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Analysis course of the University of Oxford

Past exam of the analysis course of the University of Oxford

2011

The normalization gives
$$
\widehat{e_n}(m)=\frac1{2\pi}\int_{-\pi}^{\pi}e^{i(n-m)t}\,dt=\delta_{nm}.
$$
Thus $\alpha_ne_n$ has exactly the coefficients prescribed for $Te_n$. Uniqueness yields
$$
\boxed{Te_n=\alpha_ne_n\quad(n\in\mathbb Z).}
$$
These are <eigenvectors>, including the zero-<eigenvalue> cases.

Solution

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

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

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

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