# How to prove this function is concave?

This is the function:

$\displaystyle f(\vec x) = \log \frac{\exp(x_1)}{\sum_{i=1}^n \exp(x_i)}$

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Please do not cross post between mathoverflow and math.stackexchange. – Thomas Rot Jun 22 '11 at 12:27
Irrelevant parts of the question were removed. – Łukasz Lew Jun 22 '11 at 13:54

Pick arbitrary $x,y$ and $\alpha,\beta \in (0,1),\ \alpha+\beta=1$. You want to prove that $f(\alpha x +\beta y) \geq \alpha f(x)+\beta f(y)$. This is equivalent to
$$\log \frac{e^{\alpha x_1+\beta y_1}}{\sum e^{\alpha x_i+\beta y_i}} \geq \log \frac{e^{\alpha x_1+ \beta y_1}}{(\sum e^{x_i})^\alpha (\sum e^{y_i})^\beta}$$
$$\sum_{i=1}^n e^{\alpha x_i+\beta y_i}\leq (\sum_{i=1}^n e^{x_i})^\alpha (\sum_{i=1}^n e^{y_i})^\beta$$
which is exactly Holder's inequality in the discrete case (note that $\alpha +\beta=1$.
I don't think this solution works for $g$. Maybe if you come up with some relations between $a_i$'s and $b_i$'s then you may apply something similar. – Beni Bogosel Jun 22 '11 at 13:06