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You should be able to...

Name: Anonymous 2008-12-27 21:29

Let [IEQ]X \subseteq \Re ^n[/IEQ] be a convex subset. Let [IEQ]f: X \rightarrow \Re [/IEQ] be a convex function. Prove for any [IEQ]a \in \Re[/IEQ], [IEQ]Z = \left\{x\in X \mid f(x) \leq a  \right\} [/IEQ] is a convex subset of [IEQ]\Re ^n[/IEQ]

Name: The Silent Wind of Doom 2008-12-28 16:42

>>5
Well, if n=1, a set in \mathbb{R}^n = \mathbb{R} is convex iff it is a single interval.  So the non-convex function f(x) = \sqrt{x} defined on X=\mathbb{R}^{+} is not convex, but for any a, \{x|\sqrt{x} \le a\} = (0,\sqrt{a}] is convex.

>>6
f(P) \le a and f(Q) \le a since both are in Z. Therefore

tf(P) \le ta
(1-t)f(Q) \le (1-t)a
f(P)+(1−t)f(Q) \le ta+(1−t)a = a

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