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This inequality is incorrect for an essential reason. Assume it is true for smooth functions, then using an approximation by convolution we would conclude that it is true for Sobolev spaces and hence …

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 …

This is too long for a comment so I am posting it as an answer. I believe, I have seen a statement in the literature that $W^{1,p}(\mathcal{M},\mathcal{N})$ is a Banach manifold if and if $p> \dim\mat …

The following argument is used in my recent paper with D. Azagra and A. Cappello (work in progress). The answer is yes. I asked about whether the proof of Theorem 1 can be modified so to cover Theorem …

The following result is due to Asplund [A, p.235]. Theorem. If $\varnothing\neq E\subset\mathbb{R}^n$ is closed, and $d(x)=\operatorname{dist}(x,E)$, then $f:\mathbb{R}^n\to\mathbb{R}$ defined by $f( …

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 …

Tends to 0 as n tend to infinity, which can be derived from the definition. Derived how? There is no fault in your proof, just non-trivial detailsare missing. The formula is however, true. The best …

You actually do not need to assume that the mappings are Lipschitz as it is true for general $W^{1,n}$ mappings Theorem. If $\mathcal{M}$ and $\mathcal{N}$ are smooth compact and oriented manifolds, …

There are plenty of examples of discontinuous Sobolev function in $W^{1,n}(\mathbb{R}^n)$. For example $f(x)=\log|\log|x||$ defined in a neighborhood of zero. Now take $n=2$ and restrict the function …

If the annulus is small, then it is basically an Euclidean annulus and there is an extension operator for Sobolev spaces. However, if the annulus is large is may happen that it goes around a "neck" in …

Your argument is not correct. If a property $P$ fails for $Y$ and $X\subset Y$, it does not follow that it fails for $X$. For example $X=\{0\}\subset\mathbb{R}=Y$ but there are many properties true fo …

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 …

In fact a weaker version of Aleksandrov's theorem is true for subharmonic functions. Since $\Delta u$ is a Radon measure, the following result follows from Proposition 4.4 in 1. Theorem. If $u:\Omega …

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 $u=0$ …

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 …