , then the same sequence approximates the extension by zero in H^1 (\tilde{\Omega}) . 2 If f_n \in C^\infty_c(\Omega) satisfies$$\|f_n - u\|_{H^1(\Omega)} \to 0,$$then \overline f_n \in C^\infty_c(\widetilde \Omega) satisfies$$\| \overline f_n - \overline u \|_{H^1(\widetilde \Omega)} \to 0$$Thus \overline u \in H^1_0(\widetilde \Omega). 1 The following two theorems (see Partial Differential Equations (chapter 5) by Evans) can answer your question: 4 The vanishing integral over the small ball is enough to get a Poincaré-type estimate. Let B = B(0,1) and \Omega = B(0,r). We define$$\|u\|_\star := \big|\int_B u \, \mathrm dx\big| + \|\nabla u\|_{L^p(\Omega)}.$$It is clear that \|\cdot\|_\star is a norm on W^{1,p}(\Omega) and that \|u\|_\star \le C \, \|u\|_{W^{1,p}(\Omega)} for some C > 0... 1 The answer is no in general: Take for instance u(x)= \eta(x) |x|^{-n/p}, where \eta is a smooth function that is zero in a neighborhood of the origin and is identically one outside of some big ball centered at the origin. You can check by polar coordinates that \nabla u\in L^p, but u\notin L^p (to check that test functions approximate u you can use ... 1 I believe @PhoemueX might have hit the target with his suggestion of Orlicz spaces. The vowels all match, the H I wrote was just a guess, the consonants match except the final. And the embedding holds. And after hearing Riesz (which should be pronounce Reece, or, for the German-speaking, Rieß) pronounce Rits by the same professor, hearing "cz" pronounced "ts"... 1 Here are some ideas to get you started: If you were using the space W^{1,p}(\newcommand{\R}{\mathbb R} \R^+) you would define the reflection \bar u(x) = u(-x) for x < 0 and \bar u(x) = u(x) for x > 0. This gives you continuity of \bar u at x = 0 from which the absolute continuity of \bar u follows. Suppose instead you are using the ... 1 The first question follows from the fact that u_m \rightarrow \overline{u} in W^{1,p}(\mathbb{R}^n) implies ||u_m-u_l||_{L^{p^{\ast}}(\mathbb{R}^n)} \leq C ||Du_m-Du_l||_{L^p(\mathbb{R}^n)} \leq C[||Du_m-D\overline{u}||_{L^p(\mathbb{R}^n)} + ||D\overline{u}-Du_l||_{L^p(\mathbb{R}^n)}] \rightarrow 0 as m,j \rightarrow \infty, and \lbrace u_m \... 1 The first question follows by definition, H_m(\mathbb{R}^n) is the completion of C_{c}^\infty(\mathbb{R}^n) iff (by definition) H_m(\mathbb{R}^n) = \overline{C_{c}^\infty(\mathbb{R}^n)}^{|| \cdot ||_m}, i.e. \forall u \in H_m(\mathbb{R}^n) exists \lbrace u_k \rbrace \subset C_{c}^\infty(\mathbb{R}^n) such that u_k \rightarrow u in the H_m-norm. ... 1 The equality$$ \int_{\Omega \setminus F} (\dots) \, dx = \int_{\Omega^{(i)}} \int_{ \{x_i \in \mathbb{R} \, : \, (x_1, \dots, x_i, \dots, x_N) \in \Omega \setminus F \}} (\dots) \, dx_i \, \dots \, dx_{i-1} \, d{x_{i+1}}  is Fubini's theorem. It has nothing to do with the functions being integrated, or with the assumption on $F$. It's just doing ...

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