Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 318 (Approximation Theory)

The trigonometric Chebyshev alternation theorem says that $p_n^{\ast }\in {\mathcal{T}}_n$ is best exactly when its error has at least $2n+2$ cyclically ordered extrema of equal magnitude and alternating sign. Let $5^m\le n<5^{m+1}$ and set $x_j=j\pi /5^{m+1}$. For every $k\ge m+1$,ThusThere are $2\cdot 5^{m+1}\ge 2n+2$ such extrema, while the triangle inequality bounds the tail by their common magnitude. Hence

The inverse theorem for trigonometric approximation givesSumming ${\nu }^{-\alpha }$ provesFor $c_k=a^{-k}$, part (a) gives $E_n(g)\asymp a^{-m}\asymp n^{-{\log }_{5}a}$. If $a<5$, an increment $h=\pi /5^{m+2}\le 1/n$ at $x=0$ has nonnegative summands and its $k=m+2$ term is $\asymp n^{-{\log }_{5}a}$. Part (c) handles $a=5$. Therefore

Since $5^k\equiv 1(\mathrm{mod}4)$,Choose $5^m\le n<5^{m+1}$. For $k\le m$, $\sin (5^k/n)\ge c{5}^k/n$, so each of these $m+1$ same-sign terms has magnitude at least $c/n$. The remaining tail is $O(5^{-m})=O(n^{-1})$. HenceThus $E_n(f)=O(n^{-1})$ does not imply $\omega (f,n^{-1})=O(n^{-1})$; the logarithmic gap prevents a characterization of that approximation class by this first modulus alone.

Substituting the Fourier coefficients into the Fourier partial sum and summing the finite geometric series giveswhere

The Fejér kernel is nonnegative, and preservation of constants givesBy evenness,Put $\delta =n^{-1/2}$. Use ${\omega }_2(f,|t|)\le (|t|/\delta +1{)}^2{\omega }_2(f,\delta )$ together with $F_n(t)\le n/2$ and $F_n(t)\le C/(n{t}^2)$. Splitting at $1/n$ and $\delta$ shows that the remaining weighted integral is uniformly bounded, so

Twice applying the fundamental theorem of calculus giveshence $\boxed{{\omega }_2(f,t)\le t^2\Vert f^{\prime \prime }{\Vert }_{\infty }}$. Part (b) gives $\Vert {\sigma }_n(f)-f{\Vert }_{\infty }=O(n^{-1})$. This cannot be little-$o$ for every $C^2$ function: for $f_0(x)=\cos x$,

The explicit divided difference formula makes $M_i(t)$ a finite linear combination of $(t_j-t{)}_{+}^{k-1}$. It is therefore polynomial of degree at most $k-1$ between knots and globally $C^{k-2}$. For $t<t_i$, the nodal data come from a polynomial of degree $k-1$, whose order-$k$ divided difference vanishes; for $t\ge t_{i+k}$ all truncated powers vanish. Thusand normalization does not change the degree, smoothness, knots, or support.

Apply the Leibniz rule for divided differences to $(x-t{)}_{+}^{k-1}=(x-t)(x-t{)}_{+}^{k-2}$. Only the zeroth and first divided differences of the linear factor survive. After applying the normalization, this gives the Cox-de Boor recursion formula

Induct on $k$, using the recursive divided-difference formula. After substituting the two induction hypotheses, useand the analogous identity for $g$; adjacent terms telescope. This proves

The Korovkin theorem states that positive linear operators $L_n:C[0,1]\to C[0,1]$ converge uniformly to the identity on every continuous function if they do so on the three test functions $1,t,t^2$.

Let $e_m$ denote the $m$th elementary symmetric polynomial of $t_{i+1},\dots ,t_{i+k-1}$. Expanding both sides of the Marsden identity in powers of $x$ and equating the coefficient of $x^{k-1-m}$ givesThus $a_{0,i}=1$, while $a_{1,i}$ and $a_{2,i}$ are respectively the means of the interior knots and of their pairwise products.

The Schoenberg spline operator is positive and $V_{n}1=1$. If ${\xi }_i=a_{1,i}$, then both ${\tau }_i$ and ${\xi }_i$ lie in $[t_i,t_{i+k}]$, soBecause $a_{2,i}$ averages products of knots in the same interval and all points lie in $[0,1]$,The Korovkin theorem now provesfor every $f\in C[0,1]$.

An orthonormal wavelet is $\psi \in L^2(\mathbb{R})$ such that $\{2^{j/2}\psi (2^{j}x-k){\}}_{j,k\in \mathbb{Z}}$ is an orthonormal basis. A multiresolution analysis is a nested family of closed spaces $V_j$ with trivial intersection, dense union, dyadic scaling, integer-translation invariance of $V_0$, and a generator $\varphi$ whose integer translates form an orthonormal basis of $V_0$. The Meyer-Mallat theorem says every such analysis has an orthonormal wavelet whose translates span $V_{j+1}\ominus V_j$.

Fourier transforming the refinement equation and changing variables givesBy Parseval identity, orthonormality of the translates is equivalent toThese are precisely the Fourier coefficients of the periodization $P(t)={\sum }_k|f(t+2\pi k){|}^2$. Therefore

For $f=1_{[-\pi ,\pi )}$, its $2\pi$-periodization is one almost everywhere. The refinement mask is the periodic function equal to one on $[-\pi /2,\pi /2)$ and zero on the rest of $[-\pi ,\pi )$. Fourier inversion gives the Shannon scaling functionSince $m(t)=\frac{1}{2}{\sum }_n{a}_n{e}^{-int}$, its Fourier coefficients give