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This is a step at which I am stuck in proving a theorem.

Suppose $\Omega$ is a $\mathcal C^1$ open connected and bounded subset of $\mathbb R^2$. (By $\mathcal C^1$, I mean that the boundary is defined by a $\mathcal C^1$ function). Let $\nu(x)$ be the outward normal at a boundary point $x \in \partial \Omega$. \Is it possible that $\langle x, \nu(x) \rangle = 0$ for every every $x \in \partial \Omega$?

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Since the domain is bounded and has a smooth boundary, we could apply divergence theorem which says: $$ \int_{\Omega} \mathrm{div}F = \int_{\partial \Omega} F\cdot \nu \,dS $$ Now the $F = x$, if an open, simply-connected, and bounded $\Omega$ exists such that $x\cdot \nu(x) = 0$ pointwisely, the right side is zero, while the left side is double the area of $\Omega$.


Another way to visualize the vector field $F = (x_1,x_2)$ on the plane, at ever point on the plane $F$ is pointing from the origin to that point, like this: enter image description here

and you want to find an $\Omega$ such that the normal of $\partial\Omega$ is perpendicular to this vector field everywhere, or rather the tangential direction of the boundary is parallel to $F$, such $\Omega$ cannot be bounded.

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Your proof using the divergence theorem seems correct and elegant. I will award you the bounty unless something really cool comes up. –  Abhishek Parab May 9 '12 at 15:47
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I think this is much simpler. You said that $\Omega$ is bounded and that your boundary is $C^1$. Then your boundary is a $1$-dimensional manifold and it is easy to maximize $\|x\|$ over it. Let $x_0$ be a point where $\|x\|$ attains maximal value, then we have (this is obvious, but if you have to be very formal then the method of Lagrange multipliers should do) that $\partial \Omega \perp x_0$, and therefore $x_0\ \|\ \nu(x_0)$, so $\langle x_0, \nu(x_0)\rangle \neq 0$ unless $x_0 = 0$.

Hope that helps ;-)

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