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I'm looking for a 1-homogeneous function $\pi \colon \mathbb{R}^n_{\geq 0} \to \mathbb{R}$ satisfying the following properties:

1) $\pi$ is not concave. This is equivalent to the fact that there exist $x, y \in\mathbb{R}^n_{\geq 0}$ such that $\pi(x+y) < \pi(x) + \pi(y)$.

2) For all $v = (v_1, \cdots , v_n) \in\mathbb{R}^n_{\geq 0}$ the quantity $\frac{i}{n}! \frac{\pi(v)^n}{\prod_{i}v_i}$ is bounded.

Any ideas?

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  • $\begingroup$ What is $\frac{i}{n}!$? (And what is $i$ here?) Also, when $v$ approaches a wall, do you mean some kind of limit in that quantity that you want bounded? $\endgroup$ – Alex Degtyarev Jan 21 '15 at 10:45
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    $\begingroup$ It looks like what you want is a function on $S:=S^{n-1}\cap\Bbb R^n_{\ge0}$. You can take $(\prod_iv_i)^{1/n}$, which is OK form the point of view of the boundedness, but it may be concave. Now, take any compact $K\subset S^{n-1}\cap\Bbb R_{>0}$ and change your function to any continuous function; you can do that to destroy the concavity. Since $\prod_iv_i$ is continuous and positive on $K$, the resulting ratio will still be continuous, hence bounded, just because of the compactness. $\endgroup$ – Alex Degtyarev Jan 21 '15 at 10:52
  • $\begingroup$ Thanks! I am sorry about the i. Of course is i=1. And yes, I want a limit when $v$ appraches a wall. $\endgroup$ – cata Jan 21 '15 at 11:12
  • $\begingroup$ So, judging by the $n!$ factor, you want a bound universal for all $n$? $\endgroup$ – Alex Degtyarev Jan 21 '15 at 11:15
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    $\begingroup$ Then, you should state it more carefully. $\Bbb Q$ valued function on $\Bbb R^n$ obviously cannot be homogeneous. $\endgroup$ – Alex Degtyarev Jan 21 '15 at 16:17

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