Google 翻译生成的粗略草稿，质量较低。术语依照术语表，但句子可能有误；之后会由仔细的译文取代。
§ 2。非紧区间上的积分
1. 非紧区间上的积分的定义
设I为*延长线*R ‾ \overline{\mathbf{R}} R [ a , b ] [a,\ b] [ a , b ] a a a b b b f f f [ a , b [ [a,\ b[ [ a , b [ R \mathbf{R} R g g g [ a , b ] [a,\ b] [ a , b ] f f f [ a , b ] [a,\ b] [ a , b ] a a a b b b [ a , b [ [a,\ b[ [ a , b [ f ( x ) f(x) f ( x )
我们将把自己限制在以下情况：存在I = [ a , b ] I = [a,\ b] I = [ a , b ] ( c i ) 0 ≤ i ≤ n (c_i)_{0 \leq i \leq n} ( c i ) 0 ≤ i ≤ n c 0 = a , c n = b c_0 = a,\ c_n = b c 0 = a , c n = b f f f [ c i , c i + 1 [ [c_i,\ c_{i+1}[ [ c i , c i + 1 [ c i c_i c i I I I [ a , b [ [a,\ b[ [ a , b [ [ a , b [ [a,\ b[ [ a , b [ n = 1 n = 1 n = 1
如果f f f g g g c c c [ c i , c i + 1 [ ( 0 ≤ i ≤ n − 1 ) [c_i,\ c_{i+1}[\ (0 \leq i \leq n - 1) [ c i , c i + 1 [ ( 0 ≤ i ≤ n − 1 ) x x x g ( x ) − g ( c ) = ∫ c x f ( t ) d t g(x) - g(c) = \int_c^x f(t)\,dt g ( x ) − g ( c ) = ∫ c x f ( t ) d t g g g I I I x x x c i c_i c i x x x c i + 1 c_{i+1} c i + 1 E E E ∫ c x f ( t ) d t \int_c^x f(t)\,dt ∫ c x f ( t ) d t t t t g i g_i g i f f f [ c i , c i + 1 [ ( 0 ≤ i ≤ n − 1 ) [c_i,\ c_{i+1}[\ (0 \leq i \leq n - 1) [ c i , c i + 1 [ ( 0 ≤ i ≤ n − 1 ) g g g c i c_i c i I I I 0 ≤ i ≤ n − 1 0 \leq i \leq n - 1 0 ≤ i ≤ n − 1 [ c i , c i + 1 [ [c_i,\ c_{i+1}[ [ c i , c i + 1 [ g i ( x ) + ∑ k = 1 i ( g k − 1 ( c k − ) − g k ( c k + ) ) g_i(x) + \sum_{k=1}^i (g_{k-1}(c_k-) - g_k(c_k+)) g i ( x ) + ∑ k = 1 i ( g k − 1 ( c k − ) − g k ( c k + )) I I I c i c_i c i f f f I I I f f f g + a g + a g + a E E E a a a
有人说向量函数*f f f [ a , b [ [a,\ b[ [ a , b [ R ‾ \overline{\mathbf{R}} R f f f [ a , b [ [a,\ b[ [ a , b [ g g g 是 f f f 在 [ a , b [ [a,\ b[ [ a , b [ x 0 x_0 x 0 和 x x x 是 [ a , b [ [a,\ b[ [ a , b [ g ( x ) − g ( x 0 ) g(x) - g(x_0) g ( x ) − g ( x 0 ) 的积分 f f f 从 x 0 x_0 x 0 到 x x x ∫ x 0 x f ( t ) d t \int_{x_0}^x f(t)\,dt ∫ x 0 x f ( t ) d t
这个概念显然与当区间[ x 0 , x ] [x_0,\ x] [ x 0 , x ] c i c_i c i
注位于 def 之前。图 1 表明，对于 f f f [ a , b [ [a,\ b[ [ a , b [ [ c i , c i + 1 [ [c_i,\ c_{i+1}[ [ c i , c i + 1 [ f f f I ⊂ R I \subset \mathbf{R} I ⊂ R a , b a,\ b a , b a < b a < b a < b a , b a,\ b a , b f f f a , b a,\ b a , b f f f I I I ( x , y ) (x,\ y) ( x , y ) ( a , b ) ∈ R ‾ 2 (a,\ b) \in \overline{\mathbf{R}}^2 ( a , b ) ∈ R 2 I × I I \times I I × I ∫ a x f ( t ) d t \int_a^x f(t)\,dt ∫ a x f ( t ) d t ∫ a b f ( t ) d t \int_a^b f(t)\,dt ∫ a b f ( t ) d t ∫ a b f ( t ) d t \int_a^b f(t)\,dt ∫ a b f ( t ) d t 收敛 （或 收敛 ），而不是说 f f f I I I
i) 整的 ∫ 1 + ∞ d t / t 2 \int_1^{+\infty} dt/t^2 ∫ 1 + ∞ d t / t 2
∫ 1 x d t t 2 = 1 − 1 x .
\int_1^x \frac{dt}{t^2} = 1 - \frac{1}{x}.
∫ 1 x t 2 d t = 1 − x 1 . 2) 整的∫ 0 1 d t / t \int_0^1 dt / \sqrt{t} ∫ 0 1 d t / t
∫ 1 x d t t = 2 ( 1 − x ) for x > 0.
\int_1^x \frac{dt}{\sqrt{t}} = 2(1 - \sqrt{x}) \quad \text{for} \quad x > 0.
∫ 1 x t d t = 2 ( 1 − x ) for x > 0. 3) 令( u n ) n ≥ 1 (\mathbf{u}_n)_{n \geq 1} ( u n ) n ≥ 1 f \mathbf{f} f [ 1 , + ∞ [ [1, +\infty[ [ 1 , + ∞ [ n ≤ x < n + 1 n \leq x < n+1 n ≤ x < n + 1 f ( x ) = u n \mathbf{f}(x) = \mathbf{u}_n f ( x ) = u n ∫ 1 + ∞ f ( t ) d t \int_1^{+\infty} \mathbf{f}(t)\,dt ∫ 1 + ∞ f ( t ) d t u n \mathbf{u}_n u n
∫ 1 n f ( t ) d t = ∑ p = 1 n − 1 u p ,
\int_1^n \mathbf{f}(t)\,dt = \sum_{p=1}^{n-1} \mathbf{u}_p,
∫ 1 n f ( t ) d t = p = 1 ∑ n − 1 u p , 所以条件是必要的；反之，若级数与一般项
u n \mathbf{u}_n u n 收敛于E，则
lim n → ∞ u n = 0 \lim_{n \to \infty} \mathbf{u}_n = 0 lim n → ∞ u n = 0 ；现在，如果
n ≤ x ≤ n + 1 n \leq x \leq n+1 n ≤ x ≤ n + 1 ，则有
∫ 1 n f ( t ) d t = ∑ p = 1 n − 1 u p + u n ( x − n ) \int_1^n \mathbf{f}(t)\,dt = \sum_{p=1}^{n-1} \mathbf{u}_p + \mathbf{u}_n(x-n) ∫ 1 n f ( t ) d t = ∑ p = 1 n − 1 u p + u n ( x − n ) 。所以当
x x x 趋于
+ ∞ +\infty + ∞ 时，这个积分有极限
∑ n = 1 ∞ u n \sum_{n=1}^{\infty} \mathbf{u}_n ∑ n = 1 ∞ u n 。
直接的，如果分段调节函数f \mathbf{f} f f \mathbf{f} f g \mathbf{g} g ] a , b [ ]a,\ b[ ] a , b [ f ′ \mathbf{f}' f ′ g ′ \mathbf{g}' g ′ [ f . g ] ∣ x y [\mathbf{f}.\mathbf{g}]|_x^y [ f . g ] ∣ x y [ f . g ] ∣ a b [\mathbf{f}.\mathbf{g}]|_a^b [ f . g ] ∣ a b ( x , y ) (x,\ y) ( x , y ) ( a , b ) (a,\ b) ( a , b ) a < x ≤ y < b a < x \leq y < b a < x ≤ y < b [ f . g ] ∣ a b [\mathbf{f}.\mathbf{g}]|_a^b [ f . g ] ∣ a b ∫ a b [ f ( t ) . g ′ ( t ) ] d t \int_a^b [\mathbf{f}(t).\mathbf{g}'(t)]\,dt ∫ a b [ f ( t ) . g ′ ( t )] d t ∫ a b [ f ′ ( t ) . g ( t ) ] d t \int_a^b [\mathbf{f}'(t).\mathbf{g}(t)]\,dt ∫ a b [ f ′ ( t ) . g ( t )] d t
最后，令f f f I = ] a , b [ I = ]a,\ b[ I = ] a , b [ ] a , b [ ]a,\ b[ ] a , b [ f ′ f' f ′ g g g f ( I ) f(I) f ( I ) g ( f ( x ) ) f ′ ( x ) g(f(x))f'(x) g ( f ( x )) f ′ ( x ) f f f a a a b b b g g g f ( a + ) f(a+) f ( a + ) f ( b − ) f(b-) f ( b − )
∫ a b g ( f ( t ) ) f ′ ( t ) d t = ∫ f ( a + ) f ( b − ) g ( u ) d u .
\int_a^b g(f(t))f'(t)\,dt = \int_{f(a+)}^{f(b-)} g(u)\,du.
∫ a b g ( f ( t )) f ′ ( t ) d t = ∫ f ( a + ) f ( b − ) g ( u ) d u . 事实上，如果
( x , y ) (x,\ y) ( x , y ) 趋向于
( a , b ) (a,\ b) ( a , b ) ，那么
( f ( x ) , f ( y ) ) (f(x),\ f(y)) ( f ( x ) , f ( y )) 趋向于
( f ( a + ) , f ( b − ) ) (f(a+),\ f(b-)) ( f ( a + ) , f ( b − )) 假设；只需在
x x x 和
y y y 之间应用II、p. 60的公式(12)，并传递到极限即可获得(1)。
给定非紧区间 I ⊂ R I \subset \mathbf{R} I ⊂ R f \mathbf{f} f a a a b b b a < b a < b a < b f \mathbf{f} f J ⊂ I J \subset I J ⊂ I 有向集合 K ( I ) \mathscr{K}(I) K ( I ) ⊂ 1 \subset ^1 ⊂ 1 [ α , β ] [ \alpha , \beta ] [ α , β ] [ γ , δ ] [ \gamma , \delta ] [ γ , δ ] I I I λ = min ( α , γ ) \lambda = \min(\alpha, \gamma) λ = min ( α , γ ) μ = max ( β , δ ) \mu = \max(\beta, \delta) μ = max ( β , δ ) [ λ , μ ] [ \lambda , \mu ] [ λ , μ ] I I I I I I J = [ α , β ] J = [ \alpha , \beta ] J = [ α , β ]
∫ J f ( t ) d t = ∫ α β f ( t ) d t ;
\int_J f(t) \, dt = \int_\alpha^\beta f(t) \, dt;
∫ J f ( t ) d t = ∫ α β f ( t ) d t ; 为了使f f f I I I J ↦ ∫ J f ( t ) d t J \mapsto \int_J f(t) \, dt J ↦ ∫ J f ( t ) d t K ( I ) \mathfrak{K}(I) K ( I ) ∫ a b f ( t ) d t \int_a^b f(t) \, dt ∫ a b f ( t ) d t ∫ I f ( t ) d t \int_I f(t) \, dt ∫ I f ( t ) d t
引用了 1 引用了 II § 1. 。原语和积分
命题 1（柯西积分准则） 088H
令 f f f I ⊂ R I \subset \mathbf{R} I ⊂ R a a a b b b a < b a < b a < b ∫ a b f ( t ) d t \int_a^b f(t) \, dt ∫ a b f ( t ) d t ε > 0 \varepsilon > 0 ε > 0 I I I J 0 = [ α , β ] J_0 = [\alpha, \beta] J 0 = [ α , β ] I I I J 0 J_0 J 0 K = [ x , y ] K = [x, y] K = [ x , y ] ∥ ∫ K f ( t ) d t ∥ ≤ ε \| \int_K f(t) \, dt \| \leq \varepsilon ∥ ∫ K f ( t ) d t ∥ ≤ ε
事实上，由于E E E ∫ I f ( t ) d t \int_I f(t) \, dt ∫ I f ( t ) d t ε > 0 \varepsilon > 0 ε > 0 J 0 = [ α , β ] J_0 = [\alpha, \beta] J 0 = [ α , β ] J J J J 0 ⊂ J ⊂ I J_0 \subset J \subset I J 0 ⊂ J ⊂ I ∥ ∫ J f ( t ) d t − ∫ J 0 f ( t ) d t ∥ ≤ ε \| \int_J f(t) \, dt - \int_{J_0} f(t) \, dt \| \leq \varepsilon ∥ ∫ J f ( t ) d t − ∫ J 0 f ( t ) d t ∥ ≤ ε
设J 0 = [ α , β ] J_0 = [\alpha, \beta] J 0 = [ α , β ] I I I ∥ ∫ J f ( t ) d t − ∫ J ′ f ( t ) d t ∥ ≤ ε \| \int_J f(t) \, dt - \int_{J'} f(t) \, dt \| \leq \varepsilon ∥ ∫ J f ( t ) d t − ∫ J ′ f ( t ) d t ∥ ≤ ε I I I J ′ J' J ′ J , J ′ J, J' J , J ′ ∥ ∫ K f ( t ) d t ∥ ≤ ε \| \int_K f(t) \, dt \| \leq \varepsilon ∥ ∫ K f ( t ) d t ∥ ≤ ε ∥ ∫ K f ( t ) d t ∥ ≤ ε / 2 \| \int_K f(t) \, dt \| \leq \varepsilon / 2 ∥ ∫ K f ( t ) d t ∥ ≤ ε /2 K K K I I I J 0 J_0 J 0
引用了 3 引用了 E III § 1. 序关系。 有序集 , TG I § 7. Limits , TG IV § 5. 实值函数
事实上，如果对于 J 0 ⊂ J ⊂ I J_0 \subset J \subset I J 0 ⊂ J ⊂ I J 0 ⊂ J ′ ⊂ I J_0 \subset J' \subset I J 0 ⊂ J ′ ⊂ I
∥ ∫ J f ( t ) d t − ∫ J ′ f ( t ) d t ∥ ≤ ε
\left\| \int_J f(t) \, dt - \int_{J'} f(t) \, dt \right\| \leq \varepsilon
∫ J f ( t ) d t − ∫ J ′ f ( t ) d t ≤ ε 特别是，对于x ≤ y ≤ α x \leq y \leq \alpha x ≤ y ≤ α β ≤ x ≤ y \beta \leq x \leq y β ≤ x ≤ y I I I x x x y y y ∥ ∫ x y f ( t ) d t ∥ ≤ ε \| \int_x^y f(t) \, dt \| \leq \varepsilon ∥ ∫ x y f ( t ) d t ∥ ≤ ε ∥ ∫ K f ( t ) d t ∥ ≤ ε / 2 \| \int_K f(t) \, dt \| \leq \varepsilon / 2 ∥ ∫ K f ( t ) d t ∥ ≤ ε /2 K ⊂ I K \subset I K ⊂ I K ∩ J 0 = ∅ K \cap J_0 = \emptyset K ∩ J 0 = ∅ J = [ x , y ] J = [x, y] J = [ x , y ] J ′ = [ z , t ] J' = [z, t] J ′ = [ z , t ] I I I J 0 J_0 J 0
1 回想一下（*集合理论*，III，p. 144），I I I F \mathfrak{F} F ⊂ \subset ⊂ X ∈ F X \in \mathfrak{F} X ∈ F Y ∈ F Y \in \mathfrak{F} Y ∈ F Z ∈ F Z \in \mathfrak{F} Z ∈ F X ⊂ Z X \subset Z X ⊂ Z Y ⊂ Z Y \subset Z Y ⊂ Z S ( X ) S(X) S ( X ) U ∈ F U \in \mathfrak{F} U ∈ F F \mathfrak{F} F U ⊃ X U \supset X U ⊃ X F \mathfrak{F} F S ( X ) S(X) S ( X ) F \mathfrak{F} F F \mathfrak{F} F f f f F \mathfrak{F} F F \mathfrak{F} F Gen.Top. 、IV、p. 348）。
∥ ∫ J f ( t ) d t − ∫ J ′ f ( t ) d t ∥ = ∥ ∫ x z f ( t ) d t + ∫ t y f ( t ) d t ∥ ≤ ε ,
\left\| \int_J f(t)\,dt - \int_{J'} f(t)\,dt \right\| = \left\| \int_x^z f(t)\,dt + \int_t^y f(t)\,dt \right\| \leq \varepsilon,
∫ J f ( t ) d t − ∫ J ′ f ( t ) d t = ∫ x z f ( t ) d t + ∫ t y f ( t ) d t ≤ ε , 自从
x ≤ α ≤ β ≤ y and z ≤ α ≤ β ≤ t .
x \leq \alpha \leq \beta \leq y \quad \text{and} \quad z \leq \alpha \leq \beta \leq t.
x ≤ α ≤ β ≤ y and z ≤ α ≤ β ≤ t .
如果区间I是*有界*，并且如果f f f ∫ I f ( t ) d t \int_I f(t)\,dt ∫ I f ( t ) d t y ≤ α ≤ β ≤ z y \leq \alpha \leq \beta \leq z y ≤ α ≤ β ≤ z
∥ ∫ y α f ( t ) d t ∥ ≤ ( α − a ) sup t ∈ I ∥ f ( v ) ∥ , ∥ ∫ β z f ( t ) d t ∥ ≤ ( b − β ) sup t ∈ I ∥ f ( x ) ∥
\left\| \int_y^\alpha f(t)\,dt \right\| \leq (\alpha - a) \sup_{t \in I} \|f(v)\| , \quad \left\| \int_\beta^z f(t)\,dt \right\| \leq (b - \beta) \sup_{t \in I} \|f(x)\|
∫ y α f ( t ) d t ≤ ( α − a ) t ∈ I sup ∥ f ( v ) ∥ , ∫ β z f ( t ) d t ≤ ( b − β ) t ∈ I sup ∥ f ( x ) ∥ 只要
α − a \alpha - a α − a 和
b − β b - \beta b − β 足够小就可以满足柯西准则。
人们可能注，在这种情况下，I上的f f f f f f
2. 非紧区间上的正函数积分
令 f f f I ⊂ R I \subset \mathbf{R} I ⊂ R ≥ 0 \geq 0 ≥ 0 a a a b b b a < b a < b a < b ∫ a b f ( t ) d t \int_a^b f(t)\,dt ∫ a b f ( t ) d t ∫ J f ( t ) d t \int_J f(t)\,dt ∫ J f ( t ) d t ∫ a b f ( t ) d t \int_a^b f(t)\,dt ∫ a b f ( t ) d t ∫ J f ( t ) d t \int_J f(t)\,dt ∫ J f ( t ) d t
引用了 2 引用了 TG IV § 5. 实值函数 , II § 1. 。原语和积分
事实上，由于 f ≥ 0 f \geq 0 f ≥ 0 J ⊂ J ′ J \subset J' J ⊂ J ′
∫ J f d t ≤ ∫ J ′ f d t ;
\int_J f\,dt \leq \int_{J'} f\,dt;
∫ J f d t ≤ ∫ J ′ f d t ; 因此，映射
J ↦ ∫ J f d t J \mapsto \int_J f\,dt J ↦ ∫ J f d t 是递增，命题源自单调极限定理（*Gen.Top.*，IV，p. 349，第2个）。
当映射J ↦ ∫ J f ( t ) d t J \mapsto \int_J f(t)\,dt J ↦ ∫ J f ( t ) d t K ( I ) \mathscr{K}(I) K ( I ) + ∞ +\infty + ∞ ∫ a b f ( t ) d t \int_a^b f(t)\,dt ∫ a b f ( t ) d t + ∞ +\infty + ∞ ≥ 0 \geq 0 ≥ 0
命题 3（比较原理）。 让f f f g g g I ⊂ R I \subset \mathbf{R} I ⊂ R 0 ≤ f ( x ) ≤ g ( x ) 0 \leq f(x) \leq g(x) 0 ≤ f ( x ) ≤ g ( x ) f f f g g g cf.* II、p. 61，第 6 条）。如果g g g I I I f f f ∫ I f ( t ) d t ≤ ∫ I g ( t ) d t \int_I f(t)\,dt \leq \int_I g(t)\,dt ∫ I f ( t ) d t ≤ ∫ I g ( t ) d t I I I f ( x ) = g ( x ) f(x) = g(x) f ( x ) = g ( x ) f f f g g g
现在对于每个紧凑区间 J ⊂ I J \subset I J ⊂ I
∫ J f ( t ) d t ≤ ∫ J g ( t ) d t ;
\int_J f(t)\,dt \leq \int_J g(t)\,dt;
∫ J f ( t ) d t ≤ ∫ J g ( t ) d t ; 由于∫ J g ( t ) d t ≤ ∫ I g ( t ) d t \int_J g(t)\,dt \leq \int_I g(t)\,dt ∫ J g ( t ) d t ≤ ∫ I g ( t ) d t ∫ J f d t \int_J f\,dt ∫ J f d t ∫ I f ( t ) d t \int_I f(t)\,dt ∫ I f ( t ) d t ∫ I f ( t ) d t ≤ ∫ I g ( t ) d t \int_I f(t)\,dt \leq \int_I g(t)\,dt ∫ I f ( t ) d t ≤ ∫ I g ( t ) d t f ( x ) < g ( x ) f(x) < g(x) f ( x ) < g ( x ) x ∈ I x \in I x ∈ I f f f g g g I I I [ c , d ] [c,\ d] [ c , d ] x ∈ [ c , d ] x \in [c,\ d] x ∈ [ c , d ] ∫ c d f ( t ) d t < ∫ c d g ( t ) d t \int_c^d f(t)\,dt < \int_c^d g(t)\,dt ∫ c d f ( t ) d t < ∫ c d g ( t ) d t ∫ a c f ( t ) d t ≤ ∫ a c g ( t ) d t \int_a^c f(t)\,dt \leq \int_a^c g(t)\,dt ∫ a c f ( t ) d t ≤ ∫ a c g ( t ) d t ∫ d b f ( t ) d t ≤ ∫ d b g ( t ) d t \int_d^b f(t)\,dt \leq \int_d^b g(t)\,dt ∫ d b f ( t ) d t ≤ ∫ d b g ( t ) d t ∫ a b f ( t ) d t < ∫ a b g ( t ) d t \int_a^b f(t)\,dt < \int_a^b g(t)\,dt ∫ a b f ( t ) d t < ∫ a b g ( t ) d t
这个命题提供了最常用的方法来判断函数f ≥ 0 f \geq 0 f ≥ 0 f f f g ≥ 0 g \geq 0 g ≥ 0
3. 绝对收敛积分
有人说，如果正数函数 ∥ f ( x ) ∥ \| \mathbf{f}(x) \| ∥ f ( x ) ∥ f \mathbf{f} f I ⊂ R I \subset \mathbf{R} I ⊂ R
∣ ∫ I f ( t ) d t ∣ ≤ ∫ I ∥ f ( t ) ∥ d t . (2)
\left| \int_I \mathbf{f}(t)\,dt \right| \leq \int_I \| \mathbf{f}(t) \| \,dt.\tag{2}
∫ I f ( t ) d t ≤ ∫ I ∥ f ( t ) ∥ d t . ( 2 ) 事实上，对于每个紧凑区间 J ⊂ I J \subset I J ⊂ I
∣ ∫ J f ( t ) d t ∣ ≤ ∫ J ∥ f ( t ) ∥ d t . (3)
\left| \int_J \mathbf{f}(t)\,dt \right| \leq \int_J \| \mathbf{f}(t) \| \,dt.\tag{3}
∫ J f ( t ) d t ≤ ∫ J ∥ f ( t ) ∥ d t . ( 3 ) 如果正函数 ∥ f ( x ) ∥ \| \mathbf{f}(x) \| ∥ f ( x ) ∥ ε > 0 \varepsilon > 0 ε > 0 I I I [ α , β ] [\alpha,\ \beta] [ α , β ] I I I [ λ , γ ] [\lambda,\ \gamma] [ λ , γ ] [ α , β ] [\alpha,\ \beta] [ α , β ] ∫ λ γ ∥ f ( t ) d t ∥ ≤ ε \int_{\lambda}^{\gamma} \| \mathbf{f}(t) \, dt \| \leq \varepsilon ∫ λ γ ∥ f ( t ) d t ∥ ≤ ε ∣ ∫ λ γ f ( t ) d t ∣ ≤ ε \left| \int_{\lambda}^{\gamma} \mathbf{f}(t)\,dt \right| \leq \varepsilon ∫ λ γ f ( t ) d t ≤ ε I I I
令E , F , G E, F, G E , F , G R \mathbf{R} R ( x , y ) ↦ x . y (x, y) \mapsto x.y ( x , y ) ↦ x . y E × F E \times F E × F G G G f , g \mathbf{f}, \mathbf{g} f , g I I I E E E F F F f \mathbf{f} f I I I
事实上，存在数 h > 0，使得 ∥ [ x . y ] ∥ ≤ h ∥ x ∥ . ∥ y ∥ \| [x.y] \| \leq h \| x \| . \| y \| ∥ [ x . y ] ∥ ≤ h ∥ x ∥.∥ y ∥ k = sup x ∈ I ∥ f ( x ) ∥ k = \sup_{x \in I} \| f(x) \| k = sup x ∈ I ∥ f ( x ) ∥ ∥ [ f ( x ) . g ( x ) ] ∥ ≤ h k ∥ g ( x ) ∥ \| [f(x).g(x)] \| \leq h k \| g(x) \| ∥ [ f ( x ) . g ( x )] ∥ ≤ hk ∥ g ( x ) ∥
∥ ∫ I [ f ( t ) . g ( t ) ] d t ∥ ≤ h k ∫ I ∥ g ( t ) ∥ d t .
\left\| \int_I [f(t).g(t)] \, dt \right\| \leq h k \int_I \| g(t) \| \, dt.
∫ I [ f ( t ) . g ( t )] d t ≤ hk ∫ I ∥ g ( t ) ∥ d t .
习题
参见习题为 § 2 。