# What is the norm of this operator?

Consider the Banach space $B=C[-1,1]$, with $\sup$ norm, for $f\in B$ define $\tilde f(x)=f(|x|)$, $T:B\rightarrow B, T(f)=\tilde f$ we need to show $T$ is a bounded linear operator on $B$, what is $||T||?$

$T(cf+g)= (f+g)(|x|)=cf(|x|)+ g(|x|)=c\tilde f+\tilde g$ so $T$ is linear as we know continous functions over compact set is bounded so clearly $T$ is bounded?

I am not able to determine $||T||$.

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The domain of $T$ is $B$ not $[-1,1]$. –  Rudy the Reindeer Nov 2 '12 at 8:31

$$\|T\| = \sup_{\|f\|_\infty = 1, f \in B} \|Tf\|_\infty = \sup_{\|f\|_\infty = 1, f \in C[0,1]} \|f\|_\infty = 1$$
Hence $T$ is bounded.
just to recall myself, $||T||=sup_{||x||=1}||T(x)||$? –  La Belle Noiseuse Nov 2 '12 at 8:40