A BRIEF INTRODUCTION TO UNIFORM CONVERGENCE. In the study of Fourier series, several questions arise naturally, such as: c n e int


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1 A BRIEF INTRODUCTION TO UNIFORM CONVERGENCE HANS RINGSTRÖM. Questions nd exmples In the study of Fourier series, severl questions rise nturlly, such s: () (2) re there conditions on c n, n Z, which ensure tht c n e int converges for ll t, nd, similrly, re there conditions on n, n = 0,,..., nd b n, n =, 2,..., which ensure tht ( n cos nt + b n sin nt) n= converges for ll t? re there conditions which ensure tht () nd (2) define continuous functions, continuously differentible functions etc.? when is the opertion d dt c n e int = d dt (c ne int ) = (nd the similr opertion for (2)) llowed? when is the opertion ( b ) c n e int dt = (nd the similr opertion for (2)) llowed? b c n e int dt inc n e int In order to ddress these questions, let us consider them in somewht greter generlity. Sy tht we hve sequence of (complex or rel vlued) functions f n, n =, 2,... or n = 0,,... (the cse n Z cn in prctice be reduced to two sequences of this form). We re interested in the sum f n. n= In order to illustrte tht the nswers to the bove questions re not quite s strightforwrd s might be expected, let us give n exmple. Exmple.. Let f n (t) = t 2 ( + t 2 ) n
2 2 HANS RINGSTRÖM for n = 0,,.... Compute f n (t). Solution: If t = 0, then f n (t) = 0 for ll n. In other words, s(0) = 0. For t 0, we hve t 2 ( ) n f n (t) = ( + t 2 ) n = t2 + t 2. The sum ppering on the right hnd side is the sum of geometric series. In fct, let α = + t 2. Note tht, since t 0, we hve 0 < α <. Compute ( + t 2 Thus, for t 0, To conclude ) n = t 2 α n = α = = + t2 + t +t 2 = + t2 t 2. 2 ( ) n + t 2 = t 2 + t2 t 2 = + t 2. { 0 t = 0, + t 2 t 0. In other words, s is not continuous function, since s(0) = 0 nd s(t) > for t 0. On the other hnd, ll the functions f n cn be differentited n rbitrry number of times. To conclude: even though the sum f n (t) exists for ll t, nd even though the prtil sums N re continuous (they re in fct k times continuously differentible for ny positive integer k), the sum itself is not even continuous. Exercise.2. Compute nd sketch the grph. Exercise.3. Compute nd sketch the grph. f n t ( + t 2 ) n t 3 ( + t 2 ) n
3 A BRIEF INTRODUCTION TO UNIFORM CONVERGENCE 3 It is possible to construct exmples which illustrte tht chnging the order of integrtion nd summtion is lso problemtic etc. We refer the reder interested in more exmples to Wlter Rudin s book Principles of Mthemticl Anlysis, Chpter Pointwise nd uniform convergence The reson we obtin the behviour exhibited in Exmple. is tht the sum converges pointwise but not uniformly. In the present section we wish to clrify the mening of these concepts. Since we re interested in Fourier series, we re interested in sums of the form f n (t). n= However, in the vrious definitions we shll crry out, it is convenient to note tht s cn be considered to be limit of sequence of functions s opposed to sum of sequence of functions. In fct, let n s n (t) = f k (t). Then lim s n(t). n Let us from now on consider sequence of functions s n, n =, 2,..., nd consider the limit of this sequence s n. Let us define wht is ment by pointwise convergence. Definition 2.. Let s n, n =, 2,..., be sequence of functions defined on n intervl I. If the sequence of numbers s n (t), n =, 2,..., converges for every t I, then the sequence of functions s n, n =, 2,..., is sid to converge pointwise on I. Remrk 2.2. If sequence of functions s n, n =, 2,..., converges pointwise on I, then we cn define the function s on I by lim n s n(t), nd we shll sy tht the sequence s n, n =, 2,..., converges to s pointwise on I. Exmple 2.3. One sequence which converges pointwise is s n, n =, 2,..., where n t 2 s n (t) = ( + t 2 ) k ; in Exmple. we proved tht this sequence converges pointwise on R. Exmple 2.4. Let s n (t) = t n. Then the sequence s n, n =, 2,..., converges pointwise on I = [0, ], since k=0 if 0 t < nd s n () =. If we let s n (t) = t n 0 lim n s n(t),
4 4 HANS RINGSTRÖM we thus hve { 0 t [0, ), t =. In other words, the function s hs one point of discontinuity, nmely t =. Exercise 2.5. Construct sequence of functions s n, n =, 2,..., on I = [0, ] such tht the functions s n re ll continuous, the sequence of functions s n, n =, 2,..., converges pointwise to function s on I, s hs two points of discontinuity in I. Cn you construct sequence s in the bove exercise such tht s hs three, four etc. points of discontinuity? Exercise 2.6. (This exercise is difficult). Construct sequence of functions s n, n =, 2,..., on I = [0, ] such tht the functions s n re ll continuous, the sequence of functions s n, n =, 2,..., converges pointwise to function s on I, s hs n infinite number of points of discontinuity in I. The concept of pointwise convergence should be contrsted with the concept of uniform convergence. Definition 2.7. Let s n, n =, 2,..., be sequence of functions defined on n intervl I. Then the sequence s n, n =, 2,..., is sid to converge uniformly on I to function s if, for every ɛ > 0, there is n N such tht n N implies tht for ll t I. s n (t) s(t) < ɛ Remrk 2.8. If the sequence s n, n =, 2,..., converges uniformly on I to function s, it converges to s pointwise on I (prove this). Let us demonstrte tht the sequence considered in Exmple 2.4 does not converge uniformly. In order to do so, let us ssume tht the sequence s n, n =, 2,..., converges uniformly on I = [0, ] to function S (if we cn deduce contrdiction from this ssumption, we re llowed to conclude tht the convergence is not uniform). Due to Remrk 2.8, the function S hs to coincide with the function s ppering in Exmple 2.4. Due to the definition of uniform convergence, we re llowed to first fix ɛ > 0 (let us ssume ɛ < ). Given this ɛ, there is then n N such tht for n N, we hve s n (t) s(t) < ɛ for ll t I. In the cse considered in Exmple 2.4, 0 for t [0, ). We thus hve s n (t) < ɛ for every t [0, ). Since the functions s n re continuous, this mens tht s n () ɛ, but we know tht s n () =. Since ɛ <, we hve contrdiction. Thus the convergence is not uniform.
5 A BRIEF INTRODUCTION TO UNIFORM CONVERGENCE 5 Exercise 2.9. Drw picture to illustrte the bove rgument. In order to mke the distinction between pointwise nd uniform convergence clerer, let us write down the relevnt questions to sk in order to check whether one hs pointwise or uniform convergence. Pointwise convergence: first fix t I nd then sk if, for every ɛ > 0, there is n N such tht for n N, s n (t) s(t) < ɛ (here N depends on ɛ nd t). Uniform convergence: sk if, for every ɛ > 0, there is n N such tht for n N, s n (t) s(t) < ɛ for ll t (here N only depends on ɛ). Let us, finlly, define uniform convergence explicitly for series. Definition 2.0. Let f k, k =, 2,..., be sequence of functions defined on n intervl I. Then the sum is sid to converge uniformly on I to s if the the prtil sums s n, n =, 2,..., where n s n = f k, converge uniformly on I to s. f k 3. Consequences of uniform convergence The question remins: wht is the use of uniform convergence? Let us quote few theorems tht we shll need. Let us strt with criterion which ensures tht the sum is continuous. Theorem 3.. Let f k, k =, 2,..., be sequence of continuous functions defined on n intervl I. Assume tht the sum converges uniformly on I to s. Then s is continuous on I. As ws noted in the introduction, it is of interest to find criterion which llows us to chnge the order of integrtion nd summtion. Theorem 3.2. Let f k, k =, 2,..., be sequence of Riemnn integrble functions on n intervl I = [, b] (where < < b < ). Assume tht the sum converges uniformly on I to s. Then s is Riemnn integrble on I nd b b b s(t)dt = f k (t)dt = f k (t)dt. Finlly, let us ddress the issue of differentition. f k f k
6 6 HANS RINGSTRÖM Theorem 3.3. Let f k, k =, 2,..., be sequence of continuous nd continuously differentible functions defined on n intervl I. Assume tht the sums f k, converge uniformly on I to s nd s respectively. Then s is continuous nd continuously differentible on I, nd ( ) s = f k = on I. We refer the reder interested in proof of these sttements to Wlter Rudin s book Principles of Mthemticl Anlysis, Chpter 7. f k f k 4. Weierstrss Mtest None of the theorems mentioned in the previous section re very useful unless we hve good criterion which ensures tht series converges uniformly. Such criterion is provided by the Weierstrss Mtest. Theorem 4. (Weierstrss M test). Let f k, k =, 2,..., be sequence of functions defined on n intervl I nd ssume tht there re numbers M k, k =, 2,..., such tht f k (t) M k, for t I nd k =, 2,..., nd such tht is convergent. Then the sum M k f k (t) is well defined for ll t I nd f k converges uniformly on I to s. Exmple 4.2. Consider k 2 sin kt. + We wish to use the bove results to demonstrte tht this sum defines continuous function. We hve f k (t) = k 2 sin kt, + k =, 2,.... In order to be llowed to pply the Weierstrss Mtest, we wish to estimte f k (t). We hve Let us choose f k (t) = k 2 sin kt + k 2 +. M k = k 2 +,
7 A BRIEF INTRODUCTION TO UNIFORM CONVERGENCE 7 k =, 2,.... Then M k is convergent nd Theorem 4. pplies. Thus the sum k 2 sin kt + is well defined for ll t R nd f k converges uniformly on R to s. Since the functions f k re continuous, Theorem 3. pplies in order to yield the conclusion tht s is continuous. Exmple 4.3. More generlly, if n, n = 0,, 2,..., nd b n, n =, 2,..., re complex numbers such tht ( n + b n ) <, then n= ( n cos nt + b n sin nt) n= is well defined for ll t R. Moreover, we obtin uniform convergence due to the Weierstrss Mtest nd continuity of s due to Theorem 3. (when integrting, we cn lso chnge the order of summtion nd integrtion due to Theorem 3.2). Exmple 4.4. By n rgument similr to the one presented in Exmple 4.2, it is possible to prove tht if c n, n Z, is sequence of complex numbers such tht c n <, then c n e int, conveges uniformly on R to function s. Moreover, s is continuous nd, due to Theorem 3.2, we re llowed to chnge the order of summtion nd integrtion in order to conclude tht 2π π π π s(t)e imt dt = 2π = π ( π c n 2π the second equlity is justified by Theorem 3.2. Let us turn to differentibility. π c n e int imt ) dt e int imt dt = c m ; Exmple 4.5. Consider ( + k 2 ) 2 eikt.
8 8 HANS RINGSTRÖM In this cse Since nd f k (t) = ( + k 2 ) 2 eikt, f k(t) ik = ( + k 2 ) 2 eikt. f k (t) ( + k 2 ) 2 ( + k 2 ) 2 <, we cn pply the Weierstrss Mtest in order to conclude tht conveges uniformly on R to function s. Since nd f k(t) f k k ( + k 2 ) 2 k ( + k 2 ) 2 <, we cn lso pply the Weierstrss Mtest in order to conclude tht conveges uniformly on R to function s. Due to Theorem 3.3 we thus conclude tht s is continuous nd continuously differentible nd tht s (t) = ik ( + k 2 ) 2 eikt. Exmple 4.6. By n rgument similr to tht given in the previous exmple we hve the following conclusion. If m is n integer nd n m c n <, n Z then c n e int n Z is mtimes continuously differentible nd Similrly, if then s (t) = n Z f k inc n e int,..., s (m) (t) = n Z(in) m c n e int. n m ( n + b n ) <, n= ( n cos nt + b n sin nt) n= is m times continuously differentible nd we re llowed to differentite under the summtion sign.
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