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, p, q > 1, 1 p + 1 q = 1 d5 n k =1 |a k + b k | p 1 p ≤ n k =1 |a k | p 1 p + n k =1 |b k | p 1 p , p ≥ 1 \5 ## ## n k =1 a k · b k ## ## ≤ . n k =1 a 2 k · . n k =1 b 2 k R n *5 ## ## n k =1 a k · b k ## ## ≤ n k =1 a 2 k · n k =1 b 2 k '5 n k =1 |a k + b k | p 1 p ≤ n k =1 |a k | p 1 p + n k =1 |b k | p 1 p , p ≥ 1 d5 ## ## n k =1 a k · b k ## ## 2 ≤ n k =1 a 2 k · n k =1 b 2 k \5 n k =1 |a k · b k | ≤ n k =1 |a k | p 1 p · n k =1 |b k | q 1 q , p, q > 1, 1 p + 1 q = 1 R n Y *5 ## ## n k =1 a k · b k ## ## 2 ≤ n k =1 a 2 k · n k =1 b 2 k '5 ## ## n k =1 a k · b k ## ## ≤ n k =1 a 2 k · n k =1 b 2 k d5 n k =1 |a k · b k | ≤ n k =1 |a k | p 1 p · n k =1 |b k | q 1 q , p, q > 1, 1 p + 1 q = 1 \5 n k =1 |a k + b k | p 1 p ≤ n k =1 |a k | p 1 p + n k =1 |b k | p 1 p , p ≥ 1 _ & & & c *5 R n '5 C [a, b] d5 2 \5 m $ R n & *5 1 & ! ! '5 1 & ! & d5 1 & ! \5 1 & ! % R n & *5 1 & ! ! '5 1 & ! & d5 1 & ! \5 1 & ! & 1 *5 C [a, b] '5 C 1 [a, b] d5 C 2 [a, b] \5 C 3 [a, b] ' 2 & + & *5 R n , C [a, b], 2 , m '5 R n 1 , C [a, b], 2 , c 0 d5 R n ∞ , C [a, b], 2 , m \5 C n , R n p , C 2 [a, b], m . 2 & *5 R n '5 C [a, b] d5 m \5 2 / ' & & c *5 R n '5 L 2 [0, 1] d5 2 \5 m # - 5 * * 0 9 "0 ,+ * x = (1, 1), y = (2, 2) ρ 2 (x, y) = |1 − 1| + |2 − 2| = 0 x = y. \ ρ 2 ! #" + x = (2, 3) , y = (3, 2) " ! ρ 3 (x, y) = |2 − 2| + |3 − 3| = 0 ρ 4 (x, y) = |2 · 3 − 3 · 2| = 0 + \ ρ 3 ρ 4 ! #" + $ % ρ 2 ρ 1 , ρ 3 , ρ 4 % ρ : X ×X → R ! $A % /A " `" + &A .A A " #" + '" %" + 2. $ 6. % 5. & %> ' 0, 5. . √ 2. $/ 2√π. $ % $ # $ 3. $ √ 3. $$ 2 √ 2. $% √ π. $& b $' 2π. 9 ! 2- * # * ,- "- 0 "0 "0 $ 5 5 _ !5 5 % C (1) [0, 1] C 1 [0, 1] & y n (t) = t n − t 2n . ' k ∈ N ! [0, 1] f (k) 1 , f (k) 2 , . . . , f (k) k - f (k) i (0) = 1 f (k) i (x) = ⎧ ⎪ ⎨ ⎪ ⎩ 1, agar i − 1 k < x ≤ i k , 0, agar x ∈ (0, 1]\ i − 1 k ; i k ' { g n } " { g n } " C 1 [0, 1] / x n (t) = t n − t n +1 . x n , y n , z n , e n " ! ! u n " c 0 , c m ! α p > 1 p ! 1 ! % 9 # 2 * ) 2 & 5 5 !5 & #`# ! 5 5 !5 "$"# 5 5 !5 ! #`%"! 5 5 !5 "$&# 5 5 !5 #``"! 5 5 !5 "$$# 5 B (0, 1) = {0} B [0, 1] " & 0x 1 0x 2 " S [0, 1] (0, 1) ! & ' 5 B (−∞, r) = B (∞, r) = ∅ 5 B (−∞, r) = (−∞, −(1 − r)/r) , B (∞, r) = = ((1 − r)/r, ∞.) . 5 \ ! ! diam B (x 0 , 1) = 0 < 2 · 1 !5 \ 0 = diam B(x 0 , 1) < diam B[x 0 , 1] = 1. \ & / 1 4 1 4 = ∞ i =1 2 3 2i ! ' # & \ 0, 25 ∈ K 9 "0 ,+ . _ / C [a, b]. 5 f " ( 5 f " ( lim x →±∞ f (x) = ±∞ 1 X = , − π 2 , π 2 - & (Φ, ρ 1 ) ∼ = 1 (Φ, ρ 2 ) ∼ = m. $ 5 (P, ρ 1 ) = C (2) [0, 1], 5 P, ρ 2 ) = C (1) [0, 1], !5 (P, ρ 3 ) = {x ∈ C[−1, 1] e> + }. ' R & (5, 6) ! - {x k } = (5, 6) ∩ Q. < ! & - A 1 = ∞ k =1 x k − 1 10 · 2 k , x k + 1 10 · 2 k . * & A = ((0, 1) ∩ Q) ∪ (2, 3) ∪ {4} ∪ A 1 , 0 A = (2, 3) ∪ A 1 , 0 A ⊃ [2, 3] ∪ [5, 6], 0 A = (0, 1) ∪ (2, 3) ∪ (5, 6) 0 A = [0, 1] ∪ [2, 3] ∪ [5, 6], 0 0 A ⊃ (2, 3) ∪ [5, 6]. & X = R A = Q [0, 1] Z \ X 1 ( - M = ∞ ∪ n =1 N 2n , N = ∞ ∪ n =1 N 2n−1 . ' N 2n N 2n−1 X ! ! & , M ∪ N = ∞ ∪ n =1 N n \ M ∪ N #" & % F r [a, b] = F r(a, b) = {a, b} F r Z = Z F r Q = R F r [a, ∞) = {a} F r ∅ = ∅ %% {x n } " [a, b] ! f ∈ C[a, b] b $ a f (x)dx = lim n →∞ b − a n n k =1 f (x k ) (α, β) ⊂ [a, b] ! χ (α, β) (x) " ! 1 b − a b $ a χ (α, β) (x)dx = lim n →∞ 1 n n k =1 χ (α, β) (x n ) = lim n →∞ n (α, β) n = β − α b − a 0, 1, 1 2 , 1 3 , 2 3 , 1 4 , 2 4 , 3 4 , . . . , 1 k , 2 k , . . . , k − 1 k , . . . " " {x n } , n = k(k + 1) : 2 S k = 1 k k −1 =1 f k S k = 1 k k =1 f k ! lim k →∞ |S k − S k | = lim k →∞ ## ## f (1) k ## ## = 0 2 ! lim k →∞ S k = $ 1 0 f (x)dx [ n = k (k − 1) 2 ! 0 1 1 n n =1 f (x ) = 1 n S 1 + 2 n S 2 + · · · + k n S k 1 & lim n →∞ ( 1 n S 1 + 2 n S 2 + · · · + k n S k ) = lim k →∞ S k = $ 1 0 f (x)dx [ k (k − 1) 2 < n < k (k − 1) 2 , 1 n n =1 f (x ) = 1 n k(k−1) 2 =1 f (x ) + 1 n k(k+1) 2 = k(k−1) 2 +1 f (x ) = 1 n k(k+1) 2 = k(k−1) 2 +1 f (x ) ≤ M k k (k−1) 2 = 2M k − 1 −→ 0, k → ∞. ' Download 1.57 Mb. Do'stlaringiz bilan baham: 
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