 Solution

For the diagonal operator on sequence space,Equality holds at the first coordinate vector, so $\Vert K\Vert =1$ and $K$ is a bounded linear operator. Every diagonal entry is nonzero, giving the null space $\mathcal{N}(K)=\{0\}$. A datum has a preimage exactly when its weighted coordinates belong to the l2 sequence space. HenceFinite sequences lie in the range and are dense in ${\ell }^2$, so $\mathcal{R}(K{)}^{\perp }=\{0\}$ andThe sequence $f^{(n)}=e_n/n$ converges to zero in the data norm, but $K^{†}{f}^{(n)}=e_n$ has norm one. Thus $K^{†}$ is unbounded and discontinuous. In particular the dense range is not closed; the datum $f_j=1/j$ also proves that it is a proper subset of ${\ell }^2$.