MATH 564: Fourier Series Hung Phan, UMass Lowell September, 8 Power Series A power series in the variable x is a series of the form a + a x + a x + = where the coefficients a, a,... are real or complex numbers. a k x k, Theorem. For a power series k= a kx k, there are three possibilities (i) k a kx k diverges for all x. (ii) k a kx k converges for all x. (iii) There exists R > such that k a kx k converges for all x < R and diverges for all x > R. Theorem. Given a power series k= a kx k, and suppose that either () K = lim a k+ or K = lim k ak k a k k exists. Then the following are true k= (i) If K =, then k= a kx k converges for all values of x. (ii) If K >, then the radius of convergence is R = K. If either of the limits in () fails to exist, then k= a kx k diverges for all values x. Pointwise Convergence Definition. (pointwise convergence) Suppose {f n } n N is a sequence of functions defined on an interval I. We say that f n (x) converges pointwise to the function f(x) if f n (x) n f(x) for each x I. Exercise. Prove that f n (x) = x x n converges pointwise to f(x) = x on R. Exercise.3 Prove that f n (x) = e nx converges pointwise to f(x) = {, x =, x on [, +).
3 Uniform Convergence Definition 3. A sequence of function f n (x) is said to converge uniformly to f(x) on I if f n f = sup f n (x) f(x). x I Exercise 3. Prove that f n (x) = e nx on [, 3] converges to uniformly. Example 3.3 f n (x) = xe nx on [, ) converges to uniformly. Proof. For each n, we have f n(x) = ( nx)e nx. So f n(x) = implies x = n. We then verify that f n (x) attains its maximum at x = n. We have sup f n (x) = f n ( n ) = n e as n. x [,+ So f n (x) uniformly. Exercise 3.4 Uniform convergence implies pointwise convergence. Theorem 3.5 Suppose f n (x) is a sequence of continuous functions on I that converges uniformly to f(x) on I. Then the limit function f is also continuous on I. Theorem 3.6 Suppose f n (x) is a sequence of continuous functions that converges uniformly to a continuous function f(x) on a bounded interval [a, b]. Then we have b lim n a f n (x)dx = b a lim f n(x)dx = n b a f(x)dx. Theorem 3.7 (Weierstrass Majorant Theorem) Suppose sup f k M k and k= M k < +. Then k= f k converges uniformly to some f. In addition, if each f k is continuous then f is also continuous. 4 Inner product spaces Throughout, F denotes the scalar field R (real numbers) or C (complex numbers). Example 4. R n is an inner product space with x, y = y = (y,..., y n ) R n. Example 4. C n is an inner product space with x, y = y = (y,..., y n ) C n. n x k y k k= n x k y k k= for x = (x,..., x n ) R n and for x = (x,..., x n ) C n and Example 4.3 C([a, b]) denotes the space of F-valued continuous functions on [a, b] with inner product f, g = π b a f(t)g(t)dx.
Theorem 4.4 (Cauchy-Bunyakovski-Schwarz inequality) For every u, v V, u, v u v. Equality happens if and only if u = αv or v = αu for some α F. Theorem 4.5 (generalized Pythagoras Theorem) Let {u,..., u N } be an orthonormal system, and let v span {u,..., u N }. Then v = v, u k u k and v = k= v, u k. Theorem 4.6 (Bessel s inequality) Let {u, u...} be an orthonormal system in V. Then for every v V, v, u k < v. Proof. For each N, define k= S N := v, u k uk. Clearly, (v S N ) u k for k =,..., N. So (v S N ) S N, which implies k= k= v = v S N + S N = v S N + v, u k. k= The conclusion then follows. We also have an immediate corollary Corollary 4.7 (Riemann-Lebesgue Lemma) Let {u, u...} be an orthonormal system. Then for every v V, v, u k as k. Proof. The Bessell s inequality implies v, u k < v. k= So, the term in the summand v, u k must go to zero. 5 The space L ([, π]) Definition 5. L ([, π]) is the space of all complex valued functions on [, π] satisfying f(t) dt < +. 3
On this space, the inner product is defined by f, g = π f(t)g(t)dx. The induced norm is f = π f(t) π dt. Theorem 5. (a special orthogonal system in L ([, π])) The system {, cos kt, sin kt, k =,,... } is an orthogonal system in L ([, π]). Proof. First, each vector belongs to L ([, π]): dt = π, Next, they are mutually orthogonal: cos (kt) dt = π, sin (kt) dt = π., cos kt =, sin kt =, k; cos kt, sin mt =, k, m; cos kt, cos mt =, k m; sin kt, sin mt =, k m. The proof is complete. By normalizing the above system, we obtain the following Theorem 5.3 (a special orthonormal system in L ([, π])) The system { } cos kx sin kx,,, k =,,... π π π is an orthonormal system in L ([, π]). 6 Fourier series Let f : R R be a periodic function with period T = π. The Fourier series of f is defined as the series () S(x) := a + a k cos kx + b k sin kx, where the Fourier coefficients are given by a = π f(t)dt, a k = π k= f(t) cos ktdt, b k = π 4 f(t) sin ktdt.
The complex Fourier series of f is defined as the series c k e ikx where c k = f(t)e ikt dt. π Remark. If f is only defined on [, π), we can always extend f to a periodic function on R by defining f(x) := f(y + kt ) for some y [, π) and k Z. Therefore, if suffices to consider only f : [, π] R. 7 Convergence of the Fourier series Recall that a function f(t) is said to be piecewise continuous on an interval I if (i) it is continuous on I except for a finite number of points. (ii) if a I is a point of discontinuity, then f(a + ) = lim t a + f(t) and f(a ) = lim t a f(t) exist. Let f be a piecewise continuous function on I. Recall that the right-hand derivative of f at a I is D + f(a) = lim h + f(a + h) f(a + ) h (if f(a + ) exists), and the left-hand derivative of f at a I is D f(a) = lim h f(a + h) f(a ) h (if f(a ) exists). We will use the following functional spaces Denote E the space of all piecewise continuous functions on [, π]. Denote E the space of all functions in E such that the left derivate exists for all x (, π] and the right derivative exists for all x [, π). Let f E, our goal is to prove convergence properties of the Fourier series () for f. Lemma 7. a k cos kx + b k sin kx = π f(x + t) cos ktdt. 5
Proof. We have a k cos kx + b k sin kx = π = π = π = π = π x f(t) cos k(t x)dt = π x f(s + x) cos ksds + π π x f(s + x) cos ksds + π f(t) [ cos kt cos kx + sin kt sin kx ] dt x x x f(s + x) cos ksds. f(s + x) cos ksds f(s + x) cos ksds f(s + x) cos ksds Lemma 7. We have for all N and all t, N + cos kt = sin ( N + ) t sin t. This expression is called the Dirichlet s kernel, denoted by D N (t). Proof. We have D N (t) = + k= k= (e ikt + e ikt ) = [ + e ikt + k= k= e ikt] = [ + eit e i(n+)t e it + e it e i(n+)t ] e it = [ + e it e i(n+ )t it e e i(n+ )t ] + e it e it e it e it = [ + e it e i(n+ )t it e e i(n+ )t ] i sin t + i sin t = it e e [ it + i sin t + ei(n+ )t e i(n+ )t ] i sin t = [ + i sin t i sin t + i sin(n + )t ] i sin t = sin(n + )t sin t. Notice also that D N (t) (N + ) as t. So we can define D N as follows sin(n+ )t, t, D N (t) = sin t N +, t =. Lemma 7.3 D N (t)dt = D N (t)dt = π. Proof. We have D N (t)dt = ( N + cos kt )dt = π N + k= k= cos ktdt = π. 6
Now let f E. Fix some x and define (3) g(t) := f(x + t) f(x +) sin t and h(t) := f(x + t) f(x ) sin t. Then f(x + t) f(x + ) lim g(t) = lim t + t + t t sin t So, g(t) is piecewise continuous on (, π]. Thus, we have the following = D + f(x). Lemma 7.4 Let f E and let g, h be given by (3). The functions { { g(t), t (, π], h(t), t [, ), G(t) = and H(t) =, t [, ],, t [, π], are piecewise continuous on [, π]. Hence, G, H L ([, π]). Lemma 7.5 Let f E. Then lim N π f(x + t)d N (t)dt = f(x +) and lim N π f(x + t)d N (t)dt = f(x ). Proof. We have: π f(x + t)d N (t)dt f(x +) π = π = π = π = π = f(x + t)d N (t)dt π [ f(x + t) f(x+ ) ] D N (t)dt [ f(x + t) f(x+ ) sin t f(x + )D N (t)dt ] sin ( N + ) tdt G(t) sin ( N + ) tdt (by Lemma 7.4) Notice that the system {u N = sin ( N + ) t} N= is an orthogonal system in L [, π] and u N = π Thus, by Corollary 4.7, G, u N = The second limit is proved similarly. sin ( N + Combining the above, we obtain the main result ) tdt = π [ cos(n + )t]dt =. G(t) sin ( N + ) tdt as N Theorem 7.6 (pointwise convergence of Fourier series) Let f E with the Fourier series a + a k cos kx + b k sin kx. k= 7
Then for every x [, π], the series converges pointwise to f(x +)+f(x ). In particular, f(x) = a + a k cos kx + b k sin kx k= if x is a point of continuity. Proof. For N >, define S N (t) := a N + a k cos kx + b k sin kx. We have, S N (t) = = π = π [ π k= ] f(x + t)dt + π k= [ N f(x + t) + ] cos kx dt f(x + t)d N (t)dt f(x +) + f(x ) f(x + t) cos ktdt (by Lemma 7.) (by Lemma 7.5). Theorem 7.7 (uniform convergence of Fourier series) Suppose that f is continuous on [, π], that f() = f(π), and that f E. Then its Fourier series converges uniformly to f on [, π]. a + a k cos kx + b k sin kx. k= Proof. Clearly, we can extend f to a continuous periodic function with period π. Assume that f has the Fourier series α + α k cos kx + β k sin kx. We can check that α = π α k = π Similarly, β k = ka k. By Bessel s inequality (Theorem 4.6) k= πf (t)dt =. Now using integration by parts, we have f (x) cos kxdx = π cos kxf(x) x=π x= π α k + β k f < +. k= We have a k + b k = k αk + β k. So f(x)( k) sin kxdx = kb k. ak + b k = k= k= k α ( N k + β k k k= ) α k + β k k= < + Since a k + b k a k + b k. It follows that a k + b k ak + b k < +. k= k= 8
Then by the Weierstrass Majorant Theorem (Theorem 3.7), we conclude that a + a k cos kx + b k sin kx k= converges uniformly to some (continuous) function g on [, π]. On the other hand, the series converges pointwise to f. Thus, we must have f g. 8 Fourier Series for General Interval [, T ] Let f : R R be a periodic function with period T. The Fourier series of f is defined as the series a + k= where the Fourier coefficients are given by a k cos kπ T x + b k sin kπ T x, a = T f(t)dt, a k = T f(t) cos kπ T tdt, b k = T f(t) sin kπ T tdt. To derive the above formulation, notice that h(x) = f( T π x) has period π. Thus f(x) = h( πx T ) = a + k= a k cos kπ T x + b k sin kπ T x, where a = π a k = π b k = π h(t) dt = π h(t) cos kt dt = T h(t) sin kt dt = T f( T π t) dt = π f(t) cos kπ T t dt, f(t) sin kπ T t dt. f(x) π T dx = T f(x)dx, 9