# Norm could be achieved f=1?

• April 23rd 2010, 05:15 AM
veljko
Norm could be achieved f=1?
$f:C[0,1] -> C$
$f(x)=\sum_{j=1}^{\infty} \frac{(-1)^j}{2^j} x\Bigl(\frac{1}{j}\Bigr)$.

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 $||f|| = 1$ ?
• April 23rd 2010, 08:31 AM
Focus
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.