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funksional analiz misol va masalalar yechish 1 qism
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f (x) dF (x) = L ∗ (f) = L ∗ (f). [ g : A → R ! , F : A → R Φ : A → R ( g (x) = F (x) − Φ(x) /&7, f : A → R g : A → R 1 ( " $ A f (x)dg(x) := $ A f (x)dF (x) − $ A f (x)dΦ(x). / f (x) = x 2 + 3, [0, 1] ( ! "# [0, 1] δ > 0 {(a k , b k )} n k =1 3#>#5 - n k =1 |f(b k ) − f(a k )| = n k =1 |b 2 k + 3 − a 2 k − 3| = n k =1 (b k − a k )(b k + a k ) < 2δ. ' b k + a k ≤ 2 ! b k , a k ∈ [0, 1]. [ ε > 0 ! δ = ε/2 , 3#>#5 \ f / f (x) = 2x 2 +5, x ∈ [−1, 3] ! "# f (x) = v(x) − ϕ(x), v(x) = ∨ x a [f], ϕ(x) = v(x) − f(x) ' f (x) = 2x 2 + 5, x ∈ [−1, 3] ( " ! . " ! \ v (x) = V x −1 [f] ϕ (x) = v(x) − f(x) & f v ϕ " ! f (x) = v(x)− ϕ(x) ' [−1, 0] ! [0, 3] ! ! $`a $`b" - v (x) = ⎧ ⎨ ⎩ 2 − 2x 2 , x ∈ [−1, 0] 2 + 2x 2 , x ∈ (0, 3], ϕ (x) = ⎧ ⎨ ⎩ −3 − 4x 2 , x ∈ [−1, 0] −3, x ∈ (0, 3]. / = & K 3a$a" 5 [0, 1] " ! "# = & K / ! / ! (; δ > 0 ! {(a k , b k )} n k =1 + + - K ⊂ n k =1 (a k , b k ), n k =1 (b k − a k ) < δ . (10.3) ! μ K ([0, 1]\K) = 0 μ K ([0, 1]) = K(1) − K(0) = 1. ' μ K = & / " 2 ! ' ! μ K (K) = 1. [ ! 3#>`5 - n k =1 (K(b k ) − K(a k )) = μ K n k =1 (a k , b k ) ≥ μ K (K) = 1. \ = & − K (" ; ( / + [0,∞) 2 −x dF (x) / "2 ' A = [0, ∞)− F (x) = [x] x ! "# ( F (x) = [x] " A = [0, ∞) ! 2 ! #>b`" 33#>a5 5 $ [0,∞) 2 −x dF (x) = ∞ n =1 2 −n (F (n) − F (n − 0)) * ! n ∈ N ! F (n) − F (n − 0) = 1 ( ' ! b 1 = 1/2 + q = 1 2 ! ! & 2 ! $ [0,∞) 1 2 x dF (x) = ∞ n =1 1 2 n = 1. /$ < / "2 - $ [0,3] (x + 1)dF (x). ' A = [0, 3] F (x) = x 2 + 3. ! "# ( F (x) = x 2 + 3 ! 3#>bb" 5 3#>75 - $ [0,3] (x + 1) dF (x) = $ [0,3] (x + 1) · 2x dx 2 %8 2 $ [0,3] (x + 1) dF (x) = 27. /% + [0, 1] x d K(x) / "2 ! "# ' $ [0, 1] xd K(x) = xK(x)| 1 0 − $ [0, 1] K(x) dx = 1 − 0 − 1 2 = 1 2 . ' + [0, 1] K(x) dx = 0, 5 ( 878 5" + 2 ( / 2 ! (; $ [0, 1] xd K(x) = $ [0, 1/3] xd K(x) + $ [1/3, 2/3] xd K(x) + $ [2/3, 1] xd K(x) (10.4) * K(x) = 1/2, x ∈ [1/3, 2/3] [1/3, 2/3] 3#>b5 [1/3, 2/3] &" ! 3#>ba" 5 " $ [0, 1] xd K(x) = $ [0, 1/3] xd K(x) + $ [2/3, 1] xd K(x). '! 3x = t ! x = 2 3 + t 3 + $ [0, 1] xd K(x) = 1 3 $ [0, 1] td K t 3 + 1 3 $ [0, 1] (2 + t)dK 2 3 + t 3 [ = #>`b" $ [0, 1] xd K(x) = 1 6 $ [0, 1] td K(t) + 1 6 $ [0, 1] (2 + t)dK(t) = 1 3 $ [0, 1] td K(t) + 1 3 ' $ [0, 1] xd K(x) = 0, 5 /& f (x) = K (x) F (x) = K (x) ! / "2 $ [a, b] f (x) dF (x) ! "# #>b8" 33#>85 5 - $ [0, 1] K(x) dK(x) = K(1)·K(1)−K(0)·K(0)− $ [0, 1] K(x) dK(x) = 1− $ [0, 1] K(x) dK(x). ' + [0, 1] K(x) dK(x) = 0, 5 () * +, * *+- + # - #>W"#>#7" ( /' f (x) = 3x + 1, [0, 2]. /. f (x) = 2x 2 + 5, [−1, 3]. // f (x) = sin x, [0, π]. / f (x) = 2| cos x|, [−π, π]. / f (x) = tg x 4 , [−π, π]. / f (x) = x ln(1 + x), [0, e]. / f (x) = 2 x + 5x, [−1, 3]. /$ f (x) = " |x|, [−1, 1]. /% f (x) = 3|x − 1| + 4, [0, 2]. /& * /' * & /. * f g [a, b] f ·g f : g (g(x) = 0, ∀x ∈ [a, b]) [a, b] / / * f [a, b] [a, b] ! / " / #>W"#>#7" 3 v (x) = ∨ x a [f], ϕ (x) = v(x) − f(x) 5 " / * f : [a, b] → R ! F (x) = $ [a,x] f (t) dμ [a, b] / [a, b] F F (x) = f (x) ! ! x ∈ [a, b] ! F (x) − F (a) = $ [a,x] f (t) dμ / $ * f − f (x) = 0 ! x ! f / % ! f ! f (x) = f d (x) + f s (x) + f ac (x). ' f d f s f ac / & * f [a, b] /& f [a, b] / ' * f [a, b] ϕ : R → R /& ϕ (f(x)), x ∈ [a, b] [a, b] / . * f [a, b] A ⊂ [a, b] " ! & μ (f(A)) = 0 / / 2 f : [0, 1] → [0, 1] A ⊂ [0, 1] & + - #5 μ (A) = 1, %5 μ (f(A)) = 0. / 2 f : [0, 1] → [0, 1] A ⊂ [0, 1] & + - #5 μ (A) = 0, %5 μ (f(A)) = 1. / [a, b] f [a, b] / * f : [a, b] → R $ b a f (t) dt = f(b) − f(a) f / f (x) = x + K(x), x ∈ [0 Download 1.57 Mb. Do'stlaringiz bilan baham: 
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