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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 $C > 0$...

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 ...

3

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 $$\int\limits_{\mathbb{R}} f(x) \phi''(x) d x=-\int\limits_{\mathbb{R}} f'(x) \phi '(x) d x$$

2

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 $$for some A, and the smallest such A is comparable to the \... 2 The following two theorems (see Partial Differential Equations (chapter 5) by Evans) can answer your question: 2 This is clear if you look at the Fourier transform. If f,f''\in L^2 then$$\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,$$because |\xi|^2\le\max(1,|\xi|^4). 2 If f_n \in C^\infty_c(\Omega) satisfies$$\|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$$Thus \overline u \in H^1_0(\widetilde \Omega). 2 If you use (u, v)_{H_0^1} = (\nabla u, \nabla v)_{L^2}, you get R = -\Delta, as you have shown. And there is another scalar product on H_0^1, such that R = I - \Delta. Which one? 1 You can find the general statement and proof in Chapter 6 of the book "Sobolev Spaces" by Robert A. Adams and John. J. F. Fournier. 1 A function u \in W^{1,2}(\Omega) might be unbounded for n \ge 2. Hence, it is not possible by continuous functions. In the case p > 2, I think one can develop an argument along the following lines: There is a function u \in W^{1,2}(\Omega), which has a singularity "which is stronger as the space W^{1,p}(\Omega) allows". Hence, the one-sided ... 1 Why not L^1: In studying elliptic equation, it is most convenient to consider L^p space for 1<p<\infty. The reason is that one does not have nice L^p-estimates$$\|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 . $$Clearly, compactly supported continuous functions are dense in L^2((1+t^2)^{-1}\, dt), and if ... 1 The Gauss-Green theorem (the basis of the Green identities) states that if u,v are sufficiently smooth on a nice domain \Omega then$$\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$$where \nu_j is the jth component of the external normal unit vector and ... 1 Definition: a distribution T is of order r if r is the smallest integer such that$$|T(\phi)| \le C\sum_{j=0}^{r} \sup_\Omega |\phi^{(j)}|$$holds with C independent of \phi. Example: if f\in W^{1,1}(\Omega), then both f and f' are of order 0. Property 1: if T is of order r, then T' is of order \le r+1. This follows directly ... 1 I believe @PhoemueX might have hit the target with his suggestion of Orlicz spaces. The vowels all match, the H I wrote was just a guess, the consonants match except the final. And the embedding holds. And after hearing Riesz (which should be pronounce Reece, or, for the German-speaking, Rieß) pronounce Rits by the same professor, hearing "cz" pronounced "ts"... 1 One doesn't use translations for a general domain. One uses a diffeomorphism between a part of the domain, and half-space, and then translates in the half-space. This is what Brezis does later in the text (part C); you quoted some of it in Concerning the proof of regularity of the weak solution for the laplacian problem given in Brezis. 1 Any bounded sequence in a separable Hilbert space (which is reflexive) has a weakly convergent subsequence. Added on edit: See Theorem 3.18 of the same book. 1 The better behavior of elliptic PDE with measurable coefficients in two dimensions is explained by their relation with quasiconformal maps. I know two book sources that develop this relation. Elliptic Partial Differential Equations and Quasiconformal Mappings in the Plane by Kari Astala, Tadeusz Iwaniec & Gaven Martin. Chapter 12 of Elliptic Partial ... 1 TLDR: degenerate is short for degenerate elliptic. Consider a second-order differential operator$$\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 ...

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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 $H^1 (\tilde{\Omega})$.

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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 ...

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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 $\lbrace u_m \... 1 The answer is no in general: Take for instance$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 ...

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