Sequences of Functions

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1 Sequences of Functions Uniform convergence 9. Assume that f n f uniformly on S and that each f n is bounded on S. Prove that {f n } is uniformly bounded on S. Proof: Since f n f uniformly on S, then given ε =, there exists a positive integer n such that as n n, we have f n (x) f (x) for all x S. (*) Hence, f (x) is bounded on S by the following f (x) f n (x) + M (n ) + for all x S. (**) where f n (x) M (n ) for all x S. Let f (x) M (),..., f n (x) M (n ) for all x S, then by (*) and (**), So, f n (x) + f (x) M (n ) + 2 for all n n. f n (x) M for all x S and for all n where M = max (M (),..., M (n ), M (n ) + 2). Remark: () In the proof, we also shows that the limit function f is bounded on S. (2) There is another proof. We give it as a reference. Proof: Since Since f n f uniformly on S, then given ε =, there exists a positive integer n such that as n n, we have f n (x) f n+k (x) for all x S and k =, 2,... So, for all x S, and k =, 2,... f n +k (x) + f n (x) M (n ) + (*) where f n (x) M (n ) for all x S. Let f (x) M (),..., f n (x) M (n ) for all x S, then by (*), f n (x) M for all x S and for all n

2 where M = max (M (),..., M (n ), M (n ) + ). 9.2 Define two sequences {f n } and {g n } as follows: ( f n (x) = x + ) if x R, n =, 2,..., n g n (x) = { n Let h n (x) = f n (x) g n (x). if x = or if x is irrational, b + n if x is rational, say x = a b, b >. (a) Prove that both {f n } and {g n } converges uniformly on every bounded interval. and Proof: Note that it is clear that lim f n (x) = f (x) = x, for all x R n { if x = or if x is irrational, lim g n (x) = g (x) = n b if x is ratonal, say x = a, b >. b In addition, in order to show that {f n } and {g n } converges uniformly on every bounded interval, it suffices to consider the case of any compact interval [ M, M], M >. Given ε >, there exists a positive integer N such that as n N, we have M n < ε and n < ε. Hence, for this ε, we have as n N f n (x) f (x) = x M < ε for all x [ M, M] n n and g n (x) g (x) n < ε for all x [ M, M]. That is, we have proved that {f n } and {g n } converges uniformly on every bounded interval. Remark: In the proof, we use the easy result directly from definition of uniform convergence as follows. If f n f uniformly on S, then f n f uniformly on T for every subset T of S. 2

3 (b) Prove that h n (x) does not converges uniformly on any bounded interval. Proof: Write { h n (x) = a + a n x n ( + ( n) if x = or x is irrational + + ) b bn if x is rational, say x = a b Then { if x = or x is irrational lim h n (x) = h (x) = n a if x is rational, say x = a. b Hence, if h n (x) converges uniformly on any bounded interval I, then h n (x) converges uniformly on [c, d] I. So, given ε = max ( c, d ) >, there is a positive integer N such that as n N, we have max ( c, d ) > h n (x) h (x) { ( ) x = n + n = x n + ( n if x Q c [c, d] or x = a n + + ) b bn if x Q [c, d], x = a b which implies that (x [c, d] Q c or x = ) max ( c, d ) > x n + n x n max ( c, d ) n which is absurb. So, h n (x) does not converges uniformly on any bounded interval. 9.3 Assume that f n f uniformly on S, g n f uniformly on S. (a) Prove that f n + g n f + g uniformly on S. Proof: Since f n f uniformly on S, and g n f uniformly on S, then given ε >, there is a positive integer N such that as n N, we have. and f n (x) f (x) < ε 2 g n (x) g (x) < ε 2 for all x S for all x S. Hence, for this ε, we have as n N, f n (x) + g n (x) f (x) g (x) f n (x) f (x) + g n (x) g (x) < ε for all x S. 3

4 That is, f n + g n f + g uniformly on S. Remark: There is a similar result. We write it as follows. If f n f uniformly on S, then cf n cf uniformly on S for any real c. Since the proof is easy, we omit the proof. (b) Let h n (x) = f n (x) g n (x), h (x) = f (x) g (x), if x S. Exercise 9.2 shows that the assertion h n h uniformly on S is, in general, incorrect. Prove that it is correct if each f n and each g n is bounded on S. Proof: Since f n f uniformly on S and each f n is bounded on S, then f is bounded on S by Remark () in the Exercise 9.. In addition, since g n g uniformly on S and each g n is bounded on S, then g n is uniformly bounded on S by Exercise 9.. Say f (x) M for all x S, and g n (x) M 2 for all x and all n. Then given ε >, there exists a positive integer N such that as n N, we have and f n (x) f (x) < g n (x) g (x) < which implies that as n N, we have ε 2 (M 2 + ) ε 2 (M + ) h n (x) h (x) = f n (x) g n (x) f (x) g (x) for all x S for all x S = [f n (x) f (x)] [g n (x)] + [f (x)] [g n (x) g (x)] f n (x) f (x) g n (x) + f (x) g n (x) g (x) ε < 2 (M 2 + ) M ε 2 + M 2 (M + ) < ε 2 + ε 2 = ε for all x S. So, h n h uniformly on S. 9.4 Assume that f n f uniformly on S and suppose there is a constant M > such that f n (x) M for all x in S and all n. Let g be continuous on the closure of the disk B (; M) and define h n (x) = g [f n (x)], h (x) = g [f (x)], if x S. Prove that h n h uniformly on S. 4

5 Proof: Since g is continuous on a compact disk B (; M), g is uniformly continuous on B (; M). Given ε >, there exists a δ > such that as x y < δ, where x, y S, we have g (x) g (y) < ε. (*) For this δ >, since f n f uniformly on S, then there exists a positive integer N such that as n N, we have f n (x) f (x) < δ for all x S. (**) Hence, by (*) and (**), we conclude that given ε >, there exists a positive integer N such that as n N, we have Hence, h n h uniformly on S. g (f n (x)) g (f (x)) < ε for all x S. 9.5 (a) Let f n (x) = / (nx + ) if < x <, n =, 2,... Prove that {f n } converges pointwise but not uniformly on (, ). Proof: First, it is clear that lim n f n (x) = for all x (, ). Supppos that {f n } converges uniformly on (, ). Then given ε = /2, there exists a positive integer N such that as n N, we have f n (x) f (x) = + nx < /2 for all x (, ). So, the inequality holds for all x (, ). It leads us to get a contradiction since + Nx < 2 for all x (, ) lim x + + Nx = < /2. That is, {f n } converges NOT uniformly on (, ). (b) Let g n (x) = x/ (nx + ) if < x <, n =, 2,... Prove that g n uniformly on (, ). Proof: First, it is clear that lim n g n (x) = for all x (, ). Given ε >, there exists a positive integer N such that as n N, we have /n < ε 5

6 which implies that g n (x) g = x + nx = + n < n < ε. x So, g n uniformly on (, ). 9.6 Let f n (x) = x n. The sequence {f n (x)} converges pointwise but not uniformly on [, ]. Let g be continuous on [, ] with g () =. Prove that the sequence {g (x) x n } converges uniformly on [, ]. Proof: It is clear that f n (x) = x n converges NOT uniformly on [, ] since each term of {f n (x)} is continuous on [, ] and its limit function { if x [, ) f = if x =. is not a continuous function on [, ] by Theorem 9.2. In order to show {g (x) x n } converges uniformly on [, ], it suffices to shows that {g (x) x n } converges uniformly on [, ). Note that lim g (x) n xn = for all x [, ). We partition the interval [, ) into two subintervals: [, δ] and ( δ, ). As x [, δ] : Let M = max x [,] g (x), then given ε >, there is a positive integer N such that as n N, we have M ( δ) n < ε which implies that for all x [, δ], g (x) x n M x n M ( δ) n < ε. Hence, {g (x) x n } converges uniformly on [, δ]. As x ( δ, ) : Since g is continuous at, given ε >, there exists a δ > such that as x < δ, where x [, ], we have g (x) g () = g (x) = g (x) < ε which implies that for all x ( δ, ), g (x) x n g (x) < ε. 6

7 Hence, {g (x) x n } converges uniformly on ( δ, ). So, from above sayings, we have proved that the sequence of functions {g (x) x n } converges uniformly on [, ]. Remark: It is easy to show the followings by definition. So, we omit the proof. () Suppose that for all x S, the limit function f exists. If f n f uniformly on S ( S), then f n f uniformly on S, where # (S S ) < +. (2) Suppose that f n f uniformly on S and on T. Then f n f uniformly on S T. 9.7 Assume that f n f uniformly on S and each f n is continuous on S. If x S, let {x n } be a sequence of points in S such that x n x. Prove that f n (x n ) f (x). Proof: Since f n f uniformly on S and each f n is continuous on S, by Theorem 9.2, the limit function f is also continuous on S. So, given ε >, there is a δ > such that as y x < δ, where y S, we have f (y) f (x) < ε 2. For this δ >, there exists a positive integer N such that as n N, we have x n x < δ. Hence, as n N, we have f (x n ) f (x) < ε 2. (*) In addition, since f n f uniformly on S, given ε >, there exists a positive integer N N such that as n N, we have f n (x) f (x) < ε 2 for all x S which implies that f n (x n ) f (x n ) < ε 2. (**) 7

8 By (*) and (**), we obtain that given ε >, there exists a positie integer N such that as n N, we have f n (x n ) f (x) = f n (x n ) f (x n ) + f (x n ) f (x) < ε 2 + ε 2 = ε. That is, we have proved that f n (x n ) f (x). 9.8 Let {f n } be a seuqnece of continuous functions defined on a compact set S and assume that {f n } converges pointwise on S to a limit function f. Prove that f n f uniformly on S if, and only if, the following two conditions hold.: (i) The limit function f is continuous on S. (ii) For every ε >, there exists an m > and a δ >, such that n > m and f k (x) f (x) < δ implies f k+n (x) f (x) < ε for all x in S and all k =, 2,... Hint. To prove the sufficiency of (i) and (ii), show that for each x in S there is a neighborhood of B (x ) and an integer k (depending on x ) such that f k (x) f (x) < δ if x B (x ). By compactness, a finite set of integers, say A = {k,..., k r }, has the property that, for each x in S, some k in A satisfies f k (x) f (x) < δ. Uniform convergence is an easy consequences of this fact. Proof: ( ) Suppose that f n f uniformly on S, then by Theorem 9.2, the limit function f is continuous on S. In addition, given ε >, there exists a positive integer N such that as n N, we have f n (x) f (x) < ε for all x S Let m = N, and δ = ε, then (ii) holds. ( ) Suppose that (i) and (ii) holds. We prove f k f uniformly on S as follows. By (ii), given ε >, there exists an m > and a δ >, such that n > m and f k (x) f (x) < δ implies f k+n (x) f (x) < ε for all x in S and all k =, 2,... Consider f k(x ) (x ) f (x ) < δ, then there exists a B (x ) such that as x B (x ) S, we have fk(x ) (x) f (x) < δ 8

9 by continuity of f k(x ) (x) f (x). Hence, by (ii) as n > m f k(x )+n (x) f (x) < ε if x B (x ) S. (*) Note that S is compact and S = x S (B (x) S), then S = p (B (x k) S). So, let N = max p i= (k (x p) + m), as n > N, we have f n (x) f (x) < ε for all x S with help of (*). That is, f n f uniformly on S. 9.9 (a) Use Exercise 9.8 to prove the following theorem of Dini: If {f n } is a sequence of real-valued continuous functions converginf pointwise to a continuous limit function f on a compact set S, and if f n (x) f n+ (x) for each x in S and every n =, 2,..., then f n f uniformly on S. Proof: By Exercise 9.8, in order to show that f n f uniformly on S, it suffices to show that (ii) holds. Since f n (x) f (x) and f n+ (x) f n (x) on S, then fixed x S, and given ε >, there exists a positive integer N (x) = N such that as n N, we have f n (x) f (x) < ε. Choose m = and δ = ε, then by f n+ (x) f n (x), then (ii) holds. We complete it. Remark: () Dini s Theorem is important in Analysis; we suggest the reader to keep it in mind. (2) There is another proof by using Cantor Intersection Theorem. We give it as follows. Proof: Let g n = f n f, then g n is continuous on S, g n pointwise on S, and g n (x) g n+ (x) on S. If we can show g n uniformly on S, then we have proved that f n f uniformly on S. Given ε >, and consider S n := {x : g n (x) ε}. Since each g n (x) is continuous on a compact set S, we obtain that S n is compact. In addition, S n+ S n since g n (x) g n+ (x) on S. Then S n φ (*) 9

10 if each S n is non-empty by Cantor Intersection Theorem. However (*) contradicts to g n pointwise on S. Hence, we know that there exists a positive integer N such that as n N, S n = φ. That is, given ε >, there exists a positive integer N such that as n N, we have g n (x) < ε. So, g n uniformly on S. (b) Use he sequence in Exercise 9.5(a) to show that compactness of S is essential in Dini s Theorem. Proof: Let f n (x) =, where x (, ). Then it is clear that each +nx f n (x) is continuous on (, ), the limit function f (x) = is continuous on (, ), and f n+ (x) f n (x) for all x (, ). However, f n f not uniformly on (, ) by Exercise 9.5 (a). Hence, compactness of S is essential in Dini s Theorem. 9. Let f n (x) = n c x ( x 2 ) n for x real and n. Prove that {f n } converges pointwsie on [, ] for every real c. Determine those c for which the convergence is uniform on [, ] and those for which term-by-term integration on [, ] leads to a correct result. Proof: It is clear that f n () and f n (). Consider x (, ), then x 2 := r <, then lim f n (x) = lim n c r n x = for any real c. n n Hence, f n pointwise on [, ]. Consider f n (x) = n ( c x 2) ( ) n (2n ) 2n x2, then each f n has the absolute maximum at x n = 2n. As c < /2, we obtain that f n (x) f n (x n ) ( n c = ) n 2n 2n [ ( = n c n 2 ) n ] 2n 2n as n. (*)

11 In addition, as c /2, if f n uniformly on [, ], then given ε >, there exists a positive integer N such that as n N, we have which implies that as n N, which contradicts to f n (x) < ε for all x [, ] f n (x n ) < ε { lim f n (x n ) = 2e if c = /2 n if c > /2. (**) From (*) and (**), we conclude that only as c < /2, the seqences of functions converges uniformly on [, ]. In order to determine those c for which term-by-term integration on [, ], we consider n c f n (x) dx = 2 (n + ) and f (x) dx = dx =. Hence, only as c <, we can integrate it term-by-term. 9. Prove that x n ( x) converges pointwise but not uniformly on [, ], whereas ( ) n x n ( x) converges uniformly on [, ]. This illustrates that uniform convergence of f n (x) along with pointwise convergence of f n (x) does not necessarily imply uniform convergence of f n (x). Proof: Let s n (x) = n k= xk ( x) = x n+, then { if x [, ) s n (x). if x = Hence, x n ( x) converges pointwise but not uniformly on [, ] by Theorem 9.2 since each s n is continuous on [, ]. Let g n (x) = x n ( x), then it is clear that g n (x) g n+ (x) for all x [, ], and g n (x) uniformly on [, ] by Exercise 9.6. Hence, by Dirichlet s Test for uniform convergence, we have proved that ( ) n x n ( x) converges uniformly on [, ].

12 9.2 Assume that g n+ (x) g n (x) for each x in T and each n =, 2,..., and suppose that g n uniformly on T. Prove that ( ) n+ g n (x) converges uniformly on T. Proof: It is clear by Dirichlet s Test for uniform convergence. 9.3 Prove Abel s test for uniform convergence: Let {g n } be a sequence of real-valued functions such that g n+ (x) g n (x) for each x in T and for every n =, 2,... If {g n } is uniformly bounded on T and if f n (x) converges uniformly on T, then f n (x) g n (x) also converges uniformly on T. Proof: Let F n (x) = n f k (x). Then s n (x) = n f k (x) g k (x) = F n g (x)+ and hence if n > m, we can write n (F n (x) F k (x)) (g k+ (x) g k (x)) s n (x) s m (x) = (F n (x) F m (x)) g m+ (x)+ n k=m+ Hence, if M is an uniform bound for {g n }, we have s n (x) s m (x) M F n (x) F m (x) + 2M (F n (x) F k (x)) (g k+ (x) g k (x)) n k=m+ F n (x) F k (x). (*) Since f n (x) converges uniformly on T, given ε >, there exists a positive integer N such that as n > m N, we have F n (x) F m (x) < ε M + By (*) and (**), we have proved that as n > m N, s n (x) s m (x) < ε for all x T. Hence, f n (x) g n (x) also converges uniformly on T. for all x T (**) Remark: In the proof, we establish the lemma as follows. We write it as a reference. 2

13 (Lemma) If {a n } and {b n } are two sequences of complex numbers, define A n = n a k. Then we have the identity n n a k b k = A n b n+ A k (b k+ b k ) (i) = A n b + n (A n A k ) (b k+ b k ). (ii) Proof: The identity (i) comes from Theorem In order to show (ii), it suffices to consider b n+ = b + n b k+ b k. 9.4 Let f n (x) = x/ ( + nx 2 ) if x R, n =, 2,... Find the limit function f of the sequence {f n } and the limit function g of the sequence {f n}. (a) Prove that f (x) exists for every x but that f () g (). For what values of x is f (x) = g (x)? Proof: It is easy to show that the limit function f =, and by f n (x) = nx 2 (+nx 2 ) 2, we have { if x = lim f n (x) = g (x) = n if x. Hence, f (x) exists for every x and f () = g () =. In addition, it is clear that as x, we have f (x) = g (x). (b) In what subintervals of R does f n f uniformly? Proof: Note that + nx 2 2 n x 3

14 by A.P. G.P. for all real x. Hence, x + nx 2 2 n which implies that f n f uniformly on R. (c) In what subintervals of R does f n g uniformly? Proof: Since each f n = nx2 is continuous on R, and the limit function (+nx 2 ) 2 g is continuous on R {}, then by Theorem 9.2, the interval I that we consider does not contains. Claim that f n g uniformly on such interval I = [a, b] which does not contain as follows. Consider nx 2 ( + nx 2 ) 2 + nx 2 na, 2 so we know that f n g uniformly on such interval I = [a, b] which does not contain. 9.5 Let f n (x) = (/n) e n2 x 2 if x R, n =, 2,... Prove that f n uniformly on R, that f n pointwise on R, but that the convergence of {f n} is not uniform on any interval containing the origin. Proof: It is clear that f n uniformly on R, that f n pointwise on R. Assume that f n uniformly on [a, b] that contains. We will prove that it is impossible as follows. We may assume that (a, b) since other cases are similar. Given ε =, e then there exists a positive integer N such that as n max ( ) N, b := N ( b), we have N f n (x) < e for all x [a, b] which implies that 2 Nx e (Nx)2 < e for all x [a, b] which implies that, let x = N, 2 e < e which is absurb. So, the convergence of {f n} is not uniform on any interval containing the origin. 4

15 9.6 Let {f n } be a sequence of real-valued continuous functions defined on [, ] and assume that f n f uniformly on [, ]. Prove or disprove /n lim n f n (x) dx = f (x) dx. Proof: By Theorem 9.8, we have lim n f n (x) dx = f (x) dx. (*) Note that {f n } is uniform bound, say f n (x) M for all x [, ] and all n by Exercise 9.. Hence, f n (x) dx M. (**) n /n Hence, by (*) and (**), we have /n lim n f n (x) dx = f (x) dx. 9.7 Mathematicinas from Slobbovia decided that the Riemann integral was too complicated so that they replaced it by Slobbovian integral, defined as follows: If f is a function defined on the set Q of rational numbers in [, ], the Slobbovian integral of f, denoted by S (f), is defined to be the limit S (f) = lim n n n f ( ) k, n whenever the limit exists. Let {f n } be a sequence of functions such that S (f n ) exists for each n and such that f n f uniformly on Q. Prove that {S (f n )} converges, that S (f) exists, and S (f n ) S (f) as n. Proof: f n f uniformly on Q, then given ε >, there exists a positive integer N such that as n > m N, we have f n (x) f (x) < ε/3 () 5

16 and So, if n > m N, S (f n ) S (f m ) = lim k k k j= k = lim k k f n (x) f m (x) < ε/2. (2) lim k k = ε/2 < ε j= ( ( ) ( )) j j f n f m k k ( ( ) ( )) j j f n f m k k k ε/2 by (2) j= which implies that {S (f n )} converges since it is a Cauchy sequence. Say its limit S. Consider, by () as n N, k k j= [ ( ) ] j f n ε/3 k k which implies that [ k ( ) ] j f n ε/3 k k k j= which implies that, let k k f j= k f j= S (f n ) ε/3 lim k sup k ( ) j k k ( ) [ j k k k f j= k j= [ ( ) ] j f n + ε/3 k k ( ) ] j f n + ε/3 k j= ( ) j S (f n ) + ε/3 (3) k and S (f n ) ε/3 lim k inf k k f j= ( ) j S (f n ) + ε/3 (4) k 6

17 which implies that lim sup k f k k j= lim sup k f k k j= 2ε by (3) and (4) 3 < ε. ( ) j lim k ( ) j k inf k ( ) j f k k k j= S (f n ) + lim inf k k k f j= ( ) j k S (f n ) (5) Note that (3)-(5) imply that the existence of S (f). Also, (3) or (4) implies that S (f) = S. So, we complete the proof. 9.8 Let f n (x) = / ( + n 2 x 2 ) if x, n =, 2,... Prove that {f n } converges pointwise but not uniformly on [, ]. Is term-by term integration permissible? Proof: It is clear that lim f n (x) = n for all x [, ]. If {f n } converges uniformly on [, ], then given ε = /3, there exists a positive integer N such that as n N, we have which implies that f n (x) < /3 for all x [, ] ( ) f N = N 2 < 3 which is impossible. So, {f n } converges pointwise but not uniformly on [, ]. Since {f n (x)} is clearly uniformly bounded on [, ], i.e., f n (x) for all x [, ] and n. Hence, by Arzela s Theorem, we know that the sequence of functions can be integrated term by term. 9.9 Prove that n= x/nα ( + nx 2 ) converges uniformly on every finite interval in R if α > /2. Is the convergence uniform on R? Proof: By A.P. G.P., we have x n α ( + nx 2 ) 7 2n α+ 2 for all x.

18 So, by Weierstrass M-test, we have proved that n= x/nα ( + nx 2 ) converges uniformly on R if α > /2. Hence, n= x/nα ( + nx 2 ) converges uniformly on every finite interval in R if α > / Prove that the series n= (( )n / n) sin ( + (x/n)) converges uniformly on every compact subset of R. Proof: It suffices to show that the series n= (( )n / n) sin ( + (x/n)) converges uniformly on [, a]. Choose n large enough so that a/n /2, and therefore sin ( + ( )) ( x n+ sin + x n) for all x [, a]. So, if we let fn (x) = ( ) n / n and g n (x) = sin ( + n) x, then by Abel s test for uniform convergence, we have proved that the series n= (( )n / n) sin ( + (x/n)) converges uniformly on [, a]. Remark: In the proof, we metion something to make the reader get more. () since a compact set K is a bounded set, say K [ a, a], if we can show the series converges uniformly on [ a, a], then we have proved it. (2) The interval that we consider is [, a] since [ a, ] is similar. (3) Abel s test for uniform convergence holds for n N, where N is a fixed positive integer. 9.2 Prove that the series (x2n+ / (2n + ) x n+ / (2n + 2)) converges pointwise but not uniformly on [, ]. Proof: We show that the series converges pointwise on [, ] by considering two cases: () x [, ) and (2) x =. Hence, it is trivial. Define f (x) = (x2n+ / (2n + ) x n+ / (2n + 2)), if the series converges uniformly on [, ], then by Theorem 9.2, f (x) is continuous on [, ]. However, f (x) = { 2 log ( + x) if x [, ) log 2 if x = Hence, the series converges not uniformly on [, ]. Remark: The function f (x) is found by the following. Given x [, ), then both t 2n = t and t n = ( t) converges uniformly on [, x] by Theorem 9.4. So, by Theorem 9.8, we. 8

19 have x t 2n 2 t n = = x x t 2 2 ( t) dt ( 2 t + ) + t = log ( + x). 2 2 ( ) dt t And as x =, ( x 2n+ / (2n + ) x n+ / (2n + 2) ) = = 2n + 2n ( ) n+ n + by Theorem8.4. = log 2 by Abel s Limit Theorem Prove that a n sin nx and b n cos nx are uniformly convergent on R if a n converges. Proof: It is trivial by Weierstrass M-test Let {a n } be a decreasing sequence of positive terms. Prove that the series a n sin nx converges uniformly on R if, and only if, na n as n. Proof: ( ) Suppose that the series a n sin nx converges uniformly on R, then given ε >, there exists a positive integer N such that as n N, we have 2n a k sin kx < ε. (*) Choose x = 2n, then sin 2 k=n sin kx sin. Hence, as n N, we always 9

20 have, by (*) 2n 2n (ε >) a k sin kx = a k sin kx k=n = k=n 2n a 2n sin 2 since a k > and a k k=n ( 2 sin ) (2na 2n ). 2 That is, we have proved that 2na 2n as n. Similarly, we also have (2n ) a 2n as n. So, we have proved that na n as n. ( ) Suppose that na n as n, then given ε >, there exists a positive integer n such that as n n, we have ε na n = na n < 2 (π + ). (*) In order to show the uniform convergence of n= a n sin nx on R, it suffices to show the uniform convergence of n= a n sin nx on [, π]. So, if we can show that as n n n+p a k sin kx < ε for all x [, π], and all p N k=n+ then we complete [ ] it. We consider two cases as follows. (n n ) π As x,, then n+p n+p k=n+ n+p a k sin kx = = k=n+ n+p k=n+ n+p k=n+ a k sin kx a k kx by sin kx kx if x (ka k ) x ε pπ by (*) 2 (π + ) n + p < ε. 2

21 [ ] π And as x, π, then n+p n+p k=n+ a k sin kx m k=n+ a k sin kx + n+p k=m+ [ π ] a k sin kx, where m = x m a k kx + 2a m+ sin x by Summation by parts k=n+ 2 ε 2 (π + ) (m n) x + 2a m+ sin x 2 ε π 2 (π + ) mx + 2a m+ x by 2x [, π sin x if x π 2 ε 2 (π + ) π + 2a m+ (m + ) < ε ε 2 (π + ) < ε. Hence, n= a n sin nx converges uniformly on R. Remark: () In the proof ( ), if we can make sure that na n, then we can use the supplement on the convergnce of series in Ch8, (C)- (6) to show the uniform convergence of n= a n sin nx = n= (na n) ( ) sin nx n by Dirichlet s test for uniform convergence. (2)There are similar results; we write it as references. (a) Suppose a n, then for each α (, π 2 ), n= a n cos nx and n= a n sin nx converges uniformly on [α, 2π α]. Proof: The proof follows from (2) and (3) in Theorem 8.3 and Dirichlet s test for uniform convergence. So, we omit it. The reader can see the textbook, example in pp 23. (b) Let {a n } be a decreasing sequence of positive terms. n= a n cos nx uniformly converges on R if and only if n= a n converges. Proof: ( ) Suppose that n= a n cos nx uniformly converges on R, then let x =, then we have n= a n converges. ( ) Suppose that n= a n converges, then by Weierstrass M-test, we have proved that n= a n cos nx uniformly converges on R. ] 2

22 9.24 Given a convergent series n= a n. Prove that the Dirichlet series n= a nn s converges uniformly on the half-infinite interval s < +. Use this to prove that lim s + n= a nn s = n= a n. Proof: Let f n (s) = n a k and g n (s) = n s, then by Abel s test for uniform convergence, we have proved that the Dirichlet series n= a nn s converges uniformly on the half-infinite interval s < +. Then by Theorem 9.2, we know that lim s + n= a nn s = n= a n Prove that the series ζ (s) = n= n s converges uniformly on every half-infinite interval + h s < +, where h >. Show that the equation ζ (s) = n= log n n s is valid for each s > and obtain a similar formula for the kth derivative ζ (k) (s). Proof: Since n s n (+h) for all s [ + h, ), we know that ζ (s) = n= n s converges uniformly on every half-infinite interval +h s < + by Weierstrass M-test. Define T n (s) = n k s, then it is clear that And. For each n, T n (s) is differentiable on [ + h, ), 3. T n (s) = n 2. lim n T n (2) = π2 6. log k k s converges uniformly on [ + h, ) by Weierstrass M-test. Hence, we have proved that ζ (s) = n= log n n s by Theorem 9.3. By Mathematical Induction, we know that ζ (k) (s) = ( ) k n= (log n) k n s. 22

23 . Supplement on some results on Weierstrass M- test.. In the textbook, pp , there is a surprising result called Spacefilling curve. In addition, note the proof is related with Cantor set in exercise in the textbook. 2. There exists a continuous function defined on R which is nowhere differentiable. The reader can see the book, Principles of Mathematical Analysis by Walter Rudin, pp 54. Remark: The first example comes from Bolzano in 834, however, he did NOT give a proof. In fact, he only found the function f : D R that he constructed is not differentiable on D ( D) where D is countable and dense in D. Although the function f is the example, but he did not find the fact. In 86, Riemann gave g (x) = n= sin (n 2 πx) n 2 as an example. However, Reimann did NOT give a proof in his life until 96, the proof is given by G. Hardy. In 86, Weierstrass gave h (x) = a n cos (b n πx), b is odd, < a <, and ab > + 3π 2, n= until 875, he gave the proof. The fact surprises the world of Math, and produces many examples. There are many researches related with it until now 23. Mean Convergence 9.26 Let f n (x) = n 3/2 xe n2 x 2. Prove that {f n } converges pointwise to on [, ] but that l.i.m. n f n on [, ]. Proof: It is clear that {f n } converges pointwise to on [, ], so it 23

24 remains to show that l.i.m. n f n on [, ]. Consider f 2 n (x) dx = 2 n 3 x 2 e 2n2 x 2 dx since f 2 n (x) is an even function on [, ] = 2n y 2 e y2 dy by Change of Variable, let y = 2nx 2 = 2 2 2n yd (e y2) [ = 2n 2 ye y2 2 π 4 2 since 2n e x2 dx = e y2 dy π 2 ] by Exercise So, l.i.m. n f n on [, ] Assume that {f n } converges pointwise to f on [a, b] and that l.i.m. n f n = g on [a, b]. Prove that f = g if both f and g are continuous on [a, b]. Proof: Since l.i.m. n f n = g on [a, b], given ε k = 2 k, there exists a n k such that Define then b a f nk (x) g (x) p dx 2 k, where p > h m (x) = m x a. h m (x) as x b. h m (x) h m+ (x) a f nk (t) g (t) p dt, c. h m (x) for all m and all x. So, we obtain h m (x) h (x) as m, h (x) as x, and h (x) h m (x) = k=m+ x a f nk (t) g (t) p dt as x 24

25 which implies that h (x + t) h (x) t h m (x + t) h m (x) t for all m. (*) Since h and h m are increasing, we have h and h m exists a.e. on [a, b]. Hence, by (*) m h m (x) = f nk (t) g (t) p h (x) a.e. on [a, b] which implies that f nk (t) g (t) p exists a.e. on [a, b]. So, f nk (t) g (t) a.e. on [a, b]. In addition, f n f on [a, b]. Then we conclude that f = g a.e. on [a, b]. Since f and g are continuous on [a, b], we have b a f g dx = which implies that f = g on [a, b]. In particular, as p = 2, we have f = g. Remark: () A property is said to hold almost everywhere on a set S (written: a.e. on S) if it holds everywhere on S except for a set of measurer zero. Also, see the textbook, pp 254. (2) In this proof, we use the theorem which states: A monotonic function h defined on [a, b], then h is differentiable a.e. on [a, b]. The reader can see the book, The reader can see the book, Measure and Integral (An Introduction to Real Analysis) written by Richard L. Wheeden and Antoni Zygmund, pp 3. (3) There is another proof by using Fatou s lemma: Let {f k } be a measruable function defined on a measure set E. If f k φ a.e. on E and φ L (E), then lim k lim inf f k. E k k E Proof: It suffices to show that f nk (t) g (t) a.e. on [a, b]. Since l.i.m. n f n = g on [a, b], and given ε >, there exists a n k such that b a f nk g 2 dx < 2 k 25

26 which implies that b m a f nk g 2 dx < which implies that, by Fatou s lemma, m 2 k b a lim inf m b m f nk g 2 dx lim inf f nk g 2 dx m m a b = f nk g 2 dx <. a That is, which implies that b a f nk g 2 dx < f nk g 2 < a.e. on [a, b] which implies that f nk g a.e. on [a, b]. Note: The reader can see the book, Measure and Integral (An Introduction to Real Analysis) written by Richard L. Wheeden and Antoni Zygmund, pp 75. (4) There is another proof by using Egorov s Theorem: Let {f k } be a measurable functions defined on a finite measurable set E with finite limit function f. Then given ε >, there exists a closed set F ( E), where E F < ε such that f k f uniformly on F. Proof: If f g on [a, b], then h := f g on [a, b]. By continuity of h, there exists a compact subinterval [c, d] such that f g. So, there exists m > such that h = f g m > on [c, d]. Since b a f n g 2 dx as n, 26

27 we have d c f n g 2 dx as n. then by Egorov s Theorem, given ε >, there exists a closed susbet F of [c, d], where [c, d] F < ε such that which implies that f n f uniformly on F = lim f n g n F 2 dx = lim f n g 2 dx F n = f g 2 dx m 2 F F which implies that F =. If we choose ε < d c, then we get a contradiction. Therefore, f = g on [a, b]. Note: The reader can see the book, Measure and Integral (An Introduction to Real Analysis) written by Richard L. Wheeden and Antoni Zygmund, pp Let f n (x) = cos n x if x π. (a) Prove that l.i.m. n f n converge. = on [, π] but that {f n (π)} does not Proof: It is clear that {f n (π)} does not converge since f n (π) = ( ) n. It remains to show that l.i.m. n f n = on [, π]. Consider cos 2n x := g n (x) on [, π], then it is clear that {g n (x)} is boundedly convergent with limit function { if x (, π) g = if x = or π. Hence, by Arzela s Theorem, π lim n So, l.i.m. n f n = on [, π]. cos 2n xdx = 27 π g (x) dx =.

28 (b) Prove that {f n } converges pointwise but not uniformly on [, π/2]. Proof: Note that each f n (x) is continuous on [, π/2], and the limit function { if x (, π/2] f =. if x = Hence, by Theorem9.2, we know that {f n } converges pointwise but not uniformly on [, π/2] Let f n (x) = if x /n or 2/n x, and let f n (x) = n if /n < x < 2/n. Prove that {f n } converges pointwise to on [, ] but that l.i.m. n f n on [, ]. Proof: It is clear that {f n } converges pointwise to on [, ]. In order to show that l.i.m. n f n on [, ], it suffices to note that Hence, l.i.m. n f n on [, ]. Power series f n (x) dx = for all n. 9.3 If r is the radius of convergence if a n (z z ) n, where each a n, show that lim inf a n n r lim sup a n n. a n+ a n+ Proof: By Exercise 8.4, we have lim n sup a n+ r = a n lim n sup a n n lim n inf a n+. a n Since and lim n sup lim n inf a n+ a n a n+ a n = lim n inf = lim n sup a n a n+ a n a n+, 28

29 we complete it. 9.3 Given that two power series a n z n has radius of convergence 2. Find the radius convergence of each of the following series: In (a) and (b), k is a fixed positive integer. (a) ak nz n Proof: Since 2 = we know that the radius of ak nz n is, (*) /n lim n sup a n lim n sup a k n /n = (lim n sup a n /n) k = 2k. (b) a nz kn Proof: Consider which implies that lim sup an z kn /n = lim sup a n /n z k < n n z < ( lim n sup a n /n So, the radius of a nz kn is 2 /k. (c) a nz n2 ) /k = 2 /k by (*). Proof: Consider /n lim sup a n z n2 = lim sup a n /n z n n and claim that the radius of a nz n2 is as follows. If z <, it is clearly seen that the series converges. However, if z >, lim sup a n /n lim inf z n lim sup a n /n z n n n n 29

30 which impliest that lim sup a n /n z n = +. n so, the series diverges. From above, we have proved the claim Given a power series a n x n whose coefficents are related by an equation of the form a n + Aa n + Ba n 2 = (n = 2, 3,...). Show that for any x for which the series converges, its sum is a + (a + Aa ) x + Ax + Bx 2. Proof: Consider which implies that n=2 which implies that (a n + Aa n + Ba n 2 ) x n = n=2 a n x n + Ax a n x n + Bx 2 n=2 a n x n + Ax a n x n + Bx 2 which implies that n=2 a n x n = a + (a + Aa ) x + Ax + Bx 2. a n 2 x n 2 = a n x n = a + a x + Aa x Remark: We prove that for any x for which the series converges, then + Ax + Bx 2 as follows. Proof: Consider ( ) + Ax + Bx 2 a n x n = a + (a + Aa ) x, 3

31 if x = λ ( ) is a root of + Ax + Bx 2, and a nλ n exists, we have + Aλ + Bλ 2 = and a + (a + Aa ) λ =. Note that a + Aa, otherwise, a = ( a = ), and therefore, a n = for all n. Then there is nothing to prove it. So, put λ = a a +Aa into + Aλ + Bλ 2 =, we then have a 2 = a a 2. Note that a, otherwise, a = and a 2 =. Similarly, a, otherwise, we will obtain a trivial thing. Hence, we may assume that all a n for all n. So, a 2 2 = a a 3. And it is easy to check that a n = a λ n for all n N. Therefore, a n λ n = a diverges. So, for any x for whcih the series converges, we have +Ax+ Bx Let f (x) = e /x2 if x, f () =. (a) Show that f (n) () exists for all n. Proof: By Exercise 5.4, we complete it. (b) Show that the Taylor s series about generated by f converges everywhere on R but that it represents f only at the origin. Proof: The Taylor s series about generated by f is f (n) () x n = n! x n =. So, it converges everywhere on R but that it represents f only at the origin. Remark: It is an important example to tell us that even for functions f C (R), the Taylor s series about c generated by f may NOT represent f on some open interval. Also see the textbook, pp Show that the binomial series ( + x) α = (α n) x n exhibits the following behavior at the points x = ±. (a) If x =, the series converges for α and diverges for α <. 3

32 Proof: If x =, we consider three cases: (i) α <, (ii) α =, and (iii) α >. (i) As α <, then ( α n) ( ) n = ( ) n α (α ) (α n + ), n! say a n = ( ) n α(α ) (α n+) n!, then a n for all n, and a n /n = α ( α + ) ( α + n ) (n )! α > for all n. Hence, (α n) ( ) n diverges. (ii) As α =, then the series is clearly convergent. (iii) As α >, define a n = n ( ) n ( α n), then a n+ a n = n α n if n [α] +. (*) It means that a n > for all n [α] + or a n < for all n [α] +. Without loss of generality, we consider a n > for all n [α] + as follows. Note that (*) tells us that and So, m n=[α]+ a n > a n+ > lim n a n exists. a n a n+ = α ( ) n ( α n). ( ) n ( α n) = α m n=[α]+ (a n a n+ ). By Theorem 8., we have proved the convergence of the series (α n) ( ) n. (b) If x =, the series diverges for α, converges conditionally for α in the interval < α <, and converges absolutely for α. Proof: If x =, we consider four cases as follows: (i) α, (ii) < α <, (iii) α =, and (iv) α > : (i) As α, say a n = α(α ) (α n+) n!. Then a n = α ( α + ) ( α + n ) n! 32 for all n.

33 So, the series diverges. (ii) As < α <, say a n = α(α ) (α n+) n!. Then a n = ( ) n b n, where with b n = b n+ b n α ( α + ) ( α + n ) n! = n α n >. < since < α < which implies that {b n } is decreasing with limit L. So, if we can show L =, then a n converges by Theorem 8.6. Rewrite n ( b n = α + ) k and since α+ k diverges, then by Theroem 8.55, we have proved L =. In order to show the convergence is conditionally, it suffices to show the divergence of b n. The fact follows from b n /n = α ( α + ) ( α + n ) (n )! α >. (iii) As α =, it is clearly that the series converges absolutely. (iv) As α >, we consider ( α n) as follows. Define a n = ( α n), then a n+ a n = n α n + < if n [α] +. It implies that na n (n + ) a n = αa n and (n + ) a n+ < na n. So, by Theroem 8., an = nan (n + ) a n α converges since lim n na n exists. So, we have proved that the series converges absolutely Show that a n x n converges uniformly on [, ] if a n converges. Use this fact to give another proof of Abel s limit theorem. Proof: Define f n (x) = a n on [, ], then it is clear that f n (x) converges uniformly on [, ]. In addition, let g n (x) = x n, then g n (x) is unifomrly bouned with g n+ (x) g n (x). So, by Abel s test for uniform convergence, 33

34 an x n converges uniformly on [, ]. Now, we give another proof of Abel s Limit Theorem as follows. Note that each term of a n x n is continuous on [, ] and the convergence is uniformly on [, ], so by Theorem 9.2, the power series is continuous on [, ]. That is, we have proved Abel s Limit Theorem: lim x an x n = a n If each a n > and a n diverges, show that a n x n + as x. (Assume a n x n converges for x <.) Proof: Given M >, if we can find a y near from the left such that an y n M, then for y x <, we have M a n y n a n x n. That is, lim x an x n = +. Since a n diverges, there is a positive integer p such that p a k 2M > M. (*) Define f n (x) = n a kx k, then by continuity of each f n, given < ε (< M), there exists a δ n > such that as x [δ n, ), we have n a k ε < n a k x k < n a k + ε (**) By (*) and (**), we proved that as y = δ p M p a k ε < p a k y k. Hence, we have proved it If each a n > and if lim x an x n exists and equals A, prove that a n converges and has the sum A. (Compare with Theorem 9.33.) Proof: By Exercise 9.36, we have proved the part, a n converges. In order to show a n = A, we apply Abel s Limit Theorem to complete it. 34

35 9.38 For each real t, define f t (x) = xe xt / (e x ) if x R, x, f t () =. (a) Show that there is a disk B (; δ) in which f t is represented by a power series in x. Proof: First, we note that ex = x n x := p (x), then p () = (n+)!. So, by Theorem 9. 26, there exists a disk B (; δ) in which the reciprocal of p has a power series exapnsion of the form p (x) = q n x n. So, as x B (; δ) by Theorem f t (x) = xe xt / (e x ) ( ) ( (xt) n ) x n = n! (n + )! = r n (t) x n. (b) Define P (t), P (t), P 2 (t),..., by the equation and use the identity f t (x) = P n (t) xn, if x B (; δ), n! P n (t) xn n! = etx to prove that P n (t) = n k= (n k ) P k () t n k. Proof: Since f t (x) = P n () xn n! P n (t) xn n! = etx x e x, 35

36 and f (x) = So, we have the identity P n (t) xn n! = etx P n () xn n! = x e x. P n () xn n!. Use the identity with e tx = t n n! xn, then we obtain n which implies that P n (t) n! = k= = n! P n (t) = t n k P k () (n k)! k! n ( n k) P k () t n k k= n ( n k) P k () t n k. k= This shows that each function P n is a polynomial. There are the Bernoulli polynomials. The numbers B n = P n () (n =,, 2,...) are called the Bernoulli numbers. Derive the following further properties: (c) B =, B = 2, n k= (n k ) B k =, if n = 2, 3,... Proof: Since = p(x), where p (x) := x n p(x), and := (n+)! p(x) P n () xn. n! So, where = p (x) p (x) x n = (n + )! = C n x n C n = (n + )! n k= 36 P n () xn n! ( n+ ) k Pk ().

37 So, B = P () = C =, B = P () = C P () = 2 2, by C = and note that C n = for all n, we have = C n = n ( n n! k) P k () k= = n ( n n! k) B k for all n 2. k= (d) P n (t) = np n (t), if n =, 2,... Proof: Since P n (t) = n ( n k) P k () (n k) t n k k= n = ( n k) P k () (n k) t n k k= n = k= n n! (n k) k! (n k)! P k () t (n ) k (n )! = n k! (n k)! P k () t (n ) k k= n = n k= ( n ) k Pk () t (n ) k = np n (t) if n =, 2,... (e) P n (t + ) P n (t) = nt n if n =, 2,... 37

38 Proof: Consider f t+ (x) f t (x) = so as n =, 2,..., we have [P n (t + ) P n (t)] xn n! = xe xt by f t (x) = xe xt / (e x ) = (n + ) t n xn+ (n + )!, P n (t + ) P n (t) = nt n. by (b) (f) P n ( t) = ( ) n P n (t) Proof: Note that so we have Hence, P n ( t) = ( ) n P n (t). (g) B 2n+ = if n =, 2,... f t ( x) = f t (x), ( ) n P n (t) xn n! = P n ( t) xn n!. Proof: With help of (e) and (f), let t = and n = 2k +, then it is clear that B 2k+ = if k =, 2,... (h) n + 2 n (k ) n = P n+(k) P n+ () n+ (n = 2, 3,...) Proof: With help of (e), we know that P n+ (t + ) P n+ (t) n + = t n which implies that n + 2 n (k ) n = P n+ (k) P n+ () n + (n = 2, 3,...) 38

39 Remark: () The reader can see the book, Infinite Series by Chao Wen-Min, pp (Chinese Version) (2) There are some special polynomials worth studying, such as Legengre Polynomials. The reader can see the book, Essentials of Ordinary Differential Equations by Ravi P. Agarwal and Ramesh C. Gupta. pp (3) The part (h) tells us one formula to calculte the value of the finite series m kn. There is an interesting story from the mail that Fermat, pierre de (6-665) sent to Blaise Pascal ( ). Fermat used the Mathematical Induction to show that n k (k + ) (k + p) = n (n + ) (n + p + ). (*) p + 2 In terms of (*), we can obtain another formula on m kn. 39

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