# Prove this is a bounded linear operator and find its operator norm?

I have a map $$A:(C[0,1], || \centerdot ||_\infty) \rightarrow \mathbb R, Ax = x(0) \forall x \in C[0,1]$$ and need to prove it's a bounded linear operator, and find its operator norm. I've tried applying the definition but intuitively this seems wrong to me, as $sup_{x \in c[0,1]}|x|$ and $sup|x(0)|$ seem pretty much impossible to relate. What am I missing?

$|x(0)|$ is a real number which is necessarily smaller than $\sup_{t\in [0,1]}|x(t)|$.
Indeed, we have for each $t_0\in [0,1]$ that $|x(t_0)|\leqslant \sup_{t\in [0,1]}|x(t)|$ (the supremum is the least upper bound, in particular an upper bound).
We thus have $|Ax|\leqslant \lVert x\rVert_\infty$ for all $x\in C[0,1]$, which proves that the norm is smaller than $1$. There is actually equality: take $x$ such that $x(t)=1$ for each $t$.
• I'm sorry, can you please clarify why exactly $|x(0)|$ needs to be smaller than the supremum? This is the one thing that doesn't seem clear to me.
• See edit. ${}{}$ Dec 4, 2013 at 18:58