Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 61 / 3

The original PDF prints $O(n^{\alpha })$, with a positive exponent. That premise is insufficient; the intended decay rate is $O(n^{-\alpha })$. Indeed every bounded continuous periodic function has $E_n(f)\le \Vert f{\Vert }_{\infty }=O(n^{\alpha })$. For an explicit counterexample, on $[-\pi ,\pi ]$ setand extend periodically. This is continuous, but $\omega (h,\delta )\ge 1/\log (e/\delta )$ for $0<\delta <1$, which is not $O({\delta }^{\alpha })$ for any $\alpha >0$. Thus the sign correction is required in the actual printed question, not just in its converted TeX.

For real continuous $2\pi$-periodic $f$, write $E_n$ for best uniform approximation by degree-at-most-$n$ trigonometric polynomials and $\omega (f,\delta )={\sup }_{|h|\le \delta }\Vert f(\cdot +h)-f{\Vert }_{\infty }$. A quantitative inverse theorem for trigonometric approximation iswhere $C$ is universal. Here is a direct reason for the sum. Choose best approximants $p_j$ and $2^r\le n<2^{r+1}$. WriteThe Bernstein inequality for trigonometric polynomials givesFor $j\ge 2$, monotonicity of best errors bounds $2^{j+1}{E}_{2^{j-1}}$ by $8{\sum }_{\nu =2^{j-2}}^{2^{j-1}-1}{E}_{\nu }$. The first terms are bounded by a fixed multiple of $E_0$. Also $2E_{2^r}\le 4n^{-1}{\sum }_{\nu =0}^n{E}_{\nu }$. Nowproves the theorem, for example with a sufficiently large absolute constant $C=20$.

Suppose, as intended, $E_{\nu }(f)\le A{\nu }^{-\alpha }$ for $\nu \ge 1$, after enlarging $A$ to include any finite exceptional indices, where $0<\alpha <1$. Since ${\sum }_{\nu =1}^n{\nu }^{-\alpha }\le n^{1-\alpha }/(1-\alpha )$, the inverse theorem givesFor $0<\delta \le 1$ take $n=\lfloor 1/\delta \rfloor$. Then $1/(n+1)<\delta \le 1/n$, so monotonicity of the modulus of continuity and $n^{-\alpha }\le 2^{\alpha }{\delta }^{\alpha }$ give $\omega (f,\delta )\le 2^{\alpha }{C}_f{\delta }^{\alpha }$. For $\delta >1$, the separate bound $\omega (f,\delta )\le 2\Vert f{\Vert }_{\infty }\le 2\Vert f{\Vert }_{\infty }{\delta }^{\alpha }$ covers the remaining scales. Therefore the corrected conclusion isNow put $\beta =\log 2/\log 3$. The given best approximant to the Weierstrass function leaves the uniformly convergent tail ${\sum }_{k=m+1}^{\infty }{2}^{-k}\cos (3^{k}x)$. Its supremum norm is exactly $2^{-m}$, attained at $x=0$; henceThe corrected inverse theorem gives $\omega (g,\delta )=O({\delta }^{\beta })$. To show that this order is sharp, set $h_m=\pi /3^{m+1}$. Every term of $g(0)-g(h_m)$ is nonnegative, and for $k\ge m+1$ the cosine equals $-1$ because $3^{k-m-1}$ is odd. ThusGiven small $\delta >0$, select $m$ with $h_m\le \delta <3h_m$. Then $\omega (g,\delta )\ge 4(3\pi {)}^{-\beta }{\delta }^{\beta }$. Combining the estimates gives the precise small-scale orderThe exact approximation error can also be verified independently of the supplied fact: at $x_j=j\pi /3^{m+1}$ the tail has alternating values $(-1{)}^j{2}^{-m}$. There are enough such points in one period for the trigonometric Chebyshev alternation theorem, since $n<3^{m+1}$. This is the positive lacunary trigonometric series mechanism behind the best partial sums.