8. Uniform Convergence and Differentiation Suppose That (I) A

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8. Uniform Convergence and Differentiation Suppose That (I) A 54 1. THE THEORY OF CONVERGENCE 8. Uniform convergence and differentiation Suppose that (i) a series ∞ u(x)= uk(x) Xk=1 converges pointwise on some interval [a,b]; 0 (ii) the derivatives un(x) are continuous on [a,b]; 0 (iii) the series of the derivatives k un converges uniformly on [a,b]. P Let us show that under these hypotheses the sum u(x) is continuously differentiable on [a,b] and ∞ ∞ d dun(x) (8.1) u0(x)= u (x)= dx n dx Xn=1 Xn=1 In other words, the order of differentiation and summation can be changed if the series of derivatives converges uniformly. Since the series of derivatives converges uniformly and the deriva- tives are continuous, then there exists a continuous function v(x) such that ∞ 0 v(x)= un(x) Xn=1 By the theorem about integration and uniform convergence, x ∞ x ∞ 0 v(t) dt = u (t) dt = un(x) un(a) Z Z n − a Xn=1 a Xn=1 = u(x) u(a) − for any x in [a,b], where the fundamental theorem of calculus and the convergence of the series un were used. Since v is continuous, its integral is differentiable withP respect to the upper limit and, hence, u is differentiable. Moreover, taking the derivative of both sides of the relation, x u(x)= u(a)+ v(t) dt Za it is concluded that u0(x)= v(x) which proves (8.1). It turns out that differentiability of u can also be established even 0 if the derivatives un(x) exist but are not necessarily continuous. In this 8. UNIFORM CONVERGENCE AND DIFFERENTIATION 55 case, the fundamental theorem of calculus cannot be used to establish differentiability of u(x). Nevertheless, the result still holds7: Theorem 8.1. Let un be a sequence of differentiable functions on { } [a,b]. Suppose that the sequence un(x0) converges for some x0 [a,b] and the sequence of derivatives {u0 (x) }converges uniformly on∈[a,b]. { n } Then the sequence un converges uniformly to some u(x) on [a,b], the limit function u(x){is differentiable} on [a,b] and 0 0 0 u (x)= lim un(x) = lim un(x) . n→∞ n→∞ Naturally, a similar result holds for series if the above theorem is applied to the sequence of partial sums. Example. The series ∞ sin2(nx) u(x)= 2+ n3 Xn=1 converges for all real x by the comparison test: 1 1 1 un(x) 3 3 and 3 < | |≤ 2+ n ≤ n X n ∞ In fact, it converges uniformly by the above argument. Each term of the series is a continuously differentiable function everywhere. The series of the derivatives converges uniformly everywhere because the series of upper bounds 0 n sin(2nx) n 1 1 un(x) = | 3 | 3 2 and 2 < | | 2+ n ≤ 2+ n ≤ n X n ∞ Therefore u(x) is continuously differentiable everywhere and ∞ ∞ n sin(2nx) u0(x)= u0 (x)= . n 2+ n3 Xn=1 Xn=1 8.1. Differentiability of a power series. The derivative of a function of a complex variable z is defined by f(z) f(w) f 0(z)= lim − w→z z w − In particular, (zn)0 = nzn−1 7see, e.g., W. Rudin, Principles of Mathematical Analysis, Chapter 7 56 1. THE THEORY OF CONVERGENCE because zn wn =(z w)(zn−1 + zn−2w + + wn−2z + wn−1) − − ··· Let f(z) be defined by a power series ∞ n f(z)= anz , z < R | | Xn=0 where R> 0 is the radius of convergence. The series of derivatives ∞ ∞ n−1 n annz = an+1(n + 1)z Xn=1 Xn=0 is also a power series with the same radius of convergence: n n n lim sup an+1 (n +1) = lim √n + 1 lim sup an+1 n→∞ n→∞ p| | n→∞ p| | n 1 = lim sup an = n→∞ R p| | Therefore it converges uniformly on any disk z b < R. Thus, f(z) is differentiable in the disk z < R and | |≤ | | ∞ 0 n−1 f (z)= annz , z < R | | Xn=1 The same argument can now be applied to the function g(z) = f 0(z) to conclude that ∞ 0 00 n−2 g (z)= f (z)= ann(n 1)z , z < R − | | Xn=2 In fact, it follows that a function defined by convergent power series is infinitely many times differentiable, its derivative are obtained term- by-term differentiation of the power series, ∞ (k) n−k f (z)= n(n 1) (n k)anz , z < R − ··· − | | Xn=k and these power series have the same radius of convergence. 8.2. Differentiation of a trigonometric Fourier series. Let f(x)bea2π periodic function whose derivatives are continuous (up to some order p).− By Fej´er’s theorem, its Fourier series converges to f(x) pointwise every- where: ∞ π ikx 1 −ikx f(x)= Ake , Ak = f(x)e dx 2π Z kX=−∞ −π 8. UNIFORM CONVERGENCE AND DIFFERENTIATION 57 Let us investigate how fast the Fourier coefficients Ak decay with in- creasing k . Using the integration by parts p times | | π π π 1 ikx 1 ikx 1 0 ikx Ak = f(x)e dx = f(x)e f (x)e dx 2π Z−π 2πik −π − 2πik Z−π π 1 = f 0(x)eikxdx −2πik Z−π 1 p 1 π = f (p)(x)eikxdx −ik 2π Z−π Since, by assumption, the derivative f (p)(x) is continuous and periodic, it is bounded on [ π,π] and it follows that − Mp (p) Ak p , Mp = sup f | |≤ k [−π,π] | | Thus, the Fourier coefficients of a function that has at least p continu- ous derivatives fall off at least as k−p with increasing k. In particular, if the function has infinitely many derivatives, then its Fourier coeffi- cients decay faster than any reciprocal power. This observation allows to conclude that the Fourier series of a function that is infinitely many times differentiable can be differenti- ated term-by-term any number of times and (p) p ikx f (x)= (ik) Ake Xk because the series converges uniformly for any p 0. Furthermore, suppose that f(x) is a real continuous,≥ 2π periodic function, and f 0(x) satisfies the hypotheses of Fej´er’s theorem− (see Sec- 0 tion 5.2). If Bk are the Fourier coefficients of f (x), then by integration by parts (as shown above) Bk Ak = Bk = ikAk , k = 0 − ik ⇒ − 6 π 0 Assuming that B0 = 0 (or −π f (x)dx = 0), it is concluded by Fej´er’s theorem that the series obtainedR by term-by-term differentiation of the Fourier series for f(x) converges pointwise: ikx ikx 1 0 0 ikAke = Bke = lim f (t)+ lim f (t) 2 t→x+ t→x− Xk Xk So, if the derivative f 0 is not continuous (and satisfies the above hy- potheses), the Fourier series obtained by term-by-term differentiation of the Fourier series for f converges to f 0 at each point where f 0 is continuous and to the mid-point of each jump discontinuity of f 0. 58 1. THE THEORY OF CONVERGENCE 8.3. Solutions to partial differential equations represented by Fourier series. Solutions to partial differential equations (PDEs) are often sought as series over some sets of functions (e.g., orthogonal sets of functions). In order to be a solution to a PDE, the sum of such a series should have continuous partial derivatives of sufficiently high order. For example, a solution to a wave equation 1 ∂2u ∂2u = 0 c2 ∂2t − ∂2x in two variables (x,t) should be at least twice continuously differen- tiable. If a solution is represented by a series of smooth functions, then differentiability of its sum is not obvious and requires an investigation. Suppose that periodic solutions to the above wave equation are sought: u(x + 2π,t)= u(x) The equation describes small vibrations of a circular string. Think of a circle of radius 1 as an equilibrium configuration of the string. The string can vibrate in the plane in which the circle lie and each point of the string can move only in the radial direction. In this case, u(x,t) is the shape of the vibrating string at a time moment t. By Fej´er’s theorem any continuous 2π periodic function can be represented by its trigonometric Fourier series− ∞ π ikt 1 −ikx u(x,t)= vk(t)e , vk(t)= u(x,t)e dx 2π Z kX=−∞ −π In particular, the sequence of its partial sums un(x,t) converges point- wise n ikt u(x,t)= lim un(x,t)= lim vk(t)e n→∞ n→∞ kX=−n for every real t. Let us find vk(t) for which a partial sum un(x,t) satisfies the wave equation. A physical solution is real. Owing to the linearity of the wave equation, a real solution can be obtained by taking either real or imaginary parts of a complex solution. A formulation using com- plex valued functions is technically simpler. Naturally, the periodicity ikx condition is fulfilled by un(x,t). Since the functions e are linearly independent, the coefficients vk(t) must satisfy the equation 1 v00(t)+ k2v (t) = 0 c2 k k 8. UNIFORM CONVERGENCE AND DIFFERENTIATION 59 Its general solution reads ikct −ikct vk(t)= ake + bke , k = 0 , v (t)= a + tb 6 0 0 0 where ak and bk are (complex) constants so that n ik(x+ct) ik(x−ct) un(x,t)= ake + bke kX=−n If ak and bk are such that the corresponding Fourier series converges, then it defines a function u(x,t).
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