Past exam of the analysis course of the University of Oxford / 2011 / 4 / b / iii

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- Solution iii

Solution

The normalization gives

e_{n} (m) = \frac{1}{2 π} \int_{- π}^{π} e^{i (n - m) t} d t = δ_{nm} .

Thus

α_{n} e_{n}

has exactly the coefficients prescribed for

T e_{n}

. Uniqueness yields

T e_{n} = α_{n} e_{n} (n \in Z) .

These are eigenvectors, including the zero-eigenvalue cases.

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