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HOMEWEB : http://maths- minhthe.violet.vn REAL ANALYSIS I (6650) The Minh Tran Answer : 1 a) Yes Explain We can check the polynomial ( ) 2 n 2 n 1 1 x n n 1/ 1 x p (x) 1 n We need to prove converging poiwise p (x) e on [0, 1] Indeed, we need to check p (x) at 0, 1, and 0 x 1 Pr oof : + + = + → < < 2 2 2 1 1 n 1 0 1 x n n n n n n n 2 1/(1 x ) At x 0 : We see that 1 lim p (0) lim(1 ) e e lim p (x) e at x 0 n 1 So there exist a sequence p (x) 1 of polynomial co nverging (1 x )n point wise to e e at x 0 + + →∞ →∞ →∞ + + = = + = = ⇒ = = = + + = = When 0 x 1 : We have + < < 2 2 2 2 n 2 n 1 (1 x )n 1 x 1 1 x 2 n (1 x )n 2 n n n 2 1/ (1 x ) lim 1 for 0 x 1 n 1 lim 1 e for 0 x 1 (1 x )n 1 Because we see lim 1 e for 0 x 1 (1 x )n 1 So there exist a sequence p (x) 1 of polyn (1 x )n →∞ + + + →∞ + →∞ + + < < = + = < < + + = < < + = + + 2 1/(1 x ) omial converging point wise to e when 0 x 1 + < < 2 2 n 1 1 1 2 n 2n 2 1 1 n n n n 2n n 1 1 x n n n n 2 Finally,we check p (x) At x 1: We check that 1 1 lim p (1) lim(1 ) lim (1 ) e e 2n 2n 1 (Because lim(1 ) e at x 1) 2n lim p (x) e at x 1 1 So there exist a sequence p (x) 1 of (1 x )n + →∞ →∞ →∞ →∞ + →∞ + = = + = + = = + = = ⇒ = = = + + 2 1 1/(1 x ) 2 polynomial converging point wise to e e at x 1 + = = Condition : 2 n n 2 1 1/(1 x ) 2 1 There exist a sequence p (x) 1 of polynomial conv erging (1 x )n point wise to e e on [0,1] + = + + = b) Yes Explain We can check the polynomial ( ) 2 2 n 2 n 1 1 x n 1 1 x n n 1/ 1 x p (x) 1 n We need to prove converging uniformly p (x) e on [0,1] Pr oof : As the proof above, we see that p (x) converging p oinwise to e on [0,1] Now, we need to check the property of uniform of p (x) + + + = + → 2 2 2 2 2 2 1 1 x 1 n (1 x )n 1 x 1 1 x n 2 2 n n n (1 x )n 2 n 1 1 x n 2 Indeed, set p(x) e We compute 1 1 lim p (x) lim 1 lim 1 e for 0 x 1 (1 x )n (1 x )n 1 ( because lim 1 e ) (1 x )n Moreover 1 p (x) e 1 (1 x + + + + →∞ →∞ →∞ + →∞ + = = + = + = ≤ ≤ + + + = + − = + + 2 n 1 1 x n n n n 2 e 0 , for 0 x 1 )n So given 0,there is some N(which may depend on ) such that p (x) p(x) for all 0 x 1 and all n N Condition : 1 Thus, there exist a sequence p (x) 1 of polynomial converging (1 x )n uniformly + →∞ − → ≤ ≤ ε > ε − < ε ≤ ≤ ≥ = + + 2 1/(1 x ) to e on [0,1] + c) Yes 2 2 2 2 1 n (1 x )n 1 x 1 1 x n 2 2 n n n n n 2 For every real number x, we have : 1 x 1 so 1 1 lim p (x) lim 1 lim 1 e for all x (1 x )n (1 x )n 1 Thus, there exist a sequence p (x) 1 of polynomia l converging (1 x )n po + + + →∞ →∞ →∞ + ≥ = + = + = ∈ + + = + + 2 1/(1 x ) int wise to e on + d) Yes 2 2 2 2 2 1 1 x n 1 1 x 1 n (1 x )n 1 x 1 1 x n 2 2 n n n n 2 We see that p (x) converging poinwise to e on Indeed, set p(x) e We compute 1 1 lim p (x) lim 1 lim 1 e for all x (1 x )n (1 x )n Moreover 1 p (x) p(x) 1 (1 x )n + + + + + →∞ →∞ →∞ = = + = + = ∈ + + − = + + 2 2 n 1 1 x n n n 2 1 0 1 x n n e 0 , for all x 1 In particular,for n fixed p (x) 1 1 as x and (1 x )n p(x) e e 1 as x so p (x) p(x) 0 as x So given 0,there is some N(which may depend on ) such that p (x) p(x) for all x and all n N + →∞ + − → ∈ = + → → ∞ + = → = → ∞ − → → ∞ ε > ε − < ε ∈ ≥ 2 n n 2 1/(1 x ) Condition : 1 Thus, there exist a sequence p (x) 1 of polynomial converging (1 x )n uniformly to e on + = + + 2. 2 nx n 1 1 nx n nx n nx n n n n n n ne dx Set f (x) ne and we compute n lim f (x) lim ne lim 0 for all 1 x 2 e (using l'Hospital rule) So f (x) is converges point wise to f 0 on [1,2] We have f (1/ n) n / e as n Thus,f (x) does not converges un ∞ − = − − →∞ →∞ →∞ = = = = ≤ ≤ = = → ∞ → ∞ ∑ ∫ iformly to f 0 on [1,2] = Now, we can compute 2 2 nx nx n 1 n 1 1 1 nx n 2n n 1 n 1 n 2n n 1 n 1 n n 0 ne dx ne dx 2 1 1 e 1 e e 1 1 We will compute and e e 1 We may use the formula q where q 1 1 q ∞ ∞ − − = = ∞ ∞ − = = ∞ ∞ = = ∞ = = = − = − = < − ∑ ∑ ∫ ∫ ∑ ∑ ∑ ∑ ∑ n 0 n n n 0 n 1 n 1 n n n 1 n 0 n n n 1 n 1 2 n 2 2n 2 n 1 n 1 2 q q q 1 q 1 q q q 1 1 1 q 1 q 1 1 1 1 e At q we have q 1 e e e 1 1 e 1 1 1 1 e At q we have q 1 e e e 1 1 e ∞ ∞ ∞ = = = ∞ ∞ = = ∞ ∞ = = ∞ ∞ = = = + = + ⇒ = − = − = − − = = = = − − = = = = − − ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ n 2n 2 2 n 1 2 nx 2 n 1 1 1 1 1 1 e Finally, e e e 1 e 1 e 1 Conlusion : e ne dx e 1 ∞ = ∞ − = − = − = − − − = − ∑ ∑ ∫ 3. ( ) { } n n n 1 n n Let f : M, be continuous for each n and suppose tha t f (x) 0 for all x M where f (x) f (x) for all x M and n . If M is c ompact. Show that f converges uniformly to f(x) 0 on M + ρ → ∈ → ∈ ≤ ∈ ∈ = ( ) n n n n : Suppose S is a set and f :S , n . We say that the sequence f is uniformly convergent with limit f : S if 0, N s uch that, x S n N we have f (x) f(x) ∈ → ∀ ∈ → ∀ε > ∃ ∈ ∀ ∈ ∀ ≥ − < ε Recall Proof : From the problem 3 and recall, we see that ( ) S M, ≡ ρ If M is compact then ( ) M, ρ is compact metric space. We will prove on compact metric space. Indeed, Suppose that n f is continuous, for each n ∈ , n f (x) 0 → this mean that n n lim f (x) f(x) 0 →∞ = = and suppose that where n 1 n f (x) f (x), x M + ≤ ∀ ∈ { } n n n n 1 n n 1 n n 1 n n n n n 1 N Given 0 , set B x M : f (x) f(x) so B is closed set. and f (x) f (x) is decrea sing sequence so B B (because f (x) f(x) f (x) f(x) ) We have : lim f (x) f(x) 0, x M so B . Hence, N such that B , this mean tha + + + ∞ →∞ = ε > = ∈ − ≥ ε ≤ ⊆ − ≤ − = = ∀ ∈ = ∅ ∃ ∈ = ∅ ∩ N t f (x) f(x) , x M (*)− < ε ∀ ∈ n 1 n n N Therewith f (x) f (x) is decrea sing sequence so n N then f (x) f(x) f (x) f(x) , x M (according to (*) ) + ≤ ∀ ≥ − ≤ − < ε ∀ ∈ ( ) { } n n n Thus,the sequence f (x) f(x) converges uniform ly to 0 Conlution : f converges uniformly to f(x) 0 on M − = 4. Show that ( ) ( ) h : M, N, ρ → τ is uniformly continuous on M if and only if ( ) n n n n h(x ),h(y ) 0 whenever (x ,y ) 0 τ → ρ → Recall: ( ) ( ) ( ) ( ) f : M,d N, is uniformly continuous if 0, 0 which may depend on f and such that f(x),f(y) whenever x,y M satisfy d(x,y) → ρ ∀ε > ∃δ > ε ρ < ε ∈ < δ Proof : We are proving the problem 4 on two metric spaces ( ) ( ) M, , N, ρ τ ( ) ( ) ( ) ( ) n n n n n n n Suppose h is uniformly continuous on M Given 0 , 0 ( depend on 3 and f ) such that h(x),h(y) , h(y),h(y ) , h(x ),h(x) whenever x ,y ,x,y M satisfy (x ,x) / 3, (y ,y) / 3, ( x,y) / 3 Because is a metric Hence, h(x ⇒ ∀ε > ∃δ > δ ε τ < ε τ < ε τ < ε ∈ ρ < δ ρ < δ ρ < δ τ τ ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) n n n n n n n n n n n ),h(y ) h(x ),h(x) h(x),h(y) h(y),h(y ) 3 So h(x ),h(y ) 0 whenever x ,y x ,x x,y y,y (because is a metric) / 3 / 3 / 3 ( depend on 3 and f ) So x ,y 0 ≤ τ + τ + τ < ε + ε + ε = ε τ → ρ ≤ ρ + ρ + ρ ρ < δ + δ + δ = δ δ ε ρ → ( ) ( ) ( ) ( ) ( ) ( ) n n n n n n n n n n n n Suppose h(x ),h(y ) 0 whenever x ,y M satisfy x ,y 0 We will prove that : 0, 0, such that h(x ),h(y ) whenever x , y M satisfy x ,y depend on and f ⇐ τ → ∈ ρ → ∀ε > ∃δ > τ < ε ∈ ρ < δ δ ε ( ) ( ) ( ) ( ) ( ) n n n n n n n n n n Indeed, let 0, choose 0 . We have h(x ),h(y ) 0,this mean that h(x ),h(y ) whenever x ,y M satisfy x ,y 0 this mean that x ,y depend on and f Thus, h is uniformly continuous on M ε > δ > τ → τ < ε ∈ ρ → ρ < δ δ ε 5. { } ( ) ( ) n n n n n n n 1 Define M to be the of all sequences (x ) where x 0,1 for all n . Define (x ),(y ) sup x y .Is M, separable ? Explain why or why not ? ≥ ∈ ∈ τ ≡ − τ Answer : ( ) M, is not separable τ Explain : ( ) ( ) ( ) ( ) ( ) ( ) n n n n n n n 1 n n n n n n n 1 n n n n n n n n n 1 n 1 n n n n n n n n n n n n 1 n 1 n 1 First, we can prove M, is metric space We have (x ),(y ) sup x y 0, x ,y (x ),(y ) sup x y 0 x y (x ),(y ) sup x y sup y x (y ),(x ) (x ),(y ) sup x y sup x z z y sup x z z ≥ ≥ ≥ ≥ ≥ ≥ ≥ τ τ = − ≥ ∀ τ = − = ⇔ = τ = − = − = τ τ = − = − + − ≤ − + + ( ) n n n n n n 1 n 1 y sup x z sup z y ≥ ≥ − = − + − ( ) Thus, M, is metric space τ Proof : Indeed, we have infinite subset of M such as define : ( ) n n n n n n n 1 n 1 1 2 n 1 2 n Set Sup x y 0 so that (x ),(y ) Sup x y 0 for all (x ),(x ) ,. . . ,(x ) M ; (y ),(y ) ,. . . ,(y ) M ≥ ≥ δ = − > τ = − = δ > ∈ ∈ We assume that every infinite subsets of M has a limit point n n ' ' ' ' n n ' ' ' ' n n n n so we have an inf inite set of point s in M such that (x ,y ) , Now,we assume that it have inf inite number of li mit point x , y in M such that (x ,x ) / 3, (y ,y ) / 3 (x ,y ) (x ,x ) (x ,y ) (y ,y ) / 3 / 3 / 3 where 0 τ ≥ δ τ < δ τ < δ ⇒ τ ≤ τ + τ + τ < δ + δ + δ = δ δ > This is contradiction that n n (x ,y ) τ ≥ δ so M has no limit points Thus, ( ) M, τ is not separable . We assume that every infinite subsets of M has a limit point n n ' ' ' ' n n ' ' ' ' n n n n so we have an inf inite set of point s in M such that (x ,y. M, ≡ ρ If M is compact then ( ) M, ρ is compact metric space. We will prove on compact metric space. Indeed, Suppose that n f is continuous, for each n ∈ , n f (x) 0 → this mean. ) 2 polynomial converging point wise to e e at x 1 + = = Condition : 2 n n 2 1 1/(1 x ) 2 1 There exist a sequence p (x) 1 of polynomial conv erging (1 x )n point wise to e e on [0,1] +
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Xem thêm: REAL ANALYSIS I
