$\displaystyle f:C[0,1] -> C$

$\displaystyle f(x)=\sum_{j=1}^{\infty} \frac{(-1)^j}{2^j} x\Bigl(\frac{1}{j}\Bigr)$.

Now is:

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

Than: $\displaystyle ||f|| \leq 1!$

How could I prove that it can be achieved $\displaystyle ||f|| = 1$ ?