Norm could be achieved f=1?

• Apr 23rd 2010, 04:15 AM
veljko
Norm could be achieved f=1?
$\displaystyle f:C[0,1] -> C$
$\displaystyle f(x)=\sum_{j=1}^{\infty} \frac{(-1)^j}{2^j} x\Bigl(\frac{1}{j}\Bigr)$.

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 $\displaystyle ||f|| = 1$ ?
• Apr 23rd 2010, 07:31 AM
Focus
Quote:

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

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 $\displaystyle ||f|| = 1$ ?

Is the map $\displaystyle f: C[0,1] \rightarrow \mathbb{C}$?

If so, then what you want is $\displaystyle x(1/j)=(-1)^j$. 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.