参考 Fourier Analysis by Stein,《数学分析讲义》张友金.
# 前言根据之前的例子,我们需要定义三角级数
∑ n = − ∞ + ∞ A n cos n t + B n sin n t \sum^{+\infty}_{n=-\infty}A_n\cos nt+B_n\sin nt n = − ∞ ∑ + ∞ A n cos n t + B n sin n t
以及考察其收敛性,这是研究基础,也是本节的目标。
本文提及的 可积 指的是 可积或反常绝对可积 ,如果我们说 Riemann 可积 ,会用 R \mathcal R R 表示。
# Fourier 级数的定义# 指数形式定义 :设 f f f 为 [ a , b ] [a,b] [ a , b ] 上可积的复值函数, 则定义 f f f 的 Fourier 级数 为
f ( x ) ∼ ∑ n = − ∞ + ∞ f ^ ( n ) e 2 π i n x / L f(x)\sim \sum^{+\infty}_{n=-\infty}\hat f(n)e^{2\pi inx/L} f ( x ) ∼ n = − ∞ ∑ + ∞ f ^ ( n ) e 2 π i n x / L
其中 ∣ b − a ∣ = L |b-a|=L ∣ b − a ∣ = L ,且 Fourier 系数 为
f ^ ( n ) = 1 L ∫ a b f ( x ) e − 2 π i n x / L d x \hat f(n)=\dfrac1 L\displaystyle \int^b_a f(x)e^{-2\pi inx/L}\mathrm dx f ^ ( n ) = L 1 ∫ a b f ( x ) e − 2 π i n x / L d x
指数形式的书写在一些场景更加简洁。
特别地,设 f f f 为 [ a , b ] = [ − π , π ] [a,b]=[-\pi,\pi] [ a , b ] = [ − π , π ] 上可积函数,它的 Fourier 级数为
f ( x ) ∼ 1 2 π ∑ n = − ∞ + ∞ ∫ − π π f ( t ) e i n ( x − t ) d t f(x)\sim \frac 1{2\pi}\sum^{+\infty}_{n=-\infty}\int ^{\pi}_{-\pi}f(t)e^{in(x-t)}\mathrm dt f ( x ) ∼ 2 π 1 n = − ∞ ∑ + ∞ ∫ − π π f ( t ) e i n ( x − t ) d t
# 三角形式定义 :设 f f f 为 [ a , b ] [a,b] [ a , b ] 上可积的复值函数, 则定义 f f f 的 Fourier 级数 为
f ( x ) ∼ A 0 2 + ∑ n = 1 ∞ A n cos 2 π n x L + B n sin 2 π n x L f(x)\sim \dfrac {A_0}2+\sum^\infty_{n=1}A_n\cos{\dfrac {2\pi nx}L}+B_n \sin \dfrac {2\pi nx}L f ( x ) ∼ 2 A 0 + n = 1 ∑ ∞ A n cos L 2 π n x + B n sin L 2 π n x
其中 ∣ b − a ∣ = L |b-a|=L ∣ b − a ∣ = L ,且 Fourier 系数 为
A n = 2 L ∫ a b f ( x ) cos 2 π n x L d x , B n = 2 L ∫ a b f ( x ) sin 2 π n x L d x A_n=\dfrac 2L\int^b_af(x)\cos\dfrac{2\pi nx}L\mathrm dx,\quad B_n=\dfrac 2L\int^b_af(x)\sin \dfrac {2\pi nx}L\mathrm dx A n = L 2 ∫ a b f ( x ) cos L 2 π n x d x , B n = L 2 ∫ a b f ( x ) sin L 2 π n x d x
三角形式具有奇偶性,在一些计算上更加方便。并且它和指数形式是相容的。
特别地,设 f f f 为 [ a , b ] = [ − π , π ] [a,b]=[-\pi,\pi] [ a , b ] = [ − π , π ] 上可积函数,它的 Fourier 级数为
f ( x ) ∼ A 0 2 + ∑ n = 1 ∞ A n cos n x + B n sin n x f(x)\sim \dfrac{A_0}2+\sum^\infty_{n=1}A_n\cos nx+B_n\sin nx f ( x ) ∼ 2 A 0 + n = 1 ∑ ∞ A n cos n x + B n sin n x
注意,我们只给出了形式,但是并不知道这个级数是否收敛,以及是否收敛到 f f f .
# 性质f ′ ^ ( n ) = i n f ^ ( n ) , ∀ n ∈ Z \widehat {f'}(n)=in\hat f(n),\forall n\in \mathbb Z f ′ ( n ) = i n f ^ ( n ) , ∀ n ∈ Z
# Fourier 级数的部分和# 部分和定义 :给定 f f f 的 Fourier 级数,则它的 N 次部分和 为
S N ( f ) ( x ) = ∑ n = − N N f ^ ( n ) e 2 π i n x / L S_N(f)(x)=\sum ^N_{n=-N}\hat f(n)e^{2\pi inx/L} S N ( f ) ( x ) = n = − N ∑ N f ^ ( n ) e 2 π i n x / L
# Dirichlet 核定义 :定义在 [ − π , π ] [-\pi,\pi] [ − π , π ] 上的如下三角多项式,称为 Dirichlet 核
D N ( x ) = ∑ n = − N N e i n x = sin ( N + 1 / 2 ) x sin x / 2 D_N(x)=\sum^N_{n=-N}e^{inx}=\dfrac{\sin (N+1/2)x}{\sin x/2} D N ( x ) = n = − N ∑ N e i n x = sin x / 2 sin ( N + 1 / 2 ) x
右式是它的显式形式,从中可以得到 D N ( x ) D_N(x) D N ( x ) 的 Fourier 系数有一定的性质。
性质 :对于 D N ( x ) D_N(x) D N ( x ) 的 Fourier 系数 f ^ ( n ) \hat f(n) f ^ ( n ) ,有
f ^ ( n ) = 1 , ∣ n ∣ ≤ N \hat f(n)=1,\quad |n|\leq N f ^ ( n ) = 1 , ∣ n ∣ ≤ N
证明 注意到
∫ − π π sin ( n + 1 / 2 ) x sin x / 2 d x = ∫ − π π sin ( n − 1 / 2 ) x sin x / 2 d x \int^\pi_{-\pi}\dfrac{\sin(n+1/2)x}{\sin x/2}\mathrm dx=\int^\pi_{-\pi}\dfrac{\sin (n-1/2)x}{\sin x/2}\mathrm dx ∫ − π π sin x / 2 sin ( n + 1 / 2 ) x d x = ∫ − π π sin x / 2 sin ( n − 1 / 2 ) x d x
# Poisson 核定义 :定义在 [ − π , π ] [-\pi,\pi] [ − π , π ] 上的如下级数,其中参数 r ∈ [ 0 , 1 ) r\in [0,1) r ∈ [ 0 , 1 ) ,称为 Poisson 核
P r ( θ ) = ∑ n = − ∞ ∞ r ∣ n ∣ e i n θ = 1 − r 2 1 − 2 r cos θ + r 2 P_r(\theta)=\sum^\infty_{n=-\infty}r^{|n|}e^{in\theta}=\dfrac {1-r^2}{1-2r\cos \theta+r^2} P r ( θ ) = n = − ∞ ∑ ∞ r ∣ n ∣ e i n θ = 1 − 2 r cos θ + r 2 1 − r 2
右式是它的显式形式。另外,它是绝对收敛的,求和、极限可交换。
这些例子都有 Fourier 级数,但是我们仍然不知道需要什么条件才能使 Fourier 级数收敛,进一步收敛到函数本身。在讨论收敛之前,我们先讨论唯一性。
# Uniqueness of Fourier series# Uniqueness at the continuous pointsTheorem 2.2.1 If f ∈ R ( S 1 ) f\in \mathcal R(S^1) f ∈ R ( S 1 ) and ∀ n ∈ Z , f ^ ( n ) = 0 \forall n\in \mathbb Z,\hat f(n)=0 ∀ n ∈ Z , f ^ ( n ) = 0 , then
f ( θ 0 ) = 0 f(\theta_0)=0 f ( θ 0 ) = 0
whenever f f f is continuous at the point θ 0 \theta _0 θ 0
Proof Tips: Contradiction .
Without loss of generality, let θ 0 = 0 \theta_0=0 θ 0 = 0 , then we hope that there's a family of trigonometric polynomials (to guarantee the periodicity) G n G_n G n , where
∫ − δ δ G n ⋅ f ( θ ) d θ → ∞ , as n → ∞ \int^{\delta}_{-\delta} G_n\cdot f(\theta)\mathrm d\theta\to\infty, \text{ as } n\to\infty ∫ − δ δ G n ⋅ f ( θ ) d θ → ∞ , as n → ∞
also
G n → 0 , when n → ∞ , θ ∈ S 1 − [ − δ , δ ] G_n\to 0, \text{ when } n\to \infty,\theta\in S^1-[-\delta,\delta] G n → 0 , when n → ∞ , θ ∈ S 1 − [ − δ , δ ]
for instance
G n ( x ) = [ G ( x ) ] n = ( cos θ + ϵ ) n G_n(x)=[G(x)]^n=(\cos \theta +\epsilon )^n G n ( x ) = [ G ( x ) ] n = ( cos θ + ϵ ) n
as for complex-valued situation, write f ( θ ) = u ( θ ) + i v ( θ ) f(\theta)=u(\theta)+iv(\theta) f ( θ ) = u ( θ ) + i v ( θ ) .
Corollary 2.2.2 If f ∈ C ( S 1 ) f\in C(S^1) f ∈ C ( S 1 ) and ∀ n ∈ Z , f ^ ( n ) = 0 \forall n\in \mathbb Z,\hat f(n)=0 ∀ n ∈ Z , f ^ ( n ) = 0 , then f ≡ 0 f\equiv 0 f ≡ 0 .
Corollary 2.2.3 Suppose that f ∈ C ( S 1 ) f\in C(S^1) f ∈ C ( S 1 ) , and
∑ n = − ∞ ∞ ∣ f ^ ( n ) ∣ < ∞ \sum ^\infty_{n=-\infty}|\hat f(n)|<\infty n = − ∞ ∑ ∞ ∣ f ^ ( n ) ∣ < ∞
then
S N ( f ) ( θ ) ⇉ f ( θ ) S_N(f)(\theta)\rightrightarrows f(\theta) S N ( f ) ( θ ) ⇉ f ( θ )
Proof Tips: Consider partial sum that we can interchange the infinite sum with the integral
g ( θ ) = lim N → ∞ ∑ n = − N N f ^ ( n ) e i n θ g(\theta)=\lim_{N\to\infty}\sum^N_{n=-N}\hat f(n)e^{in\theta} g ( θ ) = N → ∞ lim n = − N ∑ N f ^ ( n ) e i n θ
So what conditions on f f f would guarantee the absolute convergence of its Fourier series?
# Decay of the Fourier coefficientsCorollary 2.2.4 Suppose that f ∈ C 2 ( S 1 ) f\in C^2(S^1) f ∈ C 2 ( S 1 ) , then
f ^ ( n ) = O ( 1 / n 2 ) , as ∣ n ∣ → ∞ \hat f(n)=O(1/n^2), \text{ as } |n|\to \infty f ^ ( n ) = O ( 1 / n 2 ) , as ∣ n ∣ → ∞
so that the Fourier series of f f f converges absolutely and uniformly to f f f
Moreover, we have more generalized version
Theorem 2.2.5 Suppose that f f f is m m m times differentiable and f ( m ) f^{(m)} f ( m ) is piecewise continuous, then the Fourier series of f f f is uniformly convergent to f f f , that is to say
S N ( f ) ( x ) ⇉ f S_N(f)(x)\rightrightarrows f S N ( f ) ( x ) ⇉ f
and
∣ f − S n ( x ) ∣ < ϵ n n m − 1 / 2 |f-S_n(x)|<\dfrac{\epsilon_n}{n^{m-1/2}} ∣ f − S n ( x ) ∣ < n m − 1 / 2 ϵ n
where lim n → ∞ ϵ n = 0 , ϵ n > 0 \displaystyle \lim_{n\to\infty}\epsilon_n=0,\ \epsilon_n>0 n → ∞ lim ϵ n = 0 , ϵ n > 0 .
Proof Tips: orthogonal
Lemma 2.2.6 For f ∈ R ( S 1 ) f\in\mathcal R(S^1) f ∈ R ( S 1 )
1 2 π ∫ − π π ∣ f ( x ) ∣ 2 d x = 1 2 π ∫ − π π ∣ f ( x ) − S N ( x ) ∣ 2 d x + ∑ k = − N N ∣ f ^ ( n ) ∣ 2 \dfrac 1{2\pi}\int^\pi_{-\pi}|f(x)|^2\mathrm dx=\dfrac 1{2\pi}\int^\pi_{-\pi}|f(x)-S_N(x)|^2\mathrm dx+\sum^{N}_{k=-N}|\hat f(n)|^2 2 π 1 ∫ − π π ∣ f ( x ) ∣ 2 d x = 2 π 1 ∫ − π π ∣ f ( x ) − S N ( x ) ∣ 2 d x + k = − N ∑ N ∣ f ^ ( n ) ∣ 2
It will be proved at 3.1.