TITLE: Bounded operator that does not attain its norm
QUESTION [7 upvotes]: What is a bounded operator on a Hilbert space that does not attain its norm? An example in $L^2$ or $l^2$ would be preferred.
All of the simple examples I have looked at (the identity operator, the shift operator) attain their respective norms.

REPLY [3 votes]: This is similar to other answers but a bit more general. Consider a sequence $c=\{c_n\}\in\ell^\infty(\mathbb{Z})$ such that $|c_n|<\sup|c_n|$, e.g. $c_n$ may be positive and growing to $1$. 
Define $T:\ell^2(\mathbb{Z})\to \ell^2(\mathbb{Z})$ by 
$$T:x=\{x_n\}\mapsto Tx = cx =\{c_nx_n\}.$$
Then, for any $x=\{x_n\}\in\ell^2$, we have
$$\|Tx\|^2 = \sum_{n\in\mathbb{Z}} |c_nx_n|^2<\sum_{n\in\mathbb{Z}} \sup|c_n|^2|x_n|^2= \sup|c_n|^2\|x\|^2\tag{1}$$
that is 
$$\|T\| \leq\sup|c_n|.$$
On the other hand the choice $x=e_n=\{\delta_{kn}\}_{k\in\mathbb{Z}}$, where $\delta_{kn}$ is the Kronecker-$\delta$ (which is 0 whenever $k\ne n$ and 1 if $k=n$) shows that 
$$\sup_{\|x\|\leq1}\|Tx\|\geq\sup_{n}\|Te_n\|=\sup_n|c_n|$$
and hence
$$\|T\| =\sup|c_n|.\tag{2}$$
Now, (1) and (2) shows that the norm is never attained.