Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f:\mathbb{R}^n\to\mathbb{R}$ be a real-valued, $n$-dimensional function defined as follows:

$$ f(\mathbf{x}) = g(\mathbf{x})h(\mathbf{x}), $$

where $g:\mathbb{R}^n\to\mathbb{R}$ is convex, and $h:\mathbb{R}^n\to\mathbb{R}$ is bounded, for instance $h(\mathbf{x})\in[a,b]$, $a,b\in\mathbb{R}$, $0<a<b$, $\forall\mathbf{x}\in\mathbb{R}^n$.

What is true about the convexity of $f$? What if $g$ is quasi-convex?

Any helpful comment would be nice! Thanks!

share|cite|improve this question

Nothing can be said in general without further conditions on $h$. For instance, if $h(\mathbf{x})=-1$ then $f$ is concave. Even in $n=1$ nothing can be said. Consider for example $g(x)=x^2$ and $h(x)=x$.

Assuming $h$ positive, $0< h(x)< 1$ for instance, does not help. Take for instance $g(x)=1$ and $h(x)=(1+a\sin x)/2$ with $0<a<1$.

share|cite|improve this answer
Yeah, let's suppose that $h$ is positive, for instance, $0<h(\mathbf{x})<1$, $\forall \mathbf{x}\in\mathbb{R}^n$. What then? – nullgeppetto Jan 17 '14 at 15:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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