New answers tagged

0

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 $\...


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 $j$th component of the external normal unit vector and ...


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 $|\xi|^2\le\max(1,|\xi|^4)$.


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


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 ...


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 $\...


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 ...



Top 50 recent answers are included