Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 58 / 5 / Solution

The original PDF's kernel has an incorrect normalization. With ordinary $dt$, its integral is $\pi$, not one. Already at $n=1$ the printed kernel is the constant $1/2$. For $f\equiv 1$, the printed operator therefore returns $\pi$, while $\omega (f,\delta )=0$ for every $\delta$. Thus the literal inequality with the printed kernel is false. The Fejér kernel representing the stated average of Fourier partial sums isThe ratio at zero is interpreted by its limit. To verify the normalization and the averaging property, use the finite geometric sum:Fourier orthogonality gives unit integral; the displayed Fourier coefficients are exactly the multipliers of the Fejér sum. This also proves nonnegativity.

For the correctly normalized Fejér sum, subtraction of $f(x)$ givesTake $0<\delta \le 1$ and split at $|t|=\delta$. The inner part is at most $\omega (f,\delta )$, because the kernel is nonnegative and has mass one. The strict inequality in the definition of the modulus of continuity causes no endpoint issue: continuity gives the same bound at distance exactly $\delta$.

On $0<|t|\le \pi$, $\sin (|t|/2)\ge |t|/\pi$, soDivide an arc of length $|t|$ into at most $1+|t|/\delta$ shorter arcs. The triangle inequality then bounds $|f(x-t)-f(x)|$ by $(1+\pi /\delta )\omega (f,\delta )$. ConsequentlyChoosing $\delta =n^{-1/2}$ proves the requested fractional-scale Fejér approximation bound for the intended normalization, for example with the absolute constant $1+\pi +{\pi }^2$:The same proof applies to complex-valued continuous functions, since the estimates use absolute values.