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

• Apr 17th 2010, 01:59 AM
surjective
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.

Thanks.
• Apr 17th 2010, 03:26 AM
Enrique2
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.