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funksional analiz misol va masalalar yechish 1 qism
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−∞, ∞)
! ! + . x ∈ [0, 1] ! - x = ∞ i =1 ε i 3 i ε i − 0, 1, 2 , x = 0, ε 1 ε 2 . . . * x ∈ [0, 1] ! n + ε n = 0 ε n +1 = ε n +2 = ... = 0 x ! ! ' x ! - x = 0, ε 1 . . . ε n 0 . . . = = 0, ε 1 . . . ε n −1 (ε n − 1)22 . . . > #>> . . . e> >%%% . . . . F 1 # ! x ∈ [0, 1] " & > #e> >%%% . . . 1 4 e> >%>%>% . . . F 1 & & & " F 1 = K & ) / 0, 25 ∈ K x ∈ K ! y ∈ K + ρ (x, y) = | x − y | 9 "0 ,+ * X " ! X X A B & * B ⊂ [A] A & B & [A] = X A & $ X & * A & ! 3 ( B ⊂ X A & B 5 A (X, ρ) x M & ρ (x, M) = inf y ∈M ρ (x, y) (X, ρ) A B & ρ (A, B) = inf x ∈A,y∈B ρ (x, y) " "" X * R 3 5 ! + 7, ! " #5 R R ∗ ' %5 R R ∗ [R] = R ∗ R ∗ R "" ; R R R ( " ! \ R C 1 [−1, 1] C 1 [−1, 1] ' ! C 1 [−1, 1] f n (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ −1, x ∈ [−1, −n −1 ] n x, x ∈ (−n −1 , n −1 ) 1, x ∈ [n −1 , 1] " 3 ; #b#"! 5 ' " C 1 [−1, 1] ! ! x ∈ [−1, 1] ! |f n (x) − f m (x)| ≤ 1 n < m "%"# ρ (f n , f m ) = $ 1 −1 |f n (x) − f m (x) |dx < $ 1/n −1/n 1dx = 2 n → 0, n → ∞. ' {f n } " C 1 [−1, 1] " f ∈ C 1 [−1, 1] ϕ (x) = ⎧ ⎨ ⎩ −1, agar x ∈ [−1, 0), 1, agar x ∈ [0, 1] = f n (x) − ϕ (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ 0, x ∈ [−1, −1/n] [1/n, 1] , n x + 1, x ∈ (−1/n, 0) , nx − 1, x ∈ [0, 1/n). ' ! x ∈ [−1, 1] ! |f n (x) − ϕ (x) | ≤ 1. 2 ! 1 $ −1 |f n (x) − ϕ(x)|dx = 1/n $ −1/n |f n (x) − ϕ(x)|dx ≤ 2 n → 0, n → ∞. (14.1) * 1 $ −1 |f(x) − ϕ(x)| dx ≤ 1 $ −1 |f(x) − f n (x)| dx + 1 $ −1 |f n (x) − ϕ(x)| dx. (14.2) [ $ 1 −1 |f (x) − ϕ (x)| dx > 0 (14.3) , + #5 f (0) ≤ 0 f δ 1 > 0 + ! x ∈ [0, δ 1 ] ! f (x) < 1/2 ' |f (x) − ϕ (x) | ≥ 1/2, x ∈ [0, δ 1 ] (14.4) ! 3#bb5 [0, δ 1 ] ! $ 1 −1 |f (x) − ϕ (x)| dx ≥ $ δ 1 0 |f (x) − ϕ (x)| dx > δ 1 2 %5 * f (0) > 0 δ 2 > 0 + ! x ∈ [−δ 2 , 0) ! |f (x) − ϕ (x) | > 1/2 ' $ 1 −1 |f (x) − ϕ (x)| dx ≥ $ 0 −δ 2 |f (x) − ϕ (x)| dx > δ 2 2 . \ 3#b`5 3#b%5 1 $ −1 |f(x) − f n (x)| dx ≥ 1 $ −1 |f(x) − ϕ(x)| dx − 1 $ −1 |f n (x) − ϕ(x)| dx (14.5) 3#b#5 3#b`5 3#ba5 ρ (f, f n ) = $ 1 −1 |f (x) − f n (x)| dx ! ( { f n } " C 1 [−1, 1] 8" " "0 ,+ ! & M & ! ! & M & 0 M A & F rA = A ∩ (X\A) "" $+ & [ & #" %" & ! " M & ! & M " * M = M M * X M & ! " ! & " M & #" * M & ! ! & M %" " Q, R\Q & R ! " x ∈ R , x n = [nx] : n " x ' [x] x \ & Q " & R ! [ & R\Q & R ! x ∈ R ! y n = [nx] n + π n " x " _ ( R\Q & R ! \ [Q] = [R \Q] = R. R Q & #" R\Q & %" & ( Q " & ! {x 1 , x 2 , x 3 , . . . , x n , . . . } ! 2 Q = ∞ ∪ n =1 M n , M n = {x n } M n , n = 1, 2, . . . R ! ! 1 ( Q #" & [ & R\Q %" & 1 R\Q #" & , #bbb" R = Q ∪ (R\Q) #" & ' ' \ R\Q %" & $ x {x} n · √ 2 , n ∈ N & [0, 1] ! , " a − {n · a} , n ∈ N & [0, 1] ! ' a {n · a}, n ∈ N & [0, 1] ! [0, 1] N {a}, {2a}, . . . , {Na}, {(N +1)a} k 1 = k 2 ! {k 1 a } = {k 2 a } (k 2 − k 1 )a = [k 2 a ] − [k 1 a ] = m a = m k 2 − k 1 ' a N + 1 {a}, {2a}, . . . , {Na}, {(N + 1)a} N 0 N , 1 N , 1 N , 2 N , . . . , N − 2 N , N − 1 N , N − 1 N , N N & ! {k 1 a }, {k 2 a } ∈ − 1 N , N , k 2 > k 1 |{k 2 a } − {k 1 a }| < 1 N 2 (; {k 2 a } − {k 1 a } = (k 2 a − [k 2 a ]) − (k 1 a − [k 1 a ]) = (k 2 − k 1 )a + [k 1 a ] − [k 2 a ] * x = y + n, n ∈ Z {x} = {y} {x} = 1 − {y} \ |{k 2 a } − {k 1 a }| = {(k 2 − k 1 ) Download 1.57 Mb. Do'stlaringiz bilan baham: 
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