Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 69 / 6

For $f\in C(\mathbb{T})$, define $E_j(f)={\inf }_{T\in {\mathcal{T}}_j}\Vert f-T{\Vert }_{\infty }$, where ${\mathcal{T}}_j$ is the space of degree-at-most-$j$ trigonometric polynomials. The first inverse theorem for trigonometric approximation states that a universal constant $C$ satisfiesFor perspective, take best approximants at dyadic degrees and telescope their differences. Each difference between consecutive approximants has supremum norm at most twice the earlier error, and the Bernstein inequality for trigonometric polynomials multiplies that bound by its degree when differentiating. The resulting dyadic sum is bounded by the displayed sum because $E_j$ decreases with $j$; the remaining approximation error is also bounded by an average of earlier errors.

Now suppose $E_j(f)\le K{j}^{-\alpha }$ for large $j$, absorbing finitely many earlier terms into a constant. If $0<\alpha <1$, comparison with an integral gives ${\sum }_{j=1}^n{j}^{-\alpha }=O(n^{1-\alpha })$. If $\alpha =1$, the same comparison gives ${\sum }_{j=1}^n{j}^{-1}=O(1+\ln n)$. Substitution proves the requested two rates:The logarithm is the endpoint contribution of the harmonic sum and cannot generally be removed, as the following Weierstrass function shows.

The Weierstrass M-test gives a continuous periodic Weierstrass function, since $\sum a^{-k}<\infty$. For $5^m\le n<5^{m+1}$, the given best approximant leaves the tailIts supremum norm is at most the sum of its coefficients, and at $x=0$ all cosine terms equal one, attaining that bound. ConsequentlyPut $\alpha =\ln a/\ln 5$. Then $a^{-m}=(5^m{)}^{-\alpha }$ and $n/5^m\in [1,5)$, giving $E_n(g)\asymp n^{-\alpha }$, with constants depending only on $a$.

For $1<a<5$, part (a) gives $\omega (g,1/n)\le C_a{n}^{-\alpha }$. The first Jackson theorem for periodic approximation, $E_n(g)\le C\omega (g,1/n)$, supplies the matching lower bound. Hence the modulus of continuity has the precise orderAt $a=5$, the inverse theorem for trigonometric approximation gives the upper bound $C\delta \ln (1/\delta )$, and part (c) proves the matching lower bound. Thus the critical Weierstrass modulus isTo pass from scales $1/n$ to all small $\delta$, use monotonicity of the modulus of continuity and adjacent reciprocals, which are comparable to $\delta$.

Take $a=5$ and $x=\pi /2$. Since $5^k\equiv 1(\mathrm{mod}4)$, every cosine term vanishes at $x$, andSet $h=1/n$ and $m=\lfloor {\log }_{5}n\rfloor$. For $0\le k\le m$, the sine argument lies in $[0,1]$, where $\sin u\ge (\sin 1)u$. The initial part therefore has a common sign and magnitude at least $(m+1)(\sin 1)/n$. The remaining tail has absolute value at mostThe reverse triangle inequality now givesfor sufficiently large $n$, because $m+1>\ln n/\ln 5$. This establishes the sharp logarithmic lower boundThe constant can be reduced to include the finitely many small indices: for $n\ge 5$ the displayed lower bound is positive; for $n=2,3$ the first sine exceeds the tail bound $1/4$, and for $n=4$ the first two sines contribute $\sin (1/4)+\sin (5/4)/5$, exceeding the remaining tail bound $1/20$. At $n=1$ the logarithmic right side is zero.

Here $E_n(g)\asymp 1/n$ but $g$ is not Lipschitz continuous, so a first modulus of continuity bound $O(1/n)$ is sufficient by the first Jackson theorem for periodic approximation but is not necessary. Nor is the weaker necessary bound $O(\ln n/n)$ sufficient: an explicit comparison is the logarithmic cusp with slow trigonometric approximationextended periodically. On $[0,\pi ]$ it is increasing and concave, so its largest increment over distance $\delta$ is from zero, giving $\omega (h,\delta )=\delta \ln (e\pi /\delta )$ for $0<\delta \le \pi$.

For a concrete approximation lower bound, its cosine Fourier coefficients satisfyIndeed, integration by parts once replaces $h$ by $h^{\prime }(x)=\ln (\pi /x)$; integrating that against sine with primitive $(1-\cos jx)/j$ yields the stated positive integral. Its logarithmic growth follows because ${\int }_1^A\cos u/udu$ stays bounded by the Dirichlet test. All $a_j$ are negative and absolutely summable.

Let $V_n=2{\sigma }_{2n}-{\sigma }_n$, a de la Vallée Poussin sum. Its multiplier is one for $j\le n$, $2-j/n$ for $n<j<2n$, and zero for $j\ge 2n$; the Fejér summation is a uniform-norm contraction property gives $\Vert V_n\Vert \le 3$. Thus the functional $L_n(f)=f(0)-V_{n}f(0)$ has norm at most four and vanishes on every degree-at-most-$n$ trigonometric polynomial. Because the remaining multiplier weights are nonnegative and all $a_j<0$,The first Jackson theorem for periodic approximation gives the corresponding upper bound. The same first-modulus order $\delta \ln (1/\delta )$ occurs with both $E_n\asymp 1/n$ and $E_n\asymp \ln n/n$. Therefore first-modulus rates alone do not characterize the endpoint approximation class, unlike the matched direct and inverse rates for $0<\alpha <1$.