From Boyd & Vandenberghe, Convex Optimization Ex. 3.32 (a)
If $f$ and $g$ are convex, both nondecreasing (or nonincreasing), and positive functions
on an interval, then $fg$ is convex.
Related: $f, g$ are convex and positive $\Rightarrow f(x)g(y)$ is convex?
I'm confused as to what I need to prove here: if it is to show that for $x,y \in \mathbb{R}$ and $0 \leq \theta \leq 1$,
$$f(\theta x + (1-\theta)y)\cdot g(\theta x + (1-\theta)y) \leq \theta (f(x)g(x) + (1-\theta)f(y)g(y),$$
we are not explicitly given that $\textbf{dom }f = \textbf{dom }g$ or the two domains even intersect, so it could very well be that $x,y \in \textbf{dom }f$ but $x,y \notin \textbf{dom }g$ or vice versa. So should we pick, e.g., $x_1,x_2 \in \textbf{dom }f$ and $y_1,y_2 \in \textbf{dom }g$, and try to show
$$f(\theta x_1 + (1-\theta)x_2)\cdot g(\theta y_1 + (1-\theta)y_2) \leq \theta (f(x_1)g(y_1) + (1-\theta)f(x_2)g(y_2)$$
instead?
Since we are not explicitly given that $f, g$ are twice differentiable, we also can't use the second order condition. But the nondecreasing condition seems to imply that we should use this method…?
Also, what does "postive functions on an interval" mean in the problem statement? Is is that $f : S_1 \subseteq \mathbb{R} \rightarrow (0, \infty)$ and $g : S_2 \subseteq \mathbb{R} \rightarrow (0, \infty)$?
Sorry for the load of questions, but I'm just very confused by the phrasing of this question.
Best Answer
Yes, positive means the ranges of the functions are contained in $(0,\infty)$. It is good enough to assume that the ranges are in $[0, \infty)$.
Hint: Just apply the definition of convexity to $f$ and $g$. The question reduces to the following:
$$ \theta^{2}f(x)g(x)+(1-\theta)^{2}f(y)g(y)+\theta (1-\theta) [f(x)g(y)+g(x)f(y)]$$ $$ \leq \theta (f(x)g(x)+(1-\theta) f(y)g(y).$$
Some simple algebraic manipulation reduces this to
$$f(x)g(y)+g(x)f(y) \leq f(x)g(x)+f(y)g(y).$$
This can be written as $$f(x)[g(y)-g(x)] \leq f(y)[g(y)-g(x)].$$
Can you prove this by considering the cases $x \leq y$ and $x >y$?
[In the first case use the fact that $a \leq b$, $c \leq d$ with $a,b,c,d \geq 0$ implies $ac \leq bd$ because $bd-ac =b(d-c)+c(b-a)\geq 0$].