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Math Help - Equivalent Norms

  1. #1
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    Equivalent Norms

    Dear Colleagues,

    Could you please help me in solving the following problem:

    Let X be the space of all x such that x=(x_{1}, x_{2}, ..., x_{n}), and x_{1}, x_{2}, ..., x_{n} are numbers, and ||.|| be any norm on X, and define the norm ||.||_{2} on X to be ||x||_{2}=(|x_{1}|+|x_{2}|+...+|x_{n}|)^{1/2}. We want to show directly that there is a>0 such that a||x||_{2}\leqslant ||x|| for any x\in X, but without using the fact that all norms on a finite dimensional vector space are equivalent. Instead we want to use the fact that " A continuous mapping of a compact subset M of a normed space X into the real numbers R assumes a maximum and a minimum at some points of M ".

    I think the continuous mapping in my problem will be the norm but which norms will be used and what is the compact set.


    Best Regards.

    Raed.
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  2. #2
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    Hint: It's enough to show this for x such that ||x||=1.
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  3. #3
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    Quote Originally Posted by ymar View Post
    Hint: It's enough to show this for x such that ||x||=1.
    Thank you for your reply. In fact, ||x||=1 is not closed so not compact.

    Regards.
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  4. #4
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    Quote Originally Posted by raed View Post
    In fact, ||x||=1 is not closed so not compact.
    No, the set  \{x\in X:\|x\|=1\} is closed, and ymar's hint was a good one.

    It's probably best to start with the 2-norm, and to look at the set  \{x\in X:\|x\|_2=1\}, which is closed and bounded (as a subset of n-dimensional euclidean space) and therefore compact. You need to show that the function x\mapsto \|x\| is continuous on this space and hence attains its upper and lower bounds. The tricky part is to do this without implicitly assuming the result that you are trying to prove.
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  5. #5
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    raed, why do you think that that set isn't closed?
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