Lectures On Fourier Series. S. Kesavan Institute of Mathematical Sciences Chennai , INDIA

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1 Lectures On Fourier Series S. Kesvn Institute of Mthemticl Sciences Chenni-6 113, INDIA Third Annul Foundtionl School - Prt I December 4 3, 6

2 Contents 1 Introduction 3 Orthonorml Sets 6 3 Vritions on the Theme 11 4 The Riemnn-Lebesgue Lemm 1 5 The Dirichlet, Fourier nd Fejér Kernels 15 6 Fourier Series of Continuous Functions 3 7 Fejér s Theorem 7 8 Regulrity 3 9 Pointwise Convergence 39 1 Termwise Integrtion Termwise Differentition 47 References 5

3 1 Introduction With the invention of clculus by Newton ( ) nd Leibnitz ( ), there ws surge of ctivity in vrious topics of mthemticl physics, notbly in the study of boundry vlue problems ssocited to vibrtions of strings stretched between points nd vibrtion of brs or columns of ir ssocited with mthemticl theories of musicl vibrtions. Erly contributors to the theory of vibrting strings include B. Tylor ( ), D. Bernoulli (17 178), L. Euler ( ) nd d Alembert ( ). By the middle of the eighteenth century, d Alembert, Bernoulli nd Euler hd dvnced the theory of vibrting strings to the stge where the prtil differentil eqution (now clled the wve eqution) y t = y x ws known nd solution of the boundry vlue problem hd been found from the generl solution of tht eqution. The concept of fundmentl modes of vibrtion led them to the notion of superposition of solutions nd Bernoulli proposed solution of the form y(x, t) = n=1 b n sin nπx c cos nπt c (1.1) (where c is the length of the string). The initil position of the string f(x) will then be represented by f(x) = n=1 b n sin nπx c ( x c). (1.) Lter, Euler gve the formuls for the coefficients b n. But the generl concept of function hd not been clrified nd lengthy controversy took plce over the question of representing rbitrry functions on finite intervl by series of sine functions. The French mthemticin d Alembert gve n elegnt solution in the form y(x, t) = v(t + x) v(t x), (1.3) nd he believed tht he hd solved the problem completely. 3

4 At tht time, the word function hd very restricted mening nd ws understood s something given by n nlytic expression. Euler thought tht the initil position of plucked string need not lwys be function, but some form where different prts could be expressed by different functions. In other words, function nd grph ment different things. For every function, we cn drw its grph but every grph tht cn be drwn need not come from function. Euler strongly objected to Bernoulli s clim tht every solution to the problem of plucked string could be represented in the form (1.1), on two counts. First of ll the right-hnd side of (1.1) ws periodic function, while the left-hnd side ws rbitrry. Further, the right-hnd side of (1.) ws n nlytic formul nd hence function, while the left-hnd side, f, could be ny grph. So Euler believed tht d Alembert s solution ws vlid when f ws ny grph while Bernoulli s solution ws pplicble only to very restricted clss of functions. J. B. Fourier ( ) presented mny instructive exmples of expnsions of functions in trigonometric series in connection with boundry vlue problems ssocited to the conduction of het. His book Théorie Anlytique de l Chleur (18) is clssic. Fourier never justified the convergence of his series expnsions nd this ws objected to by his contemporries Lgrnge, Legendre nd Lplce. Fourier sserted tht ny periodic function could be written s trigonometric series. Dirichlet ( ) firmly estblished in 189 (nerly seventy yers fter the controversy strted), sufficient conditions on function f so tht its Fourier series converges to its vlue t point. Since then lot of ides nd theories grew out of need to understnd wht these series ment. Amongst them re Cntor s theory of infinite sets, the rigorous notion of function, the theories of integrtion due to Riemnn nd Lebesgue nd the theories of summbility of series. In mthemticl nlysis, we lwys try to find pproximtion of objects by simpler objects. For exmple, we pproximte rel numbers by rtionls. By truncting the Tylor series of function, we pproximte the function by polynomil. However, for function to dmit Tylor series, it hs to be infinitely differentible (but this is not sufficient!) in some intervl but this is quite restrictive. Indeed Weierstrss pproximtion theorem sttes tht ny continuous function defined on finite closed intervl cn be pproximted uniformly by polynomil. Now, consider the set of ll functions {1} {cosnt, sin nt n N} on 4

5 [, π]. Given ny point t [, π], the constnt function does not vnish t t. Further if t 1 nd t re distinct points in [, π] then we cn lwys find function in the bove set such tht it tkes different vlues t t 1 nd t. Thus, the set of ll trigonometric polynomils, viz. functions of the form N f(t) = + ( n cosnt + b n sin nt) (1.4) n=1 form n lgebr (i.e. the set is closed under pointwise ddition, multipliction nd sclr multipliction) nd it does not vnish t ny point nd seprtes points. By the Stone-Weierstrss theorem (which generlizes the Weierstrss pproximtion theorem), every periodic continuous function on [, π] cn be pproximted uniformly by trigonometric polynomils. A trigonometric polynomil of the form (1.4) cn lso be written in exponentil form: N f(x) = c n exp(inx). (1.5) n= N It is esy to see tht = c, b = nd tht or, equivlently, n = c n + c n ; b n = i(c n c n ) c n = n ib n ; c n = n + ib n. If n is non-zero integer, then exp(inx) is the derivtive of exp(inx)/in, which lso hs period π. Thus { 1 π 1 if n = exp(inx) dx = (1.6) π if n. Mulitplying (1.5) by exp( imx) nd integrting over [, π], we get, in view of (1.6), c m = 1 f(x) exp( imx) dx. (1.7) π This gives us, for ny positive integer m, m = 1 f(x) cosmxdx π b m = 1 (1.8) f(x) sin mxdx. π 5

6 It is common prctice to replce by /, so tht (1.8) is vlid for s well. We now generlize this to define the trigonometric series n= exp(inx), or, equivlently, + ( n cosnx + b n sin nx). n=1 Given π-periodic function f on [, π], we define n (n ) nd b n (n 1) by (1.8) nd the resulting series is clled the Fourier series of the function f. The n nd b n re clled the Fourier coefficients of f. The bsic question now is when does the Fourier series of function converge? If it converges, does it converge to the vlue of f t the given point? In other words, to wht extent does the Fourier series of function represent the function itself? In the sequel we will try nd nswer some of these questions. To begin with, we will look t n bstrct sitution suggested by the reltions (1.6). Orthonorml Sets Let H be Hilbert Spce (over R or C). We denote the inner product in H by (, ) nd the norm it genertes by. Definition.1. Let S = {u i i I}, where I is n indexing set, be collection of elements in H. The set S is sid to be orthonorml if } u i = 1 for ll i I (.1) (u i, u j ) = for ll i j, i, j I. Exmple.1. The stndrd bsis {e i } 1 i n in R n, where e i hs 1 in the i th -coordinte nd zero elsewhere, is orthonorml in R n with the usul innerproduct nd the eucliden norm. 6

7 Exmple.. Consider the spce of squre summble sequences l, i.e. { } l = x = (x i ) x i <. Then gin, the set of sequences {e n } n=1 where e n hs 1 s the n th entry nd zero t ll other plces, is orthonorml in l. Exmple.3. The sequence { sin nπx} is orthonorml in L (, 1). Exmple.4. The sequence { 1 π } { sinnx cos nx π, π n N} is orthonorml in L (, π). Proposition.1. Let H be seprble Hilbert spce. Then ny orthonorml set is t most countble. Proof. Let {x n } be countble dense set in H. If u nd v re elements in n orthonorml set, we hve u v =. Thus ech of the blls B n = B(x n ; /4) cn contin t most one element of n orthonorml set. Since the {x n } form dense set, every member of H must belong to one such bll. Hence the result. i=1 Henceforth, we will ssume tht H is seprble Hilbert spce over R. Proposition.. Let {e 1,...,e n } be finite orthonorml set in Hilbert spce H. Then, for ny x H, we hve n (x, e i ) x. (.) i=1 Proof. We hve x n i=1 (x, e i)e i. Expnding this, we get, using the fct tht the {e i } re orthonorml, which proves (.). x + n (x, e i ) i=1 n (x, e i ) Theorem.1 (Bessel s Inequlity). If {e i } is n orthonorml set in Hilbert Spce H, then (x, e i ) x. (.3) i 7 i=1

8 Proof. Since H is seprble, {e i } is t most countble. The result, in the finite cse, hs lredy been shown. If {e i } is countbly infinite, then for ech n, we hve tht (.) is vlid. Thus, since the result is true for ll prtil sums, it is true for the series s well nd we get (.3). Corollry.1. If {e n } is n orthonorml sequence in H, then for every x H, (x, e n ) s n. It is immedite to see tht the elements of n orthonorml set re linerly independent. Further, given set of linerly independent elements {x 1,...,x n } in H, we cn produce n orthonorml set {e 1,...,e n } such tht the liner spns of {x 1,..., x k } nd {e 1,...,e k } coincide for ll 1 k n. Indeed, set e 1 = x 1 / x 1. Define e = x (x, e 1 )e 1 x 1 (x, e 1 )e 1. (Notice, by the liner independence of x 1 nd x, the vector x (x, e 1 )e 1 cnnot be zero.) It is esy to see tht e = 1 nd tht (e 1, e ) =. In generl, ssume tht we hve constructed e 1,...,e k such tht (i) ech e i (1 i k) is liner combintion of x 1,..., x i nd (ii) the {e i } re orthonorml. Now define e k+1 = x k+1 k i=1 (x k+1, e i )e i x k+1 k i=1 (x k+1, e i )e i. Thus, inductively we obtin {e 1,...,e n }. This procedure is clled the Grm- Schmidt orthogonliztion procedure. Thus, if H is finite dimensionl spce, we cn construct n orthonorml bsis for H. Henceforth, we will ssume tht H is infinite dimensionl nd seprble. Definition.. An orthonorml set is complete if it is mximl with respect to the prtil ordering on orthonorml sets in H induced by inclusion. Let {e n } be n orthonorml sequence in n infinite dimensionl (seprble) Hilbert spce. Let x H. Define y n = n (x, e n )e n. i=1 8

9 Then, for m > n, y n y m = m i=n+1 (x, e i ) nd the right-hnd side tends to zero, by Bessel s inequlity. Thus, {y n } is Cuchy sequence nd so converges in H. We define the limit to be (x, e i )e i. i=1 Proposition.3. The vector x j=1 (x, e j)e j is orthogonl to ech e i. Further, x (x, e i )e i = x (x, e i ). (.4) i=1 Proof. Given ny n, set y n = n i=1 (x, e j)e j. Then y n i=1 (x, e i)e i in H. Now, if 1 i n, clerly (x y n, e i ) =. Fix i, nd the bove reltion holds for ll n i. Thus, i=1 (x j (x, e j )e j, e i ) =. Now, x y n = x + = x n (x, e i ) i=1 n (x, e i ). i=1 Thus, pssing to the limit s n, we get (.4). n (x, e i ) Theorem.. Let H be Hilbert spce nd {e i } n orthonorml set in H. The following re equivlent. (i) {e i } is complete. (ii) If x H such tht (x, e i ) = for ll i, then x =. i=1 (iii) If x H, then x = j (x, e j )e j. (.5) 9

10 (iv) (Prsevl s Identity) If x H, then x = i (x, e i ). (.6) { Proof. (i) (ii) If (x, e i ) = for ll i nd x, then {e i } x x } will be n orthonorml set contrdicting the mximlity of {e i }. (ii) (iii) By Proposition.3, x j (x, e j)e j is orthogonl to ech e i nd so, by (ii) we get (.5). (iii) (iv) This is n immedite consequence of (.4). (iv) (i) If {e i } were not mximl, there exists e H such tht e = 1 nd (e, e i ) = for ll i. This contrdicts (.6). Corollry.. An orthonorml set {e i } in Hilbert spce H is complete, if, nd only if, the liner spn of the {e i }, i.e. the spce of ll (finite) liner combintions of the {e i }, is dense in H. Proof. If {e i } is complete, then by (.5) we get tht ech x H is such tht x = lim n n (x, e i )e i. i=1 Thus the liner spn of {e i } is dense in H. Conversely, if the liner spn is dense, then, if x is orthogonl to ll the e i, it follows tht x =. Thus {e i } is complete. Remrk.1. In view of (.5), complete orthonorml set is lso clled n orthonorml bsis. Exmple.5. Consider the sequence {e n } in l (cf. Exmple.). This sequence is complete in l since x = x i = i=1 (x, e i ). i=1 Exmple.6. Consider the Hilbert spce L (, π). It is known tht continuous functions with compct support re dense in this spce. Such functions re periodic (they vnish t nd π) nd we sw in 1 tht, by virtue of the Stone-Weierstrss theorem, they cn be uniformly pproximted by 1

11 trigonometric polynomils. It then follows, fortiori, tht they cn lso be pproximted in the L -norm. Thus, the trignometric polynomils re dense in L (, π) nd so, by the preceding corollry, the functions { 1 π } { cosnt sin nt, n N π π } form complete orthonorml set in L (, π). In prticulr, if f L (, π), we hve, by Prsevl s identity, f dx = 1 ( f(t) dt) + π [ ( 1 π ) + f(t) cosnt dt + 1 ( ) ] f(t) sin nt dt π π which yields n=1 1 π f dx = π + ( n + b n ) (.7) n=1 where n, b n re the Fourier coefficients of f given by (1.8). By nlogy with the bove exmple, if {e i } is n orthonorml bsis for seprble Hilbert spce H, we sy tht the Fourier series of x is the series i=1 (x, e i)e i nd the quntities (x, e i ) re clled the Fourier coefficients. 3 Vritions on the Theme Let f be π-periodic function defined on [, π]. The Fourier coefficients of f re given by the formuls (1.8). These formuls lso mke sense if f L 1 (, π). By chnging the vlues of function t finite number of points, we do not chnge the vlues of the Fourier coefficients. (Indeed, recll tht the spces L p re only equivlence clsses of functions, under the equivlence reltion given by f g if f = g.e.; thus, it is meningless to tlk of the vlue of function in L p t prticulr point). Given function h such tht h(π) h(), we cn redefine it so tht h(π) = h() nd then extend it periodiclly s π-periodic function over R. We cn lso declre the function to be undefined t or π. Thus we re forced to consider the Fourier series of functions with jump discontinuities. 11

12 If function belongs to L 1 (, π) then we cn extend it either s n odd function or s n even function to L 1 (, π). In the former cse, ll the coefficients n will be zero nd we get series only involving the functions sin nt. This is clled the Fourier sine series of the given function. Similrly, in the ltter cse, only the coefficients n will be non-zero nd the resulting series, which involves only the functions cos nt, is clled the Fourier cosine series of the function. We cn lso rewrite the Fourier series of function in the mplitude-phse form. Indeed, let + ( n cosnt + b n sin nt) (3.1) n=1 be the Fourier series of function. Set In other words, n = d n cosφ n, b n = d n sin φ n. d n = n + b n, φ n = cos 1 ( n /d n ). Then the series (3.1) cn be rewritten s + n d n cos(nt φ n ). (3.) 4 The Riemnn-Lebesgue Lemm We hve seen erlier tht if {e n } were n orthonorml sequence in Hilbert spce H, then, for ny x H, we hve lim (x, e n) =. n In prticulr, if f L (, π) we deduce tht lim n f(t) cosnt dt = lim f(t) sinnt dt =. n This result is one of the forms of wht is clled the Riemnn-Lebesgue lemm. We now prove very useful generliztion of this. 1

13 Theorem 4.1 (Generlized Riemnn-Lebesgue Lemm). Let f L 1 (, b), where < b +. Let h be bounded mesurble function defined on R, such tht Then 1 lim c ± c lim ω c h(t) dt =. (4.1) f(t)h(ωt) dt =. (4.) Proof. We extend f by zero outside (, b) so tht we cn consider f L 1 (R). Let us consider n intervl [c, d] (, ). Then χ [c,d] h(ωt) dt = d = 1 ω h(ωt) dt = 1 ω c dω h(t) dt 1 ω dω cω cω h(t) dt h(t) dt nd, by hypothesis, both integrls tend to zero s ω. The result now follows, by linerity, to ll step functions mde up of chrcteristic functions of intervls. However such functions re dense in L 1 (, ). (Indeed continuous functions with compct support re dense in L 1 (, ); such functions re uniformly continuous nd by prtitioning the intervl contining the support, we cn pproximte continuous functions with compct support by step functions uniformly nd hence, fortiori, in L 1 (, ).) Let h M. Thus, if f L 1 (, ), then find step function g such tht f g dx < ε, for given ε >. Now, M f(t)h(ωt) dt f(t) g(t) h(ωt) dt + g(t)h(ωt) dt ε + g(t)h(ωt) dt nd for ω lrge enough the second term cn lso be mde to be less thn ε. A similr rgument holds for f(t)h(ωt) dt nd this completes the proof. Corollry 4.1. If f L 1 (, b) then lim n f(t) cosnt dt = lim n 13 f(t) sin nt dt =. (4.3)

14 Proof. 1 c 1 c costdt c = 1 (sin c) c 1 c sin t dt = 1 (1 cos c) c c c nd both tend to zero s c. We now give n immedite ppliction of this result. Given convergent series n=1 α n, we know tht α n s n. We now sk if trigonometric series + ( n cosnt + b n sin nt) n=1 is convergent, whether n nd b n s n. Theorem 4. (Cntor-Lebesgue Theorem). If trigonometric series + n=1 ( n cosnt + b n sin nt) converges on set E whose (Lebesgue) mesure is positive, then n nd b n. Proof. Without loss of generlity, we my ssume tht E hs finite mesure. We rewrite the trigonometric series s in (3.) where d n = n + b n nd φ n = cos 1 ( n /d n ). Since the series converges, it follows tht for ll t E, d n cos(nt φ n ) s n. Assume tht {d n } does not converge to zero. Then, there exists ε > nd subsequence {n k } such tht d nk ε >, for ll k. Then, it follows tht cos(n k t φ nk ) s k for ll t E. Since E is of finite mesure, it follows, from the dominted convergence theorem, tht cos (n k t φ nk ) dt. (4.4) E 14

15 Now cos (n k t φ nk ) = 1 [1 + cos (n kt φ nk )]. But then χ E L 1 (R) nd so cos (n k t φ nk ) dt = χ E (t) cos (n k t φ nk ) dt E R = cos φ nk χ E (t) cos n k t dt + R + sin φ nk χ E (t) sin n k t dt nd both the integrls on the right-hnd side tend to zero by Corollry 4.1. It then follows tht cos (n k t φ nk ) dt µ(e) > E which contrdicts (4.4). This shows tht d n nd so n nd b n. 5 The Dirichlet, Fourier nd Fejér Kernels Let f L 1 (, π) with Fourier series given by f(t) + ( n cosnt + b n sin nt) (5.1) n=1 where n, b n re the Fourier coefficients defined s in (1.8). We wish to exmine the convergence of this series. In prticulr, we will be interested, in the sequel, to nswers (positive or negtive) to the following questions. Does the Fourier series converge t ll points t [, π]? If it converges t t [, π], does it converge to f(t)? If it converges t ll t [, π], is the convergence uniform? To discuss the convergence, pointwise or uniform, of the Fourier series, we need to discuss the convergence of the sequence {s n } of prtil sums. We hve s n (t) = n + ( k coskt + b k sin kt). (5.) k=1 15 R

16 Equivlently, in exponentil form, s n (t) = n k= n c k exp(ikt). Using the expression for the c k (cf. (1.7)) we get where s n (t) = n 1 π k= n = 1 π D n (t) = f(x) exp(ik(t x)) dx f(x)d n (t x) dx n k= n exp(ikt). (5.3) The π-periodic function D n (t) is clled the Dirichlet Kernel. Proposition 5.1. Let {s n } be the sequence of prtil sums of the Fourier series of f L 1 (, π) which is π-periodic. Then s n (t) = 1 π = 1 π = 1 π = 1 π = 1 π f(x)d n (t x) dx f(t x)d n (x) dx f(x)d n (x t) dx f(t + x)d n (x) dx (f(t + x) + f(t x))d n (x) dx (5.4) (5.4b) (5.4c) (5.4d) (5.4e) Proof. We hve lredy estblished (5.4). By chnge of vrible t x = y we get s n (t) = 1 +t f(t y)d n (y) dy. π +t By the π-periodicity, it follows tht the integrl does not chnge s long s the length of the intervl of integrtion is π. This proves (5.4b). The reltion 16

17 (5.4c) follows from (5.4) since D n is esily seen to be n even function. Reltion (5.4d) follows from (5.4c), gin by chnge of vrible y = x t nd the fct tht the integrls do not chnge s long s the length of the intervl is π. Finlly, we split the integrl in (5.4d) s the sum of integrls over [, ] nd [, π]. Now, f(t + x)d n (x) dx = f(t y)d n (y) dy using the chnge of vrible y = x nd the evenness of D n. This proves (5.4e). Proposition 5.. Let n be n integer. Then sin(n + 1)t D n (t) = sin t t kπ, k N {}. n + 1 t = kπ, k N {}. (5.5) Further, 1 π D n (t) dt = 1. (5.6) π Proof. When n = kπ, clerly D n (t) = n+1. Assume n kπ, k N {}. Then (exp(it) 1)D n (t) = exp(i(n + 1)t) exp( int) Multiplying both sides by exp( it/) we immeditely deduce (5.5). The reltion (5.6) immeditely follows from the definition of D n (t) (cf. (5.3)) nd the reltions (1.6). We now introduce two other functions which, together with the Dirichlet kernel, will ply n importnt role in the study of the convergence of Fourier series. Definition 5.1. The continuous Fourier kernel is given by sin ωt, t Φ(ω, t) = t ω, t = (5.7) 17

18 where ω nd t re rel numbers. The ssocited discrete Fourier kernel is given by sin(n + 1)t t Φ n (t) = t (5.8) n + 1 t =, where t R nd n is non-negtive integer. Remrk 5.1. Clerly D n is π-periodic. It is esy to see tht sin(n+1/)t sin(t/) n+1 s t. Thus D n is continuous. Similrly, it is esy to see tht the continuous nd discrete Fourier kernels re lso continuous. Definition 5.. The Fejér kernel is defined by K n (t) = 1 n + 1 n D k (t). (5.9) k= Proposition 5.3. Let n be n integer. Then K n (t) = 1 n cos(n + 1)t 1 cos t Further, K n nd if < δ t π, we hve (5.1) 1 π K n (t) dt = 1. (5.11) π K n (t) Proof. As before, observe tht (n + 1)(1 cosδ). (5.1) (n + 1)K n (t) ( exp(it) 1 )( exp( it) 1 ) = ( exp( it) 1 ) m ( exp ( i(k + 1)t ) ) exp( ikt) k= = exp ( i(n + 1)t ) exp ( i(n + 1)t ) from which we deduce (5.1). The reltion (5.11) follows directly from the definition (cf. (5.9)) nd the reltion (5.6). Tht K n is non-negtive follows immeditely from (5.1). So does reltion (5.1). 18

19 We now derive some estimtes for integrls of the Dirichlet nd Fourier kernels. First we need technicl result. Lemm 5.1. Let {A k } be sequence of rel numbers such tht A k 1 > nd A k < nd such tht A k+1 < A k for ll k N. Then, for every k N, we hve < A A k < A 1. (5.13) Proof. Let k be odd. Then A 1 + (A + A 3 ) + (A 4 + A 5 ) (A k 1 + A k ) is such tht ech term in prentheses is negtive. Thus the sum is less thn A 1. Agin (A 1 + A ) + (A 3 + A 4 ) (A k + A k 1 ) + A k is such tht ech term in prentheses is greter thn nd A k >. Thus the sum is greter thn. This proves (5.13) in the cse k is odd. The proof when k is even is similr. Proposition 5.4. Let < b π. Let n be n integer. Then sin(n + 1/)t dt sin(t/) 4π. (5.14) Proof. Let A k = kπ/(n+1/) (k 1)π/(n+1/) sin(n + 1/)t sin(t/) dt, 1 k n + 1. Then in ech such intervl, the numertor of the integrnd vries like sin t between (k 1)π nd kπ. On the other hnd sin(t/) is positive nd increses. Thus clerly [ A 1 > nd A) k lterntes [ in sign nd decreses in bsolute vlue. Now let (k 1)π, kπ nπ or ]., π n+1/ n+1/ n+1/ ( nπ If is in the interior of ny of these intervls (or, if ],, π when n+1/ k = n + 1), we hve (k 1)π/(n+1/) sin(n + 1/)t sin(t/) 19 dt

20 is either greter thn (becuse the integrnd is greter thn, when k is odd) or is less thn (k even) nd sin(n + 1/)t dt sin(t/) < A k. (k 1)π/(n+1/) Thus sin(n + 1/)t sin(t/) dt = A A k 1 + (k 1)π/(n+1/) sin(n + 1/)t sin(t/) dt nd it follows tht < sin(n + 1/)t sin(t/) dt < A 1. Now, in the intervl [, π/(n + 1/)], the integrnd is positive nd decresing with mximum vlue (n + 1) t t =. Thus π A 1 < (n + 1) (n + 1/) = π. Now sin(n + 1/)t sin(t/) dt = 4π sin(n + 1/)t sin(t/) dt sin(n + 1/)t sin(t/) dt Proposition 5.5. Let < b. Then sin ωt dt t π (5.15) for ω >. Proof. Define A k = kπ/ω (k 1)π/ω sin ωt t dt = kπ (k 1)π sin t t dt.

21 Then, gin, {A k } hs lternting signs nd decreses in bsolute vlue. As in the previous lemm, we get Since sint t < sin ωt dt < A 1. t 1, we get A1 π nd so sin ωt b dt t = sin ωt dt t sin ωt t dt π. The discrete Fourier kernel is very good pproximtion of the Dirichlet kernel. Proposition 5.6. Let f L 1 (, π). Let < r π. Then r lim n f(t) sin(n + 1 )t sin t whenever either limit exists. Proof. By L Hospitl s rule, we get Define lim t r dt = lim n ( 1 sin t 1 t f(t) sin(n + 1 )t t ) =. 1 g(t) = sin(t/) 1 t t/ t =. dt (5.16) Then g is continuous. Thus it is bounded in [, r] for < r π nd the function f(t)g(t)χ [,r] (t) is integrble. So re the functions nd φ(t) = f(t)g(t) cos(t/)χ [,r] (t) ψ(t) = f(t)g(t) sin(t/)χ [,r] (t). Thus, r f(t) sin(n+ 1 [ )t 1 sin(t/) 1 ] dt = t/ φ(t) sin nt dt+ ψ(t) cosnt dt nd, by the Riemnn-Lebesgue lemm (cf. Corollry 4.1), both the integrls on the right-hnd side tend to zero s n nd the result follows. 1

22 Both the functions D n (t) nd Φ n (t) enjoy the Riemnn-Lebesgue property. More precisely, we hve the following result. Proposition 5.7. Let f L 1 (, π) nd let < r π. Then lim n r Proof. Since r >, the functions f(t)d n (t) dt = lim f(t)φ n (t) dt =. (5.17) r r φ(t) = f(t) sin(t/)g(t)χ [r,π] (t) ψ(t) = f(t) cos(t/)g(t)χ [r,π] (t) where g(t) = 1 or 1, re integrble nd the result follows, once gin, sin(t/) t/ from the Riemnn-Lebesgue lemm (Corollry 4.1). Theorem 5.1 (The loclistion principle). Let f L 1 (, π). Let < r < π. Then, for x [, π], ( r ) (f(x + t)dn (t) ) dt =. lim n r Proof. Fix x [, π]. Then, by the preceding proposition, we hve Now, r r f(x t)d n (t) dt s n. f(x t)d n (t) dt = r f(x + t)d n (t) dt using the evenness of D n nd so once gin this integrl lso tends to zero. This completes the proof. Note tht the Fourier coefficients depend on the vlues of function f throughout the intervl [, π]. However, if f nd g re in L 1 (, π) nd for some t [, π] nd r >, we hve f g in (t r, t + r) it follows from the bove theorem tht the Fourier series of f will converge t t if, nd only if, the Fourier series of g converges t t nd in this cse the sums of the Fourier series re the sme. Thus the behviour of the Fourier series t point t depends only on the vlues of the function in neighbourhood of t. This is in strong contrst with the behviour of power series. If two power series coincide in n open intervl, then they re identicl throughout their common domin of convergence.

23 6 Fourier Series of Continuous Functions A bsic question tht cn be sked is the following: does the Fourier series of continuous π-periodic function, f, converge to f(t) t every point t [, π]? Unfortuntely, the nswer is No! nd we will study this now. Proposition 6.1. We hve lim n Proof. For t R, we hve sin t t nd so D n (t) dt 4 D n (t) dt = +. (6.1) = 4 > 4 > 4 = 8 π from which (6.1) follows immeditely. sin(n + 1)t dt t (n+ 1 )π n kπ k=1 (k 1)π kπ n k=1 n k=1 (k 1)π 1 k sin t t dt sin t t dt sin t kπ dt Proposition 6.. Let V = C per [, π], the spce of continuous π-periodic functions with the usul sup-norm (denoted ) nd define φ n : V R by φ n (f) = s n (f)() where s n (f) is the n th -prtil sum of the Fourier series of f. Then φ n is continuous liner functionl on V nd φ n = 1 π D n (t) dt. (6.) 3

24 Proof. On one hnd, φ n (f) = 1 f(t)d n (t) dt π (cf. (5.4d)). Thus, nd so 1 π φ n (f) f D n (t) dt π φ n 1 D n (t) dt. π Now, let E n = {t [, π] D n (t) }. Define f m (t) = 1 m d(t, E n) 1 + m d(t, E n ) where d(t, A) = inf{ t s s A} is the distnce of t from set A. Since d(t, A) is continuous function (in fct, d(t, A) d(s, A) t s ) f m C per [, π], (it is periodic since D n is even nd so E n is symmetric set bout the origin). Also f m 1 nd f m (t) 1 if t E n while f m (t) 1 if t En. c By the dominted convergence theorem, it now follows tht from which (6.) follows. φ n (f m ) 1 π D n (t) dt Let us now recll few results from topology nd functionl nlysis. Theorem 6.1 (Bire). If X is complete metric spce, the intersection of every countble collection of dense open sets of X is dense in X. Equivlently, Bire s theorem lso sttes tht complete metric spce cnnot be the countble union of nowhere dense sets. One of the importnt consequences of Bire s theorem is the Bnch- Steinhus theorem lso known s the uniform boundedness principle. 4

25 Theorem 6. (Bnch-Steinhus). Let X be Bnch spce nd Y normed liner spce nd {Λ α } α A collection of bounded liner trnsformtions from X into Y, where α rnges over some indexing set A. Then either there exists M > such tht or, for ll x belonging to some dense G δ -set in X. Λ α M, for ll α A (6.3) sup Λ α x = (6.4) α A (Recll tht G δ -set is set which is the countble intersection of open sets). Now consider X = V s defined in Proposition 6. nd Y = R. Let A = N nd set Λ n = φ n, gin defined in bove mentioned proposition. Since, by Proposition 6.1 nd 6., we hve φ n s n, it follows from the Bnch-Steinhus theorem, tht there exists dense G δ -set (of continuous π-periodic functions) in V such tht the Fourier series of ll these functions diverge t t =. We could hve very well delt with ny other point in the intervl [, π] in the sme mnner. By nother ppliction of Bire s theorem, we cn strengthen this further. Let E x be the dense G δ -set of continuous π-periodic functions in V such tht the Fourier series of these functions diverge t x. Let {x i } be countble set of points in [, π] nd let E = n i=1 E x i V. (6.5) Then, by Bire s theorem E is lso dense G δ -set. (Ech E xi is the countble intersection of dense open sets nd so, the sme is true for E). Thus for ech f E, the Fourier series of f diverges t x i for ll 1 i. Define s (f; x) = sup s n (f)(x). n Then s (f, ) is lower semi-continuous function. Hence {x s (f; x) = } is G δ -set in (, π) for ech f. If we choose the x i bove so tht {x i } is dense (tke ll rtionls, for instnce in (, π)) then we hve the following result. 5

26 Proposition 6.3. The set E V is dense G δ -set such tht for ll f E, the set Q f (, π) where its Fourier series diverges, is dense G δ -set in (, π). Proposition 6.4. In complete metric spce, which hs no isolted points, no countble dense set cn be G δ. Proof. Let E = {x 1,...,x n,...} be countble dense set. Assume E is G δ. Thus E = n=1 W n, W n open nd dense. Then, by hypothesis, W n \ n i=1{x i } = V n is lso open nd dense. But n=1v n =, contrdicting Bire s theorem. Thus, there exists uncountbly mny π-periodic continuous functions on [, π] whose Fourier series diverge on dense G δ -set of (, π). Hving nswered our first generl question negtively, let us now prove positive result. Proposition 6.5. Let f be π-periodic function on [, π] which is uniformly Lipschitz continuous, i.e. there exists K > such tht f(x) f(y) K x y for ll x, y. Then, the Fourier series of f converges to f on [, π]. Proof. Choose < r < π such tht 1 π r r t/ sin(t/) dt < ε K. t/ This is possible since sin(t/) is bounded continuous function, nd hence integrble on [, π]. If C is n upper bound for this function, we need only choose r such tht rc < επ. Let x [, π]. K s n (x) = 1 π f(x t)d n (t) dt. Thus s n (x) f(x) = 1 π ( ) f(x t) f(x) Dn (t) dt. 6

27 If r is chosen s bove, then, by the locliztion principle (Theorem 5.1), we hve, for n lrge enough ( 1 r ) (f(x ) π + t) f(x) Dn (t) dt < ε. r On the other hnd, 1 r ( ) f(x t) f(x) Dn (t) dt π r K π K π r r r r t sin(n + 1/)t dt sin t/ t / sin(t/) dt < ε. Thus for n lrge we hve s n (x) f(x) < ε which completes the proof. Corollry 6.1. If f C 1 [, π] is π-periodic, then the Fourier series of f converges to f on [, π]. In lter section, we will relx these conditions on f nd study the pointwise convergence of Fourier series. Remrk 6.1. The convergences in Proposition 6.5 nd Corollry 6.1 bove re, in fct, uniform over [, π], cf. Exercise 3. Remrk 6.. If f is Lipschitz continuous in neighbourhood of t [, π], then we cn show, using identicl rguments, tht the Fourier series of f converges to f(t) t t. 7 Fejér s Theorem In the previous section, we sw tht there exist n uncountble number of continuous π-periodic functions whose Fourier series diverge over dense set of points. Nevertheless, s we hve observed erlier, such functions cn be pproximted uniformly by mens of trigonometric polynomils over [, π], thnks to the Stone-Weierstrss theorem. Indeed, we cn now prove the following result. 7

28 Theorem 7.1 (Fejér). Let f be continuous π-periodic function on [, π]. Let {s n } n be the sequence of prtil sums of its Fourier series. Define σ n (x) = s (x) s n (x). n + 1 Then σ n f uniformly over [, π]. Proof. It is immedite, from the definition of the Fejér kernel K n, to see tht Thus σ n (x) = 1 π σ n (x) f(x) = 1 π f(x t)k n (t) dt. ( f(x t) f(x) ) Kn (t) dt, in view of (5.11). Now, since f is continuous on [, π], it is bounded nd uniformly continuous. Let f(x) M for ll x [, π] nd, given ε >, let δ > such tht x y < δ implies f(x) f(y) < ε/. Now, choose N lrge enough such tht, for n N, K n (t) ε/4m for ll δ < t π. This is possible becuse of the estimte (5.1). Thus, since K n, δ ( ) f(x t) f(x) Kn (t) dt ε K n (t) dt = πε, δ gin by (5.11). On the other hnd, ( δ ) (f(x ) + t) f(x) Kn (t) dt M ε 4M δ if n N. It now follows tht for ll x [, π], we hve for n N which completes the proof. σ n (x) f(x) < ε π = πε Corollry 7.1. Let f nd g be two π-periodic nd continuous functions on [, π]. If they hve the sme Fourier series, then f g. Proof. If the two functions hve the sme Fourier series, then the σ n will be the sme for both functions nd we know tht σ n f nd σ n g uniformly on [, π]. Hence the result. 8

29 Given series n n, we sy tht it is Cesàro summble or (C, 1) summble to if σ n s n, where σ n = s s n, n s k being the prtil sums of the series. Thus the Fourier series of continuous π-periodic function is lwys Cesàro summble to the function. Thus if f is continuous π-periodic function whose Fourier series is given by + ( n cosnt + b n sin nt), n=1 then f is uniformly pproximted over [, π] by the trigonometric polynomils σ n (x) = n ( + 1 k ) (k cos kt + b k sin kt ). (7.1) n + 1 k=1 Strting from this, we cn deduce Weierstrss pproximtion theorem. Indeed, let f C[ 1, 1]. Define g(t) = f(cost), for t [, π]. Then g is π-periodic nd continuous. Further, g is n even function nd hence its Fourier series will only consist of cosine terms. Let the Fourier series of g be given by + k coskt. Then k=1 σ n (t) = n ( + 1 k ) k coskt. n + 1 Thus, given ε >, there exists N such tht, for ll n N, f(cost) n ( 1 k ) k coskt n + 1 < ε k=1 k=1 for ll t [, π]. This is the sme s f(t) n ( 1 k ) k cos ( k(cos 1 t) ) < ε n + 1 k=1 9

30 for ll t [ 1, 1]. Now it only remins to show tht P k (t) = cos(k cos 1 t) is polynomil in t for every non-negtive integer k. Indeed, P (t) 1 nd P 1 (t) t. Assume tht P k (t) is polynomil in t, of degree k, for every 1 k n 1, n. Then cosns = cos ns + cos(n )s cos(n )s = cos ( (n 1)s + s ) + cos ( (n 1)s s ) cos(n )s = cos(n 1)s coss cos(n )s = cossp n 1 (coss) P n (cos s). It then follows tht P n (t) = tp n 1 (t) P n (t) (7.) nd so P n (t) is polynomil of degree n in t. The polynomils {P n } defined recursively vi (7.) where P 1 nd P 1 (t) = t, re clled the Chebychev polynomils nd ply n importnt role in numericl nlysis, especilly in numericl qudrture. 8 Regulrity In this section, we will briefly discuss vrious regulrity ssumptions to be mde on functions when discussing the pointwise convergence of Fourier series. Let f : [, b] R be given function which is differentible on (, b). Assume tht f (+) = lim t f (t) exists. Then f is bounded in subintervl (, +δ) (where δ > ) nd so, by the men-vlue theorem, f is uniformly continuous on (, +δ). Thus it cn be continuously extended to [, + δ]. Consequently f(+) = lim t f(t), exists. Agin by the men-vlue theorem, pplied to f where f() = f(+) nd f(t) = f(t) for < t + δ, we hve f(t) = f(+) + f ( + θ(t ))(t ) 3

31 where < θ < 1. Thus f(t) f(+) lim = f (+). (8.1) t t Notice tht f (+) is different from the right-sided derivtive of f t (if it exists), which is given by We lso define, f (b ) = lim t b f (t). D + f(t) f() f() = lim. t t Definition 8.1. We sy tht function f : [, b] R is piecewise smooth if there exist finite number of points = < 1 <... < n = b such tht f is continuously differentible in ech subintervl ( k, k+1 ), k n 1 nd f (c+) nd f (c ) exist t ll points c [, b] except t, where f (+) exists, nd t b, where f (b ) exists. Let f : [, b] R (or C) be given function. Let P = { = x < x 1 <... < x n = b} be ny prtition of [, b]. Define V (P; f) = n f(x i ) f(x i 1 ) i=1 nd V (f;, b) = sup V (P; f) P the supremum being tken over ll possible prtitions of [, b]. The quntity V (f;, b) is clled the totl vrition of f over the intervl [, b]. Definition 8.. A rel (or complex) vlued function defined on [, b] is sid to be of bounded vrition on [, b] if V (f;, b) <. Exmple 8.1. Any monotonic function defined on [, b] is of bounded vrition. In this cse V (f;, b) = f(b) f(). 31

32 Exmple 8.. If f is uniformly Lipschitz continuous, then it is of bounded vrition. For, f(x i ) f(x i 1 ) L (x i x i 1 ) = L(b ) < i i Exmple 8.3. Define f(x) = { x sin 1 x, < x < 1, x =. Then, f is not of bounded vrition. To see this, choose prtition of [, 1] s follows: { } {, 1} π(j + 1) j n. f(x j ) f(x j 1 ) = Thus for ll n, V (f;, 1) π n j=1 π(j + 1) + π(j 1) = 4j π 4j 1. 4j π 4j = πj. 1 j nd so V (f;, 1) =. Given rel number r, define r + = mx{r, } nd r = min{r, }. Then r = r + r nd r = r + + r. Thus, if f : [, b] R, nd P = { = x <... < x n = b} is ny prtition, set nd p(p; f) = n(p; f) = n ( f(xi ) f(x i 1 ) ) + i=1 n ( f(xi ) f(x i 1 ) ). i=1 Then V (P; f) = p(p; f) + n(p; f) nd Define P(f;, b) = sup P f(b) f() = p(p; f) n(p; f). p(p; f) nd N(f;, b) = sup n(p; f); P where, gin, the supremum re tken over ll possible prtitions of [, b]. 3

33 Proposition 8.1. Let f be of bounded vrition on [, b]. Then V (f;, b) = P(f;, b) + N(f;, b) (8.) f(b) f() = P(f;, b) N(f;, b). (8.3) Proof. We know tht, for ny prtition P, p(p; f) = n(p; f) + f(b) f() N(f;, b) + f(b) f(). Then, tking the supremum over ll possible prtitions, we get P(f;, b) N(f;, b) + f(b) f(). (8.4) Similrly n(p; f) = p(p; f) + f() f(b) yields N(f;, b) P(f;, b) + f() f(b) (8.5) Reltions (8.4) nd (8.5) yield (8.3), since V (f;, b) < implies tht P(f;, b) < nd N(f;, b) <. Now V (P; f) = p(p; f) + n(p; f) gives us V (f;, b) P(f;, b) + N(f;, b). (8.6) On the other hnd, V (f;, b) p(p; f) + n(p; f) = p(p; f) ( f(b) f() ) = p(p; f) + N(f;, b) P(f;, b) using (8.3). Agin, tking the supremum over ll prtitions, we get V (f;, b) P(f;, b) + N(f;, b) P(f;, b) Reltions (8.6) nd (8.7) yield (8.). = P(f;, b) + N(f;, b) (8.7) 33

34 Proposition 8.. If f nd g re of bounded vrition on [, b], then f + g is of bounded vrition nd Proof. If P is ny prtition, V (f + g;, b) V (f;, b) + V (g;, b). (8.8) V (P; f + g) V (P; f) + V (P; g) by the tringle inequlity nd the result follows. We now hve n importnt chrcteriztion of functions of bounded vrition. Theorem 8.1. A function f : [, b] R is of bounded vrition if, nd only if, f is the difference of two monotonic incresing functions. Proof. If f is the difference of two monotonic incresing functions, since ech of these is of bounded vrition (cf. Exmple 8.1), it follows from the preceding proposition tht f is of bounded vrition. Conversely, if f is of bounded vrition on [, b], by Proposition 8.1, we hve, for ny x [, b], f(x) f() = P(f;, x) N(f;, x). Let g(x) = P(f;, x) nd h(x) = N(f;, x) f(). Clerly both functions re monotonic incresing nd f(x) = g(x) h(x). This completes the proof. From the properties of monotonic functions, we cn now deduce the following result. Corollry 8.1. If f : [, b] R is of bounded vrition, then f is differentible.e. Further, for ll x (, b), f(x+) nd f(x ) exist. So do f(+) nd f(b ). The function f hs t most countble number of jump discontinuities. The fundmentl theorem of Lebesgue integrtion sttes tht if f is integrble on [, b] nd if F(x) = c + x f(t) dt, then F = f.e. on [, b]. We now sk the question s to when function cn be expressed s n indefinite integrl of n integrble function. 34

35 Proposition 8.3. Let f be integrble on [, b]. Then set F(x) = x Then F is of bounded vrition on [, b]. f(t) dt. Proof. If P = { = x < x 1 <... < x n = b} is ny prtition of [, b], then Thus n F(x i ) F(x i 1 ) i=1 Hence the result. n i=1 V (F;, b) xi x i 1 f(t) dt = f(t) dt <. f(t) dt <. Thus function must be t lest of bounded vrition on [, b] for it to be expressed s n indefinite integrl of n integrble function. But this is not enough. Proposition 8.4. Let f : [, b] R be integrble. Given ε >, there exists δ > such tht if E [, b] is mesurble set with µ(e) < δ, then f(x) dx < ε. Proof. If f M on [, b], the result is trivilly true, since f dx Mµ(E), where µ(e) is the Lebesgue mesure of E. In the generl cse, define { f(x), if f(x) n f n (x) = n, if f(x) > n. E E Then f n is bounded nd f n f pointwise. Further {f n } is n incresing sequence of non-negtive functions. By the monotone convergence theorem, lim n f n dx = 35 f dx <.

36 Given ε >, choose N such tht ( f f N ) dx < ε/. Now choose δ > such tht µ(e) < δ implies tht f N dx < ε/. E Then E f dx E f dx ( f f N) dx+ E f N dx < ε. This completes the proof. Definition 8.3. A function f : [, b] R is sid to be bsolutely continuous on [, b] if for every ε >, there exists δ > such tht whenever we hve finite collection of disjoint intervls {(x i, x i )}n i=1 stisfying we hve n (x i x i) < δ i=1 n f(x i ) f(x i) < ε. i=1 Clerly, ny bsolutely continuous function is uniformly continuous on [, b]. Exmple 8.4. Any uniformly Lipschitz continuous function is bsolutely continuous, since f(x i ) f(x i ) L (x i x i) < Lδ. Exmple 8.5. Any indefinite integrl of n integrble function is bsolutely continuous by virtue of Proposition 8.4. We will show presently tht function cn be written s n indefinite integrl if, nd only if, it is bsolutely continuous. Proposition 8.5. An bsolutely continuous function is of bounded vrition. 36

37 Proof. Let δ correspond to ε = 1 in the definition of bsolute continuity. Let K be the integrl prt of 1 + (b )/δ, where [, b] is the given intervl. Given ny prtition P of [, b], we cn refine it to prtition consisting of K sets of sub-intervls ech of totl length less thn δ. Thus V (P; f) K nd so V (f;, b) K. Consequently, ny bsolutely continuous function is differentible.e.. on [, b]. Proposition 8.6. Let f : [, b] R be bsolutely continuous. Assume tht f =.e. in [, b]. Then f is constnt function. Proof. Let c (, b). Let E = {x (, c) f (x) = } Then µ(e) = c. Let ε, η > be rbitrry. If x E, there exists sufficiently smll h > such tht [x, x+h] [, c] nd f(x+h) f(x) < ηh. By the Vitli covering lemm, there exists finite disjoint collection of such intervls which cover ll of E except possibly subset of mesure less thn δ, where δ corresponds to ε in the definition of bsolute continuity. We lbel these intervls [x k, y k ], with x k incresing. Thus Then, Now, Also y = x 1 < y 1 x < y... y n = c = x n+1. n x k+1 y k < δ. k= n f(y k ) f(x k ) < η (y k x k ) < η(c ). k=1 n f(x k+1 ) f(y k ) < ε k=1 by bsolute continuity. Thus f(c) f() ε + η(c ) or f(c) = f() for ll c (, b). Hence the result. 37

38 Theorem 8.. A function cn be expressed s n indefinite integrl of n integrble function if, nd only if, it is bsolutely continuous. The derivtive of this function (which exists.e.) is equl.e. to the integrnd. Proof. If f were n indefinite integrl, it is bsolutely continuous (cf. Exmple 8.5). Conversely, let F : [, b] R be bsolutely continuous. Then it is of bounded vrition nd F(x) = F 1 (x) F (x) where F i (x), i = 1, re monotonic incresing. Thus F = F 1 F F 1, F. Thus,.e. nd F dx = F 1 dx + F dx F 1 dx + F dx F 1 (b) F 1 () + F (b) F () <. Thus F is integrble. Now let G(x) = x F (t) dt. Then G, being n indefinite integrl of n integrble function, is bsolutely continuous nd G = F.e. Thus G F is bsolutely continuous nd (G F) =.e.. Thus, by Proposition 8.6, (G F)(x) = (G F)() for ll x [, b]. Hence, This completes the proof. F(x) = F() + x F (t) dt. Absolutely continuous functions shre mny properties of continuously differentible functions. In prticulr, we hve the following result. Theorem 8.3 (Integrtion by prts). Let f, g be bsolutely continuous on [, b]. Then f(t)g (t) dt = f(b)g(b) f()g() f (t)g(t) dt. (8.9) 38

39 Proof. Consider the function f (x)g (y) on [, b] [, b]. Consider the integrl x f (x)g (y) dy dx. It is routine verifiction to check tht Fubini s theorem pplies nd so x f (x)g (y) dy dx = The left-hnd side gives us ( x ) g (y) dy f (x) dx = = y f (x)g (y) dxdy. ( g(x) g() ) f (x) dx g(x)f (x) dx g() [ f(b) f() ] by repeted use of Theorem 8.. The right-hnd side gives ( ) ( ) f (x) dx g (y) dy = f(b) f(y) g (y) dy y = f(b) [ g(b) g() ] Equting the two, we deduce (8.9). This completes the proof. 9 Pointwise Convergence f(y)g (y) dy. In this section, we will prove the convergence theorems of Dirichlet nd Jordn. Proposition 9.1. Let f L 1 (, π) nd ssume tht f (+) exists. Then 1 π lim f(t) sin(n + 1)t n π t dt = 1 f(+). (9.1) Proof. Adding nd subtrcting f(+) in the integrnd, we get f(t) sin(n + 1 )t t dt = (f(t) f(+)) sin(n + 1 )t t sin(n + 1 dt+f(+) )t dt. t 39

40 The second integrl on the right-hnd side becomes (fter chnge of vrible), (n+ 1 )π sin t dt t which converges to π/ s n. Thus, 1 π lim n π f(+) sin(n + 1 )t o t dt = 1 f(+). Hence we need to show tht the first term tends to zero s n. Let ε > be n rbitrrily smll number. Choose < r < π such tht for < t r, f(t) f(+) f (+) t < ε ( cf. (8.1)). Then, (f(t) f(+)) sin(n + 1 )t t dt = r (f(t) f(+))sin(n+1 )t t dt + (f(t) r f(+))sin(n+1 )t t dt. The second term tends to zero, by the Riemnn-Lebesgue property for Φ n (cf. Proposition 5.7), since f(t) f(+) L 1 (, π). Now, the first term bove cn be written s r ( ) f(t) f(+) f (+) sin(n + 1 r t )t dt + f (+) sin(n + 1 )t dt. But, nd r r ( f(t) f(+) t ) f (+) sin(n + 1 )t dt < rε < πε sin(n + 1 r )t dt = sin t r cosnt dt + cos t sin nt dt nd both these integrls tend to zero, s n, by the Riemnn-Lebesgue lemm (cf. Corollry 4.1). This completes the proof. 4

41 Theorem 9.1. (Dirichlet) Let f L 1 (, π) be piecewise smooth function. Then its Fourier series converges to 1 (f(t+) + f(t )) t ll points t [, π]. In prticulr, if f is continuous t point t, then its Fourier series converges to f(t) t tht point. Proof. Recll tht (cf. (5.4e)) s n (t) = 1 π Now, by (9.1), we hve 1 π lim f(t + x) sin(n + 1 )x n π x (f(t + x) + f(t x))d n (x) dx. dx = 1 f(t+). Thus, lim n 1 π π Hence, by Proposition 5.6, we get f(t + x) sin(n + 1 )x x dx = 1 f(t+). Similrly, lim n lim n from which the result follows. 1 π f(t + x) sin(n + 1)x π sin x 1 π f(t x) sin(n + 1)x π sin x dx = 1 f(t+). dx = 1 f(t ) Exmple 9.1. Consider the function 1, t < / f(t) =, / t π/ 1, π/ < t π. This function is piecewise smooth nd is odd. Thus, its Fourier series consists only of sine functions. Now, b n = 1 π f(t) sin nt dt = π π sin nt dt. 41

42 Thus, b n =, n odd nπ, n = 4k, k N 4 nπ, n = 4k +, k N At t = π/, the series must thus converge to 1/. Thus, 1 = [1 + + ] π which yields the well known Gregory series π 4 = In order to prove the next result, we recll version of the men vlue theorem for integrls. Proposition 9.. Let g : [, b] R be continuous nd let f : [, b] R be non-negtive nd monotonic incresing. Then, there exists c [, b] such tht f(t)g(t) dt = f(b) c g(t) dt. (9.) Remrk 9.1. The result is flse if f is not non-negtive. To see this, tke f(t) = g(t) = t on [ 1, 1]. Then the left-hnd side will be The right-hnd side is f(1) c t dt = 3. t dt = 1 c nd we cn never hve c [ 1, 1] such tht the two re equl. Remrk 9.. In the men vlue theorem of differentil clculus, the point c will be in the interior of the intervl. In cse of men vlue theorems for integrls, this need not be necessrily the cse. For instnce, if f 1 on [, b] nd if g is strictly positive on tht intervl, then we cnnot hve (9.) with c (, b). 4

43 We re now in position to prove convergence theorem for lrger clss of functions. We need preliminry result nlogous to Proposition 9.1. Proposition 9.3. Let f be of bounded vrition on [, π]. Then 1 lim n π f(t) sin(n + 1 )t t dt = 1 f(+). (9.3) Proof. Since f is of bounded vrition, it cn be written s the difference of two monotonic incresing functions. Hence we my ssume, without loss of generlity, tht f is monotonic incresing. Now, f(t) sin(n + 1 )t t dt = (f(t) f(+))sin(n+1 )t t + f(+) sin(n+ 1 )t t As lredy shown in the proof of Proposition 9.1, the second integrl converges to π/. Thus, s before, it is enough to show tht the first integrl tends to zero s n. Let ε > be n rbitrrily smll number. Choose < r π such tht f(t) f(+) < ε/4, < t r. (9.4) Then, splitting the first integrl over [, r] nd [r, π], we see tht the integrl on [r, π] tends to zero by the Riemnn-Lebesgue property for Φ n (cf. Proposition 5.7) since f(t) f(+) is integrble over tht intervl. Thus, for n sufficiently lrge, we hve 1 π r (f(t) f(+)) sin(n + 1 )t t dt < ε. On the intervl [, r], we redefine the function f s f(+) t t =, so tht the function f(t) f(+) remins non-negtive nd monotonic incresing without ltering the vlue of the integrl over tht intervl. Hence, by the men vlue theorem (cf. Proposition 9.), there exists c [, r] such tht r (f(t) f(+)) sin(n + 1 )t t dt = (f(r) f(+)) r Thus, by Proposition 5.5 nd by (9.4), it follows tht 1 r (f(t) f(+)) sin(n + 1 )t dt π t < ε. This completes the proof. 43 c dt. dt sin(n + 1 )t t dt.

44 Theorem 9.. (Jordn) Let f be function of bounded vrition on [, π]. Then, its Fourier series t ny point t in this intervl converges to 1 (f(t+)+ f(t )). In cse f is continuous t t, then the series converges to f(t). Proof. The proof is identicl to tht of Theorem 9.1, except tht we ppel to Proposition 9. in plce of Proposition 9.1. Remrk 9.3. Dirichlet proved his theorem on the convergence of Fourier series in 189. Jordn s result ws proved in In 194, Fejér showed tht if f L 1 (, π), nd if f(t+) nd f(t ) exist t t [, π], then the Fourier series is (C, 1)-summble t t to (f(t+) + f(t ))/. One of the gretest triumphs in the history of Fourier series is the result of Crlesson (1966) tht the Fourier series of ny function in L (, π) converges to the vlue of tht function.e..this ws extended by Hunt in 1968 to ll functions in L p (, π) for 1 < p <. 1 Termwise Integrtion In generl, when we hve series f(x) = n f n(x), we cn integrte the series term-by-term if the series is uniformly convergent. However, we hve seen tht Fourier series (under pproprite hypotheses) converge to (f(t+) + f(t ))/ t discontinuity. Thus, if the function is discontinuous, uniform convergence is ruled out nd the bove principle does not pply. However, Fourier series re specil nd enjoy specil properties. Vis-à-vis integrtion, we hve the following result. Theorem 1.1. Let f L 1 (, π) be extended periodiclly over R nd hve the Fourier series + ( n cosnt + b n sin nt). n=1 Then, (i) the series obtined by termwise integrtion, viz. x + ( n n sin x b ) n n cosx + C where n=1 C = b n n, n=1 44

45 converges to x f(t) dt. (ii) This convergence is uniform if f L (, π). Proof. Let F(x) = x ( f(t) ) dt. Since f is integrble, it follows tht F is bsolutely continuous. It is lso π-periodic. For, F(x + π) = x ( f(t) ) x+π ( dt + x f(t) ) dt = F(x) + ( ) f(t) dt since, by π-periodicity, the integrtion cn be done over ny intervl of length π without ffecting its vlue. But the lst integrl vnishes by the definition of. Thus F(x + π) = F(x). Since it is bsolutely continuous, it is continuous nd of bounded vition nd so its Fourier series converges to its vlue t ech point. Let F(t) = c + (c n cosnt + d n sin nt). n=1 We cn lso use integrtion by prts (vilble for bsolutely continuous functions (cf. (8.9)) to obtin c n = 1 π F(t) cosnt dt = 1 F(t) sin nt π 1 π ( nπ nπ f(t) ) sin nt dt. The boundry terms vnish by periodicity. So does the integrl of sin nt. Thus we deduce tht c n = b n /n. Similrly, d n = n /n. Thus, Evluting this t t =, we get F(t) = c + ( n n sin nt b ) n n cosnt. n=1 c = 45 n=1 b n n.