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UNIFORM CONVERGENCE 1. Pointwise Convergence

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UNIFORM CONVERGENCE 1. Pointwise Convergence UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE 1. Pointwise Convergence of a Sequence Let E be a set and Y be a metric space. Consider functions fn : E Y ! for n = 1; 2; : : : : We say that the sequence (fn) converges pointwise on E if there is a function f : E Y such that fn(p) f(p) for every p E. Clearly, such a function f is unique! and it is called !the pointwise limit2 of (fn) on E. We then write fn f on E. For simplicity, we shall assume Y = R with the usual metric. ! Let fn f on E. We ask the following questions: ! (i) If each fn is bounded on E, must f be bounded on E? If so, must sup 2 fn(p) sup 2 f(p) ? p E j j ! p E j j (ii) If E is a metric space and each fn is continuous on E, must f be continuous on E? (iii) If E is an interval in R and each fn is differentiable on E, must f be differentiable on E? If so, must f 0 f 0 on E? n ! (iv) If E = [a; b] and each fn is Riemann integrable on E, must f be b b Riemann integrable on E? If so, must fn(x)dx f(x)dx? a ! a These questions involve interchange of two Rprocesses (oneR of which is `taking the limit as n ') as shown below. ! 1 (i) lim sup fn(p) = sup lim fn(p) : n!1 p2E j j p2E n!1 (ii) For p E and pk p in E, lim lim fn(pk) = lim lim fn(pk): 2 ! k!1 n!1 n!1 k!1 d d (iii) lim (fn) = lim fn : n!1 dx dx n!1 b b (iv) lim f (x)dx = lim f (x) dx: !1 n !1 n n Za Za n Answers to these questions are all negative. Examples 1.1. (i) Let E := (0; 1] and define fn : E R by ! 0 if 0 < x 1=n; fn(x) := ≤ 1=x if 1=n x 1: ≤ ≤ Then fn(x) n for all x E; fn f on E; where f(x) := 1=x. j j ≤ 2 ! Thus each fn is bounded on E, but f is not bounded on E. (ii) Let E := [0; 1] and define fn : E R by fn(x) := 1=(nx + 1). Then ! each fn is continuous on E, fn f on E, where f(0) := 1 and f(x) := 0 if 0 < x 1. Clearly, f!is not continuous on E. ≤ 2 (iii) (a) Let E := ( 1; 1) and define fn : E R by fn(x) := 1=(nx + 1). − ! Then each fn is differentiable on E and fn f on ( 1; 1), 1 ! − 2 MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE where f(0) := 1 and f(x) := 0 if 0 < x < 1. Clearly, f is not differentiable on E. j j (b) Let E := R and define fn : R R by fn(x) := (sin nx)=n. ! Then each fn is differentiable, fn f on R, where f 0. But 0 R 0 ! ≡ fn(x) = cos nx for x , and (fn) does not converge pointwise R 2 0 on . For example, (fn(π)) is not a convergent sequence. (c) Let E := ( 1; 1) and define − [2 (1 + x)n]=n if 1 < x < 0; fn(x) := − − (1 x)n=n if 0 x < 1: − ≤ Then each fn is differentiable on E. (In particular, we have 0 fn(0) = 1 by L'H^opital's Rule.) Also, fn f on ( 1; 1), where f −0. Also, ! − ≡ n−1 0 (1 + x) if 1 < x < 0; f (x) = − − − n (1 x)n 1 if 0 x < 1: − − ≤ 0 Further, fn g on ( 1; 1); where g(0) := 1 and g(x) := 0 for 0 < x <!1. Clearly−, f 0 = g: − j j 6 (iv) (a) Let E := [0; 1] and define fn : [0; 1] R by ! 1 if x = 0; 1=n!; 2=n!; : : : ; n!=n! = 1; fn(x) := 0 otherwise. Then each fn is Riemann integrable on [0; 1] since it is discon- tinuous only at a finite number of points. 1 if x is rational; Also, fn f on [0; 1], where f(x) := ! 0 if x is irrational. For if x = p=q with p 0; 1; 2; : : : ; q N, then for all n q, 2 f g ⊂ ≥ we have n!x 0; 1; 2; : : : and so fn(x) = 1; while if x is an 2 f g irrational number, then fn(x) = 0 for all n N. We have seen that the Dirichlet function f is not Riemann2 integrable. 3 −nx (b) Let E := [0; 1]; and define fn : [0; 1] R by fn(x) := n xe . ! Then each fn is Riemann integrable and fn f on [0; 1]; where f 0. (Use L'H^opital's Rule repeatedly!to show that 3≡ st limt!1 t =e = 0 for any s R with s > 0.) However, using Integration by Parts, we hav2e 1 −nx 1 1 1 N xe dx = 2 2 n n for each n ; Z0 n − n e − ne 2 1 n 2 n so that 0 fn(x)dx = n (n=e ) (n =e ) . − − ! 1 2 −nx (c) Let E :=R [0; 1] and define fn : [0; 1] R by fn(x) := n xe . ! As above, each fn is Riemann integrable, and fn f on [0; 1]; 1 n n! where f 0, and fn(x)dx = 1 (1=e ) (n=e ) 1, which ≡ 0 − − ! 1R is not equal to 0 f(x)dx = 0: R 2. Uniform Convergence of a Sequence In an attempt to obtain affirmative answers to the questions posed at the beginning of the previous section, we introduce a stronger concept of convergence. UNIFORM CONVERGENCE 3 Let E be a set and consider functions fn : E R for n = 1; 2; : : : : We ! say that the sequence (fn) of functions converges uniformly on E if there is a function f : E R such that for every > 0, there is n N satisfying ! 0 2 n n ; p E = fn(p) f(p) < . ≥ 0 2 ) j − j Note that the natural number n0 mentioned in the above definition may depend upon the given sequence (fn) of functions and on the given positive number , but it is independent of p E. Clearly, such a function f is 2 unique and it is called the uniform limit of (fn) on E. We then write fn ⇒ f on E. Obviously, fn ⇒ f on E = fn f on E, but the converse ) ! is not true : Let E := (0; 1] and define fn(x) := 1=(nx + 1) for 0 < x 1. If ≤ f(x) := 0 for x (0; 1], then fn f on (0; 1], but fn ⇒ f on (0; 1]. To see this, let := 1=2,2 note that there!is no n N satisfying6 0 2 1 1 fn(x) f(x) = < for all n n and for all x (0; 1]; j − j nx + 1 2 ≥ 0 2 since 1=(nx + 1) = 1=2 when x = 1=n; n N. 2 A sequence (fn) of real-valued functions defined on a set E is said to be uniformly Cauchy on E if for every > 0, there is n N satisfying 0 2 m; n n ; p E = fm(p) fn(p) < . ≥ 0 2 ) j − j Proposition 2.1. (Cauchy Criterion for Uniform Convergence of a Sequence) Let (fn) be a sequence of real-valued functions defined on a set E. Then (fn) is uniformly convergent on E if and only if (fn) is uniformly Cauchy on E. Proof. = ) Let fn ⇒ f. For all m; n N and p E, we have ) 2 2 fm(p) fn(p) fm(p) f(p) + f(p) fn(p) : j − j ≤ j − j j − j =) For each p E, (fn(p)) is a Cauchy sequence in R, and so it converges ( 2 to a real number which we denote by f(p). Let > 0. There is n0 N satisfying 2 m; n n ; p E = fm(p) fn(p) < . ≥ 0 2 ) j − j For any m n and p E, letting n , we have fm(p) f(p) . ≥ 0 2 ! 1 j − j ≤ We have the following useful test for checking the uniform convergence of (fn) when its pointwise limit is known. Proposition 2.2. (Test for Uniform Convergence of a Sequence) Let fn and f be real-valued functions defined on a set E. If fn f on ! E, and if there is a sequence (an) of real numbers such that an 0 and ! fn(p) f(p) an for all p E, then fn ⇒ f on E. j − j ≤ 2 Proof. Let > 0. Since an 0, there is n N such that n n = an < , ! 0 2 ≥ 0 ) and so fn(p) f(p) < for all p E. j − j 2 n Example 2.3. Let 0 < r < 1 and fn(x) := x for x [ r; r]. Thenfn(x) n 2 −n ! 0 for each x [ r; r]. Since r 0 and fn(x) 0 r for all x [ r; r], 2 − ! j − j ≤ 2 − (fn) is uniformly convergent on [ r; r]. − Let us now pose the four questions stated in the last section with `con- vergence' replaced by `uniform convergence'. We shall answer them one by one, but not necessarily in the same order. 4 MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE Uniform Convergence and Boundedness.
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