Convexity of the product of two functions in higher dimensions

Solution 1:

Yes a generalization is possible. Here is an elementary approach to the convexity of the product of two nonnegative convex functions defined on a convex domain of $\mathbb{R}^n$.

Choose $x$ and $y$ in the domain and $t$ in $[0,1]$. Your aim is to prove that $\Delta\ge0$ with $$\Delta=t(fg)(x)+(1-t)(fg)(y)-(fg)(tx+(1-t)y). $$ But $f$ and $g$ are nonnegative and convex, hence $$ (fg)(tx+(1-t)y)\le(tf(x)+(1-t)f(y))(tg(x)+(1-t)g(y)). $$ Using this and some easy algebraic manipulations, one sees that $\Delta\ge t(1-t)D(x,y)$ with $$ D(x,y)=f(x)g(x)+f(y)g(y)-f(x)g(y)-f(y)g(x), $$ that is, $$ D(x,y)=(f(x)-f(y))(g(x)-g(y)). $$ This proves a generalization of the result you quoted to any product of convex nonnegative functions $f$ and $g$ such that $D(x,y)\ge0$ for every $x$ and $y$ in the domain of $f$ and $g$.

In particular, if $f$ and $g$ are differentiable at a point $x$ in the domain, one asks that their gradients $\nabla f(x)$ and $\nabla g(x)$ are such that $z^*M(x)z\ge0$ for every $n\times 1$ vector $z$, where $M(x)$ is the $n\times n$ matrix $$ M(x)=\nabla f(x)\cdot(\nabla g(x))^*. $$