Logarithmically convex function

Let be a convex subset of a real vector space, and let be a function taking non-negative values. Then is: • Logarithmically convex if {\log} \circ f is convex, and • Strictly logarithmically convex if {\log} \circ f is strictly convex. Here we interpret \log 0 as -\infty. Explicitly, is logarithmically convex if and only if, for all and all , the two following equivalent conditions hold: :\begin{align} \log f(tx_1 + (1 - t)x_2) &\le t\log f(x_1) + (1 - t)\log f(x_2), \\ f(tx_1 + (1 - t)x_2) &\le f(x_1)^tf(x_2)^{1-t}. \end{align} Similarly, is strictly logarithmically convex if and only if, in the above two expressions, strict inequality holds for all . The above definition permits to be zero, but if is logarithmically convex and vanishes anywhere in , then it vanishes everywhere in the interior of . Equivalent conditions If is a differentiable function defined on an interval , then is logarithmically convex if and only if the following condition holds for all and in : :\log f(x) \ge \log f(y) + \frac{f'(y)}{f(y)}(x - y). This is equivalent to the condition that, whenever and are in and , :\left(\frac{f(x)}{f(y)}\right)^{\frac{1}{x - y}} \ge \exp\left(\frac{f'(y)}{f(y)}\right). Moreover, is strictly logarithmically convex if and only if these inequalities are always strict. If is twice differentiable, then it is logarithmically convex if and only if, for all in , :f''(x)f(x) \ge f'(x)^2. If the inequality is always strict, then is strictly logarithmically convex. However, the converse is false: It is possible that is strictly logarithmically convex and that, for some , we have f''(x)f(x) = f'(x)^2. For example, if f(x) = \exp(x^4), then is strictly logarithmically convex, but f''(0)f(0) = 0 = f'(0)^2. Furthermore, f\colon I \to (0, \infty) is logarithmically convex if and only if e^{\alpha x}f(x) is convex for all \alpha\in\mathbb R. ==Sufficient conditions==

If f_1, \ldots, f_n are logarithmically convex, and if w_1, \ldots, w_n are non-negative real numbers, then f_1^{w_1} \cdots f_n^{w_n} is logarithmically convex. If \{f_i\}_{i \in I} is any family of logarithmically convex functions, then g = \sup_{i \in I} f_i is logarithmically convex. If f \colon X \to I \subseteq \mathbf{R} is convex and g \colon I \to \mathbf{R}_{\ge 0} is logarithmically convex and non-decreasing, then g \circ f is logarithmically convex. ==Properties==

A logarithmically convex function f is a convex function since it is the composite of the increasing convex function \exp and the function \log\circ f, which is by definition convex. However, being logarithmically convex is a strictly stronger property than being convex. For example, the squaring function f(x) = x^2 is convex, but its logarithm \log f(x) = 2\log |x| is not. Therefore the squaring function is not logarithmically convex. ==Examples==

• f(x) = \exp(|x|^p) is logarithmically convex when p \ge 1 and strictly logarithmically convex when p > 1. • f(x) = \frac{1}{x^p} is strictly logarithmically convex on (0,\infty) for all p>0. • Euler's gamma function is strictly logarithmically convex when restricted to the positive real numbers. In fact, by the Bohr–Mollerup theorem, this property can be used to characterize Euler's gamma function among the possible extensions of the factorial function to real arguments. ==See also==