muliplication of two convex and inceasing functions is convex

Question

Suppose that we have two functions $f$ and $g$. if $f>0 \: and \: g>0$ are both increasing and convex functions, how to prove that $f(x)g(x)$ is also convex?

Answer 1

I think the easiest way would be to show that the directional derivative $$(fg)'_+(x) = \lim_{h\to 0^+}\frac{(fg)(x + h) - (fg)(x)}{h}$$ is increasing. We have, \begin{align*} \frac{(fg)(x + h) - (fg)(x)}{h} &= \frac{(fg)(x + h) - f(x + h)g(x) + f(x + h)g(x) - (fg)(x)}{h} \\ &= f(x + h)\frac{g(x + h) - g(x)}{h} + g(x)\frac{f(x + h) - f(x)}{h} \\ &\to f(x) g'_+(x) + g(x)f'_+(x), \end{align*} as $h \to 0^+$.

Because $f$ and $g'_+$ are both positive and increasing (note that $g$ is convex), we have $f \cdot g'_+$ is increasing. The same is true for $g \cdot f'_+$, hence for $(fg)'_+$. Thus, $fg$ is convex.

Answer 2

This convexity also can be verified by the definition.

First, wlog., suppose $x<y$, due to $f$ and $g$ are both monotonic, we have \begin{equation} [f(x) - f(y)][g(x)-g(y)] = f(x)g(x) - f(x)g(y) - f(y)g(x) + f(y)g(y) \ge 0 \end{equation} Thus \begin{equation} f(x)g(x) + f(y)g(y) \ge f(x)g(y) + f(y)g(x). \end{equation}

Next, let $h(x) = f(x)g(x)$, then \begin{equation} \begin{aligned} h(\lambda x + (1-\lambda)y) &= f(\lambda x + (1-\lambda)y) g(\lambda x + (1-\lambda)y) \\ & \leq [\lambda f(x) + (1-\lambda)f(y)] [\lambda g(x) + (1-\lambda)g(y)] \\ &= \lambda^2 f(x)g(x) + \lambda(1-\lambda)[ f(x)g(y) + f(y)g(x)] + (1-\lambda)^2 g(x)g(y) \\ &\leq \lambda^2 f(x)g(x) + (1-\lambda)^2 g(x)g(y) + \lambda(1-\lambda)[f(x)g(x) + f(y)g(y)] \\ &= \lambda f(x)g(x) + (1-\lambda) f(y)g(y) = \lambda h(x) + (1-\lambda)h(y) \end{aligned} \end{equation}