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 coe cient 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= X 1 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 coe cients 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 coe cient of a function uniquely determines 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 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 coe cients 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| X 1 | 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 coe cients 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 coe cients. 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. ✓ | | | | Proof. First, we will prove that for any 2⇡-periodic and integrable function f, ⇡ 1 inx fˆ(n)= f(x + ⇡/n)e dx, 2⇡ ⇡ Z and thus ⇡ 1 inx fˆ(n)= [f(x) f(x + ⇡/n)]e dx. 4⇡ ⇡ Z By definition, we have ⇡ 1 iny fˆ(n)= f(y)e dx. 2⇡ ⇡ Z 6 SOPHIA XUE
Since f is periodic and we are integrating over the period of f, we can substitute x + ⇡n for y without changing the bounds of integration. Hence, ⇡ 1 in(x+⇡/n) fˆ(n)= f(x + ⇡/n)e dx 2⇡ ⇡ Z ⇡ 1 inx ⇡i = f(x + ⇡/n)e e dx 2⇡ ⇡ Z ⇡ 1 inx = f(x + ⇡/n)e dx. 2⇡ ⇡ Z The last equality follows from applying Euler’s Identity. Since ⇡ ⇡ 1 inx inx 2fˆ(n)= ( f(x)e dx f(x + ⇡/n)e dx), 2⇡ ⇡ ⇡ Z Z a little algebraic manipulation gives the desired result, that ⇡ 1 inx fˆ(n)= [f(x) f(x + ⇡/n)]e dx. 4⇡ ⇡ Z It follows ⇡ 1 inx fˆ(n) = [f(x) f(x + ⇡/n)]e dx | | 4⇡ | ⇡ | Z⇡ 1 inx f(x) f(x + ⇡/n) e dx. 4⇡ ⇡ | || | Z Apply H¨older and we get ⇡ 1 ↵ inx fˆ(n) C ⇡/n e dx. | | 4⇡ | ⇡ | | | | Z ⇡ inx Since e dx =2⇡, ⇡ | | C⇡↵ R fˆ(n) , | | 2 n ↵ | | which proves the theorem. ⇤
3. Convolutions Definition 3.1. Given two 2⇡-periodic integrable functions f and g on R,their convolution f g on [ ⇡,⇡] is given by ⇤ 1 ⇡ (f g)(x)= f(y)g(x y)dy. ⇤ 2⇡ ⇡ Z Note that since the product of two integrable functions is again integrable, the above integral makes sense for all x. Furthermore, since the functions are periodic, we can change variables to see that 1 ⇡ (3.2) (f g)(x)= f(x y)g(y)dy. ⇤ 2⇡ ⇡ Z The notion of convolution plays a fundamental role in Fourier analysis. It re- duces the problem of understanding SN (f) to understanding the convolution of two functions as we will see in the following example. CONVERGENCE OF FOURIER SERIES 7
th Example 3.3. DN is the N Dirichlet kernel given by
N inx DN (x)= e . n= N X For any 2⇡-periodic integrable function f, its convolution with the N th Dirichlet kernel is
⇡ N 1 in(x y) (f D )(x)= f(y)( e )dy ⇤ N 2⇡ ⇡ n= N Z X N ⇡ 1 iny inx = ( f(y)e dy)e 2⇡ n= N ⇡ X Z N = fˆ(n)einx n= N X = SN (f)(x). As we have shown above, Question 2.4 can be again reformulated as following: Question 3.4. In what sense does (f D )(x) converge to f as N ? ⇤ N !1 But before we dive in, let’s look at some nice properties of convolution. Proposition 3.5. Suppose that f, g,andh are 2⇡-periodic integrable functions. Then: (i) f (g + h)=(f g)+(f h) ⇤ ⇤ ⇤
(ii) (cf) g = c(f g)=f (cg) for any c C ⇤ ⇤ ⇤ 2 (iii) f g = g f ⇤ ⇤ (iv) (f g) h = f (g h) ⇤ ⇤ ⇤ ⇤ (v) f g is continuous ⇤
(vi) f[g(n)=fˆ(n)ˆg(n). ⇤ Part (i) and (ii) give linearity. Part (iii) gives commutativity while part (iv) demonstrates associativity. Part (v) shows that convolution is a smoothing opera- tion in the sense that even though f and g are merely integrable, f g is continuous. Part (vi) converts convolution to multiplication, which is key in the⇤ study of Fourier series. Now we move on to the proof of Proposition 3.5. Part (i) and (ii) follow directly from the linearity of integration. Part (iii) follows from Equation (3.2). Part (iv) can be easily proven through interchanging two integral signs and changing variables. Though Part (v) and (vi) are easily deduced if f and g are continuous, to prove these properties under the condition that f and g are merely integrable, we are going to need the following approximation lemma. 8 SOPHIA XUE
Lemma 3.6. Suppose f is integrable on the circle and bounded by B. Then there exists a sequence f 1 of continuous functions on the circle so that { k}k=1 sup fk(x) B x [ pi,⇡] | | 2 for all k =1, 2,...,and ⇡ f(x) fk(x) dx 0 as k . ⇡ | | ! !1 Z Proof of Lemma 3.6. When f is real, given ✏>0, a partition P of the interval [ ⇡,⇡] such that U(f,P) L(f,P) <✏since f is integrable.9 Now we define a step function g(x)= sup f(y) xj 1 y xj if x [xj 1,xj) for j [1,N]. We can see that g is bounded by B and 2 2 ⇡ ⇡ g(x) f(x) dx = (g(x) f(x))dx < ✏. ⇡ | | ⇡ Z Z To modify the step function g and make it continuous, we will take small >0 and define g*(x)=g(x) when the distance between x and any partition point in P is at least . When the distance between x and any partition point in P is more than ,defineg*(x) as the linear function that connects g(x ) and g(x + ). It is easy to see that g* is continuous. Now let g*(x)=0whenx [ ⇡ , ⇡ + ]andx [⇡ ,⇡ + ], which extends g*toa2⇡-periodic function.2 Since g* only di↵ers from2 g in N intervals of length s and is also bounded by B,wehave ⇡ g(x) g*(x) dx 4BN . ⇡ | | Z With su ciently small, we have ⇡ g(x) g*(x) dx < ✏. ⇡ | | Z By triangle inequality, ⇡ ⇡ ⇡ f(x) g*(x) dx g(x) f(x) dx + g(x) g*(x) dx < 2✏. ⇡ | | ⇡ | | ⇡ | | Z Z Z 1 Let 2✏ = k and denote g*byfk,thesequence fk is what we are looking for. When f is complex, we can apply the above{ proof} to the real part and the imaginary part separately to get the same result. ⇤ Applying Lemma 3.6, we can complete the proof of part (v) and (vi), the idea here is to substitute fk and fˆk for f and fˆ respectively and use the continuity of fk to complete the proof.
Proof of Part (v) and (vi). Let’s apply Lemma 3.6to2⇡-periodic integrable func- tions f and g. We get sequences fk and gk of approximating continuous func- tions. Note that with a bit of algebraic{ } manipulation,{ } we have f g f g =(f f ) g + f (g g ). ⇤ k ⇤ k k ⇤ k k CONVERGENCE OF FOURIER SERIES 9
Moreover, (f f ) g uniformly converges to 0 as k . k ⇤ !1 1 ⇡ (f fk) g(x) f(x y) fk(x y) g(y) dy | ⇤ 2⇡ ⇡ | || | Z 1 ⇡ sup g(y) f(y) fk(y) dy 2⇡ y | | ⇡ | | Z 0 as k . ! !1 Similarly, f (g g ) 0 uniformly in x. Therefore, as k , f g uniformly k ⇤ k ! !1 k ⇤ k converges to f g. By continuity of each fk gk, f g is also continuous, which proves Part (v).⇤ ⇤ ⇤ For each integer n,sincef g uniformly converges to f g as k ,wehave k ⇤ k ⇤ !1 f\g (n) f[g(n) as k . k ⇤ k ! ⇤ !1 By continuity of fk and gk,wehave ⇡ 1 inx f\k gk(n)= (fk gk)(x)e dx ⇤ 2⇡ ⇡ ⇤ Z ⇡ ⇡ 1 1 inx = ( fk(y)gk(x y)dy)e dx 2⇡ ⇡ 2⇡ ⇡ Z ⇡ Z ⇡ 1 iny 1 in(x y) = fk(y)e ( gk(x y)e dx)dy 2⇡ ⇡ 2⇡ ⇡ Z ⇡ Z ⇡ 1 iny 1 inx = fk(y)e ( gk(x)e dx)dy 2⇡ ⇡ 2⇡ ⇡ Z Z = fˆ(n)ˆg(n). Hence, ⇡ 1 inx fˆ(n) fˆk(n) = (f(x) fk(x))e dx | | 2⇡ | ⇡ | Z 1 ⇡ f(x) fk(x) dx 2⇡ ⇡ | | Z 0 as k , ! !1 which means fˆk(n) fˆ(n) as k . Again, similarly,g ˆk(n) gˆ(n) as k . ˆ ! !1 ! !1 By substituting fk(n) andg ˆk(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 introduce the concept of good kernels.
Definition 3.7. A family of good kernels is a sequence of functions Kn(x) n1=1 that satisfies the following properties: { }
(a) For all n 1, 1 ⇡ Kn(x)dx =1. 2⇡ ⇡ Z (b)There exists M>0 such that for all n 1, ⇡ Kn(x) dx M. ⇡ | | Z 10 SOPHIA XUE
(c) For every >0,
Kn(x) dx 0, as n . x ⇡ | | ! !1 Z | | The concept of good kernels is very important in the study of Fourier series because of the nice property good kernels have, which is stated in the following theorem.
Theorem 3.8 (Approximation to the Identity). Let Kn(x) n1=1 be a family of good kernels, and f an integrable function on the circle.{ Then }
lim (f Kn)(x)=f(x) n !1 ⇤ whenever f is continuous at x.Iff is continuous everywhere, then the above limit is uniform. Proof. If f is continuous at x, given ✏>0, such that if y <