# Lecture 1. Functional series. Pointwise and uniform convergence.

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1 1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is defined s the series ( k cos kx + b k sin kx), k=1 where the coefficients k, b k re defined s k = 1 π π π f(x) cos kx dx, b k = 1 π with k = 0, 1,... (note tht this mens tht b 0 = 0). π π f(x) sin kx dx, This is n exmple of functionl series, which is series whose terms re functions:. As usul with series, we define the bove infinite sum s limit: = lim N N, providing the limit exists. Note tht different vlues of x will, in generl, give different limits, if they exist. In this lecture we shll look t functionl series, nd functionl sequences, nd we shll consider first the question of convergence. To del with this, we consider two types of convergence: pointwise convergence nd uniform convergence. There re three min results: the first one is tht uniform convergence of sequence of continuous functions gives us continuous function s limit. The second min result is Weierstrss Mjornt Theorem, which gives condition tht gurntees tht functionl series converges to continuous function. The third result is tht integrls of sequence of functions which converges uniformly to limit function f(x) lso converge with the limit being the integrl of f(x). These results re not only good for your mentl helth, they re lso importnt tools in our lter discussion of Fourier series, nd tht is the reson for looking t them. 1

2 2 Power series. A power series in the vrible x is series of the form x + 2 x x 3 + = k x k, where the coefficients 0, 1, 2,... re rel or complex numbers. We re (in principle) llowed to put the vrible x equl to ny number we wish. For instnce, with the power series k x k we my put x = 2 nd we obtin the numericl series Or, with the power series k 2 k = (note tht in this cse ll the coefficients re equl to 1) we obtin x k 2 k = if we put x = 2. However, we hve problem: how do we know tht the numericl series we obtin by putting x = 2 in the power series k x k is convergent? We cn lwys put x = 2 nd then investigte the convergence of the numericl series, but this is rther inefficient wy of deciding if prticulr vlue of x gives us convergent series. So we come to the following question: Given the power series k x k, with given choice of coefficients 0, 1, 2,..., wht vlues of x give us convergent series, nd which vlues give divergent series? There re two very useful results which help us in exmining this question. The first one is the following: 2

3 Theorem 2.1 For power series there re three possibilities: k x k 1. The power series 2. The power series k x k diverges for ll x 0 k x k converges for ll vlues of x 3. There is positive number R such tht k x k converges for ll vlues of x with x < R nd diverges for ll vlues of x with x > R. At first sight, this looks like very useless result, becuse it doesn t nswer the question of which vlues of x re llowed. However, it is very useful result: it tells us wht sort of behviour we cn expect, nd wht to look for in power series. In prticulr, it tells us wht is the decisive fctor in our subsequent investigtion: we need to find the number R, which is clled the rdius of convergence of the power series. One word of wrning: the theorem tells us tht, when we hve found the rdius of convergence R, then the series converges for x < R nd diverges whenever x > R, but it doesn t sy nything bout the cse when x = R, tht is, when x = R nd x = R. The theorem is silent on this mtter. In fct, we must investigte the cses x = ±R seprtely (tht is, we put in x = R or x = R nd we investigte the convergence of the series tht rise). This is smll price to py when we know wht R is. So, given our theorem, how do we go bout clculting R? One result is the following: Theorem 2.2 Given the power series exist: Then the following is true: k x k, suppose tht one of the following limits K = lim k+1 k k, K = lim k k. k 1. If K = 0 then the power series k x k converges for ll vlues of x; 2. If K > 0, then the rdius of convergence R of the power series R = 1 K. k x k is 3

4 If either of the limits lim k fils to exist, then the power series k+1 k, lim k k k. k x k diverges for ll vlues of x 0. This theorem is proved using the following result (the proof is given in Mts Neymrk s Kompendium om konvergens): Theorem 2.3 Suppose limits exists k is numericl series nd suppose tht one of the following K = lim k+1 k k, K = lim k k. k If 0 K < 1 then the series converges bsolutely. If K > 1 then the series diverges. If K = 1 then convergence or divergence of the series must be investigted using some other method. 3 Pointwise Convergence. Consider the power series x k = 1 1 x for x < 1. This just sys tht for ech x ] 1, 1[ the power series in the left-hnd side converges to the number 1/(1 x). If we put then we cn rephrse this s f n (x) = n x k, f(x) = 1 1 x, f n (x) f(x) s n for ech x ] 1, 1[. We give this type of convergence nme: pointwise convergence. Note tht we hve first defined sequence of functions f n by putting for ech n = 0, 1, 2,.... f n (x) = n x k 4

5 Definition 3.1 (Pointwise convergence.) Suppose {f n (x) : n = 0, 1, 2,... } is sequence of functions defined on n intervl I. We sy tht f n (x) converges pointwise to the function f(x) on the intervl I if f n (x) f(x), s n, for ech x I. We cll the function f(x) the limit function. Exmple 3.1 f n (x) = x 1 n. Then f n(x) converges pointwise to x for ech x R: f n (x) x = 1 n 0 s n. Exmple 3.2 f n (x) = e nx on [1, 3]. For ech x [1, 3] we hve nx s n nd therefore f n (x) 0 s n for ech x [1, 3]. Thus f n (x) converges pointwise to f(x) = 0 for ech x [1, 3]. Exmple 3.3 f n (x) = e nx on [0, 3]. For ech 0 < x 3 we hve nx s n nd therefore f n (x) 0 s n for ech 0 < x 3. However, t x = 0 we hve f n (0) = 1 for ll n. Thus f n (x) converges pointwise to the function f(x) defined by f(0) = 1, f(x) = 0 for ech 0 < x 3. This is not continuous function, despite the fct tht ech function f n (x) is continuous. The lst exmple shows wht cn hppen with pointwise convergence: the limit function my fil to be continuous, even though ll functions in the sequence re continuous. Exmple 3.4 Let the sequence f n be defined s n 2 x f n (x) = (nx + 1) 3 Then f n (0) = 0 nd for ech fixed x > 0 x [0, [. n 2 x f n (x) = (nx + 1) 3 n 2 x = n 3 (x + 1 n )3 = 1 x n (x s n. )3 n So tht f n (x) f(x) = 0 pointwise on [0, [. Then for x 0 we hve f n(x) = n2 (1 2nx) (nx + 1) 4 nd we see tht for x > 0 we hve f n(x) 0 s n wheres f n(0) = n 2. Here we see tht f n f only on for x > 0. This shows tht differentibility is not lwys respected by pointwise convergence. 5

6 The lst two exmples then led us to pose the question: wht extr condition (other thn just pointwise convergence) cn gurntee tht the limit function is lso continuous or differentible? The nswer to this is given by the concept of uniform convergence. 4 Uniform convergence We define for rel-vlued (or complex-vlued) function f on non-empty set I the supremum norm of f on the set I: f I = sup f(x). x I Note tht if f is bounded function on I then sup f(x) = sup{ f(x) : x I} x I exists, by the so-clled supremum xiom. Observe tht f(x) f I for ll x I, nd tht f(x) tkes on vlues which re rbitrrily ner f I. In prticulr f I = the lrgest vlue of f(x) whenever such vlue exists (such s when I is closed, bounded intervl nd f(x) is continuous function on I). The supremum norm hs the following properties for functions f nd g on set I: f I 0 nd f I = 0 f(x) = 0 for ll x I cf = c f I for ny constnt c f + g I f I + g I (tringle inequlity) f J f I when J is subset of I. The proof of these properties is left s n exercise for the interested reder. Now we come to the definition of uniform convergence: Definition 4.1 A sequence of functions f n (x) defined on n set I is sid to converge uniformly to f(x) on I if We write this s f n f I 0 s n. or s lim f n = f n uniformly on I 6

7 f n f uniformly on I s n. Uniform convergence implies pointwise convergence, however there re sequences which converge pointwise but not uniformly. Indeed we hve so tht f n (x) f(x) sup f n (x) f(x) = f n f I, x I f n f uniformly on I s n = f n (x) f(x) 0 for ech x I = f n f pointwise on I. We record this s result: Lemm 4.1 If the sequence of functions f n (x) converges uniformly to f(x) on the intervl I, then f n (x) converges pointwise to f(x). This Lemm sys tht the limit function obtined through uniform convergence (if this occurs) is the sme s the limit function obtined from pointwise convergence. Or: if f n (x) converges to f(x) uniformly, then it must converge to f(x) pointwise. This then tells us how to go bout testing for uniform convergence: first, obtin the pointwise limit f(x) nd then see if we hve uniform convergence to f(x). Exmple 4.1 f n (x) = e nx on [1, 3]. We hve seen bove tht f n (x) converges pointwise to f(x) = 0 for ech x [1, 3]. Then we hve f n (x) f(x) = f n (x) nd we then hve f n f = sup f n (x) x [1,3] = sup e nx x [1,3] = sup e nx x [1,3] = e n 0 s n. Thus we hve uniform convergence in this cse. Note tht the lst step follows from the observtion tht e nx is strictly decresing for x 0 with n 0, so tht e n e nx for ll x 1. 7

8 Exmple 4.2 f n (x) = xe nx on I = [0, [. Here the intervl is unbounded. First we look t pointwise convergence: f n (0) = 0 nd for x > 0 we hve tht f n (x) 0 s n. Thus f n (x) 0 pointwise on I. We now need to investigte uniform convergence. Since the limit function f(x) = 0 we hve f n f = sup xe nx x [0, [ = sup xe nx x [0, [ becuse f n (x) 0 for x 0. Now, we hve f n(x) = (1 nx)e nx for x > 0 (observe tht you should never differentite on closed intervls), nd we see tht f n(x) = 0 when x = 1/n. Further, f n(x) > 0 for 0 < x < 1/n, nd f n(x) < 0 for x > 1/n, so we conclude tht f n (x) hs mximum t x = 1/n nd hence f n f = sup xe nx x [0, [ = f n ( 1 n ) = 1 ne 0 s n. So we see tht the sequence of functions f n (x) = xe nx converges uniformly to 0 on the intervl I = [0, [. 5 Uniform convergence nd continuity. We now come to two importnt results. The first is the following. Theorem 5.1 Suppose f n (x) is sequence of continuous functions on n intervl I nd suppose lso tht f n (x) converges uniformly to f(x) on the intervl I. Then the limit function f(x) is lso continuous. Proof: First note tht for x, I we my write f(x) f() = [f(x) f n (x)] + [f n (x) f n ()] + [f n () f()] from which we obtin (using the tringle inequlity) f(x) f() f(x) f n (x) + f n (x) f n () + f n () f() (5.1) Then note tht, becuse sup f(x) f n (x) 0 s n we hve for ny given choice x I of ɛ > 0 nturl number N such tht sup f(x) f n (x) < ɛ x I 3 8

9 for ll n N. We lso hve f() f n () < ɛ 3 since f() f n () sup f(x) f n (x) < ɛ x I 3. Using this in eqution (5.1), we hve, for given ɛ > 0, nturl number N so tht f(x) f() 2ɛ 3 + f n(x) f n () for n N. Fix the choice of n N, sy n = N. We now use continuity of the f n (x): for ech ɛ > 0 there exists δ > 0 such tht f n (x) f n () < ɛ 3 whenever x < δ. Consequently, for x < δ we hve f(x) f() 2ɛ 3 + f n(x) f n () < 2ɛ 3 + ɛ 3 = ɛ. This mens tht for ny ɛ > 0 there exists δ > 0 such tht f(x) f() < ɛ whenever x < δ nd this mens tht f(x) f() s x. Tht is, f(x) is continuous t ech I. Remrk: This result is very useful s quick test for the bsence of uniform convergence: if (i) f n (x), n = 0, 1, 2,... is sequence of continuous functions (on some intervl); (ii) f n (x) converges pointwise to f(x); (iii) f(x) is not continuous; (iv) Then f n (x) does not converge uniformly to f(x). Exmple 5.1 f n (x) = e nx on [0, 3] is sequence of continuous functions, converging pointwise to f(x) defined by { 1 for x = 0 f(x) = 0 for 0 < x 3, which is not continuous, nd so, by Theorem 4.1, the sequence does not converge uniformly to f(x). The second result we mention is the following, which is of gret use in integrting series: 9

10 Theorem 5.2 Suppose tht f n (x) is sequence of continuous functions which converges uniformly to continuous function f(x) on bounded intervl [, b]. Then we hve lim n f n (x)dx = Proof: The proof is quite simple: Thus: f n (x)dx lim f n(x)dx = n b f(x)dx = (f n (x) f(x))dx = f n f f n (x) f(x) dx f n f dx 1 dx f(x)dx. = f n f (b ) 0 s n. f n (x)dx f(x)dx s n if f n f uniformly on I, which is wht we wnted to prove. Remrk 5.1 Theorem 5.2 is proved here for continuous functions so tht the integrls exist. It is however possible to replce the word continuous by the word integrble, nd the theorem is still true. We now give result bout uniform convergence nd differentibility: it tells us under which conditions the limit function f(x) is differentible whenever the functions of the sequence f n (x) re differentible. Theorem 5.3 Suppose tht {f n (x); n = 0, 1, 2,... } is sequence of functions on n intervl I nd stisfying the following conditions: (i) f n (x) is differentible on I for ech n = 0, 1, 2,... (ii) f n (x)converges pointwise to f(x) on I (iii) f n(x) is continuous for ech n nd f n g converges uniformly on I where g(x) is continuous function on I. Then the limit function f(x) is differentible nd f (x) = g(x). 10

11 This result is very useful, s we shll see, in exmining the differentibility of functionl series. Proof: Becuse of differentibility we hve (for, x I) f n (x) = f n () + x f n(t)dt Furthermore, since f n(t) g(t) uniformly on I, we know from Theorem 5.2 tht x f n(t)dt x g(t)dt when n. Also, f n (x) f(x) pointwise s n. From this it follows tht f(x) = f() + x g(t)dt on letting n. Since g(x) is continuous (it is the uniform limit of sequence of continuous functions, so it is continuous by Theorem 5.1), the integrl exists nd is primitive function of g(x). Differentiting this lst eqution, we obtin This concludes the proof. f (x) = g(x). 6 Applictions to functionl series. Definition 6.1 A functionl series is series where ech term of the series is function on n intervl I. We cn lso define pointwise convergence for functionl series: Definition 6.2 The functionl series is pointwise convergent for ech x I if the limit exists for ech x I. = lim N N 11

12 Thus, we lwys define sequence of prtil sums S N (x) given s so tht S N (x) = N S 0 (x) = u 0 (x), S 1 (x) = u 0 (x) + u 1 (x), S 2 (x) = u 0 (x) + u 1 (x) + u 2 (x),... nd if lim S N(x) N exists for x then we sy tht the series = lim N S N(x) converges t x. It converges pointwise on the intervl I if lim S N(x) N exists for ech x I. With these definitions, we deduce from Theorem 4.1 tht if the functions re ll continuous on I nd if the sequence of prtil sums S N (x) converges uniformly to S(x) on I, then S(x) is continuous. However, we would like n efficient wy of deciding if functionl series converges uniformly to (unique) limit. It is not t ll esy to pply the definition of uniform convergence to n infinite sum of functions, so nother method is desirble. The pproprite result is Weierstrss Mjornt Theorem: Theorem 6.1 Suppose tht the functionl series is defined on n intervl I nd tht there is sequence of positive constnts M k so tht for ll x I. If M k, k = 0, 1, 2,... converges, then converges uniformly on I. M k 12

13 Proof: If the conditions re fullfilled then we immeditely hve, from the Comprison Theorems for Positive Series, tht, for ech x I, the series is convergent, so tht is bsolutely convergent, nd therefore convergent. This mens tht is pointwise convergent on I, nd we denote the limit by S(x). We now show tht the prtil sums S N (x) = N converges uniformly to S(x) on I under the conditions of the theorem. We hve S(x) S N (x) = k=n+1 (ll we do is subtrct the first N terms from the series). Then it follows tht S(x) S N (x) k=n+1 k=n+1 for ech x I, since M k for ech x I ccording to our ssumption. Then S S N I k=n+1 We lso know (by ssumption) tht M k converges, so we must hve tht k=n+1 M k 0 s N. Consequently, M k. M k nd our result is proved. S S N I 0 s N, Corollry 6.1 If (i) the functionl series S(x) = converges uniformly on intervl I, 13

14 (ii) is continuous function on I for ech k = 0, 1, 2,..., then S(x) is continuous on I. Proof: Becuse finite sum of continuous functions is gin continuous function, it follows tht the prtil sums S N (x) = N re continuous functions for N = 0, 1, 2,.... Then by Theorem 5.1, we hve tht S(x) = lim N S N (x) is continuous function. Exmple 6.1 Tke the functionl series We hve k=1 = sin kx k 2 sin kx k 2. = sin kx k 2 1 k 2 since sin t 1 for ll rel t. We know (stndrd positive series) tht 1 converges (series of the form 1/k α converge for α > 1 nd diverge for α 1). Hence, by Weierstrss Mjornt Theorem, k=1 1 k 2 sin kx k 2 converges uniformly for ll x, nd by Corollry 5.1 this series is continuous function of x for ll x R. Remrk 6.1 One dvntge of Weierstrss Mjornt Theorem is tht we do not hve to clculte the vlue of the series t ech x I in order to decide if we hve uniform convergence. However, drwbck is tht the conditions of the theorem re only sufficient to estblish uniform convergence, they re not bsolutely necessry for uniform convergence. In the finl section of these lecture notes we give necessry nd sufficient condition for uniform convergence. Remrk 6.2 In our sttement of Weierstrss Mjornt Theorem, we hve not sid nything bout how to find the constnts M k. Usully we tke M k = sup, x I but this is not strictly necessry: ny sequence (of constnts) will do provided tht M k converges. 14

15 Another result of interest is the following: Theorem 6.2 If (i) the functionl series converges uniformly on the intervl I (ii) is continuous on I for ech k = 0, 1, 2,..., then x ( ) u k (t) dt = ( x ) u k (t)dt for ll, x I. In other words, if the series of continuous functions converges uniformly on I, then the integrl of the sum is the sum of the integrls of the functions, just s in the cse of finite sum. Proof: See Kompendium om Konvergens. We cn lso sy something bout the differentibility of the series, using Theorem 5.3 In this cse, s in the previous two theorems, we replce f n (x) by S N (t) nd f(x) by S(t). Thus, we wnt the following: S N (x) S(x) pointwise on I S N(x) G(x) uniformly on I S N (x) is continuously differentible for ech N nd then we my conclude tht S(x) is continuously differentible with S (x) = G(x). All we need is to formulte these requirements nd result s follows: Theorem 6.3 Suppose tht stisfies the following conditions: converges pointwise on I u k(x) converges uniformly on I is continuously differentible for ech k Then is continuously differentible nd ( ) d = dx Proof: See Kompendium om Konvergens. 15 u k(x).

16 7 APPENDIX. 7.1 Supremum nd Infimum: recpitultion. Definition 7.1 Let A R. Then the supremum of A, denoted by sup A, is defined s the smllest number R with the property tht x for ll x A. In mthemticl shorthnd we hve sup A = min{ R : x for ll x A}. Similrly, the infimum of A, denoted by inf A, is defined s the lrgest number b R with the property tht x b for ll x A. In mthemticl shorthnd we hve inf A = mx{b R : x b for ll x A}. Remrk 7.1 Note tht in these definitions neither the supremum nor the infimum need belong to the set A. Exmple 7.1 (i) A = [ 1, 3]. Here we hve sup A = 3, inf A = 1, nd both these belong to A. (ii) A =] 1, 3]. Here sup A = 3, inf A = 1, but only inf A belongs to A. (iii) A =] 1, 3[. Here sup A = 3, inf A = 1, nd both re not in A. (iv) A = [ 1, [. Here inf A = 1 wheres sup A does not exist. Definition 7.2 Let f : R R be function. Then the supremum of f(x) over A is defined s the smllest number R with the property tht f(x) for ll x A. In mthemticl shorthnd we hve sup f(x) = min{ R : f(x) for ll x A}. x A Similrly, the infimum of f(x) over A is defined s the lrgest number b R with the property tht f(x) b for ll x A. In mthemticl shorthnd we hve inf = mx{b R : f(x) b for ll x A}. x A Exmple 7.2 (i) f(x) = x 3, A = [ 1, 3] Then, since f(x) is strictly incresing function, we hve sup f(x) = 27, x A inf f(x) = 1. x A 16

17 (ii) f(x) = x 2, A = [ 1, 3] Then note tht f(x) = x 2 is not strictly incresing on this intervl: it is decresing on [ 1, 0] nd then strictly incresing on [0, 3]. So we hve nd sup f(x) = 1, x [ 1,0] inf f(x) = 0 x [ 1,0] sup f(x) = 9, x [0,3] Combining these two observtions, we find tht inf f(x) = 0. x [0,3] sup f(x) = 9, x [ 1,3] inf = 0. x [ 1,3] (iii) f(x) = rctn x, A = R. Here we hve strictly incresing function, nd we hve sup f(x) = π x R 2, inf f(x) = π x R 2. It is tempting to tke the lrgest vlue of function on n intervl s the supremum, nd the lest vlue for the infimum. The lst exmple shows tht the neither the supremum nor the infimum need be ttinble vlues of function. However, we hve the following simple but useful result: Lemm 7.1 Suppose tht f(x) is rel-vlued continuous function on the closed, bounded intervl [, b]. Then sup f(x) = mx{f(x) : x [, b]}, x [,b] inf f(x) = min{f(x) : x [, b]}. x [,b] Tht is, the supremum of continuous function over closed, bounded intervl is equl to its lrgest vlue over tht intervl, nd the infimum is the lest vlue of the function over the intervl. Proof: Since f(x) is continuous nd the intervl is closed, then f(x) hs lrgest vlue nd lest vlue on the intervl: there exist x 1, x 2 [, b] so tht f(x 1 ) f(x) f(x 2 ) for ll x [, b], nd we now see tht nd the result is proved. sup f(x) = f(x 2 ), x [,b] inf f(x) = f(x 1), x [,b] 17

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