 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-2/1/i/b/solution

Let

L y_{n} = λ_{n} y_{n}

. By self-adjointness,

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} .

Taking

m = n

first shows that each eigenvalue is real:

⟨ y_{n}, y_{n} ⟩_{w} > 0

for a nonzero eigenfunction, since

w > 0

. Therefore

(λ_{n} - λ_{m}) ⟨ y_{m}, y_{n} ⟩_{w} = 0.

\int_{a}^{b} y_{m}^{*} (x) y_{n} (x) w (x) d x = 0 (λ_{m} \neq = λ_{n}) .

The weight is essential: this need not be orthogonality under the unweighted integral.

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