= Solution
The given map is the <unilateral shift operator>. Its <adjoint operator> is the <left shift operator>, since
$$
\langle Ku,f\rangle=\sum_{j\geq1}u_j\overline{f_{j+1}},\qquad K^*f=(f_2,f_3,\ldots).
$$
The shift is an <isometry> with trivial <kernel> and closed range $\{f:f_1=0\}$. Its range has <orthogonal complement> $\operatorname{span}\{e_1\}$, so \b[every square-summable datum is admissible]. Consequently,
$$
\boxed{\mathcal D(K^\dagger)=\ell^2,\qquad K^\dagger(f_1,f_2,\ldots)=(f_2,f_3,\ldots)=K^*f.}
$$
Indeed $K^\dagger K=I$, while $KK^\dagger f=f-f_1e_1$ is the required <orthogonal projection>. The discarded first component is the residual of the <least-squares solution>, not an obstruction to the existence of the <Moore–Penrose inverse of an operator>.
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