Editing Logarithmically convex function (section)

===Equivalent conditions===
If {{math|''f''}} is a differentiable function defined on an interval {{math|''I'' ⊆ '''R'''}}, then {{math|''f''}} is logarithmically convex if and only if the following condition holds for all {{math|''x''}} and {{math|''y''}} in {{math|''I''}}:
:<math>\log f(x) \ge \log f(y) + \frac{f'(y)}{f(y)}(x - y).</math>
This is equivalent to the condition that, whenever {{math|''x''}} and {{math|''y''}} are in {{math|''I''}} and {{math|''x'' > ''y''}},
:<math>\left(\frac{f(x)}{f(y)}\right)^{\frac{1}{x - y}} \ge \exp\left(\frac{f'(y)}{f(y)}\right).</math>
Moreover, {{math|''f''}} is strictly logarithmically convex if and only if these inequalities are always strict.

If {{math|''f''}} is twice differentiable, then it is logarithmically convex if and only if, for all {{math|''x''}} in {{math|''I''}},
:<math>f''(x)f(x) \ge f'(x)^2.</math>
If the inequality is always strict, then {{math|''f''}} is strictly logarithmically convex.  However, the converse is false: It is possible that {{math|''f''}} is strictly logarithmically convex and that, for some {{math|''x''}}, we have <math>f''(x)f(x) = f'(x)^2</math>.  For example, if <math>f(x) = \exp(x^4)</math>, then {{math|''f''}} is strictly logarithmically convex, but <math>f''(0)f(0) = 0 = f'(0)^2</math>.

Furthermore, <math>f\colon I \to (0, \infty)</math> is logarithmically convex if and only if <math>e^{\alpha x}f(x)</math> is convex for all <math>\alpha\in\mathbb R</math>.<ref>{{harvnb|Montel|1928}}.</ref><ref>{{harvnb|NiculescuPersson|2006|p=70}}.</ref>