When is a log-convex function convex?

Question

I consider the one-dimensional real axis. A function $f$ defined on the axis is log-convex if $f(x)f''(x)\ge (f'(x))^2$ for all $x\in\mathbb{R}$. I know that a convex function may not be log-convex, i.e., $f(x)=x$ and $f(x)=x^2$.

I also know that for log-concavity, a non-negative function that is concave is also log-concave, as $f(x)f''(x)\le 0\le (f'(x))^2$. Similarly I wish to know the conditions on a convex function that make it log-convex.

Answer 1

A log convex function is always convex. Let $g$ denote $\log(f)$ so $e^g = f$. The function $e^g$ is convex if $g$ is convex. (Show that $e^{f(x)}$ is convex.)

In the case when $\log f$ is twice differentiable it is easy to show:

$f'' = (e^g)'' = e^g(g'' + (g')^2) \geq 0$ where the inequality follows from the nonnegativity of a convex functions second derivative.

For the case when $f$ is convex we have that $g'' = \frac{f f'' - (f')^2}{f^2} \propto f f'' - (f')^2$ so $g''$ is positive whenever $f \geq \frac{(f')^2}{f''}$.

Further, $g(\lambda x + (1 - \lambda)y) \leq \log(\lambda f(x) + (1-\lambda)f(y))$ since $f$ is convex.

Thus $g$ is convex whenever $$ \log(f(\lambda x + (1-\lambda)y)) \leq \lambda \log f(x) + (1-\lambda)\log f(y) = \log f(x)^\lambda f(y)^{1 - \lambda} $$ Taking the exponent on both sides gives the condition $$ f(\lambda x + (1-\lambda)y) \leq f(x)^\lambda f(y)^{1 - \lambda}. $$

To get a simple necessity condition, we can consider the special case $\lambda = 0.5$ to get that $$ f^2(\frac{x+y}{2}) \leq f(x)f(y). $$ is a necessary but not sufficient condition. Setting $y=x+2a$ gives another interesting necessary condition: $f(x + a)f(x + a)\leq f(x)f(x+2a)$.

Answer 2

(Log-)convex functions do not need to be twice differentiable, so any proof through $\frac{d^2}{dx^2}$ lacks generality. On the other hand $e^x$ is both a convex and log-convex function, and we may wonder when the composition of two convex functions is convex. Assume that $f(x)$ is convex . Then $g(x)=e^{f(x)}$ is convex iff

$$ g(\lambda x+(1-\lambda) y) \leq \lambda g(x) + (1-\lambda) g(y)\qquad \forall \lambda\in[0,1] $$ which is equivalent to

$$ \exp\left[f(\lambda x+(1-\lambda) y)\right]\leq \lambda e^{f(x)}+(1-\lambda)e^{f(y)}\qquad \forall\lambda\in[0,1]. $$ By the convexity of $f$ we know that $f(\lambda x+(1-\lambda) y)\leq \lambda f(x)+(1-\lambda)f(y)$.
Since $\exp$ is increasing we get $$ \exp\left[f(\lambda x+(1-\lambda) y)\right]\leq \exp\left[\lambda f(x)+(1-\lambda)f(y)\right] $$ unconditionally, and since $\exp$ is convex we get $$\exp\left[\lambda f(x)+(1-\lambda)f(y)\right]\leq \lambda e^{f(x)}+(1-\lambda)e^{f(y)}$$ as wanted. In other terms, if $a(x),b(x)$ are convex functions and $a(x)$ is (weakly) increasing, then $(a\circ b)(x)$ is a convex function, as shown here, too. It follows that any log-convex function is also convex, as the exponential of a convex function.

In general nothing can be said about the composition of a concave function (like $\log$) with a convex function. If we take $f_1(x)=x^2, f_2(x)=e^{x^2}, f_3(x)=x e^{x\sqrt{x}}$, then over $\mathbb{R}^+$ we have that $f_1,f_2,f_3$ are convex but $\log f_1$ is concave, $\log f_2$ is convex and $\log f_3$ is neither convex or concave.