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Let $g(x) \colon \mathbb{R}^n_+ \to \mathbb{R}_+$ be a smooth concave and homogenuous of order 1 function: $g(\lambda x) = \lambda g(x)$ for any $\lambda > 0$. Let $B$ be the unit ball: $B = \{ x \in \mathbb{R}^n_+ \mid |x|^2 \leq 1\}$. How to show than that there exists a unique number $r \geq 0$ such that the intersection of the level set $\{ x \mid g(x) = r\}$ and of the unit ball $B$ contains a single point?

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Hint: consider the set $I = g^{-1}(1)$. It is non empty by homogeneity (in fact for each $y\in \mathbb{R}^n_+$ there is a real number $\lambda$ such that $\lambda y \in I$). Let $b:I \to \mathbb{R}$ be $b(x) = |x|^2$. By concavity and compactness of $I$, $b(x)$ attains a unique minimum on $I$. –  Willie Wong Mar 29 '12 at 14:01
(It is a lot clearer if you draw a picture. Homogeneity means that all the level sets of $g$ look like each other. Using that $g$ is concave, smooth, homogeneous, and positive, what do the level sets look like?) –  Willie Wong Mar 29 '12 at 14:03
Thank you, they look like nested cups. But $I$ is a curve in 2d (for example). What does mean 'concave' for curve? –  Nimza Mar 29 '12 at 14:16
Ah, I meant "By concavity of $g$ and compactness of $I$". In may also help to do computations in polar coordinates. That the level sets look like cups means that treated as a graph over the hyperplane $x_1 + x_2 + \cdots + x_n = 0$, the set $I$ is the graph of a convex function. On the other hand, the boundary $\partial B$ is the graph of a strictly concave function. Hence if $I$ and $\partial B$ are tangent at a point, they intersect only at that point and no other. –  Willie Wong Mar 29 '12 at 14:27
It's okay. But I don't understand why concavity of $g$ implies that $b(x)$ attains a unique minimum on compact set $g^{-1}(1)$ (I imagine this but I don't know how to show this strictly) –  Nimza Mar 29 '12 at 15:07

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