Past exam of the real and harmonic analysis course of the University of Oxford / 2007 / 4

Use normalized Haar measure $dm=dt/(2\pi )$. A homogeneous Banach space on the circle is a Banach subspace $B\subseteq L^1(\mathbb{T})$ with $\Vert f{\Vert }_1\le \Vert f{\Vert }_B$, such that translations ${\tau }_{s}f(t)=f(t-s)$ preserve $B$, satisfy $\Vert {\tau }_{s}f{\Vert }_B=\Vert f{\Vert }_B$, and obey $\Vert {\tau }_{s}f-f{\Vert }_B\to 0$ as $s\to 0$.

For continuous $g$, the map $s\mapsto g(s){\tau }_{s}f$ is continuous with values in the complete space $B$. Define its Banach-valued Riemann integral, equivalently its Bochner integral, byThe triangle inequality for Riemann sums and passage to their norm limit give the canonical estimateThe continuous inclusion into $L^1$ carries the same Riemann sums to the ordinary convolution, so this definition agrees almost everywhere with $\int f(t-s)g(s)dm(s)$. Density of continuous functions in $L^1$, completeness of $B$, and the estimate extend the operation uniquely to all $g\in L^1$. The ordinary $L^1$ convolution estimate identifies that extension with the same integral formula.

A summability kernel on the circle is a sequence of continuous periodic functions withThe frequently used stronger condition of uniform convergence to zero off each neighborhood of zero implies the last condition. For $f$ in a homogeneous Banach space,Given $\epsilon >0$, translation continuity bounds the difference inside a sufficiently small neighborhood by $\epsilon$; its integral is at most $C\epsilon$. Outside, use $2\Vert f{\Vert }_B$ and the vanishing tail condition. Hence $f\ast k_n\to f$ in $B$.

With the indexing needed in part (c), the Fejér kernel isIn particular $K_0=1$. The squared-modulus expression proves $K_n\ge 0$, and its constant Fourier coefficient proves $\int K_{n}dm=\Vert K_n{\Vert }_1=1$. The geometric series yieldsThus these are summability kernels. For $T_f:g\mapsto f\ast g$, part (a) gives $\Vert T_f\Vert \le \Vert f{\Vert }_B$. Since $\Vert K_n{\Vert }_1=1$ and $T_f{K}_n\to f$ in $B$, the reverse bound follows:

For $n\ge 1$ put $q_n(t)=2n{K}_{n-1}(t)\sin (nt)$. Using $\sin (nt)=(e^{int}-e^{-int})/(2i)$ and the Fourier coefficients of $K_{n-1}$ givesIndeed, when $1\le k\le n$ only the $e^{int}$ shift contributes, with coefficient $(-in)(1-|k-n|/n)=-ik$. For $-n\le k\le -1$ only the opposite shift contributes and gives $in(1-|k+n|/n)=-ik$; at $k=0$ both coefficients vanish.

If $P={\sum }_{|k|\le n}{a}_k{e}^{ikt}$, convolution multiplies its Fourier coefficients by those of $q_n$. Thus $-P\ast q_n=\sum ik{a}_k{e}^{ikt}=P^{\prime }$, proving the exact identity. Since $K_{n-1}\ge 0$ and has integral one,For $n=0$ the polynomial is constant and the bound is immediate. This elementary Fejér-kernel argument proves the requested estimate; the sharp Bernstein inequality for trigonometric polynomials has constant $n$.