Let $(a,b)$ be an open interval.

Let $f,g:(a,b)\rightarrow \mathbb{R}$ be positive log-convex functions.

How to prove that $f+g$ is log-convex?

I am reading a proof using quadratic forms, but I'm not really familiar with quadratic forms, so I don't get the proof. Please help.


3 Answers 3


First note that a function $f$ is log-convex is equivalent to $$f(\theta x+(1-\theta) y)\leq f(x)^\theta f(y)^{1-\theta}$$ for all $x,y\in [a,b]$ and $0\leq \theta\leq 1$. Since $f$ and $g$ are log-convex, we have the following estimate: \begin{align*} f(\theta x+(1-\theta) y)+g(\theta x+(1-\theta) y)&\leq f(x)^\theta f(y)^{1-\theta}+g(x)^\theta g(y)^{1-\theta} \end{align*} To complete the proof, we need to show $$f(x)^\theta f(y)^{1-\theta}+g(x)^\theta g(y)^{1-\theta}\leq (f(x)+g(x))^\theta(f(y)+g(y))^{1-\theta}.$$ The claim follows by combining the above two inequalities together.

Set $a=f(x),b=f(y),c=g(x),d=g(y)$, then we need to show $$a^\theta b^{1-\theta}+c^\theta d^{1-\theta}\leq (a+c)^\theta(b+d)^{1-\theta}.$$ By dividing $(a+c)^\theta$ and $(b+d)^{1-\theta}$ on both sides, we may assume that $a+c=b+d=1$. Thus it suffices to show $$a^\theta b^{1-\theta}+c^\theta d^{1-\theta}\leq 1$$ for $a+c=b+d=1$. By Young's inequality, we see that $$a^\theta b^{1-\theta}\leq \theta a+(1-\theta)b,$$ Similarily, we have $$c^\theta d^{1-\theta}\leq \theta c+(1-\theta)d.$$ Combing these two inequalities together, we obtain $$a^\theta b^{1-\theta}+c^\theta d^{1-\theta}\leq \theta(a+c)+(1-\theta)(b+d)=1.$$ The claim then follows.


Geometrically speaking, log-convexity of $f$ means: for every $x_1,x_2\in (a,b)$, with $x_1<x_2$, there exist real numbers $\alpha$ and $\beta$ such that $$f(x) \le e^{\alpha x+\beta},\quad x_1\le x\le x_2\tag{1}$$ with equality at both endpoints. This is simply the secant line description of convexity applied to $\log f$.

Keeping $x_1,x_2$ as above, record the log-convexity of $g$ as well: $$g(x) \le e^{\gamma x+\delta},\quad x_1\le x\le x_2\tag{2}$$ with equality at both endpoints.

Add (1) and (2): $$ f(x) +g (x) \le e^{\alpha x+\beta}+e^{\gamma x+\delta} ,\quad x_1\le x\le x_2 \tag{3}$$ with equality at both endpoints.

We want to replace the right hand side of (3) with a single exponential function that agrees with $f+g$ at $x_1,x_2$. To this end, we need to show that $e^{\alpha x+\beta}+e^{\gamma x+\delta}$ is log-convex. Assuming without loss of generality that $\alpha\ge \gamma$, we have $$ \frac{d}{dx}\log\left(e^{\alpha x+\beta}+e^{\gamma x+\delta}\right) =\frac{\alpha e^{\alpha x+\beta}+\gamma e^{\gamma x+\delta}}{e^{\alpha x+\beta}+e^{\gamma x+\delta}} =\gamma+\frac{\alpha -\gamma }{1+e^{(\gamma-\alpha) x+\delta-\beta}} $$ which is evidently an increasing function of $x$. $\quad\Box$


A function $f: (a, b) \to (0, \infty)$ is log-convex if and only if $$ f_\lambda(x): (a, b) \to (0, \infty), f_\lambda(x) = e^{\lambda x}f(x) $$ is convex for all $\lambda \in \Bbb R$ (see for example Characterization of log-convexity).

This immediately implies that the sum of two positive log-convex functions is again log-convex, since $(f+g)_\lambda = f_\lambda + g_\lambda$.


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