How to prove that the sum of two log-convex functions is log-convex? 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.
 A: 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: 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.
A: 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 $(f+g)_\lambda = f_\lambda + g_\lambda$.
