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。
§ 6。局部凸空间上的测度和测度
在整个截面中，仅考虑实数中超过域的向量空间。局部凸空间是指R \mathbf{R} R E ′ E' E ′ x ∈ E x \in E x ∈ E x ′ ∈ E ′ x' \in E' x ′ ∈ E ′ ⟨ x , x ′ ⟩ = x ′ ( x ) \langle x, x' \rangle = x'(x) ⟨ x , x ′ ⟩ = x ′ ( x )
1. 局部凸空间的 Promeasures
令 E 为局部凸空间。我们用有限余维的E的闭合线性子空间的F ( E ) \mathscr{F}(E) F ( E ) ⊃ \supset ⊃ V ∈ F ( E ) V \in \mathscr{F}(E) V ∈ F ( E ) p V p_V p V E / V E/V E / V F ( E ) \mathscr{F}(E) F ( E ) V ⊃ W V \supset W V ⊃ W E / W E/W E / W p V W p_{VW} p V W E / V E/V E / V Q ( E ) = ( E / V , p V W ) \mathscr{Q}(E) = (E/V, p_{VW}) Q ( E ) = ( E / V , p V W ) F ( E ) \mathscr{F}(E) F ( E )
可以证明，逆系 Q ( E ) \mathscr{Q}(E) Q ( E ) E ′ E' E ′ E ′ ∗ E'{}^* E ′ ∗ σ ( E ′ ∗ , E ′ ) \sigma(E'{}^*, E') σ ( E ′ ∗ , E ′ )
设E E E E E E ( 1 ) ^{(1)} ( 1 ) E E E
引用了 2 引用了 定义 1, IX § 4 , 命题 7
在其他字中，E E E μ \mu μ ( μ V ) V ∈ F ( E ) (\mu_V)_{V \in \mathscr{F}(E)} ( μ V ) V ∈ F ( E ) μ V \mu_V μ V E / V E/V E / V V ⊃ W V \supset W V ⊃ W μ V = p V W ( μ W ) \mu_V = p_{VW}(\mu_W) μ V = p V W ( μ W ) μ V \mu_V μ V μ \mu μ
对于E E E V V V F ( E ) \mathscr{F}(E) F ( E ) E ′ E' E ′ x 1 ′ , … , x n ′ x'_1, \ldots, x'_n x 1 ′ , … , x n ′ V V V x ∈ E x \in E x ∈ E ⟨ x , x i ′ ⟩ = 0 \langle x, x'_i \rangle = 0 ⟨ x , x i ′ ⟩ = 0 1 ≤ i ≤ n 1 \leq i \leq n 1 ≤ i ≤ n E E E E E E E ′ E' E ′
令 λ \lambda λ E E E V ∈ F ( E ) V \in \mathscr{F}(E) V ∈ F ( E ) E E E p V p_V p V E / V E/V E / V λ \lambda λ λ ~ V \widetilde{\lambda}_V λ V F ( E ) \mathscr{F}(E) F ( E ) V V V W W W p V = p V W ∘ p W p_V = p_{VW} \circ p_W p V = p V W ∘ p W V ⊃ W V \supset W V ⊃ W λ ~ = ( λ ~ V ) V ∈ F ( E ) \widetilde{\lambda} = (\widetilde{\lambda}_V)_{V \in \mathscr{F}(E)} λ = ( λ V ) V ∈ F ( E ) E E E λ ~ \widetilde{\lambda} λ λ \lambda λ λ \lambda λ λ ~ \widetilde{\lambda} λ
设E E E λ ↦ λ ~ \lambda \mapsto \widetilde{\lambda} λ ↦ λ E E E E E E ( μ V ) V ∈ F ( E ) (\mu_V)_{V \in \mathscr{F}(E)} ( μ V ) V ∈ F ( E )
对于每个 ε > 0 \varepsilon > 0 ε > 0 E E E K K K μ V ( E / V − p V ( K ) ) ≤ ε \mu_V(E/V - p_V(K)) \leq \varepsilon μ V ( E / V − p V ( K )) ≤ ε V ∈ F ( E ) V \in \mathscr{F}(E) V ∈ F ( E )
引用了 2 · 被以下引用 1 引用了 定理 1（普罗霍罗夫）, IX § 4 , 命题 2
被以下引用 定理 2（米洛斯）
已知E E E ⋂ V ∈ F ( E ) V = { 0 } \bigcap_{V \in \mathscr{F}(E)} V = \{0\} ⋂ V ∈ F ( E ) V = { 0 } ( p V ) V ∈ F ( E ) (p_V)_{V \in \mathscr{F}(E)} ( p V ) V ∈ F ( E )
特别地，映射λ ↦ λ ~ \lambda \mapsto \widetilde{\lambda} λ ↦ λ μ \mu μ E E E E E E λ \lambda λ μ = λ ~ \mu = \widetilde{\lambda} μ = λ μ \mu μ E E E μ = ( μ V ) V ∈ F ( E ) \mu = (\mu_V)_{V \in \mathscr{F}(E)} μ = ( μ V ) V ∈ F ( E ) { 0 } ∈ F ( E ) \{0\} \in \mathscr{F}(E) { 0 } ∈ F ( E ) E / { 0 } = E E/\{0\} = E E / { 0 } = E p V , { 0 } = p V p_{V,\{0\}} = p_V p V , { 0 } = p V μ V = p V ( μ { 0 } ) \mu_V = p_V(\mu_{\{0\}}) μ V = p V ( μ { 0 } ) V ∈ F ( E ) V \in \mathscr{F}(E) V ∈ F ( E ) μ = λ ~ \mu = \widetilde{\lambda} μ = λ λ = μ { 0 } \lambda = \mu_{\{0\}} λ = μ { 0 }
(1) 也称为“射影系统”。
对于每个t ∈ T t \in T t ∈ T ε t \varepsilon_t ε t f ↦ f ( t ) f \mapsto f(t) f ↦ f ( t ) ( ε t ) t ∈ T (\varepsilon_t)_{t \in T} ( ε t ) t ∈ T E ′ E' E ′ Φ \Phi Φ J ∈ Φ J \in \Phi J ∈ Φ E J E_J E J F ∈ F ( E ) F \in \mathscr{F}(E) F ∈ F ( E ) F ∘ F^\circ F ∘ E ′ E' E ′ J ∈ Φ J \in \Phi J ∈ Φ F ∘ F^\circ F ∘ ε t \varepsilon_t ε t t ∈ J t \in J t ∈ J E ′ E' E ′ F ∘ ⊂ G F^\circ \subset G F ∘ ⊂ G E J = G ∘ ⊂ F ∘ ∘ = F E_J = G^\circ \subset F^{\circ\circ} = F E J = G ∘ ⊂ F ∘∘ = F ( E J ) J ∈ Φ (E_J)_{J \in \Phi} ( E J ) J ∈ Φ F ( E ) \mathscr{F}(E) F ( E )
2. 像的优惠措施
设E和E 1 E_1 E 1 u u u E 1 E_1 E 1 F ( E ) \mathscr{F}(E) F ( E ) V 1 ∈ F ( E 1 ) V_1 \in \mathscr{F}(E_1) V 1 ∈ F ( E 1 ) V = u ‾ − 1 ( V 1 ) V = \overline{u}^{-1}(V_1) V = u − 1 ( V 1 ) u u u E / V E/V E / V u V 1 u_{V_1} u V 1 E 1 / V 1 E_1/V_1 E 1 / V 1 F ( E 1 ) \mathscr{F}(E_1) F ( E 1 ) V 1 V_1 V 1 W 1 W_1 W 1 V 1 ⊃ W 1 V_1 \supset W_1 V 1 ⊃ W 1 V = u ‾ − 1 ( V 1 ) V = \overline{u}^{-1}(V_1) V = u − 1 ( V 1 ) W = u ‾ − 1 ( W 1 ) W = \overline{u}^{-1}(W_1) W = u − 1 ( W 1 ) V ⊃ W V \supset W V ⊃ W
E → p w E / W → p v w E / V ↓ u ↓ u W 1 ↓ u V 1 E 1 → p w 1 E 1 / W 1 → p v 1 W 1 E 1 / V 1
\begin{array}{ccccc}
E & \xrightarrow{pw} & E/W & \xrightarrow{pvw} & E/V \\
\downarrow u & & \downarrow u_{W_1} & & \downarrow u_{V_1} \\
E_1 & \xrightarrow{pw_1} & E_1/W_1 & \xrightarrow{pv_{1W_1}} & E_1/V_1
\end{array}
E ↓ u E 1 pw p w 1 E / W ↓ u W 1 E 1 / W 1 p v w p v 1 W 1 E / V ↓ u V 1 E 1 / V 1 现在让 μ = ( μ V ) V ∈ F ( E ) \mu = (\mu_V)_{V \in \mathscr{F}(E)} μ = ( μ V ) V ∈ F ( E ) V 1 ∈ F ( E 1 ) V_1 \in \mathscr{F}(E_1) V 1 ∈ F ( E 1 )
(1)
ν V 1 = u V 1 ( μ u − 1 ( V 1 ) ) .
\nu_{V_1} = u_{V_1}(\mu_{u^{-1}(V_1)}).
ν V 1 = u V 1 ( μ u − 1 ( V 1 ) ) . 前面图表的交换律表明家庭ν = ( ν V 1 ) V 1 ∈ F ( E 1 ) \nu = (\nu_{V_1})_{V_1 \in \mathscr{F}(E_1)} ν = ( ν V 1 ) V 1 ∈ F ( E 1 ) E 1 E_1 E 1 ν \nu ν u u u μ \mu μ u ( μ ) u(\mu) u ( μ )
令λ \lambda λ u ( λ ) u(\lambda) u ( λ ) E 1 E_1 E 1 u u u λ \lambda λ μ \mu μ λ \lambda λ u ( μ ) u(\mu) u ( μ ) u ( λ ) u(\lambda) u ( λ )
让V ∈ F ( E ) V \in \mathscr{F}(E) V ∈ F ( E ) p V : E → E / V p_V : E \to E/V p V : E → E / V μ \mu μ E / V E/V E / V μ V \mu_V μ V
令 u 1 u_1 u 1 E 1 E_1 E 1 E 2 E_2 E 2
( u 1 ∘ u ) ( μ ) = u 1 ( u ( μ ) )
(u_1 \circ u)(\mu) = u_1(u(\mu))
( u 1 ∘ u ) ( μ ) = u 1 ( u ( μ )) （优惠措施像中的传递性） 。
3. 傅里叶变换的优惠
令 E E E μ = ( μ V ) V ∈ F ( E ) \mu = (\mu_V)_{V \in \mathscr{F}(E)} μ = ( μ V ) V ∈ F ( E ) E E E E E E x ′ x' x ′ μ x ′ \mu_{x'} μ x ′ R \mathbf{R} R E E E μ \mu μ x ′ x' x ′ μ \mu μ E ′ E' E ′ F μ \mathscr{F}\mu F μ
( F μ ) ( x ′ ) = ∫ R e i t d μ x ′ ( t ) .
(\mathscr{F}\mu)(x') = \int_{\mathbf{R}} e^{it} d\mu_{x'}(t).
( F μ ) ( x ′ ) = ∫ R e i t d μ x ′ ( t ) . 令 λ \lambda λ E E E λ \lambda λ E ′ E' E ′
( F λ ) ( x ′ ) = ∫ E e i ⟨ x , x ′ ⟩ d λ ( x ) .
(\mathscr{F}\lambda)(x') = \int_E e^{i\langle x, x' \rangle} d\lambda(x).
( F λ ) ( x ′ ) = ∫ E e i ⟨ x , x ′ ⟩ d λ ( x ) . 令 μ \mu μ λ \lambda λ x ′ ∈ E ′ x' \in E' x ′ ∈ E ′ R \mathbf{R} R μ x ′ \mu_{x'} μ x ′ E E E λ \lambda λ x ′ : E → R x' : E \to \mathbf{R} x ′ : E → R F μ = F λ \mathscr{F}\mu = \mathscr{F}\lambda F μ = F λ
令 μ \mu μ E E E u u u E E E E 1 E_1 E 1 E 1 ′ E'_1 E 1 ′ t u ^t u t u E ′ E' E ′ u u u ν \nu ν E 1 E_1 E 1 u ( μ ) u(\mu) u ( μ ) x 1 ′ ∈ E 1 ′ x'_1 \in E'_1 x 1 ′ ∈ E 1 ′ t u ( x 1 ′ ) = x 1 ′ ∘ u ^t u(x'_1) = x'_1 \circ u t u ( x 1 ′ ) = x 1 ′ ∘ u
ν x 1 ′ = x 1 ′ ( ν ) = x 1 ′ ( u ( μ ) ) = ( x 1 ′ ∘ u ) ( μ ) = μ t u ( x 1 ′ ) .
\nu_{x'_1} = x'_1(\nu) = x'_1(u(\mu)) = (x'_1 \circ u)(\mu) = \mu_{^t u(x'_1)}.
ν x 1 ′ = x 1 ′ ( ν ) = x 1 ′ ( u ( μ )) = ( x 1 ′ ∘ u ) ( μ ) = μ t u ( x 1 ′ ) . 因而，
F ( u ( μ ) ) = ( F μ ) ∘ t u .
\mathscr{F}(u(\mu)) = (\mathscr{F}\mu) \circ {}^t u.
F ( u ( μ )) = ( F μ ) ∘ t u . 特别是，让我们将E E E u u u p V p_V p V E / V E/V E / V V ∈ F ( E ) V \in \mathscr{F}(E) V ∈ F ( E ) E / V E/V E / V p V ( μ ) p_V(\mu) p V ( μ ) μ V \mu_V μ V t p V ^t p_V t p V E / V E/V E / V E ′ E' E ′ V V V V ∘ V^\circ V ∘ t p V ^t p_V t p V ( E / V ) ′ (E/V)' ( E / V ) ′ V ∘ V^\circ V ∘
( F μ ) ( x ′ ) = ∫ E ′ / V e i ⟨ x , x ′ ⟩ d μ V ( x )
(\mathscr{F}\mu)(x') = \int_{E'/V} e^{i\langle x, x' \rangle} d\mu_V(x)
( F μ ) ( x ′ ) = ∫ E ′ / V e i ⟨ x , x ′ ⟩ d μ V ( x ) 对于所有x ′ ∈ V ∘ x' \in V^\circ x ′ ∈ V ∘ E ′ = ⋃ V ∈ F ( E ) V ∘ E' = \bigcup_{V \in \mathscr{F}(E)} V^\circ E ′ = ⋃ V ∈ F ( E ) V ∘ E ′ E' E ′ F μ \mathscr{F}\mu F μ x ′ = 0 x' = 0 x ′ = 0 μ \mu μ ( F μ ) ( 0 ) (\mathscr{F}\mu)(0) ( F μ ) ( 0 )
公式 (5) 允许简化为E E E R n \mathbf{R}^n R n n ⩾ 0 n \geqslant 0 n ⩾ 0 E = R n E = \mathbf{R}^n E = R n μ \mu μ R n \mathbf{R}^n R n
∫ R n e i ⟨ x , y ⟩ d μ ( x ) = 0
\int_{\mathbf{R}^n} e^{i\langle x, y \rangle} d\mu(x) = 0
∫ R n e i ⟨ x , y ⟩ d μ ( x ) = 0 对于R n \mathbf{R}^n R n y y y μ = 0 \mu = 0 μ = 0
对于每个整数 m ⩾ 0 m \geqslant 0 m ⩾ 0 G m G_m G m R n \mathbf{R}^n R n m ⋅ Z n m \cdot \mathbf{Z}^n m ⋅ Z n R n \mathbf{R}^n R n f f f C m \mathscr{C}_m C m f ( x + a ) = f ( x ) f(x + a) = f(x) f ( x + a ) = f ( x ) x ∈ R n x \in \mathbf{R}^n x ∈ R n a ∈ G m a \in G_m a ∈ G m C m \mathscr{C}_m C m x ↦ e 2 π i ⟨ x , q ⟩ x \mapsto e^{2\pi i \langle x, q \rangle} x ↦ e 2 π i ⟨ x , q ⟩ q ∈ m − 1 ⋅ Z n q \in m^{-1} \cdot \mathbf{Z}^n q ∈ m − 1 ⋅ Z n f ∈ C m f \in \mathscr{C}_m f ∈ C m μ ( f ) = 0 \mu(f) = 0 μ ( f ) = 0
令 f f f R n \mathbf{R}^n R n m ⩾ 0 m \geqslant 0 m ⩾ 0 f m ( x ) = ∑ q ∈ G m f ( x + q ) f_m(x) = \sum_{q \in G_m} f(x + q) f m ( x ) = ∑ q ∈ G m f ( x + q ) x ∈ R n x \in \mathbf{R}^n x ∈ R n f m f_m f m C m \mathscr{C}_m C m ( f m ) (f_m) ( f m ) f f f C ⩾ 0 C \geqslant 0 C ⩾ 0 m m m ∣ f m ∣ ⩽ C |f_m| \leqslant C ∣ f m ∣ ⩽ C μ ( f ) = lim m → ∞ μ ( f m ) \mu(f) = \lim_{m \to \infty} \mu(f_m) μ ( f ) = lim m → ∞ μ ( f m ) f m ∈ C m f_m \in \mathscr{C}_m f m ∈ C m μ ( f m ) = 0 \mu(f_m) = 0 μ ( f m ) = 0 μ ( f ) = 0 \mu(f) = 0 μ ( f ) = 0 μ = 0 \mu = 0 μ = 0
4. 高斯积分的计算
∫ R ∣ x ∣ n e − x 2 / 2 d x = 2 n + 1 2 Γ ( n + 1 2 )
\int_{\mathbf{R}} |x|^n e^{-x^2/2} dx = 2^{n+1 \over 2} \Gamma \left( {n+1 \over 2} \right)
∫ R ∣ x ∣ n e − x 2 /2 d x = 2 2 n + 1 Γ ( 2 n + 1 ) (6)
∫ R x 2 n e − x 2 / 2 d x = ( 2 π ) 1 / 2 ( 2 n ) ! 2 n n !
\int_{\mathbf{R}} x^{2n} e^{-x^2/2} dx = (2\pi)^{1/2} \frac{(2n)!}{2^n n!}
∫ R x 2 n e − x 2 /2 d x = ( 2 π ) 1/2 2 n n ! ( 2 n )! (7)
∫ R x 2 n + 1 e − x 2 / 2 d x = 0.
\int_{\mathbf{R}} x^{2n+1} e^{-x^2/2} dx = 0.
∫ R x 2 n + 1 e − x 2 /2 d x = 0. (8)
回忆公式
Γ ( s ) = ∫ 0 ∞ u s − 1 e − u d u
\Gamma(s) = \int_0^\infty u^{s-1} e^{-u} du
Γ ( s ) = ∫ 0 ∞ u s − 1 e − u d u (9)
适用于每个实数 s > 0 s > 0 s > 0 x = ( 2 u ) 1 / 2 x = (2u)^{1/2} x = ( 2 u ) 1/2
∫ 0 ∞ x n e − x 2 / 2 d x = ∫ 0 ∞ ( 2 u ) n / 2 e − u 1 2 2 1 / 2 u − 1 / 2 d u = 2 n − 1 2 Γ ( n + 1 2 ) ,
\int_0^\infty x^n e^{-x^2/2} dx = \int_0^\infty (2u)^{n/2} e^{-u} {1 \over 2} 2^{1/2} u^{-1/2} du = 2^{n-1 \over 2} \Gamma \left( {n+1 \over 2} \right),
∫ 0 ∞ x n e − x 2 /2 d x = ∫ 0 ∞ ( 2 u ) n /2 e − u 2 1 2 1/2 u − 1/2 d u = 2 2 n − 1 Γ ( 2 n + 1 ) , 从而公式 (6) 自
∫ R ∣ x ∣ n e − x 2 / 2 d x = 2 ∫ 0 ∞ x n e − x 2 / 2 d x .
\int_{\mathbf{R}} |x|^n e^{-x^2/2} dx = 2 \int_0^\infty x^n e^{-x^2/2} dx.
∫ R ∣ x ∣ n e − x 2 /2 d x = 2 ∫ 0 ∞ x n e − x 2 /2 d x . 公式 (7) 由 (6) 和关系推出
Γ ( n + 1 2 ) = π 1 / 2 ( 2 n ) ! 2 2 n n ! .
\Gamma \left( n + \frac{1}{2} \right) = \pi^{1/2} \frac{(2n)!}{2^{2n} n!}.
Γ ( n + 2 1 ) = π 1/2 2 2 n n ! ( 2 n )! . (10)
对于n = 0 n = 0 n = 0 Γ ( 1 2 ) = π 1 / 2 \Gamma(\frac{1}{2}) = \pi^{1/2} Γ ( 2 1 ) = π 1/2 Γ ( x + 1 ) = x ⋅ Γ ( x ) \Gamma(x+1) = x \cdot \Gamma(x) Γ ( x + 1 ) = x ⋅ Γ ( x ) n n n
最后，公式 (8) 由函数 x ↦ x 2 n + 1 e − x 2 / 2 x \mapsto x^{2n+1} e^{-x^2/2} x ↦ x 2 n + 1 e − x 2 /2
( 2 π ) − 1 / 2 ∫ R e − x 2 / 2 e i x y d x = e − y 2 / 2 .
(2\pi)^{-1/2} \int_{\mathbf{R}} e^{-x^2/2} e^{ixy} dx = e^{-y^2/2}.
( 2 π ) − 1/2 ∫ R e − x 2 /2 e i x y d x = e − y 2 /2 . (11)
特别是，
( 2 π ) − 1 / 2 ∫ R e − x 2 / 2 d x = 1.
(2\pi)^{-1/2} \int_{\mathbf{R}} e^{-x^2/2} dx = 1.
( 2 π ) − 1/2 ∫ R e − x 2 /2 d x = 1. 变量 x ↦ − x x \mapsto -x x ↦ − x
( 2 π ) − 1 / 2 ∫ R e − x 2 / 2 e i x y d x = ( 2 π ) − 1 / 2 ∫ R e − x 2 / 2 e − i x y d x ;
(2\pi)^{-1/2} \int_{\mathbf{R}} e^{-x^2/2} e^{i xy} dx = (2\pi)^{-1/2} \int_{\mathbf{R}} e^{-x^2/2} e^{-ixy} dx;
( 2 π ) − 1/2 ∫ R e − x 2 /2 e i x y d x = ( 2 π ) − 1/2 ∫ R e − x 2 /2 e − i x y d x ; 由于 cos u = e i u + e − i u 2 \cos u = \frac{e^{iu} + e^{-iu}}{2} cos u = 2 e i u + e − i u u u u
( 12 ) ( 2 π ) − 1 / 2 ∫ R e − x 2 / 2 e i x y d x = ( 2 π ) − 1 / 2 ∫ R e − x 2 / 2 cos x y d x .
(12) \quad (2\pi)^{-1/2} \int_{\mathbf{R}} e^{-x^2/2} e^{i xy} dx = (2\pi)^{-1/2} \int_{\mathbf{R}} e^{-x^2/2} \cos xy \, dx .
( 12 ) ( 2 π ) − 1/2 ∫ R e − x 2 /2 e i x y d x = ( 2 π ) − 1/2 ∫ R e − x 2 /2 cos x y d x . 对于每个整数 n ⩾ 0 n \geqslant 0 n ⩾ 0
g n ( x ) = ( − 1 ) n ( 2 π ) − 1 / 2 ( x y ) 2 n ( 2 n ) ! e − x 2 / 2 .
g_n(x) = (-1)^n (2\pi)^{-1/2} \frac{(xy)^{2n}}{(2n)!} e^{-x^2/2}.
g n ( x ) = ( − 1 ) n ( 2 π ) − 1/2 ( 2 n )! ( x y ) 2 n e − x 2 /2 . 由(7)，
( 13 ) ∫ R ∣ g n ( x ) ∣ d x = 1 n ! ( ∣ y ∣ 2 2 ) n
(13) \quad \int_{\mathbf{R}} |g_n(x)| \, dx = \frac{1}{n!} \left( \frac{|y|^2}{2} \right)^n
( 13 ) ∫ R ∣ g n ( x ) ∣ d x = n ! 1 ( 2 ∣ y ∣ 2 ) n ( 14 ) ∫ R g n ( x ) d x = 1 n ! ( − y 2 2 ) n ,
(14) \quad \int_{\mathbf{R}} g_n(x) \, dx = \frac{1}{n!} \left( -\frac{y^2}{2} \right)^n,
( 14 ) ∫ R g n ( x ) d x = n ! 1 ( − 2 y 2 ) n , 从而
∑ n = 0 ∞ ∫ R ∣ g n ( x ) ∣ d x = e ∣ y ∣ 2 / 2 < + ∞ .
\sum_{n=0}^{\infty} \int_{\mathbf{R}} |g_n(x)| \, dx = e^{|y|^2/2} < +\infty.
n = 0 ∑ ∞ ∫ R ∣ g n ( x ) ∣ d x = e ∣ y ∣ 2 /2 < + ∞. 而且，因为，
( 2 π ) − 1 / 2 e − x 2 / 2 cos x y = ∑ n = 0 ∞ g n ( x ) ,
(2\pi)^{-1/2} e^{-x^2/2} \cos xy = \sum_{n=0}^{\infty} g_n(x),
( 2 π ) − 1/2 e − x 2 /2 cos x y = n = 0 ∑ ∞ g n ( x ) , 这个等式可以通过项、从而积分项
( 2 π ) − 1 / 2 ∫ R e − x 2 / 2 cos x y d x = ∑ n = 0 ∞ ∫ R g n ( x ) d x = e − y 2 / 2
(2\pi)^{-1/2} \int_{\mathbf{R}} e^{-x^2/2} \cos xy \, dx = \sum_{n=0}^{\infty} \int_{\mathbf{R}} g_n(x) \, dx = e^{-y^2/2}
( 2 π ) − 1/2 ∫ R e − x 2 /2 cos x y d x = n = 0 ∑ ∞ ∫ R g n ( x ) d x = e − y 2 /2 由(14)。 公式 (11) 然后从 (12) 得出。
5. 高斯预测度和测度
设 E 为局部凸空间。对于 E' 上的每个正二次方形式 Q，在 E 上存在且仅有一个 Promeasure Γ Q \Gamma_Q Γ Q F Γ Q = e − Q / 2 \mathscr{F} \Gamma_Q = e^{-Q/2} F Γ Q = e − Q /2 Γ Q \Gamma_Q Γ Q
引用了 2 · 被以下引用 2 引用了 命题 3 , 引理 2
被以下引用 命题 3, IX § 3 , § 6. 。局部凸空间上的测度和测度
Γ Q \Gamma_Q Γ Q Γ Q \Gamma_Q Γ Q ( F Γ Q ) ( 0 ) = e − Q ( 0 ) / 2 = 1 (\mathscr{F}\Gamma_Q)(0) = e^{-Q(0)/2} = 1 ( F Γ Q ) ( 0 ) = e − Q ( 0 ) /2 = 1
A) E 为有限维 n，Q 非简并。
通过 No. 4 的引理 2，具有密度 t ↦ ( 2 π ) − 1 / 2 e − t 2 / 2 t \mapsto (2\pi)^{-1/2}e^{-t^2/2} t ↦ ( 2 π ) − 1/2 e − t 2 /2 R \mathbf{R} R γ 1 \gamma_1 γ 1 γ = γ 1 ⊗ ⋯ ⊗ γ 1 \gamma = \gamma_1 \otimes \cdots \otimes \gamma_1 γ = γ 1 ⊗ ⋯ ⊗ γ 1
∫ R n e i ( a 1 t 1 + ⋯ + a n t n ) d γ ( t 1 , … , t n ) = ∏ j = 1 n ∫ R e i a j t d γ 1 ( t )
\int_{\mathbf{R}^n} e^{i(a_1 t_1 + \cdots + a_n t_n)} d\gamma(t_1, \ldots, t_n) = \prod_{j=1}^n \int_{\mathbf{R}} e^{ia_j t} d\gamma_1(t)
∫ R n e i ( a 1 t 1 + ⋯ + a n t n ) d γ ( t 1 , … , t n ) = j = 1 ∏ n ∫ R e i a j t d γ 1 ( t ) = ∏ j = 1 n ( 2 π ) − 1 / 2 ∫ R e i a j t e − t 2 / 2 d t
= \prod_{j=1}^n (2\pi)^{-1/2} \int_{\mathbf{R}} e^{ia_j t} e^{-t^2/2} dt
= j = 1 ∏ n ( 2 π ) − 1/2 ∫ R e i a j t e − t 2 /2 d t = ∏ j = 1 n e − a j 2 / 2
= \prod_{j=1}^n e^{-a_j^2/2}
= j = 1 ∏ n e − a j 2 /2 = exp ( − 1 2 ( a 1 2 + ⋯ + a n 2 ) ) .
= \exp \left( -\frac{1}{2}(a_1^2 + \cdots + a_n^2) \right).
= exp ( − 2 1 ( a 1 2 + ⋯ + a n 2 ) ) . 由于 Q 是正的且非简并，因而 Q 存在 E' 正交的基 ( e 1 ′ , … , e n ′ ) (e'_1, \ldots, e'_n) ( e 1 ′ , … , e n ′ ) R n \mathbf{R}^n R n x ↦ ( e 1 ′ ( x ) , … , e n ′ ( x ) ) x \mapsto (e'_1(x), \ldots, e'_n(x)) x ↦ ( e 1 ′ ( x ) , … , e n ′ ( x )) Γ Q \Gamma_Q Γ Q f − 1 ( γ ) f^{-1}(\gamma) f − 1 ( γ ) x ′ = a 1 e 1 ′ + ⋯ + a n e n ′ x' = a_1 e'_1 + \cdots + a_n e'_n x ′ = a 1 e 1 ′ + ⋯ + a n e n ′ x ′ ( f − 1 ( t 1 , … , t n ) ) = ∑ j = 1 n t j a j x'(f^{-1}(t_1, \ldots, t_n)) = \sum_{j=1}^n t_j a_j x ′ ( f − 1 ( t 1 , … , t n )) = ∑ j = 1 n t j a j t 1 , … , t n t_1, \ldots, t_n t 1 , … , t n
∫ E e i ⟨ x , x ′ ⟩ d Γ Q ( x ) = ∫ R n e i ( a 1 t 1 + ⋯ + a n t n ) d γ ( t 1 , … , t n )
\int_E e^{i\langle x, x' \rangle} d\Gamma_Q(x) = \int_{\mathbf{R}^n} e^{i(a_1 t_1 + \cdots + a_n t_n)} d\gamma(t_1, \ldots, t_n)
∫ E e i ⟨ x , x ′ ⟩ d Γ Q ( x ) = ∫ R n e i ( a 1 t 1 + ⋯ + a n t n ) d γ ( t 1 , … , t n ) = exp ( − 1 2 ( a 1 2 + ⋯ + a n 2 ) ) = exp ( − 1 2 Q ( x ′ ) ) .
= \exp \left( -\frac{1}{2}(a_1^2 + \cdots + a_n^2) \right) = \exp \left( -\frac{1}{2}Q(x') \right).
= exp ( − 2 1 ( a 1 2 + ⋯ + a n 2 ) ) = exp ( − 2 1 Q ( x ′ ) ) . 因而，F Γ Q = e − Q / 2 \mathscr{F}\Gamma_Q = e^{-Q/2} F Γ Q = e − Q /2
B) E 有限维的。
令 N 为由 x ′ x' x ′ Q ( x ′ ) = 0 Q(x') = 0 Q ( x ′ ) = 0 t j : E ′ → M ′ {}^t j : E' \to M' t j : E ′ → M ′ M ′ M' M ′ Q = q ∘ t j Q = q \circ {}^t j Q = q ∘ t j Γ \Gamma Γ F Γ = e − q / 2 \mathscr{F}\Gamma = e^{-q/2} F Γ = e − q /2 Γ Q = j ( Γ ) \Gamma_Q = j(\Gamma) Γ Q = j ( Γ )
F Γ Q = ( F Γ ) ∘ t j = exp ( − q ∘ t j / 2 ) = e − Q / 2
\mathscr{F}\Gamma_Q = (\mathscr{F}\Gamma) \circ {}^t j = \exp(-q \circ {}^t j/2) = e^{-Q/2}
F Γ Q = ( F Γ ) ∘ t j = exp ( − q ∘ t j /2 ) = e − Q /2 由公式 (4) 共 No. 3。
C) 一般情况。 让V ∈ F ( E ) V \in \mathscr{F}(E) V ∈ F ( E ) E / V E/V E / V E E E p V p_V p V Q V Q_V Q V ( E / V ) ′ (E/V)' ( E / V ) ′ Q ∘ t p V Q \circ {}^t p_V Q ∘ t p V μ V \mu_V μ V E / V E/V E / V e − Q V / 2 e^{-Q_V/2} e − Q V /2 W ∈ F ( E ) W \in \mathscr{F}(E) W ∈ F ( E ) V V V p V = p V W ∘ p W p_V = p_{VW} \circ p_W p V = p V W ∘ p W Q V = Q W ∘ t p V W Q_V = Q_W \circ {}^t p_{VW} Q V = Q W ∘ t p V W p V W ( μ W ) p_{VW}(\mu_W) p V W ( μ W ) ( e − Q W / 2 ) ∘ t p V W = e − Q V / 2 (e^{-Q_W/2}) \circ {}^t p_{VW} = e^{-Q_V/2} ( e − Q W /2 ) ∘ t p V W = e − Q V /2 μ V \mu_V μ V ( μ V ) V ∈ F ( E ) (\mu_V)_{V \in \mathscr{F}(E)} ( μ V ) V ∈ F ( E ) μ \mu μ E E E F μ \mathscr{F}\mu F μ e − Q / 2 e^{-Q/2} e − Q /2
设E E E E ′ E' E ′ Q Q Q e − Q / 2 e^{-Q/2} e − Q /2 E E E E E E Q Q Q Γ Q \Gamma_Q Γ Q E ′ E' E ′ Q Q Q μ = Γ Q \mu = \Gamma_Q μ = Γ Q E E E μ \mu μ
被以下引用 3 被以下引用 命题 14, IX § 1 , 命题 9, IX § 2 , 推论 2, IX § 5
通过滥用语言，如果相关的 promeasure μ ~ \widetilde{\mu} μ Γ Q \Gamma_Q Γ Q E E E μ \mu μ Q Q Q
设E和E₁为两个局部凸空间，u为E到E₁的连续线性映射。令 Q 为 E' 上的正二次方程形式，Q₁ 为 E'₁ 上的正二次方程形式 Q ∘ t u \circ {}^t u ∘ t u u ( Γ Q ) = Γ Q 1 u(\Gamma_Q) = \Gamma_{Q_1} u ( Γ Q ) = Γ Q 1
引用了 1 · 被以下引用 1 引用了 命题 3
被以下引用 § 6. 。局部凸空间上的测度和测度
集合μ = u ( Γ Q ) \mu = u(\Gamma_Q) μ = u ( Γ Q )
F μ = ( F Γ Q ) ∘ t u = e − Q / 2 ∘ t u = e − Q 1 / 2 = F Γ Q 1 ,
\mathscr{F}\mu = (\mathscr{F}\Gamma_Q) \circ {}^t u = e^{-Q/2} \circ {}^t u = e^{-Q_1/2} = \mathscr{F}\Gamma_{Q_1} ,
F μ = ( F Γ Q ) ∘ t u = e − Q /2 ∘ t u = e − Q 1 /2 = F Γ Q 1 , 从而 μ = Γ Q 1 \mu = \Gamma_{Q_1} μ = Γ Q 1
设 E 为局部凸空间，Q 为 E' 上的正二次方形式。对于每个x ′ ∈ E ′ x' \in E' x ′ ∈ E ′ x ′ x' x ′ Γ Q \Gamma_Q Γ Q R \mathbf{R} R Q ( x ′ ) Q(x') Q ( x ′ )
被以下引用 1 被以下引用 命题 6
( 16 ) ∫ E ∣ ⟨ x , x ′ ⟩ ∣ n d μ ( x ) = π − 1 / 2 2 n / 2 Γ ( n + 1 2 ) Q ( x ′ ) n / 2 ( 17 ) ∫ E ⟨ x , x ′ ⟩ 2 n d μ ( x ) = ( 2 n ) ! 2 n n ! Q ( x ′ ) n ( 18 ) ∫ E ⟨ x , x ′ ⟩ 2 n + 1 d μ ( x ) = 0.
\begin{align*}
(16) & \quad \int_E |\langle x, x' \rangle|^n d\mu(x) = \pi^{-1/2} 2^{n/2} \Gamma\left( \frac{n+1}{2} \right) Q(x')^{n/2} \\
(17) & \quad \int_E \langle x, x' \rangle^{2n} d\mu(x) = \frac{(2n)!}{2^n n!} Q(x')^n \\
(18) & \quad \int_E \langle x, x' \rangle^{2n+1} d\mu(x) = 0 .
\end{align*}
( 16 ) ( 17 ) ( 18 ) ∫ E ∣ ⟨ x , x ′ ⟩ ∣ n d μ ( x ) = π − 1/2 2 n /2 Γ ( 2 n + 1 ) Q ( x ′ ) n /2 ∫ E ⟨ x , x ′ ⟩ 2 n d μ ( x ) = 2 n n ! ( 2 n )! Q ( x ′ ) n ∫ E ⟨ x , x ′ ⟩ 2 n + 1 d μ ( x ) = 0. 特别是，
( 19 ) ∫ E ⟨ x , x ′ ⟩ 2 d μ ( x ) = Q ( x ′ ) ( x ′ ∈ E ′ ) .
(19) \quad \int_E \langle x, x' \rangle^2 d\mu(x) = Q(x') \qquad (x' \in E') .
( 19 ) ∫ E ⟨ x , x ′ ⟩ 2 d μ ( x ) = Q ( x ′ ) ( x ′ ∈ E ′ ) . 如果这些公式对于 E ′ E' E ′ x ′ x' x ′ t ⋅ x ′ t \cdot x' t ⋅ x ′ t t t Q ( x ′ ) Q(x') Q ( x ′ )
a) 假设Q ( x ′ ) = 0 Q(x') = 0 Q ( x ′ ) = 0 x ′ ( μ ) x'(\mu) x ′ ( μ ) γ 0 = ε 0 \gamma_0 = \varepsilon_0 γ 0 = ε 0 x ′ x' x ′ μ \mu μ
b) 假设Q ( x ′ ) = 1 Q(x') = 1 Q ( x ′ ) = 1 x ′ ( μ ) = γ 1 x'(\mu) = \gamma_1 x ′ ( μ ) = γ 1
∫ E ∣ ⟨ x , x ′ ⟩ ∣ n d μ ( x ) = ∫ R ∣ t ∣ n d γ 1 ( t ) = ( 2 π ) − 1 / 2 ∫ R ∣ t ∣ n e − t 2 / 2 d t
\int_E |\langle x, x' \rangle|^n d\mu(x) = \int_{\mathbf{R}} |t|^n d\gamma_1(t) = (2\pi)^{-1/2} \int_{\mathbf{R}} |t|^n e^{-t^2/2} dt
∫ E ∣ ⟨ x , x ′ ⟩ ∣ n d μ ( x ) = ∫ R ∣ t ∣ n d γ 1 ( t ) = ( 2 π ) − 1/2 ∫ R ∣ t ∣ n e − t 2 /2 d t (16) 紧随 (6) (No 4, 引理 1)。类似地，公式 (17) 和 (18) 由 (7) 和 (8) 得出。最后由(17)中的设定
n = 1 n = 1 n = 1 得到(19)。
我们现在可以证明 Cor 的逆命题。提案 5。
设 E 为局部凸空间，μ \mu μ x ′ ( μ ) x'(\mu) x ′ ( μ ) R \mathbf{R} R x ′ ∈ E ′ x' \in E' x ′ ∈ E ′ μ \mu μ
引用了 1 · 被以下引用 1 引用了 命题 6
被以下引用 定义 1
对于每个x ′ ∈ E ′ x' \in E' x ′ ∈ E ′ Q ( x ′ ) Q(x') Q ( x ′ ) R \mathbf{R} R x ′ ( μ ) x'(\mu) x ′ ( μ ) x ′ ( μ ) = γ Q ( x ′ ) x'(\mu) = \gamma_{Q(x')} x ′ ( μ ) = γ Q ( x ′ )
( F μ ) ( x ′ ) = ∫ R e i t ⋅ 1 d γ Q ( x ′ ) ( t ) = e − Q ( x ′ ) ⋅ 1 2 / 2
(\mathscr{F}\mu)(x') = \int_{\mathbf{R}} e^{it \cdot 1} d\gamma_{Q(x')}(t) = e^{-Q(x') \cdot 1^2 / 2}
( F μ ) ( x ′ ) = ∫ R e i t ⋅ 1 d γ Q ( x ′ ) ( t ) = e − Q ( x ′ ) ⋅ 1 2 /2 由
F μ \mathscr{F}\mu F μ 的定义 (No. 3, 公式 (2))。其他字、
F μ = e − Q / 2 \mathscr{F}\mu = e^{-Q/2} F μ = e − Q /2 ，还有待证明
Q Q Q 是
E ′ E' E ′ 上的正二次形式。
对于 E 与有限余维的每个闭线性子空间 V，用 E 到 E / V E/V E / V p V p_V p V μ V \mu_V μ V E / V E/V E / V p V ( μ ) p_V(\mu) p V ( μ ) Q V = Q ∘ t p V Q_V = Q \circ {}^t p_V Q V = Q ∘ t p V E ′ = ⋃ V ∈ F ( E ) Im ( t p V ) E' = \bigcup_{V \in \mathscr{F}(E)} \operatorname{Im}({}^t p_V) E ′ = ⋃ V ∈ F ( E ) Im ( t p V ) t p V {}^t p_V t p V Q V Q_V Q V ( E / V ) ′ (E/V)' ( E / V ) ′ u ∈ ( E / V ) ′ u \in (E/V)' u ∈ ( E / V ) ′ x ′ = t p V ( u ) x' = {}^t p_V(u) x ′ = t p V ( u )
u ( μ V ) = u ( p V ( μ ) ) = x ′ ( μ ) = γ Q ( x ′ ) ;
u(\mu_V) = u(p_V(\mu)) = x'(\mu) = \gamma_{Q(x')} ;
u ( μ V ) = u ( p V ( μ )) = x ′ ( μ ) = γ Q ( x ′ ) ; 那么，命题 6 意味着
Q V ( u ) = Q ( x ′ ) = ∫ R t 2 d γ Q ( x ′ ) ( t ) = ∫ E / V u ( x ) 2 d μ V ( x ) ,
Q_V(u) = Q(x') = \int_{\mathbf{R}} t^2 d\gamma_{Q(x')}(t) = \int_{E/V} u(x)^2 d\mu_V(x),
Q V ( u ) = Q ( x ′ ) = ∫ R t 2 d γ Q ( x ′ ) ( t ) = ∫ E / V u ( x ) 2 d μ V ( x ) , 因此
Q V Q_V Q V 是
( E / V ) ′ (E/V)' ( E / V ) ′ 上的正二次形式。
6. 高斯算子的例
1) 令 E 为真正的希尔伯特空间。 映射 x ′ ↦ ∥ x ′ ∥ 2 x' \mapsto \|x'\|^2 x ′ ↦ ∥ x ′ ∥ 2 E ′ E' E ′
令 A A A E E E x ′ ↦ ∥ t A ⋅ x ′ ∥ 2 x' \mapsto \|{}^tA \cdot x'\|^2 x ′ ↦ ∥ t A ⋅ x ′ ∥ 2 E ′ E' E ′ A A A μ A \mu_A μ A E E E
2) 正型的核。 设T T T T T T E = R T E = \mathbf{R}^T E = R T t ∈ T t \in T t ∈ T E E E ε t \varepsilon_t ε t f ↦ f ( t ) f \mapsto f(t) f ↦ f ( t ) ( ε t ) t ∈ T (\varepsilon_t)_{t \in T} ( ε t ) t ∈ T E ′ E' E ′
在 T T T T × T T \times T T × T K K K
(20) K ( t , t ′ ) = K ( t ′ , t ) for t , t ′ in T ,
\text{(20)} \qquad K(t, t') = K(t', t) \quad \text{for } t, t' \text{ in } T,
(20) K ( t , t ′ ) = K ( t ′ , t ) for t , t ′ in T , (21) ∑ i , j = 1 p c i c j K ( t i , t j ) ⩾ 0
\text{(21)} \qquad \sum_{i,j=1}^p c_i c_j K(t_i, t_j) \geqslant 0
(21) i , j = 1 ∑ p c i c j K ( t i , t j ) ⩾ 0 对于 T T T p p p t 1 , … , t p t_1, \ldots, t_p t 1 , … , t p c 1 , … , c p c_1, \ldots, c_p c 1 , … , c p
(22) q ( ∑ t ∈ T c t ε t ) = ∑ t , t ′ ∈ T c t c t ′ K ( t , t ′ )
\text{(22)} \qquad q\left( \sum_{t \in T} c_t \varepsilon_t \right) = \sum_{t,t' \in T} c_t c_{t'} K(t, t')
(22) q ( t ∈ T ∑ c t ε t ) = t , t ′ ∈ T ∑ c t c t ′ K ( t , t ′ ) 定义 E ′ E' E ′ q q q E ′ E' E ′
(23) K ( t , t ′ ) = 1 2 [ q ( ε t + ε t ′ ) − q ( ε t ) − q ( ε t ′ ) ]
\text{(23)} \qquad K(t, t') = \frac{1}{2} [q(\varepsilon_t + \varepsilon_{t'}) - q(\varepsilon_t) - q(\varepsilon_{t'})]
(23) K ( t , t ′ ) = 2 1 [ q ( ε t + ε t ′ ) − q ( ε t ) − q ( ε t ′ )] T T T K K K T T T E ′ E' E ′
令 K K K T T T q q q E ′ E' E ′ E E E q q q E E E K K K T T T
3) 令T T T T T T δ \delta δ
(24) δ ( t , t ′ ) = { 1 if t = t ′ 0 if t ≠ t ′ .
\text{(24)} \qquad \delta(t, t') = \begin{cases}
1 & \text{if } t = t' \\
0 & \text{if } t \neq t'.
\end{cases}
(24) δ ( t , t ′ ) = { 1 0 if t = t ′ if t = t ′ . 相应的二次形式由q ( ∑ t ∈ T c t ε t ) = ∑ t ∈ T c t 2 q\left( \sum_{t \in T} c_t \varepsilon_t \right) = \sum_{t \in T} c_t^2 q ( ∑ t ∈ T c t ε t ) = ∑ t ∈ T c t 2 t ∈ T t \in T t ∈ T μ t \mu_t μ t R \mathbf{R} R
很容易证明，R T \mathbf{R}^T R T δ \delta δ ⨂ t ∈ T μ t \bigotimes_{t \in T} \mu_t ⨂ t ∈ T μ t
4) 设n ≥ 1 n \geq 1 n ≥ 1 n n n C = ( c i j ) C = (c_{ij}) C = ( c ij ) x 1 , … , x n x_1, \ldots, x_n x 1 , … , x n ∑ i , j = 1 n c i j x i x j ≥ 0 \sum_{i,j=1}^n c_{ij} x_i x_j \geq 0 ∑ i , j = 1 n c ij x i x j ≥ 0 ( i , j ) ↦ c i j (i, j) \mapsto c_{ij} ( i , j ) ↦ c ij { 1 , 2 , … , n } \{1, 2, \ldots, n\} { 1 , 2 , … , n } R n \mathbf{R}^n R n γ C \gamma_C γ C C C C
( 25 ) ∫ R n e i ( x 1 t 1 + ⋯ + x n t n ) d γ C ( t 1 , … , t n ) = exp ( − 1 2 ∑ j , k = 1 n c j k x j x k ) ,
(25) \quad \int_{\mathbf{R}^n} e^{i(x_1 t_1 + \cdots + x_n t_n)} d\gamma_C(t_1, \ldots, t_n) = \exp \left( - \frac{1}{2} \sum_{j,k=1}^n c_{jk} x_j x_k \right),
( 25 ) ∫ R n e i ( x 1 t 1 + ⋯ + x n t n ) d γ C ( t 1 , … , t n ) = exp − 2 1 j , k = 1 ∑ n c j k x j x k , 对于 x 1 , … , x n x_1, \ldots, x_n x 1 , … , x n
( 26 ) ∫ R n t j t k d γ C ( t 1 , … , t n ) = c j k ( 1 ≤ j , k ≤ n ) .
(26) \quad \int_{\mathbf{R}^n} t_j t_k d\gamma_C(t_1, \ldots, t_n) = c_{jk} \quad (1 \leq j, k \leq n).
( 26 ) ∫ R n t j t k d γ C ( t 1 , … , t n ) = c j k ( 1 ≤ j , k ≤ n ) . 从No. 5的第5条命题，推导出公式
( 27 ) u ( γ C ) = γ U C t U ,
(27) \quad u(\gamma_C) = \gamma_{UC^t U},
( 27 ) u ( γ C ) = γ U C t U , 其中u u u R n \mathbf{R}^n R n R m \mathbf{R}^m R m U U U I n I_n I n n n n γ I n \gamma_{I_n} γ I n
( 2 π ) − n / 2 exp ( − 1 2 ( t 1 2 + ⋯ + t n 2 ) )
(2\pi)^{-n/2} \exp \left( - \frac{1}{2}(t_1^2 + \cdots + t_n^2) \right)
( 2 π ) − n /2 exp ( − 2 1 ( t 1 2 + ⋯ + t n 2 ) ) 关于R n \mathbf{R}^n R n λ n \lambda_n λ n
我们将证明，如果矩阵 C C C D = ( d j k ) D = (d_{jk}) D = ( d j k )
( 28 ) d γ C ( t 1 , … , t n ) = ( 2 π ) − n / 2 ( det D ) 1 / 2 ( exp ( − 1 2 ∑ j , k = 1 n d j k t j t k ) ) d t 1 ⋯ d t n .
(28) \quad d\gamma_C(t_1, \ldots, t_n) =
(2\pi)^{-n/2} (\det D)^{1/2} \left( \exp \left( - \frac{1}{2} \sum_{j,k=1}^n d_{jk} t_j t_k \right) \right) dt_1 \cdots dt_n.
( 28 ) d γ C ( t 1 , … , t n ) = ( 2 π ) − n /2 ( det D ) 1/2 exp − 2 1 j , k = 1 ∑ n d j k t j t k d t 1 ⋯ d t n . 因为，如果 C C C R n \mathbf{R}^n R n q q q
q ( x 1 , … , x n ) = ∑ j , k = 1 n c j k x j x k
q(x_1, \ldots, x_n) = \sum_{j,k=1}^n c_{jk} x_j x_k
q ( x 1 , … , x n ) = j , k = 1 ∑ n c j k x j x k 是非简并的。利用R n \mathbf{R}^n R n q q q n n n U U U C = U ⋅ t U C = U \cdot {}^tU C = U ⋅ t U γ C = u ( γ I n ) \gamma_C = u(\gamma_{I_n}) γ C = u ( γ I n ) u u u R n \mathbf{R}^n R n U U U Q Q Q R n \mathbf{R}^n R n
Q ( t 1 , … , t n ) = t 1 2 + ⋯ + t n 2 ;
Q(t_1, \ldots, t_n) = t_1^2 + \cdots + t_n^2;
Q ( t 1 , … , t n ) = t 1 2 + ⋯ + t n 2 ; γ I n = ( 2 π ) − n / 2 e − Q / 2 ⋅ λ n ,
\gamma_{I_n} = (2\pi)^{-n/2} e^{-Q/2} \cdot \lambda_n,
γ I n = ( 2 π ) − n /2 e − Q /2 ⋅ λ n , u ( γ I n ) = ( 2 π ) − n / 2 e − ( Q o u − 1 ) / 2 ⋅ u ( λ n ) .
u(\gamma_{I_n}) = (2\pi)^{-n/2} e^{-(Qou^{-1})/2} \cdot u(\lambda_n).
u ( γ I n ) = ( 2 π ) − n /2 e − ( Q o u − 1 ) /2 ⋅ u ( λ n ) . Q ∘ u − 1 Q \circ u^{-1} Q ∘ u − 1 R n \mathbf{R}^n R n ( t 1 , … , t n ) (t_1, \ldots, t_n) ( t 1 , … , t n ) ∑ j , k = 1 n d j k t j t k \sum_{j,k=1}^n d_{jk} t_j t_k ∑ j , k = 1 n d j k t j t k u ( λ n ) = ( det U ) − 1 ⋅ λ n = ( det D ) 1 / 2 ⋅ λ n .
u(\lambda_n) = (\det U)^{-1} \cdot \lambda_n = (\det D)^{1/2} \cdot \lambda_n.
u ( λ n ) = ( det U ) − 1 ⋅ λ n = ( det D ) 1/2 ⋅ λ n . 7. 维纳测度
在这个数中，我们用T T T R \mathbf{R} R ] 0 , 1 ] ]0, 1] ] 0 , 1 ] H \mathscr{H} H T T T ( f ∣ g ) (f|g) ( f ∣ g ) T T T C \mathscr{C} C C \mathscr{C} C ∥ f ∥ = sup t ∈ T ∣ f ( t ) ∣ \|f\| = \sup_{t \in T} |f(t)| ∥ f ∥ = sup t ∈ T ∣ f ( t ) ∣ [ 0 , 1 ] = T ∪ { 0 } [0, 1] = T \cup \{0\} [ 0 , 1 ] = T ∪ { 0 } T T T T T T C \mathscr{C} C C \mathscr{C} C T T T M 1 \mathscr{M}^1 M 1
对于每函数 f ∈ H f \in \mathscr{H} f ∈ H T T T P f Pf P f
( P f ) ( t ) = ∫ 0 t f ( x ) d x = ( f ∣ I t ) ,
(Pf)(t) = \int_0^t f(x) \, dx = (f|I_t),
( P f ) ( t ) = ∫ 0 t f ( x ) d x = ( f ∣ I t ) , I t I_t I t ] 0 , t ] ]0, t] ] 0 , t ] ∣ ( P f ) ( t ) ∣ ≤ ∥ f ∥ 2 ⋅ t 1 / 2
|(Pf)(t)| \leq \|f\|_2 \cdot t^{1/2}
∣ ( P f ) ( t ) ∣ ≤ ∥ f ∥ 2 ⋅ t 1/2 ∣ ( P f ) ( t ) − ( P f ) ( t ′ ) ∣ ≤ ∥ f ∥ 2 ⋅ ∣ t − t ′ ∣ 1 / 2 ;
|(Pf)(t) - (Pf)(t')| \leq \|f\|_2 \cdot |t - t'|^{1/2};
∣ ( P f ) ( t ) − ( P f ) ( t ′ ) ∣ ≤ ∥ f ∥ 2 ⋅ ∣ t − t ′ ∣ 1/2 ; 因而，P f Pf P f C \mathscr{C} C H \mathscr{H} H C \mathscr{C} C P P P ⩽ 1 \leqslant 1 ⩽ 1
让我们认同希尔伯特空间H \mathscr{H} H P : H → C P : \mathscr{H} \to \mathscr{C} P : H → C Π : M 1 → H \Pi : \mathscr{M}^1 \to \mathscr{H} Π : M 1 → H μ ∈ M 1 \mu \in \mathscr{M}^1 μ ∈ M 1 f ∈ H f \in \mathscr{H} f ∈ H
( Π μ ∣ f ) = μ ( P f ) = ∫ T d μ ( t ) ∫ T I t ( x ) f ( x ) d x
(\Pi \mu | f) = \mu(Pf) = \int_{\mathbf{T}} d\mu(t) \int_{\mathbf{T}} I_t(x) f(x) \, dx
( Π μ ∣ f ) = μ ( P f ) = ∫ T d μ ( t ) ∫ T I t ( x ) f ( x ) d x = ∫ T f ( x ) d x ∫ T I t ( x ) d μ ( t )
= \int_{\mathbf{T}} f(x) \, dx \int_{\mathbf{T}} I_t(x) \, d\mu(t)
= ∫ T f ( x ) d x ∫ T I t ( x ) d μ ( t ) 由勒贝格-富比尼定理。现在，
I t ( x ) = { 1 if 0 < x ≤ t ≤ 1 0 otherwise ,
I_t(x) = \begin{cases}
1 & \text{if } 0 < x \leq t \leq 1 \\
0 & \text{otherwise},
\end{cases}
I t ( x ) = { 1 0 if 0 < x ≤ t ≤ 1 otherwise , 终于从而
(34)
( Π μ ) ( x ) = μ ( [ x , 1 ] ) for x ∈ T . (\Pi \mu)(x) = \mu([x, 1]) \quad \text{for } x \in \mathbf{T}. ( Π μ ) ( x ) = μ ([ x , 1 ]) for x ∈ T . 令μ , ν \mu, \nu μ , ν M 1 \mathscr{M}^1 M 1
( Π μ ∣ Π ν ) = ∫ T Π μ ( x ) Π ν ( x ) d x = ∫ T d x ∫ T I t ( x ) d μ ( t ) ∫ T I t ′ ( x ) d ν ( t ′ )
(\Pi \mu | \Pi \nu) = \int_{\mathbf{T}} \Pi \mu(x) \Pi \nu(x) \, dx = \int_{\mathbf{T}} dx \int_{\mathbf{T}} I_t(x) \, d\mu(t) \int_{\mathbf{T}} I_{t'}(x) \, d\nu(t')
( Π μ ∣Π ν ) = ∫ T Π μ ( x ) Π ν ( x ) d x = ∫ T d x ∫ T I t ( x ) d μ ( t ) ∫ T I t ′ ( x ) d ν ( t ′ ) = ∫ T ∫ T d μ ( t ) d ν ( t ′ ) ∫ T I t ( x ) I t ′ ( x ) d x .
= \int_{\mathbf{T}} \int_{\mathbf{T}} d\mu(t) \, d\nu(t') \int_{\mathbf{T}} I_t(x) I_{t'}(x) \, dx.
= ∫ T ∫ T d μ ( t ) d ν ( t ′ ) ∫ T I t ( x ) I t ′ ( x ) d x . 现在，I t ⋅ I t ′ I_t \cdot I_{t'} I t ⋅ I t ′ ] 0 , t ] ∩ ] 0 , t ′ ] ]0, t] \cap ]0, t'] ] 0 , t ] ∩ ] 0 , t ′ ]
(35)
∫ T I t ( x ) I t ′ ( x ) d x = inf ( t , t ′ ) . \int_{\mathbf{T}} I_t(x) I_{t'}(x) \, dx = \inf(t, t'). ∫ T I t ( x ) I t ′ ( x ) d x = inf ( t , t ′ ) . 由此可见
(36)
( Π μ ∣ Π ν ) = ∫ T ∫ T inf ( t , t ′ ) d μ ( t ) d ν ( t ′ ) . (\Pi \mu | \Pi \nu) = \int_{\mathbf{T}} \int_{\mathbf{T}} \inf(t, t') \, d\mu(t) \, d\nu(t'). ( Π μ ∣Π ν ) = ∫ T ∫ T inf ( t , t ′ ) d μ ( t ) d ν ( t ′ ) . 根据前面的结果，一个定义在M 1 \mathscr{M}^1 M 1 W W W
(37)
W ( μ ) = ∫ T ∫ T inf ( t , t ′ ) d μ ( t ) d μ ( t ′ ) = ∥ Π μ ∥ 2 2 . W(\mu) = \int_{\mathbf{T}} \int_{\mathbf{T}} \inf(t, t') \, d\mu(t) \, d\mu(t') = \| \Pi \mu \|_2^2. W ( μ ) = ∫ T ∫ T inf ( t , t ′ ) d μ ( t ) d μ ( t ′ ) = ∥Π μ ∥ 2 2 . 特别是，如果 t 1 , … , t n t_1, \ldots, t_n t 1 , … , t n c 1 , … , c n c_1, \ldots, c_n c 1 , … , c n
W ( ∑ j = 1 n c j ε t j ) = ∑ j , k = 1 n c j c k inf ( t j , t k )
W \left( \sum_{j=1}^n c_j \varepsilon_{t_j} \right) = \sum_{j,k=1}^n c_j c_k \inf(t_j, t_k)
W ( j = 1 ∑ n c j ε t j ) = j , k = 1 ∑ n c j c k inf ( t j , t k ) ( t , t ′ ) ↦ inf ( t , t ′ ) (t, t') \mapsto \inf(t, t') ( t , t ′ ) ↦ inf ( t , t ′ ) 定理 1 (维纳). — 令 w 为希尔伯特空间 H \mathscr{H} H P : H → C P : \mathscr{H} \to \mathscr{C} P : H → C C \mathscr{C} C
构造、W ( μ ) = ∥ t P ( μ ) ∥ 2 2 W(\mu) = \|^tP(\mu)\|_2^2 W ( μ ) = ∥ t P ( μ ) ∥ 2 2 C \mathscr{C} C
A) 构造辅助测量空间 ^{(2)} ( Ω , m ) (\Omega, m) ( Ω , m ) n ⩾ 0 n \geqslant 0 n ⩾ 0 k / 2 n k/2^n k / 2 n k = 1 , 2 , 3 , … , 2 n k = 1, 2, 3, \ldots, 2^n k = 1 , 2 , 3 , … , 2 n D n D_n D n D = ⋃ n ⩾ 0 D n D = \bigcup_{n \geqslant 0} D_n D = ⋃ n ⩾ 0 D n Ω = R D \Omega = \mathbf{R}^D Ω = R D t ∈ D t \in D t ∈ D Ω \Omega Ω X ( t ) X(t) X ( t ) f ↦ f ( t ) f \mapsto f(t) f ↦ f ( t )
对于D中的t , t ′ t, t' t , t ′ M ( t , t ′ ) = inf ( t , t ′ ) M(t, t') = \inf(t, t') M ( t , t ′ ) = inf ( t , t ′ ) 例 2 ) 定义 Ω \Omega Ω
对于 D 中的任意 t , t ′ t, t' t , t ′
∫ Ω ∣ X ( t + t ′ 2 ) − X ( t ) + X ( t ′ ) 2 ∣ 3 d m = 1 ( 8 π ) 1 / 2 ∣ t − t ′ ∣ 3 / 2 .
\int_\Omega \left| X\left( \frac{t + t'}{2} \right) - \frac{X(t) + X(t')}{2} \right|^3 dm = \frac{1}{(8\pi)^{1/2}} |t - t'|^{3/2}.
∫ Ω X ( 2 t + t ′ ) − 2 X ( t ) + X ( t ′ ) 3 d m = ( 8 π ) 1/2 1 ∣ t − t ′ ∣ 3/2 . 注，
t + t ′ 2 \frac{t + t'}{2} 2 t + t ′ 属于到D。人们知道（No. 6，*例2*）族
( X ( t ) ) t ∈ D (X(t))_{t \in D} ( X ( t ) ) t ∈ D 是
Ω \Omega Ω 的拓扑对偶
Ω ′ \Omega' Ω ′ 的基；因此，在
Ω ′ × Ω ′ \Omega' \times \Omega' Ω ′ × Ω ′ 上存在对称双线性形式
M ^ \widehat{M} M ，其特征为
M ^ ( X ( t ) , X ( t ′ ) ) = inf ( t , t ′ ) \widehat{M}(X(t), X(t')) = \inf(t, t') M ( X ( t ) , X ( t ′ )) = inf ( t , t ′ ) 。到构造时，
Ω \Omega Ω 上的高斯测度 m 的方差是
Ω ′ \Omega' Ω ′ 上的二次方形式
ξ ↦ M ^ ( ξ , ξ ) \xi \mapsto \widehat{M}(\xi, \xi) ξ ↦ M ( ξ , ξ ) 。 集合，特别是，
ξ = X ( t + t ′ 2 ) − X ( t ) + X ( t ′ ) 2 ;
\xi = X\left( \frac{t + t'}{2} \right) - \frac{X(t) + X(t')}{2};
ξ = X ( 2 t + t ′ ) − 2 X ( t ) + X ( t ′ ) ; 简单计算得出
M ^ ( ξ , ξ ) = ∣ t − t ′ ∣ 4 .
\widehat{M}(\xi, \xi) = \frac{|t - t'|}{4}.
M ( ξ , ξ ) = 4 ∣ t − t ′ ∣ . 引用了 2 · 被以下引用 1 引用了 IX § 2. 。 运算关于措施 , 命题 6
被以下引用 引理 4
(2) Espace mesuré ：一个局部紧凑的空间，配有一个措施（第III，第 1 版，§2, No. 2，p. 52）。
根据No. 5(公式(16))第6条，
∫ Ω ∣ ξ ∣ 3 d m = π − 1 / 2 2 3 / 2 Γ ( 2 ) M ^ ( ξ , ξ ) 3 / 2 ;
\int_{\Omega} |\xi|^3 \, dm = \pi^{-1/2} 2^{3/2} \Gamma(2) \widehat{M}(\xi, \xi)^{3/2};
∫ Ω ∣ ξ ∣ 3 d m = π − 1/2 2 3/2 Γ ( 2 ) M ( ξ , ξ ) 3/2 ; 引理紧随公式 (40) 和 (41)。
B) 将Ω \Omega Ω C \mathscr{C} C 对于每个整数n ⩾ 0 n \geqslant 0 n ⩾ 0 C \mathscr{C} C E n E_n E n 1 ⩽ k ⩽ 2 n 1 \leqslant k \leqslant 2^n 1 ⩽ k ⩽ 2 n [ k − 1 2 n , k 2 n ] \left[ \frac{k-1}{2^n}, \frac{k}{2^n} \right] [ 2 n k − 1 , 2 n k ] R \mathbf{R} R
∥ f ∥ = sup 1 ⩽ k ⩽ 2 n ∣ f ( k 2 n ) ∣
\|f\| = \sup_{1 \leqslant k \leqslant 2^n} \left| f\left( \frac{k}{2^n} \right) \right|
∥ f ∥ = 1 ⩽ k ⩽ 2 n sup f ( 2 n k ) f ∈ E n f \in E_n f ∈ E n
对于每一个函数g ∈ Ω g \in \Omega g ∈ Ω n ⩾ 0 n \geqslant 0 n ⩾ 0 u n ( g ) u_n(g) u n ( g ) E n E_n E n g g g D n D_n D n T n g = u n + 1 ( g ) − u n ( g ) T_n g = u_{n+1}(g) - u_n(g) T n g = u n + 1 ( g ) − u n ( g ) D n D_n D n Ω \Omega Ω C \mathscr{C} C T n T_n T n m m m
— *对于每个整数 n ⩾ 0 n \geqslant 0 n ⩾ 0
∫ Ω ∥ T n g ∥ 3 d m ( g ) ⩽ 1 ( 8 π ) 1 / 2 2 − n / 2 .
\int_{\Omega} \|T_n g\|^3 \, dm(g) \leqslant \frac{1}{(8\pi)^{1/2}} 2^{-n/2}.
∫ Ω ∥ T n g ∥ 3 d m ( g ) ⩽ ( 8 π ) 1/2 1 2 − n /2 . 引用了 3 引用了 引理 3 , IV § 3. 。 L$^p$空间 , 命题 6
设g ∈ Ω g \in \Omega g ∈ Ω n ∈ N n \in \mathbf{N} n ∈ N E n ⊂ E n + 1 E_n \subset E_{n+1} E n ⊂ E n + 1 T n g T_n g T n g E n + 1 E_{n+1} E n + 1 D n D_n D n
∥ T n g ∥ 3 = sup 1 ⩽ k ⩽ 2 n ∣ T n g ( 2 k − 1 2 n + 1 ) ∣ 3 ⩽ ∑ k = 1 2 n ∣ T n g ( 2 k − 1 2 n + 1 ) ∣ 3 .
\|T_n g\|^3 = \sup_{1 \leqslant k \leqslant 2^n} \left| T_n g\left( \frac{2k-1}{2^{n+1}} \right) \right|^3 \leqslant \sum_{k=1}^{2^n} \left| T_n g\left( \frac{2k-1}{2^{n+1}} \right) \right|^3.
∥ T n g ∥ 3 = 1 ⩽ k ⩽ 2 n sup T n g ( 2 n + 1 2 k − 1 ) 3 ⩽ k = 1 ∑ 2 n T n g ( 2 n + 1 2 k − 1 ) 3 . 让我们约定g ( 0 ) = 0 g(0) = 0 g ( 0 ) = 0 u n ( g ) u_n(g) u n ( g ) g g g
T n g ( 2 k − 1 2 n + 1 ) = g ( 2 k − 1 2 n + 1 ) − 1 2 ( g ( k − 1 2 n ) + g ( k 2 n ) )
T_n g\left( \frac{2k-1}{2^{n+1}} \right) = g\left( \frac{2k-1}{2^{n+1}} \right) - \frac{1}{2} \left( g\left( \frac{k-1}{2^n} \right) + g\left( \frac{k}{2^n} \right) \right)
T n g ( 2 n + 1 2 k − 1 ) = g ( 2 n + 1 2 k − 1 ) − 2 1 ( g ( 2 n k − 1 ) + g ( 2 n k ) ) 1 ⩽ k ⩽ 2 n 1 \leqslant k \leqslant 2^n 1 ⩽ k ⩽ 2 n
∫ Ω ∣ T n g ( 2 k − 1 2 n + 1 ) ∣ 3 d m ( g ) = ∫ Ω ∣ X ( 2 k − 1 2 n + 1 ) − 1 2 ( X ( k − 1 2 n ) + X ( k 2 n ) ) ∣ 3 d m ;
\int_{\Omega} \left| T_n g\left( \frac{2k-1}{2^{n+1}} \right) \right|^3 \, dm(g) = \int_{\Omega} \left| X\left( \frac{2k-1}{2^{n+1}} \right) - \frac{1}{2} \left( X\left( \frac{k-1}{2^n} \right) + X\left( \frac{k}{2^n} \right) \right) \right|^3 \, dm;
∫ Ω T n g ( 2 n + 1 2 k − 1 ) 3 d m ( g ) = ∫ Ω X ( 2 n + 1 2 k − 1 ) − 2 1 ( X ( 2 n k − 1 ) + X ( 2 n k ) ) 3 d m ; !图1
图1
然后可以应用引理 3 和 t = k − 1 2 n t = \frac{k-1}{2^n} t = 2 n k − 1 t ′ = k 2 n t' = \frac{k}{2^n} t ′ = 2 n k
∫ Ω ∣ T n g ( 2 k − 1 2 n + 1 ) ∣ 3 d m ( g ) = 1 ( 8 π ) 1 / 2 2 − 3 n 2 .
\int_{\Omega} \left| T_n g \left( \frac{2k-1}{2^{n+1}} \right) \right|^3 dm(g) = \frac{1}{(8\pi)^{1/2}} 2^{-\frac{3n}{2}} .
∫ Ω T n g ( 2 n + 1 2 k − 1 ) 3 d m ( g ) = ( 8 π ) 1/2 1 2 − 2 3 n . 由(44)，我们有
∫ Ω ∥ T n g ∥ 3 d m ( g ) ≤ ∑ k = 1 2 n ∫ Ω ∣ T n g ( 2 k − 1 2 n + 1 ) ∣ 3 d m ( g ) = 1 ( 8 π ) 1 / 2 2 n ⋅ 2 − 3 n 2 ,
\int_{\Omega} \| T_n g \|^3 dm(g) \leq \sum_{k=1}^{2^n} \int_{\Omega} \left| T_n g \left( \frac{2k-1}{2^{n+1}} \right) \right|^3 dm(g) = \frac{1}{(8\pi)^{1/2}} 2^n \cdot 2^{-\frac{3n}{2}} ,
∫ Ω ∥ T n g ∥ 3 d m ( g ) ≤ k = 1 ∑ 2 n ∫ Ω T n g ( 2 n + 1 2 k − 1 ) 3 d m ( g ) = ( 8 π ) 1/2 1 2 n ⋅ 2 − 2 3 n , 从而引理。
由引理 4、Ω \Omega Ω T n T_n T n C \mathscr{C} C L C 3 ( Ω , m ) L^3_{\mathscr{C}}(\Omega, m) L C 3 ( Ω , m ) N 3 ( T n ) ≤ 1 ( 8 π ) 1 / 6 ( 2 − 1 / 6 ) n N_3(T_n) \leq \frac{1}{(8\pi)^{1/6}} (2^{-1/6})^n N 3 ( T n ) ≤ ( 8 π ) 1/6 1 ( 2 − 1/6 ) n ∑ n = 0 ∞ N 3 ( T n ) < + ∞ \sum_{n=0}^{\infty} N_3(T_n) < +\infty ∑ n = 0 ∞ N 3 ( T n ) < + ∞ Ω 0 ⊂ Ω \Omega_0 \subset \Omega Ω 0 ⊂ Ω Ω − Ω 0 \Omega - \Omega_0 Ω − Ω 0 m m m g ∈ Ω 0 g \in \Omega_0 g ∈ Ω 0 ∑ n = 0 ∞ T n ( g ) \sum_{n=0}^{\infty} T_n(g) ∑ n = 0 ∞ T n ( g ) C \mathscr{C} C Ω \Omega Ω m m m u u u C \mathscr{C} C
u ( g ) = { ∑ n = 0 ∞ T n g = lim n → ∞ u n ( g ) for g ∈ Ω 0 0 for g ∈ Ω − Ω 0 .
u(g) = \begin{cases}
\sum_{n=0}^{\infty} T_n g = \lim_{n \to \infty} u_n(g) & \text{for } g \in \Omega_0 \\
0 & \text{for } g \in \Omega - \Omega_0 .
\end{cases}
u ( g ) = { ∑ n = 0 ∞ T n g = lim n → ∞ u n ( g ) 0 for g ∈ Ω 0 for g ∈ Ω − Ω 0 . 由于0 ≤ m ≤ n 0 \leq m \leq n 0 ≤ m ≤ n D m ⊂ D n D_m \subset D_n D m ⊂ D n u n ( g ) u_n(g) u n ( g ) g g g g ∈ Ω 0 g \in \Omega_0 g ∈ Ω 0 u ( g ) u(g) u ( g ) D D D g g g
C) 构造 C \mathscr{C} C w ′ w' w ′ C \mathscr{C} C m m m u : Ω → C u : \Omega \to \mathscr{C} u : Ω → C m m m w ′ w' w ′ C \mathscr{C} C W W W w = w ′ w = w' w = w ′ D \mathscr{D} D M 1 \mathscr{M}^1 M 1 t t t ε t \varepsilon_t ε t D D D
对于每一个度量μ ∈ D \mu \in \mathscr{D} μ ∈ D
∫ C e i ⟨ f , μ ⟩ d w ′ ( f ) = e − W ( μ ) / 2 .
\int_{\mathscr{C}} e^{i \langle f, \mu \rangle} \, dw'(f) = e^{-W(\mu)/2}.
∫ C e i ⟨ f , μ ⟩ d w ′ ( f ) = e − W ( μ ) /2 . 被以下引用 1 被以下引用 引理 6
集合 μ = c 1 ε t 1 + c 2 ε t 2 + ⋯ + c n ε t n \mu = c_1 \varepsilon_{t_1} + c_2 \varepsilon_{t_2} + \cdots + c_n \varepsilon_{t_n} μ = c 1 ε t 1 + c 2 ε t 2 + ⋯ + c n ε t n t 1 , … , t n t_1, \ldots, t_n t 1 , … , t n D D D c 1 , … , c n c_1, \ldots, c_n c 1 , … , c n R \mathbf{R} R g ∈ Ω 0 g \in \Omega_0 g ∈ Ω 0 u ( g ) u(g) u ( g ) g g g D D D
⟨ u ( g ) , μ ⟩ = ∑ j = 1 n c j g ( t j ) ( g ∈ Ω 0 ) .
\langle u(g), \mu \rangle = \sum_{j=1}^n c_j g(t_j) \qquad (g \in \Omega_0).
⟨ u ( g ) , μ ⟩ = j = 1 ∑ n c j g ( t j ) ( g ∈ Ω 0 ) . 还，
W ( μ ) = ∑ j , k = 1 n c j c k inf ( t j , t k ) ,
W(\mu) = \sum_{j,k=1}^n c_j c_k \inf(t_j, t_k),
W ( μ ) = j , k = 1 ∑ n c j c k inf ( t j , t k ) , 并且，由于 m m m Ω \Omega Ω M M M Ω − Ω 0 \Omega - \Omega_0 Ω − Ω 0 m m m
∫ Ω 0 e i ∑ j = 1 n c j g ( t j ) d m ( g ) = exp ( − 1 2 ∑ j , k = 1 n c j c k inf ( t j , t k ) ) .
\int_{\Omega_0} e^{i \sum_{j=1}^n c_j g(t_j)} \, dm(g) = \exp \left( - \frac{1}{2} \sum_{j,k=1}^n c_j c_k \inf(t_j, t_k) \right).
∫ Ω 0 e i ∑ j = 1 n c j g ( t j ) d m ( g ) = exp − 2 1 j , k = 1 ∑ n c j c k inf ( t j , t k ) . 现在，Ω − Ω 0 \Omega - \Omega_0 Ω − Ω 0 m m m w ′ = u ( m ) w' = u(m) w ′ = u ( m )
∫ C e i ⟨ f , μ ⟩ d w ′ ( f ) = ∫ Ω 0 e i ⟨ u ( g ) , μ ⟩ d m ( g ) .
\int_{\mathscr{C}} e^{i \langle f, \mu \rangle} \, dw'(f) = \int_{\Omega_0} e^{i \langle u(g), \mu \rangle} \, dm(g).
∫ C e i ⟨ f , μ ⟩ d w ′ ( f ) = ∫ Ω 0 e i ⟨ u ( g ) , μ ⟩ d m ( g ) . 公式 (48) 紧随公式 (49) 到 (52)。
让μ ∈ M 1 \mu \in \mathscr{M}^1 μ ∈ M 1 μ n ∈ D \mu_n \in \mathscr{D} μ n ∈ D μ ( f ) = lim n → ∞ μ n ( f ) \mu(f) = \lim_{n \to \infty} \mu_n(f) μ ( f ) = lim n → ∞ μ n ( f ) f ∈ C f \in \mathscr{C} f ∈ C W ( μ ) = lim n → ∞ W ( μ n ) W(\mu) = \lim_{n \to \infty} W(\mu_n) W ( μ ) = lim n → ∞ W ( μ n )
引用了 8 引用了 命题 13, III § 1 , 命题 15, III § 1 , 命题 17, III § 1 , 推论 1, III § 2 , 推论 2, III § 3 , 命题 6 , 定理 2, IV § 4 , 引理 5
让I = [ 0 , 1 ] I = [0, 1] I = [ 0 , 1 ] T = ] 0 , 1 [ T = ]0, 1[ T = ] 0 , 1 [ M 1 \mathscr{M}^1 M 1 M ( I ) \mathscr{M}(I) M ( I ) 0 0 0 M ( I ) \mathscr{M}(I) M ( I )
(3) 即I I I T = I − { 0 } T = I - \{0\} T = I − { 0 }
I I I M ( I ) \mathscr{M}(I) M ( I ) t ↦ ε t t \mapsto \varepsilon_t t ↦ ε t D \mathscr{D} D D ‾ \overline{\mathscr{D}} D ν ∈ D \nu \in \mathscr{D} ν ∈ D ∥ ν ∥ ≤ ∥ μ ∥ \| \nu \| \leq \| \mu \| ∥ ν ∥ ≤ ∥ μ ∥ μ \mu μ M ( I ) \mathscr{M}(I) M ( I ) M ( I ) \mathscr{M}(I) M ( I ) M ( I ) \mathscr{M}(I) M ( I ) μ n ∈ A \mu_n \in A μ n ∈ A μ \mu μ C \mathscr{C} C f ∈ C f \in \mathscr{C} f ∈ C μ ( f ) = lim n → ∞ μ n ( f ) \mu(f) = \lim_{n \to \infty} \mu_n(f) μ ( f ) = lim n → ∞ μ n ( f ) C ( I ) ⊗ C ( I ) \mathscr{C}(I) \otimes \mathscr{C}(I) C ( I ) ⊗ C ( I ) C ( I × I ) \mathscr{C}(I \times I) C ( I × I ) lim n → ∞ μ n = μ \lim_{n \to \infty} \mu_n = \mu lim n → ∞ μ n = μ ∥ μ n ∥ ≤ ∥ μ ∥ \| \mu_n \| \leq \| \mu \| ∥ μ n ∥ ≤ ∥ μ ∥ lim n → ∞ ( μ n ⊗ μ n ) = μ ⊗ μ \lim_{n \to \infty} (\mu_n \otimes \mu_n) = \mu \otimes \mu lim n → ∞ ( μ n ⊗ μ n ) = μ ⊗ μ μ n \mu_n μ n μ \mu μ
W ( μ n ) = ∫ I ∫ I inf ( t , t ′ ) d μ n ( t ) d μ n ( t ′ ) ,
W(\mu_n) = \int_I \int_I \inf(t, t') d\mu_n(t) d\mu_n(t'),
W ( μ n ) = ∫ I ∫ I inf ( t , t ′ ) d μ n ( t ) d μ n ( t ′ ) , W ( μ ) = ∫ I ∫ I inf ( t , t ′ ) d μ ( t ) d μ ( t ′ ) ,
W(\mu) = \int_I \int_I \inf(t, t') d\mu(t) d\mu(t'),
W ( μ ) = ∫ I ∫ I inf ( t , t ′ ) d μ ( t ) d μ ( t ′ ) , 从而lim n → ∞ W ( μ n ) = W ( μ ) \lim_{n \to \infty} W(\mu_n) = W(\mu) lim n → ∞ W ( μ n ) = W ( μ )
还需要证明w ′ w' w ′ e − W / 2 e^{-W/2} e − W /2 μ ∈ M 1 \mu \in \mathscr{M}^1 μ ∈ M 1 μ n ∈ D \mu_n \in \mathscr{D} μ n ∈ D w ′ w' w ′ ∣ e i ⟨ f , μ n ⟩ ∣ = 1 |e^{i \langle f, \mu_n \rangle}| = 1 ∣ e i ⟨ f , μ n ⟩ ∣ = 1 n n n
∫ C e i ⟨ f , μ ⟩ d w ′ ( f ) = lim n → ∞ ∫ C e i ⟨ f , μ n ⟩ d w ′ ( f ) = lim n → ∞ e − W ( μ n ) / 2 = e − W ( μ ) / 2 .
\int_{\mathscr{C}} e^{i \langle f, \mu \rangle} dw'(f) = \lim_{n \to \infty} \int_{\mathscr{C}} e^{i \langle f, \mu_n \rangle} dw'(f)
= \lim_{n \to \infty} e^{-W(\mu_n)/2} = e^{-W(\mu)/2}.
∫ C e i ⟨ f , μ ⟩ d w ′ ( f ) = n → ∞ lim ∫ C e i ⟨ f , μ n ⟩ d w ′ ( f ) = n → ∞ lim e − W ( μ n ) /2 = e − W ( μ ) /2 . Q.E.D.
C \mathscr{C} C w w w e − W / 2 e^{-W/2} e − W /2 C \mathscr{C} C
8. 傅里叶变换的连续性
— 令 E 为局部凸空间，μ \mu μ Φ \Phi Φ μ \mu μ
∣ Φ ( x ′ ) ∣ ≤ Φ ( 0 )
|\Phi(x')| \leq \Phi(0)
∣Φ ( x ′ ) ∣ ≤ Φ ( 0 ) ∣ Φ ( x ′ ) − Φ ( y ′ ) ∣ 2 ≤ 2 Φ ( 0 ) ( Φ ( 0 ) − ℜ Φ ( x ′ − y ′ ) )
|\Phi(x') - \Phi(y')|^2 \leq 2\Phi(0)\left(\Phi(0) - \Re \Phi(x' - y')\right)
∣Φ ( x ′ ) − Φ ( y ′ ) ∣ 2 ≤ 2Φ ( 0 ) ( Φ ( 0 ) − ℜΦ ( x ′ − y ′ ) ) 被以下引用 5 被以下引用 § 6. 。局部凸空间上的测度和测度 , 命题 2 , 命题 3 , 定理 2（米洛斯） , 定理 3（普罗霍罗夫-萨佐诺夫）
E ′ E' E ′ x ′ , y ′ x', y' x ′ , y ′
No. 3的公式(5)允许简化为E是有限维且μ \mu μ
∣ Φ ( x ′ ) ∣ = ∣ ∫ E e i ⟨ x , x ′ ⟩ d μ ( x ) ∣ ≤ ∫ E ∣ e i ⟨ x , x ′ ⟩ ∣ d μ ( x ) = ∫ E d μ ( x ) = Φ ( 0 ) ,
|\Phi(x')| = \left| \int_E e^{i\langle x, x' \rangle} d\mu(x) \right| \leq \int_E |e^{i\langle x, x' \rangle}| d\mu(x) = \int_E d\mu(x) = \Phi(0),
∣Φ ( x ′ ) ∣ = ∫ E e i ⟨ x , x ′ ⟩ d μ ( x ) ≤ ∫ E ∣ e i ⟨ x , x ′ ⟩ ∣ d μ ( x ) = ∫ E d μ ( x ) = Φ ( 0 ) , 从而 (53)。此外，如果a a a b b b
∣ e i a − e i b ∣ 2 = ∣ e i b ∣ 2 ∣ e i ( a − b ) − 1 ∣ 2 = ( e i ( a − b ) − 1 ) ( e − i ( a − b ) − 1 ) = 2 − 2 cos ( a − b ) ;
|e^{ia} - e^{ib}|^2 = |e^{ib}|^2|e^{i(a-b)} - 1|^2 = (e^{i(a-b)} - 1)(e^{-i(a-b)} - 1) = 2 - 2\cos(a - b);
∣ e ia − e ib ∣ 2 = ∣ e ib ∣ 2 ∣ e i ( a − b ) − 1 ∣ 2 = ( e i ( a − b ) − 1 ) ( e − i ( a − b ) − 1 ) = 2 − 2 cos ( a − b ) ; 根据柯西-施瓦茨不等式，我们有
∣ Φ ( x ′ ) − Φ ( y ′ ) ∣ 2 = ∣ ∫ E ( e i ⟨ x , x ′ ⟩ − e i ⟨ x , y ′ ⟩ ) d μ ( x ) ∣ 2
|\Phi(x') - \Phi(y')|^2 = \left| \int_E (e^{i\langle x, x' \rangle} - e^{i\langle x, y' \rangle}) d\mu(x) \right|^2
∣Φ ( x ′ ) − Φ ( y ′ ) ∣ 2 = ∫ E ( e i ⟨ x , x ′ ⟩ − e i ⟨ x , y ′ ⟩ ) d μ ( x ) 2 ≤ ∫ E ∣ e i ⟨ x , x ′ ⟩ − e i ⟨ x , y ′ ⟩ ∣ 2 d μ ( x ) ∫ E 1 2 d μ ( x )
\leq \int_E |e^{i\langle x, x' \rangle} - e^{i\langle x, y' \rangle}|^2 d\mu(x) \int_E 1^2 d\mu(x)
≤ ∫ E ∣ e i ⟨ x , x ′ ⟩ − e i ⟨ x , y ′ ⟩ ∣ 2 d μ ( x ) ∫ E 1 2 d μ ( x ) = ∫ E ( 2 − 2 cos ⟨ x , x ′ − y ′ ⟩ ) ) d μ ( x ) ⋅ Φ ( 0 )
= \int_E (2 - 2\cos\langle x, x' - y' \rangle)) d\mu(x) \cdot \Phi(0)
= ∫ E ( 2 − 2 cos ⟨ x , x ′ − y ′ ⟩)) d μ ( x ) ⋅ Φ ( 0 ) = 2 Φ ( 0 ) ( Φ ( 0 ) − ℜ Φ ( x ′ − y ′ ) ) ,
= 2\Phi(0)\left(\Phi(0) - \Re \Phi(x' - y')\right),
= 2Φ ( 0 ) ( Φ ( 0 ) − ℜΦ ( x ′ − y ′ ) ) , 从而 (54)。
— 为E ′ E' E ′ Φ \Phi Φ ℜ Φ \Re \Phi ℜΦ Φ \Phi Φ
被以下引用 1 被以下引用 命题 9
这是从不等式 (54) 得出的。
令 F 为局部凸空间。我们为F的对偶F ′ F' F ′ F ′ F' F ′ F ′ F' F ′ F ′ F' F ′ μ \mu μ F μ \mathscr{F}\mu F μ
( F μ ) ( x ) = ∫ F ′ e i ⟨ x , x ′ ⟩ d μ ( x ′ ) .
(\mathscr{F}\mu)(x) = \int_{F'} e^{i\langle x, x' \rangle} d\mu(x').
( F μ ) ( x ) = ∫ F ′ e i ⟨ x , x ′ ⟩ d μ ( x ′ ) .
令μ \mu μ F ′ F' F ′ Φ \Phi Φ ε > 0 \varepsilon > 0 ε > 0 F ′ F' F ′ μ ( F ′ − K ) ≤ ε \mu(F' - K) \leq \varepsilon μ ( F ′ − K ) ≤ ε σ ( F ′ , F ) \sigma(F', F) σ ( F ′ , F ) U ∘ U^\circ U ∘ ε U \varepsilon U ε U
Φ ( 0 ) − ℜ Φ ( x ) = ∫ F ′ ( 1 − cos ⟨ x , x ′ ⟩ ) d μ ( x ′ ) .
\Phi(0) - \Re \Phi(x) = \int_{F'} (1 - \cos \langle x, x' \rangle) d\mu(x').
Φ ( 0 ) − ℜΦ ( x ) = ∫ F ′ ( 1 − cos ⟨ x , x ′ ⟩) d μ ( x ′ ) . 现在，每个x ′ ∈ F ′ − K x' \in F' - K x ′ ∈ F ′ − K 0 ≤ 1 − cos ⟨ x , x ′ ⟩ ≤ 2 0 \leq 1 - \cos \langle x, x' \rangle \leq 2 0 ≤ 1 − cos ⟨ x , x ′ ⟩ ≤ 2
1 − cos ⟨ x , x ′ ⟩ ≤ 1 2 ⟨ x , x ′ ⟩ 2 ≤ ε 2 2
1 - \cos \langle x, x' \rangle \leq \frac{1}{2} \langle x, x' \rangle^2 \leq \frac{\varepsilon^2}{2}
1 − cos ⟨ x , x ′ ⟩ ≤ 2 1 ⟨ x , x ′ ⟩ 2 ≤ 2 ε 2 x ′ ∈ K ⊂ U ∘ x' \in K \subset U^\circ x ′ ∈ K ⊂ U ∘
0 ≤ Φ ( 0 ) − ℜ Φ ( x ) ≤ 2 μ ( F ′ − K ) + ε 2 2 μ ( K ) ≤ 2 ε + ε 2 2 μ ( F ′ ) .
0 \leq \Phi(0) - \Re \Phi(x) \leq 2 \mu(F' - K) + \frac{\varepsilon^2}{2} \mu(K) \leq 2\varepsilon + \frac{\varepsilon^2}{2} \mu(F').
0 ≤ Φ ( 0 ) − ℜΦ ( x ) ≤ 2 μ ( F ′ − K ) + 2 ε 2 μ ( K ) ≤ 2 ε + 2 ε 2 μ ( F ′ ) . 这个不等式的第二个成员随着ε \varepsilon ε ℜ Φ \Re \Phi ℜΦ
9. 米洛斯的引理
令 T 为有限维向量空间且 μ \mu μ T ′ T' T ′ T ′ T' T ′ μ \mu μ Φ \Phi Φ ε > 0 \varepsilon > 0 ε > 0 r > 0 r > 0 r > 0 x ′ ∈ T ′ x' \in T' x ′ ∈ T ′ C r C_r C r ⟨ x , x ′ ⟩ 2 ≤ r 2 h ( x ) \langle x, x' \rangle^2 \leq r^2 h(x) ⟨ x , x ′ ⟩ 2 ≤ r 2 h ( x ) x ∈ T x \in T x ∈ T
— *在假设 Φ ( 0 ) − ℜ Φ ≤ ε + q \Phi(0) - \Re \Phi \leq \varepsilon + q Φ ( 0 ) − ℜΦ ≤ ε + q
μ ( T ′ − C r ) ≤ 3 ( ε + r − 2 Tr ( q / h ) )
\mu(T' - C_r) \leq 3 (\varepsilon + r^{-2} \operatorname{Tr}(q/h))
μ ( T ′ − C r ) ≤ 3 ( ε + r − 2 Tr ( q / h )) 引用了 3 · 被以下引用 2 引用了 引理 2 , 引理 1 , 命题 2
被以下引用 命题 3, IX § 3 , 定理 2（米洛斯）
对于每个 r > 0 r > 0 r > 0
q 右 h 的迹可写为 Tr ( q / h ) \operatorname{Tr}(q/h) Tr ( q / h ) Tr ( q / h ) \operatorname{Tr}(q/h) Tr ( q / h ) Tr ( q / h ) \operatorname{Tr}(q/h) Tr ( q / h ) h ( x ) = 0 h(x) = 0 h ( x ) = 0 x ∈ T x \in T x ∈ T q ( x ) = 0 q(x) = 0 q ( x ) = 0
令a 1 , … , a n a_1, \ldots, a_n a 1 , … , a n T T T D D D x ′ ∈ T ′ x' \in T' x ′ ∈ T ′ ∑ j = 1 n ⟨ a j , x ′ ⟩ 2 > 1 \sum_{j=1}^n \langle a_j, x' \rangle^2 > 1 ∑ j = 1 n ⟨ a j , x ′ ⟩ 2 > 1 t ≥ 0 t \geq 0 t ≥ 0 3 ( 1 − e − t / 2 ) ≥ 0 3(1 - e^{-t/2}) \geq 0 3 ( 1 − e − t /2 ) ≥ 0
3 ( 1 − e − t / 2 ) ≥ 3 ( 1 − e − 1 / 2 ) ≥ 3 ( 1 − ( 9 4 ) − 1 / 2 ) = 1
3(1 - e^{-t/2}) \geq 3(1 - e^{-1/2}) \geq 3\left(1 - \left(\frac{9}{4}\right)^{-1/2}\right) = 1
3 ( 1 − e − t /2 ) ≥ 3 ( 1 − e − 1/2 ) ≥ 3 ( 1 − ( 4 9 ) − 1/2 ) = 1 对于t > 1 t > 1 t > 1 e > 9 4 e > \frac{9}{4} e > 4 9 t = ∑ j = 1 n ⟨ a j , x ′ ⟩ 2 t = \sum_{j=1}^n \langle a_j, x' \rangle^2 t = ∑ j = 1 n ⟨ a j , x ′ ⟩ 2
μ ( D ) ≤ 3 ∫ T ′ ( 1 − exp ( − 1 2 ∑ j = 1 n ⟨ a j , x ′ ⟩ 2 ) ) d μ ( x ′ ) .
\mu(D) \leq 3 \int_{T'} \left(1 - \exp \left(-\frac{1}{2} \sum_{j=1}^n \langle a_j, x' \rangle^2\right)\right) d\mu(x').
μ ( D ) ≤ 3 ∫ T ′ ( 1 − exp ( − 2 1 j = 1 ∑ n ⟨ a j , x ′ ⟩ 2 ) ) d μ ( x ′ ) . 令 γ \gamma γ R \mathbf{R} R t ↦ ( 2 π ) − 1 / 2 e − t 2 / 2 t \mapsto (2\pi)^{-1/2} e^{-t^2/2} t ↦ ( 2 π ) − 1/2 e − t 2 /2
∫ R e i u t d γ ( t ) = e − u 2 / 2
\int_{\mathbf{R}} e^{iut} d\gamma(t) = e^{-u^2/2}
∫ R e i u t d γ ( t ) = e − u 2 /2 对于所有真实的u u u
1 − exp ( − 1 2 ∑ j = 1 n ⟨ a j , x ′ ⟩ 2 ) = ∫ ⋯ ∫ ( 1 − e i ∑ j = 1 n ⟨ a j , x ′ ⟩ t j ) d γ ( t 1 ) ⋯ d γ ( t n )
\begin{align*}
1 - \exp \left(-\frac{1}{2} \sum_{j=1}^n \langle a_j, x' \rangle^2\right) \\
&= \int \cdots \int \left(1 - e^{i \sum_{j=1}^n \langle a_j, x' \rangle t_j}\right) d\gamma(t_1) \cdots d\gamma(t_n)
\end{align*}
1 − exp ( − 2 1 j = 1 ∑ n ⟨ a j , x ′ ⟩ 2 ) = ∫ ⋯ ∫ ( 1 − e i ∑ j = 1 n ⟨ a j , x ′ ⟩ t j ) d γ ( t 1 ) ⋯ d γ ( t n ) 对于所有x ′ ∈ T ′ x' \in T' x ′ ∈ T ′ x ′ , t 1 , … , t n x', t_1, \ldots, t_n x ′ , t 1 , … , t n μ \mu μ γ \gamma γ d μ ( x ′ ) d\mu(x') d μ ( x ′ ) μ \mu μ γ \gamma γ
∫ T ′ ( 1 − exp ( − 1 2 ∑ j = 1 n ⟨ a j , x ′ ⟩ 2 ) ) d μ ( x ′ ) = ∫ ⋯ ∫ ( Φ ( 0 ) − Φ ( ∑ j = 1 n t j a j ) ) d γ ( t 1 ) ⋯ d γ ( t n ) .
\begin{align*}
\int_{T'} \left(1 - \exp \left(-\frac{1}{2} \sum_{j=1}^n \langle a_j, x' \rangle^2\right)\right) d\mu(x') \\
&= \int \cdots \int \left(\Phi(0) - \Phi\left(\sum_{j=1}^n t_j a_j\right)\right) d\gamma(t_1) \cdots d\gamma(t_n).
\end{align*}
∫ T ′ ( 1 − exp ( − 2 1 j = 1 ∑ n ⟨ a j , x ′ ⟩ 2 ) ) d μ ( x ′ ) = ∫ ⋯ ∫ ( Φ ( 0 ) − Φ ( j = 1 ∑ n t j a j ) ) d γ ( t 1 ) ⋯ d γ ( t n ) . 由于q q q T T T q j k q_{jk} q j k
q ( ∑ j = 1 n t j a j ) = ∑ j , k q j k t j t k
q \left( \sum_{j=1}^n t_j a_j \right) = \sum_{j,k} q_{jk} t_j t_k
q ( j = 1 ∑ n t j a j ) = j , k ∑ q j k t j t k t 1 , … , t n t_1, \ldots, t_n t 1 , … , t n 真实；特别是，
q j j = q ( a j ) q_{jj} = q(a_j) q j j = q ( a j ) 代表
1 ≤ j ≤ n 1 \leq j \leq n 1 ≤ j ≤ n 。此外，对于
n = 0 , 1 , 2 n = 0, 1, 2 n = 0 , 1 , 2 ，整的
∫ R t n d γ ( t ) \int_{\mathbf{R}} t^n d\gamma(t) ∫ R t n d γ ( t ) 分别有值 1、0、1（No. 4、引理 1）。由此，立即推断出
∫ ⋯ ∫ ( ε + q ( ∑ j = 1 n t j a j ) ) d γ ( t 1 ) … d γ ( t n ) = ε + ∑ j = 1 n q ( a j ) .
\int \cdots \int \left( \varepsilon + q \left( \sum_{j=1}^n t_j a_j \right) \right) d\gamma(t_1) \ldots d\gamma(t_n) = \varepsilon + \sum_{j=1}^n q(a_j).
∫ ⋯ ∫ ( ε + q ( j = 1 ∑ n t j a j ) ) d γ ( t 1 ) … d γ ( t n ) = ε + j = 1 ∑ n q ( a j ) . 现在，(58)和
Φ ( 0 ) \Phi(0) Φ ( 0 ) 的第一个成员是实数；因而，在（58）的第二个成员中，可以通过替换
Φ \Phi Φ 乘
ℜ Φ \Re \Phi ℜΦ 。 不等式
Φ ( 0 ) − ℜ Φ ≤ ε + q \Phi(0) - \Re \Phi \leq \varepsilon + q Φ ( 0 ) − ℜΦ ≤ ε + q 和公式 (56), (58) 和 (59) 那么意味着
μ ( D ) ≤ 3 ( ε + ∑ j = 1 n q ( a j ) ) .
\mu(D) \leq 3 \left( \varepsilon + \sum_{j=1}^n q(a_j) \right).
μ ( D ) ≤ 3 ( ε + j = 1 ∑ n q ( a j ) ) . 让我们修复数
r > 0 r > 0 r > 0 。由于二次方形式
h h h 为正，因此在 0 和
n n n 之间存在
T T T 和整数
m m m 的基
( e 1 , … , e n ) (e_1, \ldots, e_n) ( e 1 , … , e n ) ，使得
h ( ∑ j = 1 n t j e j ) = ∑ j = 1 m t j 2
h \left( \sum_{j=1}^n t_j e_j \right) = \sum_{j=1}^m t_j^2
h ( j = 1 ∑ n t j e j ) = j = 1 ∑ m t j 2 t 1 , … , t n t_1, \ldots, t_n t 1 , … , t n 实数（附件，No. 1，提案2）。那么直接的
C r C_r C r 由
x ′ ∈ T ′ x' \in T' x ′ ∈ T ′ 组成，这样
∑ j = 1 m ⟨ e j , x ′ ⟩ 2 ≤ r 2 , ∑ j = m + 1 n ⟨ e j , x ′ ⟩ 2 = 0.
\sum_{j=1}^m \langle e_j, x' \rangle^2 \leq r^2, \qquad \sum_{j=m+1}^n \langle e_j, x' \rangle^2 = 0.
j = 1 ∑ m ⟨ e j , x ′ ⟩ 2 ≤ r 2 , j = m + 1 ∑ n ⟨ e j , x ′ ⟩ 2 = 0. 对于每个整数
l ≥ 1 l \geq 1 l ≥ 1 ，令
D l D_l D l 为
x ′ ∈ T ′ x' \in T' x ′ ∈ T ′ 满足不等式中的集合
∑ j = 1 m ⟨ r − 1 e j , x ′ ⟩ 2 + ∑ j = m + 1 n ⟨ l e j , x ′ ⟩ 2 > 1.
\sum_{j=1}^m \langle r^{-1} e_j, x' \rangle^2 + \sum_{j=m+1}^n \langle l e_j, x' \rangle^2 > 1.
j = 1 ∑ m ⟨ r − 1 e j , x ′ ⟩ 2 + j = m + 1 ∑ n ⟨ l e j , x ′ ⟩ 2 > 1. 人们很容易看出序列
( D l ) l ≥ 1 (D_l)_{l \geq 1} ( D l ) l ≥ 1 是递增与并
T ′ − C r T' - C_r T ′ − C r ，从而
μ ( T ′ − C r ) = lim l → ∞ μ ( D l ) .
\mu(T' - C_r) = \lim_{l \to \infty} \mu(D_l).
μ ( T ′ − C r ) = l → ∞ lim μ ( D l ) . 但到了(60)，
μ ( D l ) ≤ 3 ( ε + ∑ j = 1 m r − 2 q ( e j ) + ∑ j = m + 1 n l 2 q ( e j ) ) ;
\mu(D_l) \leq 3 \left( \varepsilon + \sum_{j=1}^m r^{-2} q(e_j) + \sum_{j=m+1}^n l^2 q(e_j) \right);
μ ( D l ) ≤ 3 ( ε + j = 1 ∑ m r − 2 q ( e j ) + j = m + 1 ∑ n l 2 q ( e j ) ) ; 对于j = m + 1 , … , n j = m+1, \ldots, n j = m + 1 , … , n h ( e j ) = 0 h(e_j) = 0 h ( e j ) = 0 q ( e j ) = 0 q(e_j) = 0 q ( e j ) = 0 Tr ( q / h ) = ∑ j = 1 m q ( e j ) \operatorname{Tr}(q/h) = \sum_{j=1}^m q(e_j) Tr ( q / h ) = ∑ j = 1 m q ( e j )
Q.E.D.
10. 对核空间的双重措施
令 F 为局部凸空间。令T s \mathscr{T}_s T s F ′ F' F ′ σ ( F ′ , F ) \sigma(F', F) σ ( F ′ , F ) T c \mathscr{T}_c T c F ′ F' F ′ T s \mathscr{T}_s T s T c \mathscr{T}_c T c F ′ F' F ′ F ′ F' F ′ T s \mathscr{T}_s T s T c \mathscr{T}_c T c T \mathscr{T} T T \mathscr{T} T F T ′ F'_{\mathscr{T}} F T ′ F ′ F' F ′ T \mathscr{T} T F T ′ F'_{\mathscr{T}} F T ′ T \mathscr{T} T F ′ F' F ′ μ \mu μ
人们将 F 上的 Sazonov 拓扑称为由连续半范数 N 满足定义的局部凸拓扑 S \mathscr{S} S N 2 N^2 N 2 Tr ( N 2 / H ) < + ∞ \operatorname{Tr}(N^2/H) < +\infty Tr ( N 2 / H ) < + ∞ S \mathscr{S} S
定理 2（米洛斯） 09ST
— *令 F 为局部凸空间，T \mathscr{T} T F ′ F' F ′ T s \mathscr{T}_s T s T c \mathscr{T}_c T c μ \mu μ F T ′ F'_{\mathscr{T}} F T ′ μ \mu μ Φ \Phi Φ μ \mu μ F T ′ F'_{\mathscr{T}} F T ′
引用了 3 · 被以下引用 2 引用了 命题 8 , 命题 10 , 命题 1
被以下引用 推论 , 定理 3（普罗霍罗夫-萨佐诺夫）
让ε > 0 \varepsilon > 0 ε > 0 Φ \Phi Φ Tr ( Q / H ) < + ∞ \operatorname{Tr}(Q/H) < +\infty Tr ( Q / H ) < + ∞
Φ ( 0 ) − R Φ ( x ) ≤ ε / 6
\Phi(0) - \mathscr{R}\Phi(x) \leq \varepsilon/6
Φ ( 0 ) − R Φ ( x ) ≤ ε /6 对于每个x ∈ F x \in F x ∈ F Q ( x ) ≤ 1 Q(x) \leq 1 Q ( x ) ≤ 1 ∣ R Φ ( x ) ∣ ≤ Φ ( 0 ) |\mathscr{R}\Phi(x)| \leq \Phi(0) ∣ R Φ ( x ) ∣ ≤ Φ ( 0 ) x ∈ F x \in F x ∈ F
Φ ( 0 ) − R Φ ( x ) ≤ ε / 6 + 2 Φ ( 0 ) Q ( x )
\Phi(0) - \mathscr{R}\Phi(x) \leq \varepsilon/6 + 2\Phi(0)Q(x)
Φ ( 0 ) − R Φ ( x ) ≤ ε /6 + 2Φ ( 0 ) Q ( x ) 对于所有x ∈ F x \in F x ∈ F
集合 r = ( 12 Φ ( 0 ) Tr ( Q / H ) ε − 1 ) 1 / 2 r = (12\Phi(0)\operatorname{Tr}(Q/H)\varepsilon^{-1})^{1/2} r = ( 12Φ ( 0 ) Tr ( Q / H ) ε − 1 ) 1/2 x ′ ∈ F T ′ x' \in F'_\mathscr{T} x ′ ∈ F T ′ ⟨ x , x ′ ⟩ 2 ≤ r 2 H ( x ) \langle x, x' \rangle^2 \leq r^2 H(x) ⟨ x , x ′ ⟩ 2 ≤ r 2 H ( x ) x ∈ F x \in F x ∈ F H 1 / 2 H^{1/2} H 1/2 F F F F T ′ F'_\mathscr{T} F T ′ F T ′ F'_\mathscr{T} F T ′
设V是F T ′ F'_\mathscr{T} F T ′ μ V \mu_V μ V T ′ T' T ′ μ \mu μ F T ′ F'_\mathscr{T} F T ′ p V p_V p V Φ \Phi Φ p V ( K ) p_V(K) p V ( K ) x ′ ∈ T ′ x' \in T' x ′ ∈ T ′ C r C_r C r x ∈ T x \in T x ∈ T ⟨ x , x ′ ⟩ 2 ≤ r 2 H ( x ) \langle x, x' \rangle^2 \leq r^2 H(x) ⟨ x , x ′ ⟩ 2 ≤ r 2 H ( x ) T ′ T' T ′ μ V \mu_V μ V 2 Φ ( 0 ) Q 2\Phi(0)Q 2Φ ( 0 ) Q Tr ( q / h ) ≤ 2 Φ ( 0 ) Tr Q / H ) \operatorname{Tr}(q/h) \leq 2\Phi(0)\operatorname{Tr}Q/H) Tr ( q / h ) ≤ 2Φ ( 0 ) Tr Q / H )
μ V ( T ′ − C r ) ≤ 3 ( ε 6 + 2 Φ ( 0 ) Tr ( Q / H ) r − 2 ) = ε .
\mu_V(T' - C_r) \leq 3 \left( \frac{\varepsilon}{6} + 2\Phi(0)\operatorname{Tr}(Q/H)\,r^{-2} \right) = \varepsilon.
μ V ( T ′ − C r ) ≤ 3 ( 6 ε + 2Φ ( 0 ) Tr ( Q / H ) r − 2 ) = ε . 由于p V p_V p V F T ′ / V F'_\mathscr{T}/V F T ′ / V T ′ T' T ′ μ \mu μ F T ′ F'_\mathscr{T} F T ′
Q.E.D.
— *令 F 为桶状核空间，T \mathscr{T} T F ′ F' F ′ T s \mathscr{T}_s T s T c \mathscr{T}_c T c μ \mu μ F T ′ F'_\mathscr{T} F T ′ Φ \Phi Φ μ \mu μ μ \mu μ Φ \Phi Φ
引用了 2 · 被以下引用 1 引用了 命题 9 , 定理 2（米洛斯）
被以下引用 推论
必要性来自于No. 8的第9条命题，充分性来自于Th。 2.
11. 希尔伯特空间的测量
令 E 为实希尔伯特空间，其中标量乘积表示为 ( x ∣ y ) (x|y) ( x ∣ y ) E ′ E' E ′ x , y x, y x , y ⟨ x , j ( y ) ⟩ = ( x ∣ y ) \langle x, j(y) \rangle = (x|y) ⟨ x , j ( y )⟩ = ( x ∣ y ) E ′ E' E ′ μ \mu μ F μ \mathscr{F}\mu F μ μ \mu μ
( F μ ) ( x ) = ∫ E e i ( x ∣ y ) d μ ( y ) ( x ∈ E ) . (\mathscr{F}\mu)(x) = \int_{E} e^{i(x|y)} d\mu(y) \quad (x \in E). ( F μ ) ( x ) = ∫ E e i ( x ∣ y ) d μ ( y ) ( x ∈ E ) .
a) 对于 Sazonov 拓扑*，*函数 Φ 在 E 上连续。
b) 对于每个 ε > 0，E 上存在一个核正二次方程形式 Q，使得 Φ(0) − ℜ Φ \Re \Phi ℜΦ 。
c) promeasure μ 是对 Es 的测量。
b ) ⇒ a ) b) \Rightarrow a) b ) ⇒ a )
a ) ⇒ c ) a) \Rightarrow c) a ) ⇒ c )
c ) ⇒ b ) c) \Rightarrow b) c ) ⇒ b ) x ∈ E x \in E x ∈ E n ≥ 1 n \geq 1 n ≥ 1 B n B_n B n ≤ n \leq n ≤ n E s E_s E s E = ⋃ n ≥ 1 B n E = \bigcup_{n \geq 1} B_n E = ⋃ n ≥ 1 B n n ≥ 1 n \geq 1 n ≥ 1 μ ( E − B n ) < ε 2 \mu(E - B_n) < \frac{\varepsilon}{2} μ ( E − B n ) < 2 ε
Q ( x ) = 1 2 ∫ B n ( x ∣ y ) 2 d μ ( y ) Q(x) = \frac{1}{2} \int_{B_n} (x|y)^2 d\mu(y) Q ( x ) = 2 1 ∫ B n ( x ∣ y ) 2 d μ ( y ) 定义 E 上的正二次形式 Q。集合 C = n 2 2 μ ( B n ) C = \frac{n^2}{2} \mu(B_n) C = 2 n 2 μ ( B n ) ( e 1 , … , e p ) (e_1, \ldots, e_p) ( e 1 , … , e p )
∑ j = 1 p ( e j ∣ y ) 2 ≤ ∥ y ∥ 2 ≤ n 2
\sum_{j=1}^p (e_j|y)^2 \leq \|y\|^2 \leq n^2
j = 1 ∑ p ( e j ∣ y ) 2 ≤ ∥ y ∥ 2 ≤ n 2 贝塞尔的不等式每y ∈ B n y \in B_n y ∈ B n
∑ j = 1 p Q ( e j ) = 1 2 ∫ B n ∑ j = 1 p ( e j ∣ y ) 2 d μ ( y ) ≤ n 2 2 μ ( B n ) = C ,
\sum_{j=1}^p Q(e_j) = \frac{1}{2} \int_{B_n} \sum_{j=1}^p (e_j|y)^2 d\mu(y) \leq \frac{n^2}{2} \mu(B_n) = C,
j = 1 ∑ p Q ( e j ) = 2 1 ∫ B n j = 1 ∑ p ( e j ∣ y ) 2 d μ ( y ) ≤ 2 n 2 μ ( B n ) = C , 因此Q是核。
此外，每实数t1 − cos t ≤ inf ( 2 , t 2 2 ) 1 - \cos t \leq \inf \left( 2, \frac{t^2}{2} \right) 1 − cos t ≤ inf ( 2 , 2 t 2 )
Φ ( 0 ) − ℜ Φ ( x ) = ∫ E ( 1 − cos ( x ∣ y ) ) d μ ( y )
\Phi(0) - \Re \Phi(x) = \int_{E} (1 - \cos(x|y)) d\mu(y)
Φ ( 0 ) − ℜΦ ( x ) = ∫ E ( 1 − cos ( x ∣ y )) d μ ( y ) ≤ ∫ B n 1 2 ( x ∣ y ) 2 d μ ( y ) + ∫ E − B n 2 ⋅ d μ ( y )
\leq \int_{B_n} \frac{1}{2}(x|y)^2 d\mu(y) + \int_{E - B_n} 2 \cdot d\mu(y)
≤ ∫ B n 2 1 ( x ∣ y ) 2 d μ ( y ) + ∫ E − B n 2 ⋅ d μ ( y ) < Q ( x ) + ε
< Q(x) + \varepsilon
< Q ( x ) + ε 对于所有x ∈ E x \in E x ∈ E
Q.E.D.
通过本文介绍的E 1 E_1 E 1 E 2 E_2 E 2 ν \nu ν Φ ∘ u ∗ \Phi \circ u^* Φ ∘ u ∗ u ∗ u^* u ∗ u u u u ∗ u^* u ∗ E 2 E_2 E 2 E 1 E_1 E 1 E 2 E_2 E 2 y ↦ ∥ u ∗ ( y ) ∥ 2 y \mapsto \|u^*(y)\|^2 y ↦ ∥ u ∗ ( y ) ∥ 2 ( E 2 ) S (E_2)_\mathscr{S} ( E 2 ) S E 2 E_2 E 2 u ∗ u^* u ∗ ( E 2 ) S (E_2)_\mathscr{S} ( E 2 ) S E 1 E_1 E 1 F V = Φ ∘ u ∗ \mathscr{F}_V = \Phi \circ u^* F V = Φ ∘ u ∗ ( E 2 ) S (E_2)_\mathscr{S} ( E 2 ) S ν \nu ν E 2 E_2 E 2
Γ Q \Gamma_Q Γ Q Φ \Phi Φ e − Q / 2 e^{-Q/2} e − Q /2 t t t Φ ( 0 ) − ℜ Φ ≤ Q / 2 \Phi(0) - \Re \Phi \leq Q/2 Φ ( 0 ) − ℜΦ ≤ Q /2 e t ≥ 1 + t e^t \geq 1 + t e t ≥ 1 + t Γ Q \Gamma_Q Γ Q E s E_s E s
— 令 G G G ，则 G G G Φ \Phi Φ
∑ j , k = 1 p c j c k ‾ Φ ( x j x k − 1 ) ≥ 0
\sum_{j,k=1}^p c_j \overline{c_k} \Phi(x_j x_k^{-1}) \geq 0
j , k = 1 ∑ p c j c k Φ ( x j x k − 1 ) ≥ 0 被以下引用 2 被以下引用 引理, IV § 4 , 习题 5, § 3
适用于任何 x 1 , … , x p x_1, \ldots, x_p x 1 , … , x p 在 G G G 和任何复数 c 1 , … , c p c_1, \ldots, c_p c 1 , … , c p
稍后将详细研究这个概念。
— 令 E 为有限维向量空间，μ \mu μ Φ \Phi Φ μ \mu μ Φ \Phi Φ E ′ E' E ′
引用了 1 · 被以下引用 2 引用了 命题 9
被以下引用 IX § 4. 。 逆极限措施 , 定理 4
Φ \Phi Φ
让我们证明Φ \Phi Φ x 1 ′ , … , x p ′ x'_1, \ldots, x'_p x 1 ′ , … , x p ′ E ′ E' E ′ c 1 , … , c p c_1, \ldots, c_p c 1 , … , c p
∑ j , k c j c k ‾ Φ ( x j ′ − x k ′ ) = ∫ E ∑ j , k c j c k ‾ e i ⟨ x , x j ′ − x k ′ ⟩ d μ ( x )
\sum_{j,k} c_j \overline{c_k} \Phi(x'_j - x'_k) = \int_E \sum_{j,k} c_j \overline{c_k} e^{i \langle x, x'_j - x'_k \rangle} \, d\mu(x)
j , k ∑ c j c k Φ ( x j ′ − x k ′ ) = ∫ E j , k ∑ c j c k e i ⟨ x , x j ′ − x k ′ ⟩ d μ ( x ) = ∫ E ∣ ∑ j = 1 p c j e i ⟨ x , x j ′ ⟩ ∣ 2 d μ ( x ) ≥ 0.
= \int_E \left| \sum_{j=1}^p c_j e^{i \langle x, x'_j \rangle} \right|^2 \, d\mu(x) \geq 0.
= ∫ E j = 1 ∑ p c j e i ⟨ x , x j ′ ⟩ 2 d μ ( x ) ≥ 0. Q.E.D.
一个可以证明逆命题被称为 Bochner 的定理 ：正型的 E ′ E' E ′
— 令 E 为局部凸空间。傅立叶变换是 E 上的算子的集合的双射到E ′ E' E ′
引用了 2 · 被以下引用 2 引用了 命题 11 , 命题 3
被以下引用 推论 , 定理 3, VI § 3
我们知道（No. 3，命题 3）傅里叶变换是单射的。令 μ = ( μ V ) V ∈ F ( E ) \mu = (\mu_V)_{V \in \mathscr{F}(E)} μ = ( μ V ) V ∈ F ( E ) Φ \Phi Φ E ′ E' E ′ E / V E/V E / V Φ \Phi Φ Φ T \Phi_T Φ T E / V E/V E / V μ V \mu_V μ V Φ T \Phi_T Φ T Φ \Phi Φ E ′ E' E ′
相反，令Φ \Phi Φ E ′ E' E ′ E ′ E' E ′ V ∈ F ( E ) V \in \mathscr{F}(E) V ∈ F ( E ) E ′ E' E ′ E / V E/V E / V V ∘ V^\circ V ∘ Φ \Phi Φ V ∘ V^\circ V ∘ Φ V \Phi_V Φ V E / V E/V E / V μ V \mu_V μ V Φ V \Phi_V Φ V F ( E ) \mathscr{F}(E) F ( E ) W ⊂ V W \subset V W ⊂ V p V W p_{VW} p V W E / W E/W E / W E / V E/V E / V t p V W {}^t p_{VW} t p V W V ∘ V^\circ V ∘ W ∘ W^\circ W ∘
F ( p V W ( μ W ) ) = ( F μ W ) ∘ t p V W = Φ W ∘ t p V W = Φ V = F μ V ,
\mathscr{F}(p_{VW}(\mu_W)) = (\mathscr{F}\mu_W) \circ {}^t p_{VW} = \Phi_W \circ {}^t p_{VW} = \Phi_V = \mathscr{F}\mu_V,
F ( p V W ( μ W )) = ( F μ W ) ∘ t p V W = Φ W ∘ t p V W = Φ V = F μ V , () 这个问题将在即将出版的《光谱理论》一书中的章中进行研究。关于这个问题读者可以查阅L.H的书。 LOOMIS， 抽象谐波数学分析*，范诺斯特兰德，纽约，1953 年。
从而 p v w ( μ w ) = μ v pvw(\mu_w) = \mu_v p v w ( μ w ) = μ v μ = ( μ v ) v ∈ F ( E ) \mu = (\mu_v)_{v \in \mathscr{F}(E)} μ = ( μ v ) v ∈ F ( E ) E E E Φ \Phi Φ μ \mu μ
— *令 F 为桶装核空间；为 F' 配备一个局部凸拓扑 T \mathscr{T} T σ ( F ′ , F ) \sigma(F', F) σ ( F ′ , F )
引用了 2 引用了 推论 , 定理 4
这紧随定理 4 和 Cor.的 Th。 2 个No. 10。*
习题
参见习题为 § 6 。
