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?
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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 $F$.
Chappers
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Good answer, but you need to reverse all the ≥ to ≤ to be in accord with the definition of convex. As written, you have proved that under exactly the same condition (g is increasing), the result is true that the composition of two concave functions is concave! – Mathemagical Dec 29 '17 at 07:11
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@Mathemagical whoops, fixed. Thanks. – Chappers Dec 29 '17 at 14:26
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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.
Mathemagical
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