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Math Help - Prove a set is convex algebraically

  1. #1
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    Prove a set is convex algebraically

    How can I determine if a set is convex without graphing it?

    I have that a set, S, is convex if, given any two points, x_1, x_2 \in S, the line connecting them is entirely within S, or:

    \lambda x_1 + (1-\lambda) x_2 \in S \qquad \forall \lambda \in [0,1]

    Some of the sets in question:
    b. \{(x,y,z) \colon z=\left| y \right|, x \le 3 \}
    c. \{(x,y): y-3x^2=0\}
    d. \{(x,y,z): y \ge x^2, x+2y+z \le 4\}
    e. \{(x,y):x=3, \left| y \right|, \le 4 \}
    f. \{(x,y,z): z= \left| y \right|, x \le 3 \}

    I have to hand in f. I'm not good enough to draw it, but I can envision it. It looks like a "V" going from x=3 down to -\infty. Clearly, it is not convex. But how would I show that mathematically?

    Thanks.
    Last edited by MSUMathStdnt; January 30th 2011 at 02:03 PM.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by MSUMathStdnt View Post
    Some of the sets in question (I can't make the braces in math mode?):

    If you write \{a\}, you'll obtain \{a\} .


    b. { (x,y,z) \colon z=\left| y \right|, x \le 3 }

    It is not convex. Choose:

    x_1=(0,-1,1),\;x_2=(0,1,1),\;\lambda=1/2.


    Fernando Revilla
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