Let $\{e_n\}$ be the usual basis for $l^2$ and $\{\alpha_n\}$ be a bounded sequence of scalars. For all n, define $Ae_n=\alpha_n e_n$ on $l^2$. Show that $||A||=\sup\alpha_n$.
I can show $||A||\leq\sup\alpha_n$ easily. My problem is showing $||A||\geq\sup\alpha_n$. I can not find a suitable element of $l^2$ for it.
First, you mean $\|A\|=\sup_n |\alpha_n|$, not $\sup_n \alpha_n$. Some of the $\alpha_n$ could be negative.
$A e_n = \alpha_n e_n$, so $\|A\|\ge |\alpha_n|$ for all $n$. It then follows $\|A\|\ge \sup_n |\alpha_n|$. To see this, just note that if $\|A\|=\sup_n|\alpha_n|-\epsilon$ for some $\epsilon>0$, then there would be an $m$ such that $\|A\|<\alpha_m$ (by definition of $\sup$), which is a contradiction.
