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A Sobolev space is a vector space of functions equipped with a norm that is a combination of Lp-norms of the function itself and its derivatives up to a given order.
9
votes
Sobolev spaces and geometry
Unlike their Hölder cousins, most Sobolev spaces are reflexive Banach spaces. Reflexivity is a highly desirable feature for variational problems because it gives a little bit of compactness, enough …
10
votes
Sobolev spaces of differential forms and regular atlases
There is a coordinate free way of defining Sobolev spaces of sections of a vector bundle $E$ over a manifold $M$. You need to make a few choices: a metric $g$ on $M$, a metric $h$ on $E$ and a con …
2
votes
Harmonic functions vanishing on the boundary and distance function asymptotics
$\DeclareMathOperator{\dist}{dist}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\pa}{\partial}$ Suppose that $N=2$ and
$\Omega$ is is the unit disk. Choose
$$
u= -1+ar^4+br^5\in C^2(\overline{\Omeg …
4
votes
Nice way to express $H^{-1}(\mathbb{S}^1)$
Note first of all that $H^{-1}$ will contain objects that are not functions, such as the Dirac $\delta$ concentrated at a point. (Think that $2\delta_0=\frac{d^2}{dx^2}|x|$.)
The correct definitio …
6
votes
Accepted
Rellich-Kondrachov compacteness Theorem for the Euclidean space with Gaussian measure
I'll start with several known facts.
Proposition 1. Suppose that $E$ is a real Hilbert space with norm $\Vert -\Vert$ and $K: E\to E$ is a a compact, selfadjoint
positive operator. Denote …
2
votes
Coercivity for functional and complete orthonormal system
$J$ is not coercive in $W^{1,2}$ For that to happen you need to show that $\Vert u_n\Vert_{1,2}\to \infty$ implies $J(u_n)\to \infty$. Take for example the function $u_n$ which is identically $0$ …
17
votes
Accepted
Boundedness of the derivative of the trace of an H^1 function
Clearly, what you call $\newcommand{\bn}{\boldsymbol{n}}$ $\nabla u\cdot \bn $ is the normal derivative $\frac{\partial u}{\partial \bn}$. The trace theorem (see e.g. Lions and Magenes, Non-Homogene …
7
votes
Accepted
Convergence of Schwartz kernels implies convergence of operators
Here is a classical theorem. $\newcommand{\bR}{\mathbb{R}}$
Suppose that for $0< a,b<\infty$
$$\sup_x\left(\int_{\bR^n} |k(x,y)|^a dy\right)^{\frac{1}{a}}=M_1(k)<\infty, $$
$$\sup_y\left(\int_{\bR^ …
2
votes
Function extension in a Sobolev space
The answer is yes, with the caveat indicated in the comment below. Consider an arbitrary extension $\hat{\theta}\in H^2(\Omega)$. Now choose a compactly supported smooth function $\eta$ such that
…
2
votes
Sobolev embedding proof without Gagliardo–Nirenberg–Sobolev inequality or Morrey's inequality
Sobolev's original proof is different from the two approaches described by Deane. He uses certain integral formulae. You can read about this approach in the classic monograph
C. Morrey: …
2
votes
Reference request: Simple facts about vector-valued Sobolev space
If you read French then this book is the place you are looking for
Brézis, H. Opérateurs maximaux
monotones et semi-groupes de
contractions dans les espaces de
Hilbert. (French) North-Holl …