CONVERGENCE OF FOURIER SERIES

Transcription

1 CONVERGENCE OF FOURIER SERIES SOPHIA UE Abstract. The subject of Fourier analysis starts as physicist and mathematician Joseph Fourier s conviction that an arbitrary function f could be given as a series. In this expository paper, we build up from the basic defitions of Fourier analysis to answer the question what sense does the Fourier series of a function converge to the function itself. Di erent criteria for convergence will be troduced along the way. The proof of Mean Square Convergence will conclude this paper. Contents 1. Introduction to Fourier Series 1 2. Uniqueness of Fourier series 3 3. Convolutions 6 4. Mean-square convergence of Fourier series 11 Acknowledgments 15 References Introduction to Fourier Series Throughout this paper, an tegrable function should be terpreted as tegrable the Riemann sense. In this section, we will go through the basic defitions of Fourier Analysis that will appear throughout this paper. Examples and variations are supplied to aid understandg. Defition 1.1. If f is an tegrable function given on an terval [a, b] of length L (that is, b a = L), the n th Fourier coe cient of f is defed by ˆf(n) L Z b a f(x)e 2x/L dx, n 2 Z. Note that more often than not, we will be workg with functions that are tegrable on a circle, which means that the function is tegrable on every terval of length 2. Defition 1.2. The Fourier series of f is given normally by The notation 1 ˆf(n)e 2x/L. n 1 f(x) n 1 ˆf(n)e 2x/L

2 2 SOPHIA UE means that the series on the right-hand side is the Fourier series of f. Defition 1.3. The N th partial sum of the Fourier Series of f, for N a positive teger, is given by N S N (f)(x) = ˆf(n)e 2x/L. n= N The notion of N th partial sum of the Fourier Series of f is very important the study of Fourier Analysis. Usg the partial sums of the Fourier series, we can view the convergence of Fourier series as the limit of these symmetric sums as N tends to fity. Indeed, the basic question can be reformulated as follows: Question 1.4. In what sense does S N (f) converge to f as N!1? Followg are some simple examples to familiarize ourselves with above defitions. Example 1.5. Let f( ) = for apple apple. To calculate the Fourier coe cients of f, we should split it to two cases. First, when n 6=, ˆf(n) Z e d 2 e [ 2 = e + e 2 = cos(n) = ( 1)n+1. Z e d ] + e e (2)( ) + s(n) 2 When n=, Hence, the Fourier series of f is ˆf(n) Z d =. 2 f( ) ( 1) n+1 1 e n+1 s(n ) =2 ( 1). n n6= n=1 Furthermore, by the alternatg series test, we can easily see that the Fourier series of f is convergent. Example 1.6. Let f( ) = for apple apple. First, when n 6=, ˆf(n) 2 Z e d Z 1 e d. 2

3 CONVERGENCE OF FOURIER SERIES 3 Usg the result from last example, we get that ˆf(n) e [ 2 e [ 2 Z e d + e 1 ()( ) e +( ) e Z e + d ] cos( n)+is( n) cos(n) is(n) [ 2 =+ e 1 n 2 cos(n) isn 1 = n 2 = ( 1)n 1 n 2. On the other hand, when n =, Hence, ˆf(n) 2 f(n) 1 2 e + n6= 2 e + Z ( 1) n 1 n 2 e n=2k+1,k2n 1 e ()( ) ] d = 2. 4 n 2 e 2 (cos(n ) is(n )) 4 n=2k+1,k2n + e 1 1+e ] ()( ) cos(n ) 2. Uniqueness of Fourier series is(n ) n 2. If we were to believe that the Fourier series of functions f somehow converge to f, then we could fer that the Fourier coe cient of a function uniquely determes the function. In other words, for our assumption to be true, if f and g have the same Fourier coe cients, then f and g are necessarily equal. The statement can be reformulated as followg by takg the di erence f g: Proposition 2.1. If ˆf(n) =for all n 2 Z, then f =. It is obvious that this proposition cannot be true without reservation. Sce calculatg Fourier coe cients requires tegration and any two functions that are different at fitely many pots can have the same tegration, one particular Fourier series can be shared by two functions that are di erent at fitely many pots. However, we do have the followg positive result regardg contuous pots. Theorem Suppose that f is an tegrable function on the circle with ˆf(n) = for all n 2 Z. Then f( )=whenever f is contuous at the pot. This result is really nice. It shows that f vanishes for most values of. Followg is a staightforward corollary. 1 Rigorous proof of this theorem can be found Elias M. Ste and Rami Shakarchi s Fourier Analysis an Introduction (23), p39-41 followg Theorem 2.1.

4 4 SOPHIA UE Corollary 2.3. If f is contuous on the circle and ˆf(n) =for all n 2 Z, then f. Moreover, this corollary is useful answerg Question 2.4 under the condition that the Fourier series converges absolutely. Corollary 2.4. Suppose that f is a contuous function on the circle and that the Fourier series of f is absolutely convergent, P 1 n ˆf(n) < 1. Then, the Fourier series converges uniformly to f, thatis, uniformly x. lim S N (f)(x) =f(x) N!1 Proof. Let g(x) = P 1 ˆf(n)e n x. Sce P 1 n ˆf(n) < 1, g(x) isdefed everywhere. By triangle-equality, we get that S N (f)(x) g(x) = ˆf(n)e x 1 n ˆf(n)e x = n >N ˆf(n)e x apple n >N Sce the Fourier series of f is bounded, for any >, we can fd large N such that P n >N ˆf(n) <. Hence, S N (f)(x) uniformly converges to g(x) as N!1. Note that for large N, the Fourier coe cients of g(x) are ĝ(n) 2 2 = ˆf(n). Z 2 Z 2 g(x)e x dx ( napplen ˆf(n)e x )e x dx In other words, f and g have identical Fourier coe cients. Now let s defe a new function h = f g. By distributivity of the tegral, it is easy to see that ĥ(n) =. Apply Corollary 3.3 to h, we get that h = f g = or f = g. Combg this result with the uniform convergence of S N (f)(x) tog(x), we get the desired result, i.e. S N (f)(x) uniformly converges to f(x) as N!1. Now that we have proven when the Fourier series of a contuous function is absolutely convergent, its Fourier series converge uniformly to the said function, what conditions on f would guarantee the absolute convergence of its Fourier series? Corollary 2.5. Suppose that f is a twice contuously di erentiable function on the circle, then ˆf(n) =O(1/( n ) 2 as n! 1, so that the Fourier series of f converges absolutely and uniformly to f. The notation ˆf(n) =O(1/( n ) 2 as n! 1 means that the left-hand side is bounded by a constant multiple of the right-hand side, i.e. there exists C> with ˆf(n) apple C/( n ) 2 for all large n. ˆf(n).

5 CONVERGENCE OF FOURIER SERIES 5 Proof. Through tegratg by parts twice, we have ˆf(n) 2 Z 2 f( )e d 2 2 f( ) e Z 2f ( )e d 2 2 f ( ) n 2 Z 2 e + 1 Z 2f ( )e d f ( )e d. Sce f and f are both periodic, f( ) e be a bound for f.thus, 2 n 2 ˆf(n) apple Z 2 Z 1 2() 2 2f ( )e d 2 f ( )e d apple Sce B is dependent of n, we have that = and f ( ) e Z 2 ˆf(n) =O(1/( n ) 2 as n! 1 2 f ( ) d apple 2B. There are also stronger versions of Corollary 2.5 such as the followg. = Let B Corollary 2.6. If f is 2-periodic and has an tegrable derivative, then its Fourier series converges absolutely and uniformly to f. More generally, Theorem 2.7. The Fourier series of f converges absolutely (and hence uniformly to f) iff satisfies a Hölder condition of order, with >1/2, thatis sup f( + t) f( ) apple A t for all t. Proof. First, we will prove that for any 2-periodic and tegrable function f, and thus By defition, we have ˆf(n) 2 ˆf(n) 4 Z Z [f(x) f(x + /n)e x dx, f(x + /n)]e x dx. ˆf(n) Z f(y)e y dx. 2

6 6 SOPHIA UE Sce f is periodic and we are tegratg over the period of f, we can substitute x + n for y without changg the bounds of tegration. Hence, ˆf(n) Z Z Z f(x + /n)e (x+/n) dx f(x + /n)e x e i dx f(x + /n)e x dx. The last equality follows from applyg Euler s Identity. Sce 2 ˆf(n) 2 ( Z f(x)e x dx Z f(x + /n)e x dx), a little algebraic manipulation gives the desired result, that It follows ˆf(n) 4 ˆf(n) 4 Apply Hölder and we get apple 1 4 Sce R e x dx =2, which proves the theorem. Z Z Z ˆf(n) apple 1 4 [f(x) f(x + /n)]e x dx. [f(x) f(x + /n)]e x dx f(x) f(x + /n) e x dx. Z C /n e x dx. ˆf(n) apple C 2 n, 3. Convolutions Defition 3.1. Given two 2-periodic tegrable functions f and g on R, their convolution f g on [, ] is given by (f g)(x) 2 Z f(y)g(x y)dy. Note that sce the product of two tegrable functions is aga tegrable, the above tegral makes sense for all x. Furthermore, sce the functions are periodic, we can change variables to see that (3.2) (f g)(x) 2 Z f(x y)g(y)dy. The notion of convolution plays a fundamental role Fourier analysis. It reduces the problem of understandg S N (f) to understandg the convolution of two functions as we will see the followg example.

7 CONVERGENCE OF FOURIER SERIES 7 Example 3.3. D N is the N th Dirichlet kernel given by D N (x) = N n= N e x. For any 2-periodic tegrable function f, its convolution with the N th Dirichlet kernel is (f D N )(x) Z N f(y)( e (x y) )dy 2 = = N n= N N n= N ( 1 2 = S N (f)(x). Z n= N ˆf(n)e x f(y)e y dy)e x As we have shown above, Question 2.4 can be aga reformulated as followg: Question 3.4. In what sense does (f D N )(x) converge to f as N!1? But before we dive, let s look at some nice properties of convolution. Proposition 3.5. Suppose that f, g, and h are 2-periodic tegrable functions. Then: (i) f (g + h) =(f g)+(f h) (ii) (iii) (cf) g = c(f g) =f (cg) for any c 2 C f g = g f (iv) (f g) h = f (g h) (v) (vi) f g is contuous [f g(n) = ˆf(n)ĝ(n). Part (i) and (ii) give learity. Part (iii) gives commutativity while part (iv) demonstrates associativity. Part (v) shows that convolution is a smoothg operation the sense that even though f and g are merely tegrable, f g is contuous. Part (vi) converts convolution to multiplication, which is key the study of Fourier series. Now we move on to the proof of Proposition 3.5. Part (i) and (ii) follow directly from the learity of tegration. Part (iii) follows from Equation (3.2). Part (iv) can be easily proven through terchangg two tegral signs and changg variables. Though Part (v) and (vi) are easily deduced if f and g are contuous, to prove these properties under the condition that f and g are merely tegrable, we are gog to need the followg approximation lemma.

8 8 SOPHIA UE Lemma 3.6. Suppose f is tegrable on the circle and bounded by B. Then there exists a sequence {f k } 1 k=1 of contuous functions on the circle so that for all k =1, 2,...,and Z sup f k (x) apple B x2[ pi,] f(x) f k (x) dx! as k!1. Proof of Lemma 3.6. When f is real, given >, 9 a partition P of the terval [, ] such that U(f,P) L(f,P) < sce f is tegrable. Now we defe a step function if x 2 [x j g(x) = x j sup f(y) 1appleyapplex j 1,x j ) for j 2 [1,N]. We can see that g is bounded by B and Z g(x) f(x) dx = Z (g(x) f(x))dx <. To modify the step function g and make it contuous, we will take small > and defe g*(x) =g(x) when the distance between x and any partition pot P is at least. When the distance between x and any partition pot P is more than,defeg*(x) as the lear function that connects g(x ) and g(x + ). It is easy to see that g* is contuous. Now let g*(x) =whenx 2 [, + ]andx 2 [, + ], which extends g* toa2-periodic function. Sce g* only di ers from g N tervals of length s and is also bounded by B, wehave Z With su ciently small, we have By triangle equality, Z f(x) g*(x) dx apple Z g(x) g*(x) dx apple 4BN. Z g(x) g*(x) dx <. Z g(x) f(x) dx + g(x) g*(x) dx < 2. Let 2 k and denote g* byf k,thesequence{f k } is what we are lookg for. When f is complex, we can apply the above proof to the real part and the imagary part separately to get the same result. Applyg Lemma 3.6, we can complete the proof of part (v) and (vi), the idea here is to substitute f k and ˆf k for f and ˆf respectively and use the contuity of f k to complete the proof. Proof of Part (v) and (vi). Let s apply Lemma 3.6 to 2-periodic tegrable functions f and g. We get sequences {f k } and {g k } of approximatg contuous functions. Note that with a bit of algebraic manipulation, we have f g f k g k =(f f k ) g + f k (g g k ).

9 CONVERGENCE OF FOURIER SERIES 9 Moreover, (f f k ) g uniformly converges to as k!1. (f f k ) g(x) apple 1 2 Z apple 1 2 sup g(y) y! as k!1. f(x y) f k (x y) g(y) dy Z f(y) f k (y) dy Similarly, f k (g g k )! uniformly x. Therefore, as k!1, f k g k uniformly converges to f g. By contuity of each f k g k, f g is also contuous, which proves Part (v). For each teger n, scef k g k uniformly converges to f g as k!1,wehave By contuity of f k and g k,wehave Hence, f\ k g k (n) \ f k g k (n)! [f g(n) as k!1. Z Z Z Z = ˆf(n)ĝ(n). (f k g k )(x)e x dx Z 1 2 ( f k (y)g k (x f k (y)e y ( 1 2 f k (y)e y ( 1 2 ˆf(n) ˆfk (n) 2 apple 1 2 Z Z Z Z! as k!1, y)dy)e x dx g k (x y)e (x y) dx)dy g k (x)e x dx)dy (f(x) f k (x))e x dx f(x) f k (x) dx which means ˆf k (n)! ˆf(n) as k!1. Aga, similarly, ĝ k (n)! ĝ(n) as k!1. By substitutg ˆf k (n) and ĝ k (n) for f(n) and g(n), we can easily get Part (vi). Now that we have familiarized ourselves with the basic properties of convolution, we will troduce the concept of good kernels. Defition 3.7. A family of good kernels is a sequence of functions {K n (x)} 1 n=1 that satisfies the followg properties: (a) For all n 1, Z 1 K n (x)dx =1. 2 (b)there exists M>suchthat for all n 1, Z K n (x) dx apple M.

10 1 SOPHIA UE (c) For every >, Z K n (x) dx!, as n!1. apple x apple The concept of good kernels is very important the study of Fourier series because of the nice property good kernels have, which is stated the followg theorem. Theorem 3.8 (Approximation to the Identity). Let {K n (x)} 1 n=1 be a family of good kernels, and f an tegrable function on the circle. Then lim (f K n)(x) =f(x) n!1 whenever f is contuous at x. If f is contuous everywhere, then the above limit is uniform. Proof. If f is contuous at x, given >, 9 such that if y < then f(x y) f(x) <. By property (a) of good kernel, we have (f K n (x) f(x)) 2 2 Z Z K n (y)f(x y)dy f(x) K n (y)[f(x y) f(x)]dy. Thus, let B be a bound for f, by triangle equality, we get the followg: 1 (f K n )(x) f(x) = 2 apple 1 Z 2 Z + 1 K n (y) f(x 2 apple y apple y) f(y) dy R 2 apple Z 2 Z y < K n (y)[f(x y) f(x)]dy K n (y) f(x y) f(x) dy K n (y) dy + 2B Z K n (y) dy. 2 apple y apple By property (b) of good kernels, K n(y) dy apple M 2. By property (c) of R good kernels, for large n, B apple y apple K n(y) dy <. Let C>besome constant. For large n, wehave (f K n )(x) f(x) apple C., ], f is uniformly cont- If f is contuous everywhere, by compactness of [ uous, which completes the proof. Let s recall what we have shown Example 3.3, that (f D N )(x) =S N (f)(x). Havg proven the above theorem, it is natural to wonder whether or not the Dirichlet kernels are a family of good kernel. If the Dirichlet kernels are a family of good kernels, Theorem 3.8 would imply that the Fourier series of f is pot-wise convergent to f at pots of contuity.

11 CONVERGENCE OF FOURIER SERIES 11 Example 3.9 (the Dirichlet kernels is not a family of good kernel). Let s first remd ourselves that D N ( ) = Ne ik s((n +1/2) ) =. s( /2) k= N The last part of the equation is the closed form formula for the N th Dirichlet kernel. x Sce s(x) 1 for x 2 [ /2,/2], It follows then, Z s(n +1/2) D N ( ) 2 for 2 [ Z D N ( ) d 4 =4 4 s(n +1/2) d Z (N+1/2) Z N s( ) N 1 Z (k+1) =4 k= k N (k + 1) = 8 k= N 1 k= 1 k +1 = 8 log(n + 1) 8 logn, s( ) d s( ) d Z (k+1) k, pi]. s( ) d which means R D N ( ) d is not bounded and thus the Dirichlet kernel does not satisfy property (b) of good kernels. This result is dishearteng as it suggests that the pot-wise convergence of Fourier series may even fail at pots of contuity. 4. Mean-square convergence of Fourier series Although we hit a dead end at the end of last section when we saw that Dirichlet kernels are not good kernels, this section, we will show that regardg the overall behavior of a function f over the entire terval [, 2], there are still nice results. Orthogonality is at the heart of the theorem we are about to prove. We first lay out the defitions and notations that will be used throughout this section. Consider a space R of tegrable functions on the circle. The ner product of two tegrable functions f and g is defed as (f,g) 2 Z 2 f( )g( )d.

12 12 SOPHIA UE The norm f is defed as f 2 =(f,f) 2 Z 2 f( ) 2 d. Notation-wise, for convenience, we use a n to denote the Fourier coe cients of f, wheref is an tegrable function on the circle. As e would appear over and over this section, for each teger n, lete n = e. Now that we have everythg defed, let s get down to some easy examples to familiarize ourselves with the defitions and notations. Example 4.1. Consider the family {e n } n2z. When n = m, When n 6= m, (e n,e m ) =1. (e n,e m ) 2 Z 2 e e im d Z 2 e i(n m) d Z 2 Z 2 = frac12 2 1d e e im Z 2 e i(n m) d e i(n m) 2 i(n m) e i(n m)2 e 2 i(n m) i(n m) =. Therefore, we have shown that the family {e n } n2z is orthonormal. Besides the orthonormal property of the family {e n } n2z, another important observation is that the Fourier coe cients of the function f can be represented as the ner products of f with elements of the above family: (f,e n ) 2 Z 2 Moreover, it is also important to note that S N (f) = f( )e d = a n. a n e n. With these md, we are ready to prove a simple but useful lemma.

13 CONVERGENCE OF FOURIER SERIES 13 Lemma 4.2. For any complex number b n, (f a n e n )? b n e n. P Proof. We will first show that (f a ne n )?e n. (f a n e n,e n )=(f,e n ) ( a n e n,e n ) For any complex number b n, (f a n e n, b n e n )=(f = b N (f = a n a n = = =. a n e n,b N e N )+... +(f a n e n,e N ) b N (f a n e n,b N e N ) a n e n,e N ) While this lemma may not seem too terestg by itself, it has two very useful results. For the first result, let s recall the Pythagorean theorem, which states that if and Y are orthogonal, then + Y 2 = 2 + Y 2. Let b n = a n, we get f a n e n + a n e n 2 = f which can be rewritten as followg: f 2 = f S N (f) 2 + a n e n 2 + a n 2. a n e n 2, The second result is best known as the best approximation lemma, which states that Lemma 4.3 (Best approximation lemma). If f is tegrable on the circle with Fourier coe cients a n, then for any complex numbers c n f S N (f) apple f c n e n. Moreover, equality holds precisely when c n = a n for all n apple N. Proof. Let b n = a n c n 8n, f c n e n = f a n e n + b n e n.

14 14 SOPHIA UE P By Lemma 4.2, we know that (f a P ne n )?(f b ne n ). Hence, we can apply the Pythagorean theorem and get the followg result: f (f c n e n ) 2 = f S N (f) 2 + (f b n e n ) 2. When c n = a n, b n = f (f When c n 6= a n, b n 6= Hence, f c n e n ) = f S N (f). b n e n 2 Z 2 b n e n 2 d >. 2 S N (f) < f c n e n. Theorem 4.4 (Mean Square Convergence). Suppose f is tegrable on the circle. Then Z 1 2 f( ) S N (f)( ) 2 d! as N!1. 2 Proof. Apply Lemma 3.6 and choose a contuous function g on the circle such that and Sce sup x2[,] f(x) = B, sup g( ) apple sup f( ) = B 2[,2] 2[,2] Z 2 f g apple B apple C 2 f( ) g( ) d < 2. Z 2 Z 2 Z 2 f( ) g( ) 2 d f( ) g( ) f( ) g( ) d f( ) g( ) d where C = B is some constant. Moreover, we can approximate g by a trigonometric polynomial P of degree M such that g( ) P ( ) <,for all. After takg squares and tegratg, we get g P <. It follows directly that f P <C 1/2. We can then get the desired result by applyg the best approximation lemma.

15 CONVERGENCE OF FOURIER SERIES 15 Acknowledgments. It is a pleasure to thank my mentor, Mariya Sardarli and Dolores Walton, for all their help. Without their patient support and guidance, this paper would not be possible. I would also like to thank Professor Peter May for organizg this amazg program and Professor Lszl Babai for his terestg lectures that made my summer meangful. References [1] Elias M. Ste and Rami Shakarchi. Fourier analysis-an troduction, Prceton University Press. 23. [2] Maria Christa Pereyra and Lesley A. Ward. Harmonic Analysis: from Fourier to Haar