7
$\begingroup$

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.

$\endgroup$

3 Answers 3

10
$\begingroup$

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.

$\endgroup$
3
4
$\begingroup$

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 $x$. $\quad\Box$

$\endgroup$
1
$\begingroup$

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 $(f+g)_\lambda = f_\lambda + g_\lambda$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.