# Under what circumstances is the bounded set of a concave function connected?

Let $f : \mathbb{R}^n_{\ge 0} \to \mathbb{R}^n$ be a concave function ($f(\lambda x + (1 - \lambda)y) \ge \lambda f(x) + (1 - \lambda)f(y)$ when $\lambda \in [0,1]$) with $f(0) = 0$. Fix some $a \in \mathbb{R}^n_{\ge 0}$ satisfying $f(a) \ge 0$. I am interested in the set $P =\{x \ge 0 \text{ }|\text{ } 0 \le f(x) \le f(a)\}$. Clearly, this set contains $0$ and $a$. Under what circumstances can we connect $0$ and $a$ with a path contained in $P$?

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Did you mean concave or should that first inequality be reversed? Also how are you ordering $\mathbb{R}^n$? –  Tim Duff Oct 5 '12 at 18:14
Oops, you're completely right - I meant concave. Edited. My inequalities on $\mathbb{R}^n$ are pointwise: $a \ge b$ iff $a_i \ge b_i$ for each $i \in \{1, \dots, n\}$. –  GMB Oct 5 '12 at 18:16