Bounded operator that does not attain its norm 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.
 A: On $\ell^2(\mathbb{N})$, take $T\in B(\ell^2(\mathbb{N}))$ to be the map
$$
(a_n)_{n\in\mathbb{N}}\mapsto \left((1-\frac1n)a_n\right)_{n\in\mathbb{N}}
$$
Since every coefficient is multiplied by scalar less than $1$, $T$ is contractive (i.e. $\|T\|\leq1$). Also, $\|Te_n\|=(1-\frac1n)$, where $e_n$ are the elements in the canonical basis, so $\|T\|=1$. And
$$
\|T(a_n)\|^2=\sum_n|(1-\frac1n)a_n|^2<\sum_n|a_n|^2=\|(a_n)\|_2^2,
$$
so the norm is never attained. 
Edit: another easy example is as follows. Let $H=L^2[0,1]$, and let $T$ be the operator such that 
$$
(Tf)(t)=tf(t).
$$
Then it is easy to see that $\|T\|=1$ (using functions supported near $1$). And, given any $f\in L^2[0,1]$, if $f\ne0$ then 
$$
\|Tf\|_2^2=\int_0^1t^2|f(t)|^2<\int_0^1|f(t)|^2=\|f\|_2^2
$$
(the inequality has to be strict: otherwise we get $\int_0^1(1-t^2)|f(t)|^2=0$, which implies $f=0$). 
A: This is similar to other answers but a bit more general. Consider a sequence $\{c_n\}\in\ell^\infty(\mathbb{Z})$ such that $|c_n|<\sup|c_n|=c$, 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
$$\|Tx\|^2 = \sum_{n\in\mathbb{Z}} |c_nx_n|^2<\sum_{n\in\mathbb{Z}} c^2|x_n|^2= c^2\|x\|^2\tag{1}$$
That is, for non-zero $x$ we have
$$\|Tx\| < \sup|c_n|\|x\| $$
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 $k=n$) leads to
$$\sup_{\|x\|\leq1}\|Tx\|\geq\sup_{n}\|Te_n\|=\sup_n|c_n|$$
and hence
$$\|T\| =\sup|c_n|.\tag{2}$$
Now, (1) and (2) shows that the norm is never attained.
A: Let $T:l^{2}\longrightarrow l^{2}$ gven by $T(x)= \Big( \dfrac{n x_n}{n+1}\Big)_{n=1}^{\infty}$. Notice that
\begin{equation}
\|T(x)\|_{l^{2}} = \left | \sum_{n} \Big( \dfrac{n}{n+1}\Big)^{2}x_{n}^{2} \right |^{1/2} < \Big(\sum_{n} x_{n}^{2}\Big)^{1/2}= \|x_n\|_{l^{2}}.
\end{equation}
Then $T$ is bounded and if $\|x_n\| = 1, \|T(x)\|_{l^{2}} < 1$. But $\|T(e_i)\| = n/(n+1)$ . Hence $\|T\|=1$.
A: I would take a countable subset $(b_n)$ of an ONB, $n\in \mathbb N$. And then define a linear map through $b_n \mapsto \frac{n}{n+1} b_n$ for all vectors in your countable subset. All the other vectors of your ONB shall be mapped to $0$. This can be viewed as a linear map on a dense subset which is bounded by $1$, so there is a bounded continuation on the whole Hilbert space. By this continuation no vector of norm $1$ is mapped onto a vector of norm 1. This can be seen by writing this vector in ONB representation.
A: For an example in $L^2[0,1]$, consider the operator of multiplication by $x$, i.e. $(Tf)(x) = x f(x)$.  
A: In finite dimensional spaces all operator is norm attaining: we use the fact that the unit ball is compact. Then $\{Tx,\lVert x\rVert=1\}$ is bounded because it's the image of a compact set by a continuous map.  
Otherwise, it's not true. Consider $H:=\ell^2(\Bbb N)$. Define $Tx:=\sum_{n\geq 1}\left(1-\frac 1n\right)\langle x,e_n\rangle e_n$ where $e_n$ is the sequence whose $n$-th term is $1$, the others $0$. The norm of $T$ is $1$, but for $x\neq 0$, $$\lVert Tx\rVert^2=\sum_{n\geq 1}\left(1-\frac 1n\right)^2|\langle x,e_n\rangle|^2\leq \sum_{n\geq 1}|\langle x,e_n\rangle|^2=\lVert x\rVert^2.$$
The last inequality isn't an equality, since $\langle x,e_n\rangle\neq 0$ for some $n$.
