# When is $\{ x \ge 0 | f(x) \le 0\}$ path-connected?

I'm trying to determine the conditions on $f : \mathbb{R}^n_{\ge 0} \to \mathbb{R^n}$ under which $\{ x \ge 0 | f(x) \le 0 \}$ is path-connected. We can assume that $f$ is continuous and concave.

Is the co-domain of $f$ really meant to be $\mathbb{R}^n$ or that was a typo for $\mathbb{R}$? In the former case, what do you mean by concave? Note that a sublevel set of a (real valued) concave function is the complement of a convex set, so it is path-connected -at least for n≥2. –  Pietro Majer Oct 8 '12 at 19:13
Yes, the co-domain of $f$ is meant to be $\mathbb{R}^n$. By concave, I mean that it is concave on each index: for any $\lambda \in [0, 1]$, $f(\lambda x + (1 - \lambda)y) \ge \lambda f(x) + (1 - \lambda) f(y)$ (the inequality is pointwise). I think the sublevel set is only necessarily path-connected if the domain of $f$ is $\mathbb{R}^n$, not $\mathbb{R}^n_{\ge 0}$. –  user21816 Oct 8 '12 at 19:21
A bad example is $f(x)=-\langle x,v\rangle^2$, in this case the set is not connected even in $\mathbb R^n$. –  Anton Petrunin Oct 8 '12 at 19:56