Lecture 3: Fourier Series: pointwise and uniform convergence.

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1 Lecture 3: Fourier Series: pointwise and uniform convergence. 1. Introduction. At the end of the second lecture we saw that we had for each function f L ([, π]) a Fourier series f a + (a k cos kx + b k sin kx), where the coefficients a k, b k are defined as a k f(x) cos kx dx, b k f(x) sin kx dx, with k =, 1,... (note that this means that b = ). This is called the real form of the Fourier series. The two objects, f and the series, are not always equal at each point x [, π], but we noted that they are equal as vectors in L ([, π]). In this lecture we shall look at conditions which allow us to put f(x) = a + (a k cos kx + b k sin kx). To this end, we define two spaces of functions which are contained in L ([, π]). Definition 1.1. We define the space E as the space of piecewise continuous functions on [, π]. The function f(x) is piecewise continuous on an interval I if it is continuous on I except perhaps for a finite number of points, and if a I is a point of discontinuity for f(x) then f(a + ) and f(a ) exist: that is are required to exist. f(a + ) = lim x a + f(x), f(a ) = lim x a f(x) Example 1.1. The function f(x) defined as e x for x <, f(x) = x for x, x for x >. To see this, we note that the points of discontinuity are x = and x =. Then we easily calculate that f( ) = 1, f( + ) = and f( ) =, f( + ) = 4. Then we build upon this definition to define another space E as follows: 1

2 Definition 1.. The space E is defined as the space of all functions f(x) E such that the right-hand derivative D + f(x) exists for all x < π, and the left-hand derivative D f(x) exists for all < x π. Recall that f(x + h) f(x + ) D + f(x) = lim h + h f(x + h) f(x ) D f(x) = lim. h h Note that we need to know f(x + ) in order to calculate D + f(x), and f(x ) in order to calculate D f(x). Remark 1.1. Every continuous function is piecewise continuous, that is, belongs to E. Every function which is differentiable on [, π] belongs to E. All functions in E belong to L ([, π]).. Pointwise convergence of Fourier series. We shall prove that the Fourier series of a given function converges pointwise to that function provided the function is continuous and belongs to E. But first we need some notation: we write S N (x) for Our result is: Theorem.1. If f E and a N + (a k cos kx + b k sin kx). then the series converges pointwise to f a + (a k cos kx + b k sin kx) That is f(x + ) + f(x ). In particular, we have S N (x) f(x +) + f(x ) as N. f(x) = a + (a k cos kx + b k sin kx) at every point x [, π] where f(x) is continuous. Theorem.1 tells us about the pointwise convergence of the Fourier series to the value of the function at points where the function is continuous, and to the mean value of the left- and right-hand limits of the function at points of (finite) discontinuity. However, we would also like to know when we have uniform convergence. This will give us the possibility of integrating the series term-wise. To this end, we have the following result:

3 Theorem.. Suppose that f(t) is continuous on [, π] f() = f(π) f (t) E 3 Then a + (a k cos kx + b k sin kx) converges uniformly to f(x) on [, π]. We leave the proofs of these theorems to the Appendix of these notes. 3. Sine and Cosine series. In this section we take up the definitions of even and odd functions: Definition 3.1. f(t) is said to be even if f( t) = f(t) for all t R. f(t) is said to be odd if f( t) = f(t) for all t R. Elementary examples are: f(t) = cos t is even, and f(t) = sin t is odd. Not all functions are of these two types, but they are a mixture of them. More precisely: Proposition 3.1. Each function f(t) can be written as f(t) = F + (t) + F (t) where F + (t) is even and F (t) is odd. Proof: Put f(t) + f( t) f(t) f( t) F + (t) =, F (t) =. it is elementary to verify that F + (t) is even and that F (t) is odd. useful results are: Proposition 3.. (1) f(t), g(t) even = f(t)g(t) is even. () f(t), g(t) odd = f(t)g(t) is even. (3) f(t) even and g(t) odd = f(t)g(t) is odd.

4 4 The proof of this is an elementary verification. We also record the results: a a f(t)dt = if f(t) is odd (and integrable, of course), whereas we have if f(t) is even (and integrable). a a f(t)dt = a f(t)dt Note that the Fourier series of an odd function contains only sine-terms, whereas the Fourier series of an even function contains only cosine-terms. Now we come to another definition: Definition 3.. If f(t) is defined on [, π], then the function F (t) defined by { f(t) if t π F (t) = f( t) if t <, is called the even extension of f(t). The function G(t) defined by f(t) if < t π G(t) = t = f( t) if t <, is called the odd extension of f(t). As we have noted above, the Fourier series of an even function contains only cosineterms, and the Fourier series of an odd function contains only sine-terms. This then gives us the following definition: Definition 3.3. The cosine series of f(t) defined on [, π] is the Fourier series of the even extension of f(t). The sine series of f(t) defined on [, π] is the Fourier series of the odd extension of f(t). 4. Solving ordinary differential equations with Fourier series. In this section I want to show how to solve differential equations using Fourier series. First, note that there is one restriction: the function(s) in the differential equation must be periodic. In our case, we look at functions with peruiod π, but the theory holds for any other period. An important fact is that if y(t) has period T then its derivative y (t) also has period T (provided it exists). This is easy to prove: Lemma 4.1. If y(t) has period T and is differentiable, then y (t) also has period T.

5 Proof: We know that y(t + T ) = y(t) for all t R and we want to prove that y (t + T ) = y (t). We have y y(t + T + h) y(t + T ) (t + T ) = lim h h y(t + h) y(t) = lim h h = y (t). Another result that we need is the following: Lemma 4.. If y(t) has period π and is differentiable and has complex Fourier series 5 then y(t) c k e ikt y (t) ikc k e ikt. In other words, if c k, k =..., 1,, 1,,... are the Fourier coefficients of y(t), then the Fourier coefficients of y (t) are ikc k, k =..., 1,, 1,,... Proof: Suppose we have Then we have y (t) d k e ikt. because we have d k = 1 π y (t)e ikt dt = [y(t)e ikt ] π + 1 π = ikc k (ik)y(t)e ikt dt [y(t)e ikt ] π = since both y(t) and e ikt have period π and because We now look at an example. c k = 1 π y(t)e ikt dt. Example: Find all functions y(t) with period π and which satisfy the differential equation y (t) y(t + π) = cos t.

6 6 Solution: First note that y (t) exists and so both y (t) and y(t) are continuous, since if f (t) exists at t, then f(t) is continuous at t. Note also that we then have y (t) = y(t + π) + cos t so that y (t) is also continuous, because both y(t) and cos t are continuous. Since both these functions are differentiable and because their derivatives are continuous, then y (t) is differentiable and y (t) is continuous. From the above reasoning, we conclude that y(t), y (t) and y (t) all satisfy the conditions of Dirichlet s theorem, and so we may put them equal to their Fourier series. So if we have then y(t) = c k e ikt and y (t) = ikc k e ikt. Further, y (t) = (ik) c k e ikt. since y(t + π) = = = c k e ik(t+π) e ikπ c k e ikt ( 1) k c k e ikt e ikπ = (e iπ ) k = ( 1) k. Putting all this into the differential equation, we have (by rewriting cos t in complex form) which gives the equation (ik) c k e ikt ( 1) k c k e ikt = eit + e it, From this equation, it follows that [(ik) ( 1) k ]c k e ikt = eit + e it.

7 7 [(ik) ( 1) k ]c k = for all k ±1. The reasoning behind this step is that the functions {e ikt, k =, ±1, ±, ±3,... } form a basis for L ([, π]), and that coefficients of basis vecotrs in the left-hand side of an equation are equal to the corresponding coeffiients in the right-hand side of the equation. We then have that either c k = or (ik) ( 1) k = for these k. Now gives (ik) ( 1) k = k = ( 1) k and this has no integer solution: for even k we have k =, which is impossible; for odd k we have k = which is also impossible because is irrational. Thus we must have (ik) ( 1) k for all k ±1 and hence c k = for k ±1. For k = ±1 we have the following: so that and so we have 3c 1 = 1, 3c 1 = 1 c 1 = 1 6, c 1 = 1 6, y(t) = 1 cos t. 3

8 8 5. Appendix: Proofs of pointwise and uniform convergence. Theorem 5.1. If f E and then the series converges pointwise to f a + (a k cos kx + b k sin kx) That is f(x + ) + f(x ). In particular, we have S N (x) f(x +) + f(x ) as N. f(x) = a + (a k cos kx + b k sin kx) at every point x [, π] where f(x) is continuous. Proof: The proof is quite lengthy, so we break it up into a series of Lemmas. The first step is to prove the result Lemma 5.1. a k cos kx + b k sin kx f(x + t) cos kt dt Proof: We have, using the definition of the constants a k, b k, a k cos kx + b k sin kx x (π x) f(t)[cos kt cos kx + sin kt sin kx] dt f(t) cos k(t x) dt f(s + x) cos ks ds f(s + x) cos ks ds where we have first used the substitution s = t x, and then used the result that if F (s) has period π then α+π α F (s)ds = F (s)ds for all α: in fact, this result with F (s) = f(s + x) cos ks and α = (π x) gives us Thus we have 1 π x f(s + x) cos ks ds = 1 f(s + x) cos ks ds. π (π x) π

9 9 This result and the fact that a k cos kx + b k sin kx a = 1 f(x + t) dt, π f(t + x) cos kt dt. which is easy to verify, when substituted into S N (x) (as defined above) give where D N (t) is defined as S N (x) which bears the name Dirichlet s kernel. The aim is to prove that ( N ) 1 f(x + t) + cos kt dt f(x + t)d N (t) dt, D N (t) = 1 N + cos kt, We do this by proving 1 π f(x + t)d N (t) dt f(x +) + f(x ) π as N. and as N. 1 π f(x + t)d N (t) dt f(x +) π 1 f(x + t)d N (t) dt f(x ) π The next step is to prove the formula Lemma 5.. Proof: We have first that D N (t) = sin(n + 1 )t sin t for t.

10 1 Then note that N D N (t) = 1 N + cos kt q k = 1 qn+1 1 q = ( = N [ e ikt + e ikt] N e ikt + ) N e ikt. whenever q 1, so if t we put q = e it to obtain N and then we take q = e it to obtain e ikt = eit e (N+1)it 1 e it = eit e (N+1)it e it (e it e it = e it e (N+ 1 )it i sin t, N Combining these two calculations, we find e ikt = e it e (N+1)it 1 e it = e it e (N+1)it e it (e it e it = e it e (N+ 1 )it i sin t. N N e ikt + e ikt 1 [ ] = i sin t e it it e + e (N+ 1 )it + e (N+ 1 )it 1 = [ i i sin t sin it + i sin(n + 1 ] )it = 1 + sin(n + 1)it sin t and from this it follows that D N (t) = sin(n + 1 )it sin t

11 for t. We can calculate D N (): we have D N () = N + 1, and it is easy to verify that D N (t) D N () as t, so we may write D N (t) = { sin(n+ 1 )it sin t Then we define two functions g(t) and h(t) by g(t) = f(x + t) f(x +) sin t, t N + 1, t =., h(t) = f(x + t) f(x ) sin t. The function g(t) is piecewise continuous on ], π] and h(t) is piecewise continuous on [, [. We find that 11 as well as f(x + t) f(x + ) lim g(t) = lim t + t + t = D + f(x) t sin t f(x + t) f(x ) lim h(t) = lim t t + t = D f(x). t sin t The limits exist because f E. Finally, using these results, we calculate that and that 1 π lim f(x + t)d N (t) dt = f(x +) N π 1 lim N π From all this follows the result: f(x + t)d N (t) dt = f(x ). lim S N(x) = 1 f(x + t)d N (t) dt N π = f(x +) + f(x ), as claimed. The rest of the theorem follows easily. To show that these limits hold, we have: Lemma 5.3. and 1 π lim f(x + t)d N (t) dt = f(x +) N π 1 lim N π f(x + t)d N (t) dt = f(x ).

12 1 Proof: An elementary calculation using the definition of D N (t) gives Then we have D N (t)dt = π, D N (t)dt = π. 1 π f(x + t)d N (t) dt f(x +) π f(x + t)d N (t) 1 π (f(x + t) f(x + ))D N (t)dt sin t g(t)d N(t)dt ( g(t) sin N + 1 ) tdt. Now, by the Riemann-Lebesgue Lemma (see Lecture ), we find that ( g(t) sin N + 1 ) tdt as N. Consequently, f(x + )D N (t)dt or 1 π f(x + t)d N (t) dt f(x +) π as N, To show that we put 1 π lim f(x + t)d N (t) dt = f(x +) N π ( g(t) sin N + 1 ) tdt as N, G(t) = { g(t) if < t π if t. Then, since g(t) is piecewise continuous on [, π], it follows that G(t) is piecewise continuous on [, π], and hence G(t) L ([, π]) so that the Riemann-Lebesgue Lemma gives ( g(t) sin N + 1 ) ( tdt = G(t) sin N + 1 ) tdt as N. A similar calculation on 8 π, ] using the function h(t) defined above gives us 1 lim N π We leave the details to you. f(x + t)d N (t) dt = f(x ).

13 Theorem.1 tells us about the pointwise convergence of the Fourier series to the value of the function at points where the function is continuous, and to the mean value of the left- and right-hand limits of the function at points of (finite) discontinuity. However, we would also like to know when we have uniform convergence. This will give us the possibility of integrating the series term-wise. To this end, we have the following result: Theorem 5.. Suppose that f(t) is continuous on [, π] Then f() = f(π) f (t) E a + (a k cos kx + b k sin kx) converges uniformly to f(x) on [, π]. Proof: Since we have f (t) E, we have the Fourier series 13 and, with f (t) α + (α k cos kt + β k sin kt) f(t) a + (a k cos kt + b k sin kt), it is easy to calculate (using the definition of the coefficients in a Fourier expansion) that α = (this is guaranteed by f() = f(π)) and that α k = kb k, β k = ka k so we find that f (t) (kb k cos kt ka k sin kt). Now, Parseval s identity (see Lecture ) guarantees that so that ( α k + β k ) = f < +, (k a k + k b k ) converges. We shall show that the series ak + b k also converges. First consider, for each N 1, the vectors

14 14 and Then we have u N = (1, 1, 1 3,..., 1 N ) v N = ( α 1 + β 1, α + β,..., α N + β N ). N ak + b k = N 1 αk k + β k = u N v N u N v N by the Cauchy-Schwarz Theorem = N 1 N α k + β k, k and from this it follows that (on taking the limit as N ) ak + b k 1 α k + β k which then shows that k ak + b k converges. Having proved this, we use the fact that a k a k + b k, b k a k + b k in order to deduce that From this it follows that ( a k + b k ) < +. a + (a k cos kt + b k sin kt) converges uniformly on [, π], by Weierstrass Majorant Theorem because a k cos kt + b k sin kt a k + b k. This proves the result.

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