[-]
Download 1.57 Mb. Pdf ko'rish
|
funksional analiz misol va masalalar yechish 1 qism
|
A
= ∞ ∪ n =1 A n , A i ∩ A j = ∅, i = j, f A 5 A n f ∞ n =1 $ A n f (x)dμ $ A f (x)dμ = ∞ n =1 $ A n f (x)dμ. (7.8) [ ( ( 8#" ! " & "" 2 A A 1 , A 2 , . . . , A n , . . . A = ∞ ∪ n =1 A n , A i ∩ A j = ∅, i = j. ! A n f ∞ n =1 $ A n |f(x)| dμ f A 38W5 & "" $1 & ! f A (μ(A) < ∞) ε > 0 δ > 0 μ (D) < δ D ⊂ A ## ## $ D f (x)dμ ## ## < ε [ Z / " & "" ! [a, b] I = (R) $ b a f (x)dx 6 f [a, b] 1 " (L) $ [a, b] f (x)dμ = (R) $ b a f (x) dx. & = & y 1 , y 2 , . . . , y n , . . . ! f : A → R ! ! A n = { x ∈ A : f(x) = y n } & ! 8 f A & ! A n & ! 7a" ! 4 A n & ! f " ! ! A (f < c) = {x ∈ A : f(x) < c} = ∪ y n A n ! & ! ! f A ! & = & K K n & ! / ! "# K n = 2 n−1 k =1 K nk & k − K nk K y k = (2k − 1) · 2 −n , (k = 1, 2, 3, . . . , 2 n −1 ) ( A k = {x ∈ K n : K(x) = y k } = K nk , k = 1, 2, 3, . . . , 2 n −1 . ' ! k ∈ 1, 2, 3, . . . , 2 n −1 ! μ (K nk ) = 3 −n ! / (; $ K n K(x)dμ = 2 n−1 k =1 2k − 1 2 n · 1 3 n = 1 2 n · 3 n 2 n−1 k =1 (2k − 1) = = 1 2 n · 3 n · 1 + 2 n − 1 2 · 2 n −1 = 1 4 · 2 n 3 n ' & n ! S n = a 1 + a n 2 n & = & K [0, 1]\K & ! " ' K − = & ! "# [0, 1]\K = ∞ n =1 K n K n & + "+ / σ − " 38#" 38W5 5 - $ [0, 1]\K K(x)dμ = ∞ n =1 $ K n K(x)dμ = ∞ n =1 1 4 · 2 n 3 n = 1 4 · 2 3 1 − 2 3 = 1 2 . (7.9) ' + ! ! & ! S = b 1 1 − q & d ! A & ! f " ! ' ! / d5 f ! ! M > 0 ! x ∈ A |f(x)| ≤ M f A i & f i , i |f i |μ (A i ) ≤ M · i μ (A i ) = M · μ (A). \ 8`" ( f ! &$ A = (0, 1] f ! - f (x) = n, x ∈ A n = 1 2 n , 1 2 n −1 , n ∈ N. f A = (0, 1] & / ( ! c * ! ! "# ( ∞ ∪ n =1 A n = (0, 1], A n ∩ A m = ∅, n = m. A n = {x ∈ A : f(x) = n} 2 ! / (; ∞ n =1 y n μ (A n ) = ∞ n =1 n · 1 2 n (7.10) ! f A = (0, 1] ! ' \ " - lim n →∞ a n +1 a n = lim n →∞ n + 1 2 n +1 · 2 n n = 1 2 < 1. \ 38#>5 ! ' f / ( ! ! [ 38#>5 , S n ! S n = 2S n − S n = 1 + 2 2 + 3 4 + 4 8 + · · · + n 2 n −1 − 1 2 + 2 4 + 3 8 + · · · + n 2 n = = 1 + 2 2 − 1 2 + 3 4 − 2 4 + · · · + n 2 n −1 − n − 1 2 n −1 − n 2 n = 1 + 1 2 + 1 4 + 1 8 + · · · + 1 2 n −1 − n 2 n ( ' n → ∞ $ (0, 1] f (x)dμ = lim n →∞ S n = lim n →∞ ⎛ ⎜ ⎝ 1 − 1 2 n 1 − 1 2 − n 2 n ⎞ ⎟ ⎠ = 2 2 ( + ! ( Z (; ! " ! d ! Z (: / ! " ! ! (: &% . ! f : A → R A (μ(A) < ∞) & ! ! n ∈ N f but n (x) = [nf(x)] n (7.11) ! 8 f : A → R ! 8#% 8#$" n ∈ N 38##5 f but n ! < |f but n (x)| ≤ |f(x)| + 1 f but n ! ! 4 f but n n ∈ N ! {f but n } " f ! " ! x ∈ A ## f (x) − f but n (x) ## = ## ##f(x) − [nf(x)] n ## ## = ## ## nf (x) − [nf(x)] n ## ## = {nf(x)} n ≤ 1 n \ {f but n } " f 8a" ( f A & ! && / ( ! Z ( ! ! "# \ [0, 2] / Z ( ! D / - $ [0, 2] D(x)dμ = 1 · μ ([0, 2] ∩ Q) + 0 · μ ([0, 2]\Q) = 0. \ [0, 2] Z ( ! ' ! [0, 2] 0 = x 0 < x 1 < x 2 < · · · < x n −1 < x n = 2 n ( \ [x k −1 , x k ] ! ! M k ! k ∈ {1, 2, . . . , n} ! 1 \ ! ! m k 0 ' \ Ω n ω n - Ω n = 2 n n k =1 M k = 2 n n k =1 1 = 2, ω n = 2 n n k =1 m k = 2 n n k =1 0 = 0. ' lim n →∞ Ω n = 2, lim n →∞ ω n = 0 \ \ [0, 2] Z (" ! 2 ( / ! ' Z ! ' 8W"8$ 8b$"8a>" ! &' / Z ! ( ! ! Z ( ! ! "# [0, 2] \ , ! ( ! / ( ! \ [0, 2] Z ( ! ' 88" &. / Z ! _ ( ! ϕ : A → R ! f : Download 1.57 Mb. Do'stlaringiz bilan baham: 
|