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Math Help - Norm could be achieved f=1?

  1. #1
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    Norm could be achieved f=1?

    f:C[0,1] -> C
    f(x)=\sum_{j=1}^{\infty} \frac{(-1)^j}{2^j} x\Bigl(\frac{1}{j}\Bigr).

    Now is:
    |f(x)| \leq \sum_{j=1}^{\infty} \frac{1}{2^j} |x\Bigl(\frac{1}{j}\Bigr)| \leq ||x||_{\infty} \sum_{j=1}^{\infty}\frac{1}{2^j} =||x||_{\infty}.

    Than: ||f|| \leq 1!
    How could I prove that it can be achieved ||f|| = 1 ?
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  2. #2
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    Quote Originally Posted by veljko View Post
    f:C[0,1] -> C
    f(x)=\sum_{j=1}^{\infty} \frac{(-1)^j}{2^j} x\Bigl(\frac{1}{j}\Bigr).

    Now is:
    |f(x)| \leq \sum_{j=1}^{\infty} \frac{1}{2^j} |x\Bigl(\frac{1}{j}\Bigr)| \leq ||x||_{\infty} \sum_{j=1}^{\infty}\frac{1}{2^j} =||x||_{\infty}.

    Than: ||f|| \leq 1!
    How could I prove that it can be achieved ||f|| = 1 ?
    Is the map f: C[0,1] \rightarrow \mathbb{C}?

    If so, then what you want is  x(1/j)=(-1)^j. You can linearly interpolate the points in between to give you a continuous map, and when you do so the norm of x is obviously 1.
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