Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 61 / 4

For a periodic function use the modulus of continuityOn the interval, take the supremum over pairs of points whose distance is at most $\delta$. The mean value theorem applied to cosine, whose derivative has absolute value at most one, givesBoth cosine values lie in $[-1,1]$, so for $|h|\le \delta$,Taking the supremum provesThe cosine substitution for polynomial approximation is also a linear isometry in the supremum norm, since cosine maps a full period onto $[-1,1]$; its image consists of even continuous periodic functions.

The first Jackson theorem for periodic approximation asserts that a universal constant $C$ satisfiesfor every continuous $2\pi$-periodic function $g$, where the infimum is over degree-at-most-$n$ trigonometric polynomials.

Apply it to the even function $g(\theta )=f(\cos \theta )$. If $q_n$ is a trigonometric polynomial approximating $g$, its even parthas no larger error, because $g(-\theta )=g(\theta )$ and the triangle inequality bounds each averaged error by $\Vert g-q_n{\Vert }_{\infty }$. Every even trigonometric polynomial has the formHere the Chebyshev polynomials $T_j$ have algebraic degree $j$, so $p_n$ has degree at most $n$. Surjectivity of cosine givesConversely, every algebraic polynomial of degree at most $n$ yields such an even trigonometric polynomial. Taking infima therefore proves the exact identity $E_n^{\mathrm{alg}}(f)=E_n^{\mathrm{trig}}(\tilde{f})$, not merely an inequality. Combine this with the periodic theorem and the preceding modulus of continuity estimate:Symmetrization, the degree correspondence and the equality of norms justify every step in transferring the Jackson-type estimate.

For the indicated function the cosine substitution givesThe absolute-value function is Lipschitz continuous with constant one, and so is sine. Their composition therefore satisfiesUse the sharper intermediate estimate from the preceding part, rather than the ordinary interval modulus of continuity:

The usual inverse theorem for trigonometric approximation cannot hold verbatim with the ordinary interval modulus of continuity. It would implyBut comparison with the endpoint $1$ giveswhich is of order $n^{-1/2}$ and contradicts that bound as $n\to \infty$. This is an endpoint obstruction to an algebraic inverse approximation theorem. The algebraic approximation rate measures smoothness after cosine substitution; cosine compresses distances quadratically near the endpoints. A valid algebraic inverse theorem must account for that endpoint geometry rather than using the unchanged periodic formulation.