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If $E$ is a closed set such that Hausdorff measure $H^{n-p}(E)$ is $\sigma$-finite, then its capacity satisfies $Cap_p(E)=0$ and it follows that $W^{1,p}(X)=W^{1,p}(X\setminus E)$. Sobolev $W^{1,p}$ f …

Yes, that is true. This is a consequence of the classical Trace Theorem for Sobolev spaces. The proof can be found in any textbook on Sobolev spaces for example in Evans' Partial Differential Equatio …

Theorem. Multiplication by a Lipschitz function defines a bounded operator in $H^{1/2}(\partial\Omega)$. First proof. More generally, if $\Omega\subset\mathbb{R}^n$ is a bounded Lipschitz domain, …

It is well known and easy to verify that (Exercise 14 p. 309 in [1]) $$ \log\Big|\log\sqrt{x^2+y^2}\Big|\in H^1(B^2(0,e^{-1})) $$ so the trace of this function on the $x$-axis belongs to the trace spa …

Since the norm is quite specific, I am not sure if you can find it in any book. However, you can prove compactness of the embedding directly. Given a sequence $f_k\in C^{\beta}(\mathbb{R}^n)$, the com …

In your notation $H^k=W^{k,2}$, where $W^{k,p}$ represents the Sobolev space of functions whose derivatives of orders $\leq k$ are in $L^p$. A function $f\in W^{1,p}(a,b)$ if and only if there is $g …

Your function need not be Hölder continuous. Let $\Omega$ be the union of two exponential cusps with a common vertex and let $E$ be the complement of these cusps. Let $u=1$ in the upper cusp and $u=0$ …

Smooth functions $C^\infty(D)$ are always dense in $W^{1,p}(D)$ when $1\leq p<\infty$. This is a classical result of Meyers and Serrin and you can find it in any textbook on Sobolev spaces. However, t …

Yes if you choose a suitable representative of a Sobolev function. Lemma. Let $f\in W^{1,p}(\mathbb{R}^n)$, $1\leq p<\infty$. Then for every $\epsilon>0$, there is a Lipschitz function $g:\mathbb{R}^n …

Definition. If $U\subset\mathbb{R}$ is open, we say that $u\in {AC}(U)$ if $u$ is absolutely continuous on every compact interval in $U$. Let $\Omega\subset\mathbb{R}^n$. We say that $u$ is absolutely …

Here is a related example from: P. Hajlasz, Sobolev mappings: Lipschitz density is not a bi-Lipschitz invariant of the target. Geom. Funct. Anal. 17 (2007), 435-467. Theorem. There is a Lipschitz fu …

The answer presented here is copied from the paper [2]. Many similar results (sometimes with more complicated proofs) can be found in [1]. Let the space $L^{k,p}$ be defined by: $$ L^{k,p}(\mathbb{R}^ …

The restriction of $f$ to the boundary has degree zero. It is true also in higher dimensions. The proof presented below is based on the proof of density of $C^\infty(M,N)$ in $W^{1,p}(M,N)$, $p\geq \o …

The answer is no. Consider the form $$ \omega = \frac{-y}{x^2+y^2}\, dx + \frac{x}{x^2+y^2}\, dy $$ on the annulus $\Omega=\{ (x,y):\, 1<x^2+y^2<2\}$. This form is closed, but not exact, because the …

For $s>0$ the easiest way to obtain convolution estimates on manifolds is described below. First of all, by the Whitney embedding theorem (or by Nash theorem if you want to preserve the Riemannian m …