why are logarithmically convex functions convex? I know that the converse is not true; there are convex functions that are not logarithmically convex. But how can I prove that a logarithmically convex function is convex? I tried to use the definition of convex functions directly but that doesn't seem to work. Could anyone help me?
 A: The composition of a convex function $f$ with a convex, increasing function $g: \mathbb{R} \to \mathbb{R}$ is convex: since
$$ f(tx+(1-t)y) \leq t f(x) + (1-t) f(y), $$
we have
$$ g\big(f(tx+(1-t)y)\big) \leq g\big(t f(x) + (1-t) f(y)\big) $$
because $g$ is increasing, and then
$$ g\big(t f(x) + (1-t) f(y)\big) \leq t g(f(x))+(1-t)g(f(y)) $$
because $g$ is convex.
The exponential is such a function, so if $f$ is convex, so is $\exp(f)$. Equivalently, if $\log{F}$ is convex, so is $F$.
A: Note: The answer from Chappers is complete (+1).  I'd like to add an illustration for the special case where the composition $h=g \circ f$ is of differentiable functions. Then by applying the chain rule twice, $$h''(x) =g'[f(x)]f''(x)+\left[f'(x)\right]^2g''[f(x)]$$  Under the conditions that Chappers mentions ($g'>0,f''>0,g''>0$), we can immediately see that $h''>0$, showing convexity. In the particular case of g being the exponential
$$h''(x) =e^{f(x)}\left(f''(x)+\left[f'(x)\right]^2\right)$$
Always positive, showing the log-convex $h$ is convex.
