Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 318 / 1 / a / Solution

Averaging the integral representations of the Fourier partial sums gives

σ_{n} (f, x) = \frac{1}{π} \int_{T} (\frac{1}{n} \sum_{k = 0}^{n - 1} D_{k} (x - t)) f (t) d t .

The finite trigonometric sum is

\sum_{k = 0}^{n - 1} sin (k + \frac{1}{2}) u = \frac{s i n ^{2} ( n u /2 )}{s i n ( u /2 )} .

Since

D_{k} (u) = sin ((k + 1/2) u) / (2 sin (u /2))

, the Fejér kernel is therefore

F_{n} (u) = \frac{1}{n} k = 0 \sum n - 1 D_{k} (u) = \frac{1}{2 n} \frac{sin ^{2} ( n u /2 )}{sin ^{2} ( u /2 )}

and

σ_{n} (f, x) = \frac{1}{π} \int_{T} F_{n} (x - t) f (t) d t .

The displayed square shows that

F_{n} \geq 0

. Every Fourier partial sum preserves the constant function, so

σ_{n} (1) = 1

. Substituting

f = 1

in the integral formula gives

\frac{1}{π} \int_{T} F_{n} (t) d t = 1.

Nonnegativity then yields

\frac{1}{π} \int_{T} ∣ F_{n} (t) ∣ d t = 1 .

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