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funksional analiz misol va masalalar yechish 1 qism
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E
(f = a) = ∞ n =1 E a ≤ f < a + 1 n . ' E (a ≤ f < a+1/n) & %5 & ! " ! . ! & ! ! E (f = a) & ! b5 E (f ≤ a) & ! ( `5 E (f ≤ a ) = E(f < a) ∪ E(f = a) ! a5 E (f > a) & ! E (f > a) = E\E(f ≤ a) ! & ! ! %% * f g E ! {x ∈ E : f(x) > g(x)} & ! Z & Q ! " ! ( Q = {r 1 , r 2 , . . . , r n , . . . } - {x ∈ E : f(x) > g(x)} = ∞ ∪ k =1 ({x : f(x) > r k } ∩ {x : g(x) < r k }) (6.2) x 0 ∈ {x ∈ E : f(x) > g(x)} Z ! " r k ∈ Q + f (x 0 ) > r k > g (x 0 ) \ x 0 ∈ { x : f(x) > r k } ∩ { x : g(x) < r k } . ' x 0 ∈ ∞ ∪ k =1 ({ x : f(x) > r k } ∩ { x : g(x) < r k }) ! [ x 0 ∈ ∞ ∪ k =1 ({ x : f(x) > r k } ∩ { x : g(x) < r k }) x 0 & ! " ( r k ∈ Q + f (x 0 ) > r k g (x 0 ) < r k ' f (x 0 ) > g(x 0 ) x 0 ∈ {x ∈ E : f(x) > g(x)} ! 37%5 {x ∈ E : f(x) > g(x)} & ! 37%5 " ! & ! ! # * ,-0) 0 "0 "4--4 ) 2- ' " ! " % 7, E f g μ ({x ∈ E : f(x) = g(x)}) = 0 f g f ∼ g % 7, ! E E [ 7%" ( ! * " % 7, ! E {f n } f $ lim n →∞ f n (x) = f(x) E x & {f n } E f ' E & {f n } ! " f ! %$7, ! δ > 0 lim n →∞ μ ({x ∈ E : |f n (x) − f(x)| ≥ δ}) = 0 {f n } E f %"" $4 & E {f n } f 5 δ > 0 E δ ⊂ E μ (E\E δ ) < δ E δ {f n } f % "" $1 & [a, b] f ε > 0 [a, b] ϕ μ ({x ∈ [a, b] : f(x) = ϕ(x)}) < ε %& \ D 33%#5 5 Z R 33%`5 5 θ (x) ≡ 0 I (x) ≡ 1 + ! "# ( Q & ! μ ( Q) = 0 / ! " ! 3a%>" ( 5 A ⊂ Q ! μ (A) = 0. [ - {x : D(x) = θ(x)} = Q, {x : R(x) = θ(x)} = Q, {x : D(x) = R(x)} ⊂ Q, {x : D(x) = I(x)} = R\Q. ' - μ ({x : D(x) = θ(x)}) = μ ({x : R(x) = θ(x)}) = μ (Q) = 0, μ ({x : D(x) = R(x)}) = 0, μ ({x : D(x) = I(x)}) = μ (R\Q) = 0. \ D ∼ θ, R ∼ θ, D ∼ R D, R θ " I %' * E = A 1 ∪ A 2 A 1 ∩ A 2 = ∅ * f 1 : A 1 → R f 2 : A 2 → R ! f (x) = ⎧ ⎨ ⎩ f 1 (x), agar x ∈ A 1 f 2 (x), agar x ∈ A 2 E & ! c ∈ R { x ∈ E : f(x) < c } = {x ∈ A 1 : f 1 (x) < c} ∪ {x ∈ A 2 : f 2 (x) < c} & " ! {x ∈ A 1 : f 1 (x) < c} {x ∈ A 2 : f 2 (x) < c } & ! f 1 f 2 ! ! {x ∈ E : f(x) < c} ! & ! \ f E ! () * +, * *+- + # - %. * f g E & ! f + g, f − g & f · g ! * E ! x ! g (x) = 0 f : g E ! %/ A ⊂ R & 3%%$" 3%b5 5 y = χ A (x) ! ! A ! % . ! ! c A ⊂ E = [0, 1] ! & ! f (x) = χ A (x) g (x) = χ E \A (x) % . ! & ! c A ⊂ E = [0, 1] ! & f (x) = χ A (x) g (x) = χ E \A (x) % * f E ! g E ! f + g E ! c % . ! ! 7`" 37#5 f %$ . ! ! 37#5 f %% * a, b ∈ R ! 7a" #5 %5 b5 a5 & ! f E & ! %& a ∈ R ! E (f = a) & ! f E & ! ! %' A ⊂ [0, 1] ! & L : R → R ! - L(x) = ⎧ ⎨ ⎩ x, agar x ∈ A −x, agar x /∈ A. (6.3) ' ! a ∈ R {x : L(x) = a} & ! %. 37`5 L ! {x ∈ [0, 1] : L(x) < 0} & ! % / 37`5 L E = [−1, 1] & ! % f : E → R ! ! A ⊂ R ' & ! f −1 (A) ! & % K : R → R f = & K n , n = 1, 2, . . . = & n − ! K ni , ' & K −1 (K 1 ), K −1 (K 2 ), K −1 (K 3 ) K −1 (K n ) & % * f : E → R ! f E ! A ! % [−2, 2] ! % $ [−2, 2] ! ! " % % f : [a, b] → R ! f + (x) = max {f(x), 0} , f − (x) = min {f(x), 0} < 5 * f ! f + f − ! 5 * f + f − ! f ! % & 2 f g " ! ! A ⊂ E = [0, 1] ! " & f (x) = χ A (x) g (x) = χ E \A (x) % ' 2 f g & ! ! % . \ 3%`" 3%#5 5 D [0, 3] = E & ! ( % / * f E ! h (x) = sign f(x) ! % * f E & ! f + (x) = 1 2 (f(x) + |f(x)|) (6.4) ! % * f E & ! f − (x) = 1 2 (|f(x)| − f(x)) (6.5) ! % * f g Download 1.57 Mb. Do'stlaringiz bilan baham: 
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