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funksional analiz misol va masalalar yechish 1 qism
x) ≥ 0
+ A f (x)dμ ≥ 0 / + & *5 * μ (A) = 0 + A f (x)dμ = 0 '5 + A [f(x) + g(x)]dμ = + A f (x)dμ + + A g (x)dμ d5 + A k · f(x)dμ = k + A f (x)dμ \5 * f (x) ≥ 0 + A f (x)dμ ≥ 0 & *5 * μ (A) = 0 + A f (x)dμ = 0 '5 + A [f(x) + g(x)]dμ = + A f (x)dμ + + A g (x)dμ d5 + A k · f(x)dμ = k + A f (x)dμ \5 * f (x) ≥ g(x) + A f (x)dμ ≥ + A g (x)dμ / ! & a 0 = 0, a 1 = + (0, 1) sin 2 x x dμ, a 2 = + (0, 1) sin 4 x x 3 dμ, a 3 = + (0, 1) sin xdμ *5 a 1 '5 a 2 d5 a 3 \5 a 0 A = [0, 3] & f (x) = 3 + D(x) *5 3 '5 2 d5 7 \5 9 A = [0, 2] & f (x) = 4 − [x] *5 3 '5 7 d5 6 \5 4 $ [−2, 2] & *5 f (x) = cos x '5 f (x) = [x] d5 f (x) = e x 2 \5 f (x) = sin x % f (x) = [x], x ∈ [−2, 2] & " & *5 {−1; 0; 1} '5 {−1; 0; 1; 2} d5 {−1; 1} \5 {−2; −1; 0; 1} & [0, 3] & f 1 (x) = x, f 2 (x) = 1, f 3 (x) = [x] *5 f 1 , f 2 , f 3 '5 f 2 , f 3 d5 f 1 , f 2 \5 f 1 , f 3 ' [0, 1] ! f 1 (x) = − cos x, f 2 (x) = sin x, f 3 (x) = 1 + [x] *5 f 1 , f 2 , f 3 '5 f 2 , f 3 d5 f 1 , f 2 \5 f 1 , f 3 . [−1, 1] f 1 (x) = cos x, f 2 (x) = (1 − x) 2 , f 3 (x) = 1 − [x] *5 f 1 , f 2 , f 3 '5 f 2 , f 3 d5 f 1 , f 2 \5 f 1 , f 3 / [0, 1] ! f 1 (x) = sin x, f 2 (x) = [x], f 3 (x) = cos x *5 f 1 , f 2 , f 3 '5 f 2 , f 3 d5 f 1 , f 2 \5 f 3 f (x) = signx x 0 = 0 & *5 % '5 > d5 # \5 ` f (x) = [x] x 0 = 0 ! & & *5 % '5 > d5 −1 \5 # f (x) = 2x − {x} *5 x + [x] '5 x − [−x] d5 2x + [x] \5 {x} + [x] f (x) = [2x] [−2, 3] & *5 ` '5 a d5 #> \5 7 $ f (x) = sin x [−π, π] & *5 % '5 b d5 # \5 a % [−2, 2] & *5 {x} '5 [x] d5 e x 2 \5 signx & [−2, 2] & *5 x '5 [x] d5 e x 2 \5 sin πx ' [0, 1] & *5 cos x '5 R(x) d5 D(x) \5 K(x) . / "2 f (x) = x " A = [0, 3] & ! + A f (x) dF (x), F (x) = [x] *5 % '5 ` d5 b \5 7 / / "2 f (x) = 2x " A = [1, 3] & ! + A f (x) dF (x), F (x) = ln x *5 % '5 ` d5 b \5 7 # - 5 * * 0 % § # * ,-0) . ! ! ! 1 f g (f + g) (x) = 1, x ∈ [0, 1] ! . ! & ! ! 1 f g & f (x) · g(x) = 0, x ∈ [0, 1] ! f (x) = ⎧ ⎨ ⎩ 1, x ∈ A −1, x ∈ [0, 1] \A, A ⊂ [0, 1] ! & $ #b & 7#W f L(x) ' a ∈ R {x ∈ R : L(x) = a} & & & L(x 1 ) = L(x 2 ) * x 1 , x 2 ∈ A L(x 1 ) = L(x 2 ) x 1 = x 2 ! " x 1 , x 2 ∈ A L(x 1 ) = L(x 2 ) −x 1 = −x 2 x 1 = x 2 ! * x 1 ∈ A x 2 ∈ A L(x 1 ) = L(x 2 ) x 1 = −x 2 ! 2 {x ∈ R : L(x) = a} & a ∈ A \ {0} ! & a ! & 2 ! {x ∈ R : L(x) = a} ! & . {x ∈ [0, 1] : L(x) < 0} = [0, 1] \ A A ! & ! [0, 1] \ A ! / '! x ∈ [−1, 0] L(x) ≥ 0 ! {x ∈ [−1, 1] : L(x) < 0} = [0, 1] \ A 3#$" 5 [0, 1] \ A ! ! L E ! K −1 (K 1 ) = 1 2 , K −1 (K 2 ) = 1 4 , 3 4 , K −1 (K 3 ) = 1 8 , 3 8 , 5 8 , 7 8 , K −1 (K n ) = 2k − 1 2 k , k = 1, 2, . . . , 2 n −1 . A ⊂ [−2, 2] ! & χ A (x) = ⎧ ⎨ ⎩ 1, x ∈ A 0, x ∈ [−2, 2] \A. $ A ⊂ [−2, 2] ! & f (x) = ⎧ ⎨ ⎩ 1, x ∈ A −1, x ∈ [−2, 2] \A. & A ⊂ [0, 1] = E ! & f (x) = χ A (x), g(x) = χ E \A (x) ! f (x) + g(x) = 1. ' A ⊂ [0, 1] = E ! & f (x) = χ A (x), g(x) = −χ E \A (x) f (x) · g(x) ≡ 0, x ∈ E, . E (D < c) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ ∅, c ≤ 0 E \Q, 0 < c ≤ 1 E, c > 1. % * f : E → R ! {x ∈ E : f(x) < a} = {x ∈ E : f 3 (x) < a 3 } f 3 ! ! d a −∞ +∞ ! a 3 (−∞, ∞) & {x ∈ E : f 3 (x) < c} = {x ∈ E : f(x) < c 1 3 } \ f f 3 ! ! [ e ix = cos x + i sin x cos x sin x [−π, π] ! f (x) = e ix [−π, π] & ! . [0, 1] = A ∪ B A B ! & μ (A) > 0, 5; μ (B) = 0, 5. A B '" ! x ∈ [0, 1] δ > 0 ! μ (A ∩ (x − δ, x + δ)) > 0 μ (B ∩ (x − δ, x + δ)) > 0 , χ A (x) " ! $' [0, 1] n n ∈ N ! {χ [ k−1 n , k n ] (x)} n k =1 ' ! f 1 (x) = χ [0, 1] (x), f 2 (x) = χ [0, 1 2 ] (x), f 3 (x) = χ [ 1 2 , 1] (x) ν ! f ν (x) = χ [0, 1 n ] (x), f ν +1 (x) = χ [ 1 n , 2 n ] (x), . . . , f ν +n−1 (x) = χ [ n−1 n , 1] (x), . . . f n " [0, 1] ! ! " %/ E (f < c) = {x ∈ [−1, 2] : sign x < c} & ! c ∈ R ! % E δ = ε 10 3 , π − ε 10 3 π + ε 10 3 , 2π − ε 10 3 , ε ∈ 0, 1 4 . %$ ϕ (x) = 0, x ∈ [0, 1] . %% f (x) = 0, x ∈ [0, 1) , f n (x) = x n , x ∈ [0, 1) . %& f n (x) = x n , x ∈ [0, 1] " θ (x) ≡ 0 " ! ! %' %. _ &/ 5 g (x) ≡ 0, b) g(x) = π, c) g(x) = 0, d) g(x) = ln(1 + |x|). & 5 g (x, y) = x 2 , b ) g(x, y) = cos x − sin y, c) g(x, y) = xy, d ) g(x, y) = chx. & 5 g (x) ≡ 0, 5 g (x) ≡ 0, !5 g (x) ≡ 0, 5 g (x) ≡ 1, 5 g (x) ≡ 0, 5 g (x) ≡ 0. &$ 5 g 1 (x, y) ≡ 0, g 2 (x, y) = ⎧ ⎨ ⎩ 1, (x, y) ∈ Q × Q 0, (x, y) / ∈ Q × Q . 5 g 1 (x, y) ≡ 0, g 2 (x, y) = ⎧ ⎨ ⎩ 1, (x, y) ∈ Q × Q 0, (x, y) / ∈ Q × Q . !5 g 1 (x, y) ≡ 0, g 2 (x, y) = ⎧ ⎨ ⎩ 1, (x, y) ∈ R × Q 0, (x, y) / ∈ R × Q . 5 g 1 (x, y) ≡ 1, g 2 (x, y) = ⎧ ⎨ ⎩ 1, (x, y) / ∈ Q × Q 2, (x, y) ∈ Q × Q . 5 g 1 (x, y) = ⎧ ⎨ ⎩ |x|, |x| ≥ |y| |y|, |x| < |y| , g 2 (x, y) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ |x|, |x| > |y| |y|, |x| < |y| 0, |x| = |y| . 5 g 1 (x, y) = | x | + | y | , g 2 (x, y) = ⎧ ⎨ ⎩ 0, x = y |x| + |y|, x = y . 5 g 1 (x, y) = e x , g 2 (x, y) = ⎧ ⎨ ⎩ e x , x Download 1.57 Mb. Do'stlaringiz bilan baham: |
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