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Recall: A function $f:\mathbb{R}^n\rightarrow\mathbb{R}$ is called $0$-homogeneous if $f(\lambda x)= f(x)$ for every $\lambda>0$ and every $x\in \mathbb{R}^n$.

Question: Let $B$ a convex balanced and absorbent bounded domain of $\mathbb{R}^n$. Is the space of $0$-homogeneous $C^\infty(\mathbb{R}^n\setminus\{0\})$ functions dense in $H^1(\partial B)$?

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If "0-homogeneous" means $f(tx) = f(x)$ for all $t > 0$ and $\Omega$ is not convex, why would you expect it to be dense? –  Deane Yang Apr 29 '12 at 20:14
    
Do you mean $C^\infty(\mathbb{R}^n\setminus\{0\})$? The only 0-homogeneous $C^0(\mathbb{R}^n)$ functions are the constants. If you mean that, consider the two cases $\Omega$ being the unit ball centered around the origin and $\Omega$ being the unit ball centered around the point $(2,0,0,\ldots)$. –  Willie Wong May 1 '12 at 15:25
    
What Willie is hinting at is that you need to assume that $0$ is in the interior of $\Omega$. –  Deane Yang May 1 '12 at 17:24
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Do you intend $\partial \Omega$ to be "regular" in some sense -- piecewise smooth or rectifiable, for example? I guess that even if $\Omega$ is star shaped the boundary might be very rough. If it is, then how do you define $H^1(\partial \Omega)$? (This may be my ignorance -- possibly there is a general definition of $H^1(S)$ for closed subsets $S$ of $\mathbb{R}^n$ but I don't know it.) –  Jeff Schenker May 1 '12 at 18:10
    
The point is that maybe you should try to figure this out on your own. You might not be able to get the best possible result yourself, but you should be able to figure out what conditions are sufficient. Just star-shaped is not enough. I also encourage you to work it out in detail for a two-dimensional domain. The key point is that the tangent plane at any point in the boundary has to be transversal to the ray from the origin. This allows you to use the inverse function theorem to prove what you want. I concede that this must be in the literature somewhere, but I don't know where. –  Deane Yang May 1 '12 at 18:11

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