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I have seen many questions along this line, but none quite answered my question as far as I could tell.

On all of $\mathbb{R}$, is the Sobolev norm ever defined as follows $$\|f\|_{W_2^k(\mathbb{R})} := \|f\|_{L_2(\mathbb{R})}+|f|_{W_2^k(\mathbb{R})},$$ where $|f|_{W_2^k(\mathbb{R})}:=\|f^{(k)}\|_{L_2(\mathbb{R})}$ denotes the usual seminorm?

Usually you see this definition for domains satisfying the uniform cone condition or something like that.

Keaton
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  • For 1D, you can just view the cone condition as: "For any $x\in \Omega$, there exists an open neighborhood containing $x$, which has a $\geq\alpha>0$ measure". This will guarantee that the (weak) derivative can be defined. Also for the whole space, since you can view that $\partial \mathbb{R}^n = \emptyset$, so that you don't have to worry that much for domain having a corner like Gabriel's Horn or such. – Shuhao Cao Sep 16 '13 at 16:22

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Yes, this is a perfectly fine definition on $\mathbb R$. The purpose of the uniform (interior) cone condition is to make sure that all points of the domain are uniformly easy to approach "from the inside"; this enables us to control lower order derivatives globally, by integrating higher order derivatives (Poincaré's inequality). When the domain is all of $\mathbb R$ or $\mathbb R^n$, access "from the inside" isn't an issue at all. Depending on how the uniform cone condition is stated, one even may be able to say that it holds vacuously when the boundary is empty.

By the way, you can easily check that this definition gives control on all derivatives, by using the Fourier transform. The $L^2$ norms with weights $1$ and $|\xi|^k$ together control all $L^2$ norms with weights $|\xi|^j$, $0< j<k$.

Post No Bulls
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