A function is said to be quasi-convex if $f(\theta x +(1-\theta)y)\leq \max\{f(x), f(y)\}$ for all $x,y$ and $\theta \in [0,1]$. A function is said to be quasi-linear if $f$ and $-f$ are both quasiconvex (e.g $\log$ or $\tanh$, $\sqrt{x}$ are such functions). Given $f_1$ and $f_2$ is quasi-linear, is it true that $f_1+f_2$ is also quasi-linear?
I know it is not true for quasi-convex functions, but could it be for quasi-linear?
On
Your definition of "quasi-linear" is equivalent to stating that for all $x, y$ and $\theta \in [0, 1]$:
$\begin{align*} \min\{ f(x), f(y)\} &\le f(\theta x + (1 - \theta) y) \le \max\{ f(x), f(y)\} \end{align*}$
Take $f(x) = \sqrt{x}$, $g(x) = \sqrt{1 - x}$. Then $h(x) = f(x) + g(x) = \sqrt{x} + \sqrt{1 - x}$, for $x = 0, y = 1$ it is $\min\{h(0), h(1)\} = \max\{h(0), h(1)\} = 1$, but for $\theta = 1/2$ we have $h(1/2) = 2 \sqrt{1 / 2} = \sqrt{2} > 1$, $h$ isn't quasi-linear.
I think the answer is no, because if you take $f_1(x)=\sqrt{x}$ and $f_2(x)=-\log(x)$ (which are both quasi-convex) then $f_1+f_2$ is quasi-convex but $-(f_1+f_2)$ is not! Therefore it is not quasi-linear