 Solution

The exact condition for this diagonal operator on sequence space isFirst, $T$ is bounded precisely when ${\sup }_n|{\lambda }_n|<\infty$, with operator norm equal to that supremum: the upper bound follows by summing $|{\lambda }_n{x}_n{|}^2$, and the coordinate unit vectors $e_n$ give the lower bound. If the supremum is infinite, choose distinct indices with $|{\lambda }_{n_j}|\ge j$ and set $x_{n_j}=1/j$, other coordinates zero. Then $x\in {\ell }^2$ but $Tx\notin {\ell }^2$, so the displayed prescription would not even define an everywhere-defined map into ${\ell }^2$.

If $T$ is compact but ${\lambda }_n\nrightarrow 0$, infinitely many indices satisfy $|{\lambda }_n|\ge ϵ>0$. Along those indices,contradicting compactness. Conversely, if ${\lambda }_n\to 0$, truncate to $T_{N}x=({\lambda }_1{x}_1,\dots ,{\lambda }_N{x}_N,0,\dots )$. This is a finite-rank compact operator, and $\Vert T-T_N\Vert ={\sup }_{n>N}|{\lambda }_n|\to 0$. To see compactness directly, approximate the image of the unit ball by $T_N$ to within $ϵ/2$, and cover its bounded finite-dimensional image by finitely many $ϵ/2$-balls. Hence $T$'s unit-ball image is totally bounded; its closure is compact because ${\ell }^2$ is complete.