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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: Anonymous 2008-12-27 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]

Name: Anonymous 2008-12-27 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

Name: The Silent Wind of Doom 2008-12-27 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.

Name: Anonymous 2008-12-28 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

It's not that hard.

Name: Anonymous 2008-12-28 13:41

>>4
I don't see how you get the first inequality.

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

Name: Anonymous 2008-12-28 16:43

>>7

Err, should be (0,a^2] at the end of the first line.

Name: Anonymous 2008-12-28 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.

Name: The Silent Wind of Doom 2008-12-28 19:04

>>9
Same thing.

For any P,Q \in X, each i \in I, and all t \in [0,1]

a_i = ta_i + (1-t)a_i \ge tf_i(P) + (1-t)f_i(Q) \ge f_i(tP + (1-t)Q)

So tP + (1-t)Q \in \left\{ x \in X \mid f_i(x) \le a_i\right\} for every i. Therefore, etc.

Don't change these.
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