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CONVERGENCE of FOURIER SERIES Contents 1. Introduction To CONVERGENCE OF FOURIER SERIES SOPHIA XUE Abstract. The subject of Fourier analysis starts as physicist and mathemati- cian 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 definitions of Fourier analysis to answer the question in what sense does the Fourier series of a function converge to the function itself. Di↵erent criteria for convergence will be introduced 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 15 1. Introduction to Fourier Series Throughout this paper, an integrable function should be interpreted as integrable in the Riemann sense. In this section, we will go through the basic definitions of Fourier Analysis that will appear throughout this paper. Examples and variations are supplied to aid understanding. Definition 1.1. If f is an integrable function given on an interval [a, b] of length L (that is, b a = L), the nth Fourier coefficient of f is defined by − b ˆ 1 2⇡inx/L f(n)= f(x)e− dx, n Z. L 2 Za Note that more often than not, we will be working with functions that are integrable on a circle, which means that the function is integrable on every interval of length 2⇡. Definition 1.2. The Fourier series of f is given normally by 1 fˆ(n)e2⇡inx/L. n= X1 The notation 1 f(x) fˆ(n)e2⇡inx/L ⇠ n= 1 X 1 2 SOPHIA XUE means that the series on the right-hand side is the Fourier series of f. Definition 1.3. The N th partial sum of the Fourier Series of f, for N a positive integer, is given by N 2⇡inx/L SN (f)(x)= fˆ(n)e . n= N X− The notion of N th partial sum of the Fourier Series of f is very important in the study of Fourier Analysis. Using 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 infinity . Indeed, the basic question can be reformulated as follows: Question 1.4. In what sense does S (f) converge to f as N ? N !1 Following are some simple examples to familiarize ourselves with above defini- tions. Example 1.5. Let f(✓)=✓ for ⇡ ✓ ⇡. To calculate the Fourier coefficients of f, we should split it into two cases.− First, when n = 0, 6 ⇡ 1 in✓ fˆ(n)= ✓e− d✓ 2⇡ ⇡ Z− 1 e in✓ ⇡ ⇡ e in✓ = [✓ − − d✓] 2⇡ in ⇡ − ⇡ in − − Z− − in⇡ in⇡ in⇡ in⇡ e− + e e− e = + − − 2in (2⇡in)( in) − cos(n⇡) sin(n⇡) = − + in i⇡n2 ( 1)n+1 = − . in When n=0, 1 ⇡ fˆ(n)= ✓d✓ =0. 2⇡ ⇡ Z− Hence, the Fourier series of f is ( 1)n+1 1 sin(n✓) f(✓) − ein✓ =2 ( 1)n+1 . ⇠ in − n n=0 n=1 X6 X Furthermore, by the alternating series test, we can easily see that the Fourier series of f is convergent. Example 1.6. Let f(✓)= ✓ for ⇡ ✓ ⇡. First, when n = 0, | | − 6 ⇡ 0 1 in✓ 1 in✓ fˆ(n)= ✓e− d✓ ✓e− d✓. 2⇡ 0 − 2⇡ ⇡ Z Z− CONVERGENCE OF FOURIER SERIES 3 Using the result from last example, we get that 1 e in✓ ⇡ ⇡ e in✓ e in✓ 0 0 e in✓ fˆ(n)= [✓ − − d✓ ✓ − + − d✓] 2⇡ in 0 − 0 in − in ⇡ ⇡ in − Z − − − Z− − in⇡ in⇡ in⇡ in⇡ 1 e− e− 1 e 1 e− = [⇡ 0+ − 0+( ⇡) − ] 2⇡ in − (in)( in) − − in − (in)( in) − − − − in⇡ in⇡ 1 cos( n⇡)+isin( n⇡) cos(n⇡) isin(n⇡) e− 1 1+e− = [⇡ − − − − + − − ] 2⇡ in (in)( in) in⇡ − − e− 1 =0+ − ⇡n2 cos(n⇡) isinn⇡ 1 = − − ⇡n2 ( 1)n 1 = − − . ⇡n2 On the other hand, when n = 0, 1 ⇡ 1 2⇡2 ⇡ fˆ(n)= ✓ d✓ = = . 2⇡ ⇡ | | 2⇡ ⇤ 2 2 Z− Hence, 1 ( 1)n 1 f(n) ein✓ + − − ein✓ ⇠ 2⇡ ⇡n2 n=0 X6 1 4 = ein✓ + − ein✓ 2⇡ ⇡n2 n=2k+1,k X 2N 1 4 cos(n✓) isin(n✓) = (cos(n✓) isin(n✓)) − . 2⇡ − − ⇡ n2 n=2k+1,k X 2N 2. Uniqueness of Fourier series If we were to believe that the Fourier series of functions f somehow converge to f, then we could infer that the Fourier coefficient of a function uniquely determines the function. In other words, for our assumption to be true, if f and g have the same Fourier coefficients, then f and g are necessarily equal. The statement can be reformulated as following by taking the di↵erence f g: − Proposition 2.1. If fˆ(n)=0for all n Z, then f =0. 2 It is obvious that this proposition cannot be true without reservation. Since cal- culating Fourier coefficients requires integration and any two functions that are dif- ferent at finitely many points can have the same integration, one particular Fourier series can be shared by two functions that are di↵erent at finitely many points. However, we do have the following positive result regarding continuous points. Theorem 2.2. 1 Suppose that f is an integrable function on the circle with fˆ(n)=0 for all n Z. Then f(✓0)=0whenever f is continuous at the point ✓0. 2 This result is really nice. It shows that f vanishes for ”most” values of ✓. Following is a staightforward corollary. 1Rigorous proof of this theorem can be found in Elias M. Stein and Rami Shakarchi’s Fourier Analysis– an Introduction (2003), p39-41 following Theorem 2.1. 4 SOPHIA XUE Corollary 2.3. If f is continuous on the circle and fˆ(n)=0for all n Z, then f 0. 2 ⌘ Moreover, this corollary is useful in answering Question 2.4 under the condition that the Fourier series converges absolutely. Corollary 2.4. Suppose that f is a continuous function on the circle and that ˆ the Fourier series of f is absolutely convergent, n1= f(n) < . Then, the Fourier series converges uniformly to f,thatis, 1 | | 1 P lim SN (f)(x)=f(x) N !1 uniformly in x. ˆ inx ˆ Proof. Let g(x)= n1= f(n)e .Since n1= f(n) < , g(x)isdefined everywhere. 1 1 | | 1 By triangle-inequality,P we get that P 1 S (f)(x) g(x) = fˆ(n)einx fˆ(n)einx = fˆ(n)einx fˆ(n) . | N − | | − | | | | | n N n= n >N n >N | X| X1 | X| | X| Since the Fourier series of f is bounded, for any ✏>0, we can find large N such that fˆ(n) <✏. Hence, S (f)(x) uniformly converges to g(x) as N . n >N | | N !1 Note that| | for large N, the Fourier coefficients of g(x) are P 2⇡ 1 inx gˆ(n)= g(x)e− dx 2⇡ 0 Z 2⇡ 1 inx inx = ( fˆ(n)e )e− dx 2⇡ 0 n N Z X = fˆ(n). In other words, f and g have identical Fourier coefficients. Now let’s define a new function h = f g. By distributivity of the integral, it is easy to see that hˆ(n) = 0. Apply Corollary− 3.3 to h, we get that h = f g = 0 or f = g. Combining this − result with the uniform convergence of SN (f)(x)tog(x), we get the desired result, i.e. S (f)(x) uniformly converges to f(x) as N . N !1 ⇤ Now that we have proven when the Fourier series of a continuous 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 continuously di↵erentiable function on the circle, then fˆ(n)=O(1/( n )2as n , | | | |! 1 so that the Fourier series of f converges absolutely and uniformly to f. The notation fˆ(n)=O(1/( n )2as n means that the left-hand side is bounded by a | | | |! 1 constant multiple of the right-hand side, i.e. there exists C>0 with fˆ(n) C/( n )2 for all large n . | | | | | | CONVERGENCE OF FOURIER SERIES 5 Proof. Through integrating by parts twice , we have 2⇡ 1 in✓ fˆ(n)= f(✓)e− d✓ 2⇡ Z0 in✓ 2⇡ 1 e− 1 in✓ = f(✓)− + 2⇡f0(✓)e− d✓ 2⇡ in 2⇡in 0 Z0 1 in✓ = 2⇡f0(✓)e− d✓ 2⇡in Z0 in✓ 2⇡ 1 e− 1 in✓ = f 0(✓)− + 2⇡f00(✓)e− d✓ 2⇡in in 2⇡(in)2 0 Z0 2⇡ 1 in✓ = − f 00(✓)e− d✓. n2 Z0 2⇡ 2⇡ in✓ in✓ e− e− Since f and f 0 are both periodic, f(✓) − in = 0 and f 0(✓) − in = 0 Let B 0 0 be a bound for f .Thus, 00 2⇡ 2⇡ 2 in✓ 2⇡ n fˆ(n) f 00(✓)e− d✓ f 00(✓) d✓ 2⇡B. | | | || | | | Z0 Z0 Since B is independent of n, we have that fˆ(n)=O(1/( n )2as n | | | |! 1 ⇤ There are also stronger versions of Corollary 2.5 such as the following. Corollary 2.6. If f is 2⇡-periodic and has an integrable 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 H¨older condition of order ↵,with↵>1/2,thatis sup f(✓ + t) f(✓) A t ↵ for all t.
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