— For every integer 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 . Let 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 . Let us make the convention 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 ) ) for 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 ; !Figure 1
FIGURE 1
one can then apply Lemma 3 with 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 . By (44), we then have
∫ 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 , whence the lemma.
By Lemma 4, the mapping T n T_n T n Ω \Omega Ω 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 ∑ n = 0 ∞ T n ( g ) \sum_{n=0}^{\infty} T_n(g) ∑ n = 0 ∞ T n ( g ) C \mathscr{C} C g ∈ Ω 0 g \in \Omega_0 g ∈ Ω 0 m m m u u u Ω \Omega Ω 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 . Since u n ( g ) u_n(g) u n ( g ) g g g D m ⊂ D n D_m \subset D_n D m ⊂ D n 0 ≤ m ≤ n 0 \leq m \leq n 0 ≤ m ≤ n u ( g ) u(g) u ( g ) D D D g g g g ∈ Ω 0 g \in \Omega_0 g ∈ Ω 0
C) Construction of a Gaussian measure on C \mathscr{C} C w ′ w' w ′ C \mathscr{C} C m m m m m m u : Ω → C u : \Omega \to \mathscr{C} u : Ω → C 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 ε t \varepsilon_t ε t t t t D D D
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.
