 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-2/1/a/ii/solution

First, setting

f = g = y_{n}

in the self-adjoint operator identity shows that

λ_{n} ⟨ y_{n}, y_{n} ⟩_{w}

is real, so every eigenvalue is real. For two eigenfunctions,

λ_{n} ⟨ y_{m}, y_{n} ⟩_{w} = ⟨ y_{m}, L y_{n} ⟩_{w} = ⟨ L y_{m}, y_{n} ⟩_{w} = λ_{m} ⟨ y_{m}, y_{n} ⟩_{w} .

Distinct eigenvalues therefore imply

⟨ y_{m}, y_{n} ⟩_{w} = \int_{a}^{b} y_{m}^{*} y_{n} w d x = 0.

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