s_{n} (f, x) = \frac{1}{2 π} \int_{T} f (t) \sum_{k = - n}^{n} e^{ik (x - t)} d t = \frac{1}{π} \int_{T} D_{n} (x - t) f (t) d t,

where

D_{n} (u) = \frac{sin (( n + \frac{1}{2} ) u )}{2 sin ( u /2 )} .

b

Solution

The Fejér kernel is nonnegative, and preservation of constants gives

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

By evenness,

σ_{n} (f, x) - f (x) = \frac{1}{2 π} \int_{T} F_{n} (t) [f (x - t) - 2 f (x) + f (x + t)] d t .

Put

δ = n^{- 1/2}

. Use

ω_{2} (f, ∣ t ∣) \leq (∣ t ∣/ δ + 1)^{2} ω_{2} (f, δ)

together with

F_{n} (t) \leq n /2

and

F_{n} (t) \leq C / (n t^{2})

. Splitting at

1/ n

and

δ

shows that the remaining weighted integral is uniformly bounded, so

∥ σ_{n} (f) - f ∥_{\infty} \leq C ω_{2} (f, n^{- 1/2}) .

f (x + h) - 2 f (x) + f (x - h) = \int_{0}^{h} \int_{- s}^{s} f^{''} (x + u) d u d s,

hence

ω_{2} (f, t) \leq t^{2} ∥ f^{''} ∥_{\infty}

. Part (b) gives

∥ σ_{n} (f) - f ∥_{\infty} = O (n^{- 1})

. This cannot be little-

o

for every

C^{2}

f_{0} (x) = cos x

σ_{n} (f_{0}) = (1 - n^{- 1}) cos x, ∥ σ_{n} (f_{0}) - f_{0} ∥_{\infty} = n^{- 1} .

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