muliplication of two convex and inceasing functions is convex 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?
 A: 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.
A: 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}
