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Thread: Balls (don't know what else to call it)

  1. #1
    Feb 2010

    Balls (don't know what else to call it)

    Hello all,

    I am looking at a supspace W of C[0,1]. Given \delta>0 i put:

    g(x)=\delta\sum_{n=1}^{\infty}\frac{1}{n+1}x^{n} , \hspace{0,5cm} x\in[0,\frac{1}{2}]

    I would like to show that g\in B(\bf{0},\delta).

    My thought-process: I know that B(\bf{0},\delta) signifies a ball centered at (0,0) and radius \delta>0. This sets our limit, i.e. g may not exceed this limit. So what I basically think is that, if I can show that  |g(x)-g_{k}(x)|<\delta ( k\in\mathbb{N}) then I would have shown that g\in B(\bf{0},\delta) (is my understanding correct?). I have tried this but my attempt was not succesful.

    Assistance would be great.

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  2. #2
    Jun 2009
    I suppose you are talking about C[0,1/2] instead of C[0,1].

    To observe that g is in B(0,\delta) you have to show that \|g\|=\|g-0\|<\delta,
    but you can compute that for each 0\leq x\leq\frac{1}{2}

    |g(x)|\leq\delta\sum_{n=1}^{\infty}\frac{1}{n+1}\f  rac{1}{2^n}<\delta\sum_{n=1}^{\infty}\frac{1}{2^n}  =\delta. This certainly implies that the supremum of |g| over [1,1/2] (i.e. the norm of g) is smaller that \delta.

    Notice that these estimates show that the function f is actually continuous by Weierstrass M test.
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