Norm could be achieved f=1?

Quote:

Originally Posted by veljko

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

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.