= Solution
For the <diagonal operator on sequence space>,
$$
\|Ku\|_2^2=\sum_{j\geq1}\frac{|u_j|^2}{j^2}\leq\|u\|_2^2.
$$
Equality holds at the first coordinate vector, so $\|K\|=1$ 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>. Hence
$$
\boxed{\mathcal R(K)=\left\{f\in\ell^2:\sum_{j\geq1}j^2|f_j|^2<\infty\right\},\qquad (K^\dagger f)_j=jf_j.}
$$
Finite sequences lie in the range and are dense in $\ell^2$, so $\mathcal R(K)^\perp=\{0\}$ and
$$
\boxed{\mathcal D(K^\dagger)=\mathcal R(K),\qquad\mathcal N(K)=\{0\}.}
$$
The <sequence> $f^{(n)}=e_n/n$ converges to zero in the data <norm>, but $K^\dagger f^{(n)}=e_n$ has <norm> one. Thus \b[$K^\dagger$ 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 $\ell^2$.
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