Give an example of $T\in B(l^2)$ such that $\sigma (T)=S^1$ . Where $\sigma (T) =\{ \lambda\in \mathbb C : \lambda I-T \notin GL(B(l^2))\}$
I began with choosing a countable dense subset $\{a_n\}_{n=1}^\infty$ of $S^1$. Define $T:l^2\to l^2 $ as $T((x_i))= (a_i x_i)$ . Clearly $T$ is bounded operator on $l^2$ . Now I claim that $\sigma (T)= S^1$ . Given $\lambda\in S^1$ , we have $(T-\lambda I)((x_i))=((a_i-\lambda)x_i )$. For $T-\lambda I$ to be invertible , $\inf_|a_i-\lambda| $ should be non zero. Clearly $\{a_i\}$ being dense in $S^1$ , we have have $\inf |a_i-\lambda|=0 $. Hence $\lambda\in \sigma(T)$ , hence $S^1\subset\sigma(T)$. Similarly we can show other way inclusion. Am I constructing the example correct?
That's a perfectly good example.
The standard example is different: Note that $\ell_1(\Bbb Z)$ is isomorphic to $L^2(S^1)$. Define $T:L^2(S^1)\to L^2(S^1)$ by $Tf(z)=zf(z)$. (The same as a right shift on $\ell_2(\Bbb Z)$.)