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Math Help - Ball in metric space

  1. #1
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    Ball in metric space

    Let \mathbb{B}((x,y),r) on \mathbb{R}^2 be a ball in euclidean norm. We define a norm ||x|| that ball in this norm satisfes:
    B=\{ x \in \mathbb{R}^2: ||x||<1 \} = ( (-1,1) \times (-1,1) \cup \mathbb{B}((1,0),1) \cup \mathbb{B}((-1,0),1).
    Is this norm determined by a dot product?

    I think that this problem need to use parallelogram law ... but how?
    Also we know that ball in eclidean norm is determined by the dot product. Maybe any draw should solve everything?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Ball in metric space

    Quote Originally Posted by Camille91 View Post
    Let \mathbb{B}((x,y),r) on \mathbb{R}^2 be a ball in euclidean norm. We define a norm ||x|| that ball in this norm satisfes:
    B=\{ x \in \mathbb{R}^2: ||x||<1 \} = ( (-1,1) \times (-1,1) \cup \mathbb{B}((1,0),1) \cup \mathbb{B}((-1,0),1).
    Is this norm determined by a dot product?

    I think that this problem need to use parallelogram law ... but how?
    Also we know that ball in eclidean norm is determined by the dot product. Maybe any draw should solve everything?
    Right, ok, so go on. What exactly is your idea?
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  3. #3
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    Re: Ball in metric space

    Take two vectors  x, y \in B and show that  ||x-y||^2+||x+y||^2=2(||x||^2+||y||^2) ? I don't get it.
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  4. #4
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    Re: Ball in metric space

    Now I see it seems to be simple. This norm doesn't satisfe parallelogram law. Let's take e_1=(1,0) and e_2=(0,1). With so defined ball we know that: e_2, e_1-e_2, e_1+e_2 \in S so  ||e_2||=1, ||e_1+e_2||=1, ||e_1-e_2||=1 but  e_1 \notin S, ||e_1||=a, a \not= 1 . S is the sphere of this ball.
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