proof that $g\left(\boldsymbol{x}\right)=-\prod_{i=1}^{n}x_{i}^{p_{i}},p_{i}\geq0,\sum_{i}p_{i}=1$ is convex given $p_{i}\geq0,\sum_{i}p_{i}=1$

Question

I want to proof that for the function: Let $g\left(\boldsymbol{x}\right):\mathbb{R}_{++}^{n}\longrightarrow\mathbb{R}$ be $g\left(\boldsymbol{x}\right)=-\prod_{i=1}^{n}x_{i}^{p_{i}}$ and $\sum_{i=1}^{n}p_{i}=1,\forall i\in\left\{ 1,\ldots,n\right\} :p_{i}\geq0$

proof that $g\left(\boldsymbol{x}\right)$ is convex function for all $\boldsymbol{x}\in\mathbb{R}_{++}^{n}$.

what I tried so far:

1. tried to look for a transformation that preserve convexity (like log ... etc) , but $g\left(\boldsymbol{x}\right)$ is negative for all $\boldsymbol{x}\in\mathbb{R}_{++}^{n}$. so log is not a valid transformation .
2. tried to understand how can I use Jensen's inequality with another convex function f : $$f\left(\sum_{i=1}^{N}\lambda_{i}x_{i}\right)\leq\sum_{i=1}^{N}\lambda_{i}f\left(x_{i}\right)\\\sum_{i=1}^{N}\lambda_{i}=1\\\lambda_{1},\ldots,\lambda_{N}\geq0$$
3. tried to use First-order characterization: $$\forall\boldsymbol{x},\boldsymbol{y}\in\text{dom}\left(f\right):f\left(\boldsymbol{y}\right)\geq f\left(\boldsymbol{x}\right)+\nabla f\left(\boldsymbol{x}\right)^{T}\left(\boldsymbol{y}-\boldsymbol{x}\right)\text{ and }\text{dom}\left(f\right)\text{\,is convex}\iff f\text{\,is convex}$$ but without any meaningful results..
4. tried to understand how can I use second-order characterization as well but with no success.

Thanks :D

Answer 1

$$ \frac{\partial g}{\partial x_j\partial x_k}(x)=\cases{-\frac{p_jp_k}{x_jx_k}\prod_\limits{i=1}^nx_i^{p_i}& if $\ j\ne k$\\ -\frac{p_j\left(p_j-1\right)}{x_j^2}\prod_\limits{i=1}^nx_i^{p_i}& if $\ j=k$}\ . $$ Therefore \begin{align} \sum_{j=1}^n \sum_{k=1}^ny_jy_k \frac{\partial g}{\partial x_j\partial x_k}(x)=\\ -&\left(\left(\sum_{j=1}^n \frac{p_jy_j}{x_j}\right)^2-\sum _{j=1}^n \frac{p_jy_j^2}{x_j^2}\right) \prod_\limits{i=1}^nx_i^{p_i}\ . \end{align} By the Cauchy-Schwarz inequality, \begin{align} \left(\sum_{j=1}^n \frac{p_jy_j}{x_j}\right)^2 &= \left(\sum_{j=1}^n \sqrt{p_j}\frac{\sqrt{p_j} y_j}{x_j}\right)^2\\ &\le \left(\sum_{j=1}^n p_j\right)\left(\sum_{j=1}^n \frac{p_jy_j^2}{x_j^2}\right)\\ &= \sum_{j=1}^n \frac{p_jy_j^2}{x_j^2}\ , \end{align} from which it follows that $\ \displaystyle \sum_{j=1}^n \sum_{k=1}^ny_jy_k \frac{\partial g}{\partial x_j\partial x_k}(x)\ge0\ $ for all $\ x\in\mathbb{R}_{++}^n\ $ and all $\ y\in\mathbb{R}^n\ $. That is, the Hessian of $\ g(x)\ $ is non-negative semidefinite for all $\ x\in\mathbb{R}_{++}^n\ $, from which it follows that $\ g\ $ is convex on $\ \mathbb{R}_{++}^n\ $.

Answer 2

Remind that $-g$ is Cobb-Douglas function. The prove is that way:

First, $g$ is convex iff $-g$ is concave. So you can take $ln(g)=\sum_i p_i ln(x_i)$, which is clearly concave, thus quasi-concave.

Second, the increasing transformation of quasi-concave function is quasi-concave. So take $e^{ln(g)}=g$ gives quasi-concavity of $g$. Besides, it is easy to verify that g is homogenous of degree 1.

Last, it is not hard to show that a function is quasi-concave and homogenous of degree 1 is concave so g is concave.