Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 326 / 1 / ii

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- Solution ii

Solution

The given map is the unilateral shift operator. Its adjoint operator is the left shift operator, since

⟨ K u, f ⟩ = \sum_{j \geq 1} u_{j} \overline{f_{j + 1}}, K^{*} f = (f_{2}, f_{3}, \dots) .

The shift is an isometry with trivial kernel and closed range

{f : f_{1} = 0}

. Its range has orthogonal complement

span {e_{1}}

, so every square-summable datum is admissible. Consequently,

D (K^{†}) = ℓ^{2}, K^{†} (f_{1}, f_{2}, \dots) = (f_{2}, f_{3}, \dots) = K^{*} f .

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.

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