Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 61 / 1

Take $n\ge 1$. In the normalization used here, the Dirichlet kernel has the finite expansionAveraging a finite number of the integral formulas for the Fourier partial sums is legitimate by linearity of the integral. Hence the Fejér sum is convolution withTo obtain the nonnegative form of the Fejér kernel, expand a squared geometric sum:The coefficient $n-r$ counts pairs of indices whose difference is $r$. Dividing by $2n$ and summing the geometric progression givesAt $u\in 2\pi \mathbb{Z}$ the ratio has its continuous limiting value $F_n(u)=n/2$. Thus this is a continuous, nonnegative trigonometric polynomial, not a kernel with genuine singularities.

Integration over a full period kills every nonconstant cosine term, soTranslation invariance of integration over the circle consequently gives, for every $x$,The first step is the integral triangle inequality, the second uses nonnegativity, and the last uses the mass just computed. Taking the supremum norm provesIn particular, Fejér summation is a uniform-norm contraction. The factor $\pi$, rather than $2\pi$, is essential for the half-normalized Dirichlet kernel and Fejér kernel in this problem.

Let $m$ be a positive integer and let the trigonometric polynomial $t_n$ have frequencies only in $[-n,n]$. For every $j\ge n$, its Fourier partial sum is the polynomial itself: $s_j(t_n)=t_n$. Every term in the defining average of the de la Vallée Poussin sum therefore equals $t_n$, givingThis is exact reproduction of the degree-at-most-$n$ trigonometric polynomials, irrespective of the positive averaging length $m$.

The indexing of the Fejér sums givesSubtracting removes precisely the initial $n$ Fourier partial sums. ThusFor $n=0$, omit the second term, so that no undefined ${\sigma }_0$ is needed. Apply the triangle inequality and the uniform-norm contraction of Fejér summation estimate to getHence the operator norm of the de la Vallée Poussin sum is at most $1+2n/m$.

For any degree-at-most-$n$ trigonometric polynomial $t_n$, linearity and reproduction giveThe operator norm bound in the preceding part yieldsTake the infimum over all such trigonometric polynomials. By the definition of best uniform approximation,No choice of a minimizer is needed for this argument. It is an instance of the polynomial reproduction error bound: a bounded linear reproducing operator has error at most $1+\Vert P\Vert$ times the optimal error.