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funksional analiz misol va masalalar yechish 1 qism
A → R
" ! x ∈ A |f(x)| ≤ ϕ(x) f Z ( ! ! "# < - ϕ (x) ≡ 2 f (x) = ⎧ ⎨ ⎩ −1, agar x ∈ Q 1, agar x ∈ R\Q. (7.12) '! x ∈ [0, 2] ! |f(x)| ≤ ϕ(x) ϕ : [0, 2] → R [0, 2] Z ( " ! / f [0, 2] Z ( " ! ' D Z ( ! " &/ 7 ( A ! & ; ϕ c > 0 μ {x ∈ A : ϕ(x) ≥ c} ≤ 1 c $ A ϕ (x)dμ (7.13) 38#`5 7 ! "# * A c = { x ∈ A : ϕ(x) ≥ c} , $ A ϕ (x) dμ = $ A c ϕ (x) dμ + $ A \A c ϕ (x) dμ ≥ $ A c ϕ (x) dμ ≥ c · μ (A c ) . ' 38#`5 ! & f (x) = 3x 2 + 2, x ∈ [0, 1] ( X Z / " 8b" ! "# ' f (x) = 3x 2 + 2, x ∈ [0, 1] ' ! [0, 1] ! ! ! f ! ! {f n } " n ∈ N f n lim n →∞ + A f n (x) dμ 2 ! [0, 1] 0 < 1 n < 2 n < · · · < n − 1 n < n n = 1 n A k = k − 1 n , k n , k = 1, 2, . . . , n − 1, A n = n − 1 n , 1 1 & + "+ n k =1 A k = [0, 1] f n [0, 1] " ! - f n (x) = f k n = 3 k 2 n 2 + 2, x ∈ A k , k = 1, 2, . . . , n. 1 " [0, 1] f max 0≤ x ≤1 |f n (x) − f(x)| = max 1≤ k ≤n max x ∈A k |f n (x) − f(x)| = = max 1≤ k ≤n max x ∈A k |f(k/n) − f(x)| = max 1≤ k ≤n 3(2k − 1) n 2 = 3(2n − 1) n 2 . \ " [0, 1] f [ f n [0, 1] & ! / $ [0,1] f n (x) dμ = n k =1 f k n μ (A k ) = n k =1 3k 2 n 2 + 2 1 n = 3 n 3 n k =1 k 2 + + 2 n n k =1 1 = 3 n 3 n (n + 1)(2n + 1) 6 + 2 = (n + 1)(2n + 1) 2 n 2 + 2. (7.14) _ ! n ∈ N ! 1 2 + 2 2 + 3 2 + · · · + n 2 = n (n + 1)(2n + 1) 6 38#b5 n → ∞ lim n →∞ $ [0,1] f n (x) dμ = lim n →∞ (n + 1)(2n + 1) 2 n 2 + 2 = 1 + 2 = 3 . + Z / 8b" $ 1 0 (3x 2 + 2) dx = (x 3 + 2x) ## 1 0 = 1 + 2 − 0 = 3. \ ( () * +, * *+- + # - & * f : A → R ! g (x) = [f(x)] A ' [a] a & . ! A ⊂ E & y = χ A (x) E ( 8#" A n & " ! & y = sign x E = [−1, 3] ( " 8#" &$ * f 1 : A → R f 2 : E\A → R f (x) = ⎧ ⎨ ⎩ f 1 (x), x ∈ A f 2 (x), x ∈ E\A E &% = & K [0, 1]\K ' K − = & && K : K → R = ' K − = & &' < [0, 1] & ! / a ) f(x) = ⎧ ⎨ ⎩ 0, agar x ∈ K n, agar x ∈ K n , b ) g(x) = ⎧ ⎨ ⎩ 1, x ∈ K 2 −n , x ∈ K n . &. 2 & " &/ 2 & * f g α f + β g & * f : A → R g : A → R f · g & f : A → R ! ! ! " + " & = & K [0, 1] ! ! ! " &$ * f g A & ! α f + β g A & ! $ A (α f(x) + β g(x) ) dμ = α $ A f (x)dμ + β $ A g (x)dμ &% f (x) = [x], x ∈ [0, 5) = A A & ! && ! A ⊂ E ! + E χ A (x) dμ = μ(A) &' A = {x ∈ [−π, π] : sin x < 0, 5} ! + [−π, π] χ A (x) dμ " &. \ ( , A = [0, 3] & ! &/ Z A = [0, 1] & ! 8`#"8`8" f : A → R & f (x) = [2x], A = [0, 2) & f (x) = sign x, A = [−1, 3] & f (x) = χ [0,1]\Q (x), A = [−1, 3] & f (x) = [x] + sign x, A = [−1, 2] &$ f (x) = sign x + χ [1,2] (x), A = [−1, 4] &% f (x) = n, x ∈ A n = 1 3 n , 1 3 n −1 , n ∈ N, A = (0, 1]. && f (x) = 1 n , x ∈ A n = 1 (n + 1)! , 1 n ! , n ∈ N, A = (0, 1]. &' f ! A & ! " ! 38b5 + &. ' f ! 38b5 ! {f n } " & / / ' " ! $ A k · f(x) dμ = k $ A f (x) dμ, k ∈ R. & / ' ! - $ A (f(x) + g(x)) dμ = $ A f (x) dμ + $ A g (x) dμ. 8b> 8b#" 36#9 5 / & / A & ! ! f ! & / 3 5 A & " ; f (x) ≥ 0 ; & / * μ (A) = 0 f : A → R & $ * ! x ∈ A ! f (x) = g(x) $ A f (x)dμ = $ A g (x)dμ ( ' / 8b7 8b8" / " 36#9 5 & % * ϕ A & ! ! x ∈ A ! |f(x)| ≤ ϕ (x) f ! A & ! & & * f ! f |f| ! ! & ' * n ∈ N ! 38##5 ! f but n ! $ A f (x) dμ = lim n →∞ $ A f but n (x) dμ. & . / Z ! _ ( f : A → R g : A → R Z ( ! ! ! A = [0, 2] \ D(x) θ (x) = 0 &$/ / Z ! _ ( Z ( ! f : A → R " |f| Z ( " ! 38#%5 Download 1.57 Mb. Do'stlaringiz bilan baham: |
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