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Math Help - [SOLVED] Convex Sets

  1. #1
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    [SOLVED] Convex Sets

    Let u be a fixed point in R^n and let c be a fixed real number. Prove that each of the following three sets is convex:
    a.
    b.
    c.
    ================================================== ===========
    a. Let

    By definition of convex, we claim that the line segment is contained in S.
    Since by definition of a scalar product, then



    The problem I have is that in R, the set (a) would be a line greater than c, set (b) would be a point, and set (c) would be a line lessthan c.
    If in R^2, then the set (a) be various regions depending on the signs for ?
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello

    In R^2, for v(x,y), the condition can be written so the corresponding region is a half plane. (bounded below* by the straight line ) Similarly, if u_1 or u_2=0 than the region is either a half plane, either the whole plane, either the empty set : even if the regions are various, it's always the same kind of domain : a (half) plane.

    To answer the first question, you can proceed this way : For any two points v and w in S and for any real number t in ]0,1[, show that t*v+(1-t)*w is in S that is to say that .

    Does it help ?

    (*) : Is their a more suitable expression than "bounded below" that could be used in this case ?
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  3. #3
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    Hi flyingsquirrel,
    Yes, it helps a lot.
    1. First, I was curious about this notation :
    ]0,1[
    2.
    (*) : Is their a more suitable expression than "bounded below" that could be used in this case ?
    Since the plane is bounded below by a line, then the plane would be the greatest lower bound or infimum, correct?

    3. So for , I need to show that belongs in <u,v>

    that is



    which can be written as



    Then factor out the "tv+(1-t)w gives



    Divide both sides by "tv+(1-t)w yields



    Am I on the right track?
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Paperwings View Post
    1. First, I was curious about this notation :
    That's the interval (0,1). (It's a notation used in France and probably in some other countries)

    2.
    Since the plane is bounded below by a line, then the plane would be the greatest lower bound or infimum, correct?
    I don't know. I was just looking for another expression than "bounded below" which doesn't sound correct. (by the way, a greatest lower bound is a real number not a plane or a point of R^3)

    3. So for , I need to show that belongs in <u,v>
    and <u,v> are real numbers, can you say that a real number belong to another ? What you meant is "I need to show that if v and w are in S then the points tv+(1-t)w belong to S too"

    that is

    Where is u ?

    which can be written as

    If u_1,..., u_n are the coordinates of u, (that is real numbers) this is wrong. It'd be true if you had written [tv+(1-t)w]_1 * u_1 + ... + [tv+(1-t)w]_n * u_n where [tv+(1-t)w]_i is the i-th coordinate of the vector tv+(1-t)w.

    Then factor out the "tv+(1-t)w gives



    Divide both sides by "tv+(1-t)w yields
    You're dividing by a vector here ! (hopefully you did not do what you wrote in this case )



    Am I on the right track?
    I'm not sure. The only thing we know is that . How do you plan to use that ? As we don't know anything about u, what are you going to do with u_1, u_2, ... u_n ?
    Last edited by flyingsquirrel; July 28th 2008 at 08:04 AM. Reason: modified meaningless sentecne
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  5. #5
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    First, thank you for your precious time on this problem and the other problem.

    I think I got this!

    Since we know that

    For the scalar product has linearity, then for some alpha, and beta are any real numbers:



    In, this case we need to show that

    Let and

    then ( I switched v = u, and u = v to correspond with the problem)



    Since scalar products are symmetric, then by your hint , then:



    since and such that

    Similarly, this can be done for the other two sets.
    Last edited by Paperwings; July 28th 2008 at 08:04 AM.
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  6. #6
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Paperwings View Post
    I think I got this!
    I think you got it too. Well done !
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