3
$\begingroup$

It is well known that in optimization the concept of quasi-convex function is the following: $f: \Omega\subset \mathbb{R}^n \rightarrow \mathbb{R}$ is a quasi convex function if for every $\alpha\in\mathbb{R}$, we have that $level_{\alpha}(f):=\{x\in\Omega; f(x)\leq \alpha\}$ is a convex set.

Every convex function is quasi convex in this sense. Although the converse is not true, as one can see from, for example, the level sets of the Heavyside function.

My question: Which additional conditions make a quasi-convex function $f: \Omega\rightarrow \mathbb{R}$ convex?

One should ask at least that f is locally Lipschitz, since every convex function satisfies this hypothesis.

Thanks for the attention.

$\endgroup$
  • 1
    $\begingroup$ Lipschitz is not enough. Not even differentiability can save you because on $\Bbb R$ any increasing function is quasi-convex. $\endgroup$ – M. Winter Sep 18 '17 at 13:54
  • $\begingroup$ Given a non-convex increasing function $f:\Bbb R\to\Bbb R$, define $g(x)=-f(\|x\|):\Bbb R^n\to\Bbb R$, which is not convex but all level sets are circles. So my previous comment also holds in higher dimensions. I think local restrictions are not enough. $\endgroup$ – M. Winter Sep 18 '17 at 14:12
  • $\begingroup$ That was an excellent example. Thank you very much $\endgroup$ – Renan Assimos Sep 19 '17 at 11:34

Your Answer

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

Browse other questions tagged or ask your own question.