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Math Help - Show that ||x||= d(x,0) defines a norm on V

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    Show that ||x||= d(x,0) defines a norm on V

    Let V be a vector space, and let d be a metric on V satisfying d(x,y) = d(x-y,0) and  d(ax,ay) = |a|d(x,y) for every x,y \in V and every scalar a. Show that ||x||=d(x,0) defines a norm on V (that has d as it standard metric). Give an example of a metric on the vector space R that fails to be associated with a norm in this way.
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    MHF Contributor Drexel28's Avatar
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    Re: Show that ||x||= d(x,0) defines a norm on V

    Quote Originally Posted by wopashui View Post
    Let V be a vector space, and let d be a metric on V satisfying d(x,y) = d(x-y,0) and  d(ax,ay) = |a|d(x,y) for every x,y \in V and every scalar a. Show that ||x||=d(x,0) defines a norm on V (that has d as it standard metric). Give an example of a metric on the vector space R that fails to be associated with a norm in this way.
    What have you tried? Is there a particular place you are having trouble?
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    Re: Show that ||x||= d(x,0) defines a norm on V

    for a counter-example, use the discrete metric. which hypothesis does this fail to satisfy, and why does it fail to yield a norm?
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    Re: Show that ||x||= d(x,0) defines a norm on V

    Quote Originally Posted by Drexel28 View Post
    What have you tried? Is there a particular place you are having trouble?
    I have trouble starting the proof, do I just use the definition of norm to show that d(x,0)=|x-0| satisfy the properites of norm?
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    MHF Contributor Drexel28's Avatar
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    Re: Show that ||x||= d(x,0) defines a norm on V

    Quote Originally Posted by wopashui View Post
    I have trouble starting the proof, do I just use the definition of norm to show that d(x,0)=|x-0| satisfy the properites of norm?
    It seems like you are thinking this norm is given. You aren't DEFINING d you are DEFINING \|\cdot\|, right? So, you need to show that the norm \|\cdot\| defined by \|x\|=d(x,0) is really a norm. For example \|ax\|=d(ax,0)=d(ax,a0)=|a|d(x,0)=|a|\|x\|, etc.
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    Re: Show that ||x||= d(x,0) defines a norm on V

    ok, so you've proved d is positive scalable, but is it subadditive? does it separate points? i don't want any of those cheap semi-norms.
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