部分和 = 與狄利克雷核的卷積
把係數積分代入 S_N f(x)，並交換求和與積分。諸純音坍縮成一個閉式權重，即狄利克雷核 D_N。於是部分和是一個卷積：S_N f(x) = (f * D_N)(x)。傅立葉級數的收斂，實質上是核 D_N 行為如何的問題。
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.為什麼逐點收斂確實困難
由於 ‖D_N‖₁ → ∞，映射 f ↦ S_N f(0) 隨 N 增大而無界。由一致有界原理，存在一個連續函數，其傅立葉級數在某點發散。因此傅立葉級數的逐點收斂即便對連續函數也可能失敗 —— 單靠連續性不夠。正面結論需要額外的光滑性；乾淨的充分條件是下面的狄利克雷條件。
費耶爾：對部分和取平均，一切便順
良方是停止求和、改為取平均。切薩羅均值 σ_N f = (S_0 f + … + S_{N−1} f)/N 是與費耶爾核 K_N 的卷積。與 D_N 不同，費耶爾核非負 —— 僅此一處符號的改變，便使它成為一個逼近恆等：一個在 0 處聚集、卻保持均值為 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