I'm given a function $g:\mathbb{R}^n \mapsto \mathbb{R}$, $g(y) = \prod_{i\in[n]} (1+y_i\cdot c_i)$, where $c_i>0$. Let $e_a,e_b$ be two arbitrary standard vectors. It is easy to show that for any $y,a,b$ the function $z \mapsto g(y + z(e_a - e_b))$ is concave.

My problem is the following. I need to consider, for a fixed $y$, a function of all possible swaps between standard vectors, i.e., $h\left( (z_{ab})_{a,b\in[n]} \right) = g(y + \sum_{a,b} z_{ab}(e_a-e_b))$. For such a function I would like to use Jensen's inequality, and so I need to argue that the function $h\left( (z_{ab})_{a,b\in[n]} \right)$ is concave. Can I argue as follows?

  1. For any $(z_{ab})_{a,b\in[n]}$:

  2. the function is strictly concave in every standard direction, i.e., $z_{ab}$, hence

  3. since the function is $C^\infty$

  4. in appropriately small neighborhood of $(z_{ab})_{a,b\in[n]}$ the function is concave, therefore

  5. the function is concave everywhere.

  • $\begingroup$ First of all, how is it from $\mathbb{R}^n$ to $[0,1]$? $\endgroup$ – Fedor Petrov Sep 10 '15 at 7:18
  • $\begingroup$ You mean $g$? It takes $n$ numbers and outputs a product of $n$ numbers. Function $h$ is from $\mathbb{R}^{n(n-1)/2}$. $\endgroup$ – Marek Adamczyk Sep 10 '15 at 8:06
  • $\begingroup$ Yes, $g$, why do its values belong to $[0,1]$? $\endgroup$ – Fedor Petrov Sep 10 '15 at 8:48
  • $\begingroup$ Sorry, a mistake. It's from $\mathbb{R}$. The original I was considering was from $[0,1]$ because it had a normalization factor. $\endgroup$ – Marek Adamczyk Sep 10 '15 at 8:52
  • $\begingroup$ Basically, I wanted to know if this is a well established fact. I think I can come up with a tailored argument using the fact that the function is quite simple, but would prefer to rely on a general property. $\endgroup$ – Marek Adamczyk Sep 10 '15 at 8:57

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