 Solution

The given map is the unilateral shift operator. Its adjoint operator is the left shift operator, sinceThe shift is an isometry with trivial kernel and closed range $\{f:f_1=0\}$. Its range has orthogonal complement $\mathrm{span}\{e_1\}$, so every square-summable datum is admissible. Consequently,Indeed $K^{†}K=I$, while $K{K}^{†}f=f-f_1{e}_1$ 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.