Past exam of the mathematics course of the University of Cambridge 2001 iii Paper 58 5 Solution Created 2026-10-03 Updated 2026-10-14
The original PDF's kernel has an incorrect normalization. With ordinary , its integral is , not one. Already at the printed kernel is the constant . For , the printed operator therefore returns , while for every . 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 givesTake and split at . The inner part is at most , 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 .
On , , soDivide an arc of length into at most shorter arcs. The triangle inequality then bounds by . ConsequentlyChoosing proves the requested fractional-scale Fejér approximation bound for the intended normalization, for example with the absolute constant :The same proof applies to complex-valued continuous functions, since the estimates use absolute values.
Past exam of the mathematics course of the University of Cambridge 2006 iii Paper 67 3 Solution Created 2026-10-03 Updated 2026-10-14
Use the periodic modulus of continuitySubdividing a displacement into at most steps of size at most proves the chaining inequalityFor , the sharper bound holds directly.
Set , where and . Positivity and unit normalized mass of the Fejér kernel giveThe portion is at most . For , and , soApply this bound and the chaining inequality to the two tails:Let . Calculus gives , attained at , and . Thus a possible explicit constant for the fractional-scale Fejér approximation bound isThe constant depends only on , not on or . Since a continuous periodic function is uniformly continuous, the right-hand side tends to zero. The logarithmic tail estimate is what permits every ; a crude global-oscillation bound would not prove the whole requested range.