部分和 = 与狄利克雷核的卷积
把系数积分代入 S_N f(x)，并交换求和与积分。诸纯音坍缩成一个闭式权重，即狄利克雷核 D_N。于是部分和是一个卷积：S_N f(x) = (f * D_N)(x)。傅里叶级数的收敛，实质上是核 D_N 行为如何的问题。
D_N(t) = sum_{n=-N}^{N} e^{i n t}. This is a finite geometric series with ratio e^{i t}:
D_N(t) = e^{-iNt} * ( e^{i(2N+1)t} - 1 ) / ( e^{it} - 1 ).
Factor e^{i t/2} top and bottom to symmetrize:
D_N(t) = ( e^{i(N+1/2)t} - e^{-i(N+1/2)t} ) / ( e^{it/2} - e^{-it/2} )
= sin((N + 1/2) t) / sin(t/2).
Key facts:
(1) (1/2pi) integral over [-pi,pi] D_N(t) dt = 1 (the n=0 term integrates to 2pi, the rest to 0).
(2) D_N CHANGES SIGN and oscillates; it is NOT >= 0.
(3) Its L^1 size grows: (1/2pi) integral |D_N(t)| dt ~ (4/pi^2) log N -> infinity.
These 'Lebesgue constants' diverging is the seed of every pointwise pathology to come.为什么逐点收敛确实困难
由于 ‖D_N‖₁ → ∞，映射 f ↦ S_N f(0) 随 N 增大而无界。由一致有界原理，存在一个连续函数，其傅里叶级数在某点发散。因此傅里叶级数的逐点收敛即便对连续函数也可能失败 —— 单靠连续性不够。正面结论需要额外的光滑性；干净的充分条件是下面的狄利克雷条件。
费耶尔：对部分和取平均，一切便顺
良方是停止求和、改为取平均。切萨罗均值 σ_N f = (S_0 f + … + S_{N−1} f)/N 是与费耶尔核 K_N 的卷积。与 D_N 不同，费耶尔核非负 —— 仅此一处符号的改变，便使它成为一个逼近恒等：一个在 0 处聚集、却保持均值为 1 的尖峰。
Fejer kernel: K_N(t) = (1/N) sum_{m=0}^{N-1} D_m(t) = (1/N) ( sin(Nt/2) / sin(t/2) )^2 >= 0.
Three properties making {K_N} an approximate identity:
(i) K_N(t) >= 0 (a square -- positivity!)
(ii) (1/2pi) integral over [-pi,pi] K_N = 1 (mean one)
(iii) for fixed delta > 0, (1/2pi) integral_{delta <= |t| <= pi} K_N(t) dt -> 0 (mass piles up at 0)
Fejer's theorem: if f is continuous and 2pi-periodic, then sigma_N f -> f UNIFORMLY.
Proof sketch: sigma_N f(x) - f(x) = (1/2pi) integral K_N(t) [ f(x-t) - f(x) ] dt.
Split into |t| < delta and |t| >= delta.
Near piece: |f(x-t) - f(x)| < epsilon by UNIFORM CONTINUITY of f, and integral K_N <= 1, so < epsilon.
Far piece: |f(x-t)-f(x)| <= 2 max|f|, times (iii) which -> 0; choose N large.
Both bounds are independent of x => UNIFORM convergence. QED