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

I'm trying to determine the conditions on $f : \mathbb{R}^n_{\ge 0} \to \mathbb{R^n}$ under which $\{ x | f(x) \le 0\}$ is path-connected. We can assume that $f$ is continuous and concave (i.e. for any $\lambda \in [0, 1]$, $f(\lambda x + (1 - \lambda) y) \ge \lambda f(x) + (1 - \lambda) f(y)$).

Inequalities on $\mathbb{R}^n$ are pointwise: $a \ge b$ iff $a_i \ge b_i$ for each $i$.

Thanks!

Note $L_{\lambda} = \{x\in\mathbf{R}^n\;|\;f(x)\leq \lambda\}$. Asking path-connectedness of only $L_0$ is way too weak. If $f$ is of class $\mathscr{C}^1$ and is coercive then if there is a $\lambda$ such that $L_{\lambda}$ is non path-connected implied the existence of a critical point, this is the so-called "moutain pass theorem". (See for instance the nice book of Aubin and Ekeland, "Applied non-linear analysis".) If $f$ is of class $\mathscr{C}^2$, the Hessian operator of $f$ at this critical point will have at least one positive eigenvalue and at most one negative eigenvalue. (See weak and strong Morse indexes theory for instance.) So if $f$ hasn't such critical point, and is smooth enough, all $L_{\lambda}$ will be path-connected.