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Math Help - Proof (metric spaces)

  1. #1
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    Proof (metric spaces)

    Hello everybody,
    I am stuck with a proof of which I think that I have an idea of how to solve it.

    I have given two vectors x,y \in R^k with k >= 3. I know that |x-y| = d > 0 and r > 0.

    Now I am supposed to show a couple of things and what I am left to do is to prove that there is only one z \in R^k s.t. 2r = d and |z-x| = |z-y| = r.

    OK, now here is what I have done. I know from the Schwarz inequality that |x-z + z - y| = |x-z| + |z-y| is possible in general if I chose the right z. I get this same equality if I set the given definition of d into my other constraint. Also should I be able to solve for x since I have one unknown z and three requirements for z. However, whenever I solve for z I get some nonsence. I am not exactly sure about what I am doing wrong or how else I can show that there is only one z for that all these requirements hold. Can one of you help me. Even a hint would be more than enough. Thanks a ton! Cheers.
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by raphw View Post
    Hello everybody,
    I am stuck with a proof of which I think that I have an idea of how to solve it.

    I have given two vectors x,y \in R^k with k >= 3. I know that |x-y| = d > 0 and r > 0.

    Now I am supposed to show a couple of things and what I am left to do is to prove that there is only one z \in R^k s.t. 2r = d and |z-x| = |z-y| = r.

    OK, now here is what I have done. I know from the Schwarz inequality that |x-z + z - y| = |x-z| + |z-y| is possible in general if I chose the right z. I get this same equality if I set the given definition of d into my other constraint. Also should I be able to solve for x since I have one unknown z and three requirements for z. However, whenever I solve for z I get some nonsence. I am not exactly sure about what I am doing wrong or how else I can show that there is only one z for that all these requirements hold. Can one of you help me. Even a hint would be more than enough. Thanks a ton! Cheers.
    What condition needs to be satisfied for equality to hold in the Cauchy-Schwarz inequality? Use that additional information about x-z and z-y to determine z.
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  3. #3
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    I am not sure if I got your hint. However, here is what I have done.
    First, for convenience, I defined: x' = x - y and y' = y - y = 0 what should, in my opinion, only "shift" the problem but not change it.

    I know that: |z| = |z - x'| = d/2 and therefore {x'}^2 = 2zx'. Since d^2 = |x'|^2 = 2zx', I know that z = d^2/2 * x'^{-1} and can therefore not be any other value.

    Q.E.D.? I'm still confused.
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  4. #4
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    Quote Originally Posted by raphw View Post
    I am not sure if I got your hint. However, here is what I have done.
    First, for convenience, I defined: x' = x - y and y' = y - y = 0 what should, in my opinion, only "shift" the problem but not change it.

    I know that: |z| = |z - x'| = d/2 and therefore {x'}^2 = 2zx'. Since d^2 = |x'|^2 = 2zx', I know that z = d^2/2 * x'^{-1} and can therefore not be any other value.

    Q.E.D.? I'm still confused.
    No, that's no correct, because the inverse of x' x'^{-1} is not generally defined. x' is a vector. Vectors don't usually have multiplicative inverses.
    You have to use the Cauchy-Schwarz inequality here to conclude that x' and z must be collinear:
    d^2 = |x'|^2 = 2zx' => d^2/2 = zx'
    C-S inequality says that zx' \leq |z||x'| with equality iff z and x' are collinear. In this case we have equality . So there is a \lambda \in \mathbb{R}, such that x' \lambda = z.
    From this it follows that |x'| |\lambda| =|z| => d |\lambda| = d/2 => \lambda = +-\frac{1}{2}. So z= \frac{1}{2} x' is the only possible value for z.
    qed
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