Let $a_j\to 0$ and let $T:l^2 \to l^2$ be the operator defined by $ T(s_1,s_2,s_3,...)=(0,a_1s_1,a_2s_2,...)$. Compute the operator norm $||T||$.
The hint of the problem is prove that $T$ is a compact operator and compute the spec of $T$. I proved that $T$ is effectively compact, thus the spec consiste only on the eigenvalues. But $0$ is the only eigenvalue. I don't know if this is useful. At least I proved that $||T||\le \displaystyle \sup_{j\in \mathbb N} |a_j|$.
Thanks!
EDITED: ($a_j \to 0$)
Let $x\in l^2:x=(s_1,s_2,..)$. Then $\frac {\|Tx\|}{\|x\|}=(\frac {\sum_{n=1}^\infty \|a_ns_n\|^2}{\sum_{n=1}^\infty \|s_n\|^2})^{1/2}$
We have that $\frac {\sum_{n=1}^\infty \|a_ns_n\|^2}{\sum_{n=1}^\infty \|s_n\|^2}=\frac {|a_1s_1|^2+|a_2s_2|^2+...}{|s_1|^2+|s_2|^2+...}\leq sup|a_i|^2_{i\in \Bbb N}$ .Thus for $x:\|x\|\leq 1$ we have that $\|T\|\leq sup|a_i|_{i\in \Bbb N}$. If $sup|a_i|_{i\in \Bbb N}=max|a_i|_{i\in \Bbb N}=a_{i_0} $ then take $x=(0,0,...,1(i_0),0,0,...)$ with $1$ in the $i_0$ position and thus $\|T\|= sup|a_i|_{i\in \Bbb N}$
If there is no maximum then take a subsequence of $a_i$ let's say $a_{k_i}\to sup|a_i|_{i\in \Bbb N}$,(this subsequence exists because every bounded sequence has a monotone increasing subsequence which goes to supremum in this occasion). So for every $ε>0$ there is a $n_0\in \Bbb N:\|a_{k_i}- sup|a_i|_{i\in \Bbb N}\|<ε$ for every $i\geq n_0$. For this $i$ if we take his $k_i$ then we can take $x=(0,0,...,1(k_{n_0}),0,0,..)$ with $1$ in the $k_{n_0}$ position and thus $\|T\|= sup|a_i|_{i\in \Bbb N}$.