4
$\begingroup$

Let $\Omega$ be a set, e.g $\mathbb{R}^n$ for $n \in \mathbb{N}$. Let $f$ be a convex function from $\Omega$ to $\mathbb{R}$. Let $g$ be a quasi-convex function from $\Omega$ to $\mathbb{R}$.

Is $f+g$ quasi-convex? The proof is not straightforward from using the definition: Let $\lambda \in \left[0,1\right]$. Let $(x,y) \in \Omega$. Then, $\lambda (f+g)(x)+(1-\lambda)(f+g)(y) \leq max(g(x),g(y)) + \lambda f(x)+(1-\lambda)f(y)$. which does not help to conclude anything.

$\endgroup$

1 Answer 1

6
$\begingroup$

The statement is wrong for $\Omega = \mathbb{R}$.

Let $f(x)=-x$ and $g(x)=x-\frac{1}{2}|x|$. $f$ is obviously convex, and $g$ is monotonically increasing, and thus quasi-convex, but their sum $(f+g)(x)=-\frac{1}{2}|x|$ is obviously not quasi-convex.

$\endgroup$
2
  • $\begingroup$ I do not follow you when you say that $g$ is monotonically increasing, thus quasi-convex. Isn't it that the $g$ you propose is quasi concave? $\endgroup$
    – baptiste
    Mar 7, 2018 at 7:01
  • $\begingroup$ Every monotonic function is both quasi-convex and quasi-concave. $\endgroup$
    – ManfP
    Mar 7, 2018 at 10:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .