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

Table of contents
- Solution b

Solution

Because the Fejér kernel has normalized integral one,

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

For

f \in Lip α

this implies

∣ σ_{n} (f, x) - f (x) ∣ \leq \frac{2 M}{π} \int_{0}^{π} F_{n} (t) t^{α} d t .

On

0 \leq t \leq π

, the inequalities

sin (t /2) \geq t / π

and

∣ sin (n t /2) ∣ \leq n ∣ sin (t /2) ∣

give

F_{n} (t) \leq min (\frac{n}{2}, \frac{π ^{2}}{2 n t ^{2}}) .

Splitting the integral at

1/ n

gives

\int_{0}^{π} F_{n} (t) t^{α} d t \leq \frac{n}{2} \int_{0}^{1/ n} t^{α} d t + \frac{π ^{2}}{2 n} \int_{1/ n}^{π} t^{α - 2} d t .

For

0 < α < 1

, both terms are

O (n^{- α})

. For

α = 1

, the second is

O ((lo g n) / n)

. Uniformly in

x

,

∥ σ_{n} (f) - f ∥_{\infty} \leq {c_{α} n^{- α}, c_{1} (lo g n) / n, 0 < α < 1, α = 1,

where the constants absorb the Lipschitz continuity constant

M

.

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