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]
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Anonymous2008-12-27 21:30
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]
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Anonymous2008-12-27 21:32
Hopefully this is more readable.
Let X \subseteq \Re ^n be a convex subset. Let f: X \rightarrow \Re be a convex function. Prove for any a \in \Re, Z = \left\{x\in X \mid f(x) \leq a \right\} is a convex subset of \Re ^n
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The Silent Wind of Doom2008-12-27 22:14
Let P,Q \in Z. Then for any t \in [0,1]
a = ta + (1-t)a \ge tf(P) + (1-t)f(Q) \ge f(tP + (1-t)Q)
Then tP + (1-t)Q \in X since X is convex, and so is in Z, and therefore Z is convex.
Less homework plox :3.
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Anonymous2008-12-28 11:43
Something more interesting, well slightly so, find a counterexample to the converse,
ie. a function f s.t the condition that for all a { x | F(x)<=a} is convex, but f isn't a convex function
>>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
Err, should be (0,a^2] at the end of the first line.
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Anonymous2008-12-28 17:01
>>7
Ah, I get it now, thanks bro. But that question was just a warm up.
Prove also that if \left\{ f_i \right\}_{i \in I} is a collection of convex functions on a convex subset X \subseteq \Re^n and \left\{ a_i \right\}_{i \in I} is a collection of real numbers, then the set \left\{ x \in X \mid f_i(x) \leq c_i, \forall i \in I\right\} is a convex subset of \Re^n.