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Given a vector $\vec{x}={x_1,x_2,\ldots,x_n}$ ($x_i \geq 0$), why is it that in a function $U(\vec{x})=w \log \left ( \sum_{i=1}^n x_i \right ) + \varepsilon \sum_{i=1}^n \log (x_i)$ the $\varepsilon$ term is is said to ensure strictly concavity for the function $U(\vec{x})$? Why is $V(\vec{x})=w \log \left ( \sum_{i=1}^n x_i \right )$ (i.e. without the $\varepsilon$ term) not strictly concave? Is there any relation of this with the log-sum inequality, i.e. for non-negative numbers $a_1$, $a_2$, \ldots , $a_n$ and $b_1$, $b_2$, \ldots , $b_n$, we have $\sum_{i=1}^{n} a_i \log \frac{a_i}{b_i} \geq \left ( \sum_{i=1}^n a_i \right ) \log \frac{\sum_{i=1}^n a_i}{\sum_{i=1}^n b_i}$ with equality iff $a_i/b_i$ = constant?

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The sum $\sum$ is not a strictly concave function. So you might have $\vec{x}$ and $\vec{x}'\neq \vec{x}$ such that $v=\sum_i x_i=\sum x_i'$. Then for every $\lambda\in (0,1)$, $v=\sum_i \lambda x_i+(1-\lambda)x_i'$. Hence $w\log(\sum_i x_i)=w\log(\sum_i x_i')=w\log(\sum_i \lambda x_i+(1-\lambda)x_i')$. So $w\log(\sum_i\cdot)$ is not strictly concave.

But the sum of a concave function and a strictly concave function is strictly concave, so adding the $\epsilon$-term ensures strict concavity (the logarithm is clearly strictly concave).

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  • $\begingroup$ Thanks a lot, Michael! In fact I suspected the convex function property $f(\theta x_1 + (1-\theta) x_2) < \theta f(x_1) + (1-\theta) f(x_2)$ played some role here, but I simply forgot this other property you mentioned, i.e. the sum of a concave function and a strictly concave function is strictly concave, so you were of great help! $\endgroup$ – Fuad Abinader Mar 26 '12 at 17:17

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