If $J=\left\lbrace f\in B(\ell^2): f^*(e_1)=0 \right\} $, is the set $\overline{\langle \{ f^{*}_{1}f_2: f_1,f_2\in J\}\rangle } $ equal to $B(\ell^2)$?
( $\ell ^2 $ is the Hilbert space $(\ell^2, \lVert.\rVert _2) $ and $B(\ell^2) $ is the C$^*$-algebra of bounded operators on $\ell^2$ and $e_1=(1,0,0,0,\ldots )$ and the symbol $\langle .\rangle $ means the linear span of inside it )
I know $J$ is a closed right ideal and $\langle J,J \rangle $ is two sided ideal; and $ \langle J,J\rangle \supseteq J$ with the properties of approximate unit, But $\langle J,J\rangle =?$
Yes, and you don't even need to take the span nor the closure. That is, $$ J_0=\{f_1^*f_2: f_1,f_2\in J\}=B(H). $$ Let $S$ be the unilateral shift. Then $S^*e_1=0$, so $S\in J$. Thus $ST\in J$ for all $T\in B(H)$. And then $$ T=(T^*)^*S^*S=(ST^*)^*S\in J_0. $$