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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 $\displaystyle W$ of $\displaystyle C[0,1]$. Given $\displaystyle \delta>0$ i put:

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

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

    My thought-process: I know that $\displaystyle B(\bf{0},\delta)$ signifies a ball centered at $\displaystyle (0,0)$ and radius $\displaystyle \delta>0$. This sets our limit, i.e. $\displaystyle g$ may not exceed this limit. So what I basically think is that, if I can show that $\displaystyle |g(x)-g_{k}(x)|<\delta$ ($\displaystyle k\in\mathbb{N}$) then I would have shown that $\displaystyle g\in B(\bf{0},\delta)$ (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 $\displaystyle C[0,1/2]$ instead of $\displaystyle C[0,1]$.

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

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

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