Convexity proof (another)

Question

$\def\dom{\mathop{\mathrm{dom}}}$How can I prove convexity of this function? $$\ f(x_1,x_2,...,x_n)= \prod_{i=1}^n \left(\frac{x_i}{r_i}\right)^{-p_i},$$ where $$ 0\le{p_i}\le1 ,\quad\sum_{i = 1}^n {p_i}=1,\quad \forall x_i,r_i\in \mathbb{R},\quad x_i,r_i\gt0. $$ I know it is convex and I know function $f : \mathbb{R}^n → \mathbb{R}$ is convex if $\dom f$ is a convex set and if for all $x,y ∈ \dom f$, and $θ$ with $0 ≤ θ ≤ 1$, we have: $$f(θx + (1 − θ)y) ≤ θf(x) + (1 − θ)f(y).$$ It is given that $\dom f$ is a convex set but I don't know how to prove convexity of the functin $f$. Thank you!

Answer 1

What you need is to show that the Hessian $\nabla^2 f(x)$ is positive semi-definite. Here I only show for $n=2$ and the proof will be similar for $n\ge 3$. For $n=2$, $$ f(x)=\left(\frac{x_1}{r_1}\right)^{-p_1}\left(\frac{x_1}{r_2}\right)^{-p_2}. $$ Without loss of generality, assume $r_1>0,r_2>0$ and $x_1>0,x_2>0$. Note $$ \frac{\partial^2f}{\partial x_1^2}=\frac{1}{x_1^2}p_1(p_1+1)\left(\frac{x_1}{r_1}\right)^{-p_1}\left(\frac{x_1}{r_2}\right)^{-p_2},\frac{\partial^2f}{\partial x_2^2}=\frac{1}{x_2^2}p_2(p_2+1)\left(\frac{x_1}{r_1}\right)^{-p_1}\left(\frac{x_1}{r_2}\right)^{-p_2}, $$ and $$ \frac{\partial^2f}{\partial x_1\partial x_2}=\frac{1}{x_1x_2}p_1p_2\left(\frac{x_1}{r_1}\right)^{-p_1}\left(\frac{x_1}{r_2}\right)^{-p_2}. $$ Thus \begin{eqnarray} \det \nabla^2 f(x)&=&\det\left(\begin{matrix} \frac{\partial^2f}{\partial x_1^2}&\frac{\partial^2f}{\partial x_1\partial x_2}\\ \frac{\partial^2f}{\partial x_1\partial x_2}&\frac{\partial^2f}{\partial x_2^2} \end{matrix}\right)\\ &=&\frac{\partial^2f}{\partial x_1^2}\frac{\partial^2f}{\partial x_2^2}-\left(\frac{\partial^2f}{\partial x_1\partial x_2}\right)^2\\ &=&\frac{1}{x_1^2x_2^2}p_1p_2(p_1+p_2+1)\left(\frac{x_1}{r_1}\right)^{-2p_1}\left(\frac{x_1}{r_2}\right)^{-2p_2}>0, \end{eqnarray} and also $$ \frac{\partial^2f}{\partial x_1^2}>0,\frac{\partial^2f}{\partial x_2^2}>0$$ and hence $\nabla^2f(x)$ is positive semi-definite.