Fourier Series. 1. Review of Linear Algebra

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1 Fourier Series In this section we give a short introduction to Fourier Analysis. If you are interested in Fourier analysis and would like to know more detail, I highly recommend the following book: Fourier Analysis, Elias Stein and Rami Shakarchi. Princeton University Press, 3. These notes follow the book closely. Fourier Analysis studies how a function f(x) can be expressed as a series of cos x and sin x or more symmetrically as a series of e inx. This is an analogue of Taylor series which expresses a nice function as a power series. As you can see from what will be covered later, the problem of expressing a function f(x) as a series of cos x and sin x is much more difficult than the convergence of Taylor series. To prepare for the discussion, we first review Linear Algebra, in particular, inner products and norms. Then we ll give two examples of infinite dimensional vector spaces equipped with inner products. After that we introduce orthogonal and orthonormal systems and use these to define Fourier coefficients and Fourier series of a periodic function. The rest of the section is devoted to the convergence of Fourier series: global and local convergence.. Review of Linear Algebra We recall the concepts of inner product and norm as needed for our introduction to Fourier series. Definition. Let V be a vector space over R (or C). An inner product on V is a function (, ) on V V R (or C) such that the following conditions are satisfied: (a) (v, w) = (w, v)(or (v, w) = (w, v)), v, w V. (b) (av +bv, w) = a(v, w)+b(v, w), (v, cw +dw ) = c(v, w )+d(v, w )(or (v, cw +dw ) = c(v, w ) + d(v, w )), v, v, v, w, w, w V. (c) (v, v) for any v V. We define the norm of v by v = (v, v). If v = implies v = then the inner product is called strictly positive-definite. Example. Let V = R n. For any v = (v, v,, v n ) and w = (w, w,, w n ) define (v, w) = v i w i. i=

2 It is easy to see that the three conditions are satisfied. The norm of the vector v = (v, v,, v n ) is v = The inner product is strictly positive-definite. v + v + + v n. Example. Let l (Z) be the set of all (two-sided) infinite sequences of real numbers such that (, a n,, a, a, a,, a n, ) a n <. We define the addition and scalar multiplication componentwise, i.e., if n Z v = (, a n,, a, a, a,, a n, ) and w = (, b n,, b, b, b,, b n, ), define v + w = (, a n + b n,, a + b, a + b, a + b,, a n + b n, ) and if λ is any real number, define λv = (, λa n,, λa, λa, λa,, λa n, ). We show that this set is a vector space over R. If v and w are two elements in l (Z), we show that v + w is in l (Z). Write v = (, a n,, a, a, a,, a n, ) and w = (, b n,, b, b, b,, b n, ), it suffices to show that v + w converges absolutely. Since a n + b n is increasing as n. So we only need to show that a n + b n is bounded i= n above. This is obtained by using triangle inequality: i= n ( a n + b n ) ( a n ) + ( b n ) ( a n ) + ( b n ). i= n i= n i= n n Z n Z Thus a n + b n ( v + w ). i= n Hence v + w converges absolutely. It is straightforward to verify the other conditions.

3 3 Now we define an inner product as follows: for any v = (, a n,, a, a, a,, a n, ) and w = (, b n,, b, b, b,, b n, ), define (v, w) = n Z a n b n. We need to verify that this definition satisfies the conditions of inner product. In fact the only point worth mentioning is whether (v, w) is well defined, in other words, we need to show that (v, w) converges. Here the idea is same as the convergence of v + w. It suffices to show that (v, w) converges absolutely. We use Cauchy-Schwarz inequality: a n b n ( a n ) ( b n ) < v w. i= n i= n i= n So we have (v, w) v w. Thus (v, w) is bounded above. Hence it converges absolutely. This vector space is infinite-dimensional since the vectors e n = (,, ) are linear independent for all n where e n has at i-th component and everywhere else. Example 3. Let R([, ]) denote the set of complex-valued Riemann integrable functions on [, ]. We define addition and scalar multiplication of functions f and g as follows: if f and g are two Riemann integrable functions on [, ], define (f + g)(x) = f(x) + g(x), x [, ] and (λf)(x) = λf(x), λ C, x [, ]. Under these two operations, the set R([, ]) is a vector space over C. We define an inner product by The norm of f is (f, g) = f(x)g(x)dx. f = It is easy to see that this is an inner product. ( f(x) dx). This vector space is still infinite-dimensional since obviously R([, ]) contains polynomials on [, ] and all x n are linearly independent for any n.

4 4 The example and example 3 are two examples of infinite-dimensional vectors spaces with inner products. A vector space equipped with an inner product is called complete if every Cauchy sequence converges to a limit in the vector space. A complete vector space equipped with an inner product is called Hilbert space. We can show that the example is complete since every Cauchy sequence converges to a limit in the space l (Z). However the example 3 is not complete. We ll give a Cauchy sequence that doesn t converge to a limit in R([, ]). Let {, if x n f n (x) = ; ln x, if n < x. {, if x = ; It is not hard to see that f n (x) converges to the function f(x) = ln x, if < x. Now f n (x) is bounded and Riemann integrable since ln x is bounded and Riemann integrable on [ n, ]. However the function f(x) is unbounded on [, ], hence not Riemann integrable, i.e., f R([, ]). So R([, ]) is not complete. The following concept will play an important role in the study of Fourier series. Definition. Let V be a vector space equipped with an inner product. Two vectors v and w are said to be orthogonal if (v, w) =. The above three examples of vector spaces share the following three formulae. Some of these have already been used in the above examples:. The Pythagorean theorem: if v and w are two orthogonal vectors in a vector space V equipped with an inner product, then v + w = v + y.. The Cauchy-Schwarz inequality: for any v, w V, we have (v, w) v w. 3. The triangle inequality: for any v, w V, we have v + w v + w. To prove (), we use the orthogonality of v and w. Then we see that v+w = (v+w, v+w) = (v, v) + (v, w) + (w, v) + (w, w) = (v, v) + (w, w) = v + w. For (), consider, for any real number t, v + tw.

5 5 We also have v + tw = (v, v) + t(v, w) + t (w, w). When viewed as a quadratic polynomial in t, the quadratic polynomial is always bigger than, thus we have the discriminant = 4(v, w) 4(v, v)(w, w). Hence we obtain (v, w) v w. Taking square roots of both sides, we obtain (v, w) v w. For (3), we notice v+w = (v+w, v+w) = (v, w)+(v, w)+(w, v)+(w, w) v + v w + w = ( v + w ). The last inequality is obtained by using Cauchy-Schwarz inequality. Taking square roots again, we obtain v + w v + w.. Fourier Series We define the Fourier coefficients and series relative to an orthonormal sequence of periodic function on [a, b] and give examples of the Fourier series of some functions. Definition. Let {φ n } be a sequence of complex-valued functions on [a, b] such that b a φ n (x)φ m (x)dx =, n m. Then such a sequence is called an orthogonal system of function on [a, b]. If in addition, Then {φ n } is called orthonormal. b a φ n (x) dx =, n. Examples. The sequence {e n (x) = e inx } is an orthonormal sequence in the vector space R([, ]). For any e n, e m R([, ]), we compute (e n (x), e m (x)) as follows: If n m, then (e n (x), e m (x)) = If n = m, then e inx e imx dx = e i(n m)x dx = i(n m) ei(n m)x =. (e n, e n ) = e inx e inx dx = dx =.

6 6 Examples. The sequence {, cos x, sin x, cos x, sin x, } is another orthonormal system of functions on [, ]. We ll verify one case and leave other cases to you to verify. For cos nx, cos mx, n, m, we compute ( cos nx, cos mx) = = (m + n) sin(m + n)x + cos nx cos mxdx = [cos(m + n)x + cos(m n)xdx] sin(m n)x =, if m n; (m n) sin nx (x + ) =, if n = m. Definition. If f is an integrable function on [a, b], then the n-th Fourier coefficient of f is defined by f(n) = b a The Fourier series of f is given by b f(x) a f(x)e inx b a dx, n Z. n= f(n)e inx b a. We write c n = f(n). The reason that why we write the Fourier coefficient as f(n) is that the Fourier coefficients coincide with the value of the Fourier transform of f(x) at x = n. Recall that for any integrable function f(x) on R, its Fourier transform for ξ R is defined by f(ξ) = f(x)e ixξ dx. In the definition of Fourier series we didn t use equality since we don t know whether the series converges and whether f(x) is equal to the series. More generally if {φ n } is an orthonormal system on [a, b], for any integrable function f on [a, b], the number c n = b a b a f(x)φ n (x)dx, n =,, is called the n-th Fourier coefficients of f relative to {φ n }. The series The Fourier series of f. We write f(x) c n φ n (x). n= c n φ n (x) is called n=

7 7 The N-th partial sum of the Fourier series of f is given by S N (f)(x) = f(n)e inx/(b a). One of main goals of Fourier analysis is to determine whether and when the N-th partial sum converges to f, i.e. Problem: In what sense does S N (f) converge to f as N. We ll study some theorems related to the convergence problem in next section. then Example 3. Let f(x) = x on [, π]. Let s compute the Fourier series of f. First if n, If n =, then c n = xe inx dx = [( x in e inx ) π + e inx in dx] = [ π in e inπ π in einπ (in) e inx π ] = [ π in (einπ e inπ ) ( n einπ n e inπ )] = cos nπ [ in = = ( )n in Thus the Fourier series of f = x is cos nπ in n (einπ e inπ )] π xdx = x 4π π =. f(x) n e inx in = n= ( einx in e inx in ) = n= sin nx. n (π x) Example 4. Let f(x) = on [, ]. If n, then 4 c n = (π x) e inx dx = t 4 4 e in(π t) dt

8 8 after the substitution t = π x. So we have If n =, then c n = = = = = = cos nπ cos nπ cos nπ cos nπ = n t 4 e in(π t) dt t 4 eint e inπ dt t 4 eint dt ( t e int e int ) t 4in π in dt [ π 4in (einπ e inπ ) + teint n π [ ] π n (einπ e inπ ) eint in 3 π e int ] n dx Hence the Fourier series of f(x) = c = = = (π3 πx = = π π 6 (π x) 4 f(x) π + c n e inx = π + n n= (π x) dx 4 ( π 4 πx + x 4 )dx is 4 + x3 ) n (einx + e inx ) = π + Recall that we showed the uniqueness of power series in last section, i.e., if f(x) = a n (x c) n = n= b n (x c) n, n= n= cos nx n. Then we have a n = b n for any n =,,. We may ask the same question for Fourier series. It turns out that the situation for Fourier series is more subtle. The condition on the function f really matters. The following theorem shows the uniqueness if the function f is continuous.

9 9 Notice that in our discussion so far, we only assume that f is integrable. Theorem.3 Suppose that f is an integrable periodic function on [, π] with f(n) = for all n Z. Then f(x ) = whenever f is continuous at the point x. Proof. From the condition that f(n) =, for any n, it follows that f(n) = for any n. Since e inx = cos nx i sin nx, we obtain f(x) cos nxdx = f(x)e inx dx = f(x) sin nxdx = for any n. We are going to construct a sequence of trigonometric polynomials p k (x) such that as k. Thus we obtain a contradiction. f(x)p k (x)dx First we assume that f is a real-valued function. Let s assume the opposite, i.e., f(x ). We may assume that x = and f(x ) >. Then there exists a δ > such that Let p(x) = cos x + ɛ where ɛ is chosen so that and Summing up, we have p(x) = estimate the integral f(x) > f(), for x < δ. p(x) < ɛ, for δ < x < π p(x) > + ɛ for x < η < δ. < ɛ, δ < x < π; + ɛ, if x < η;, if η x δ. f(x)p k (x)dx. We have f(x)p k (x)dx = ( δ< x π We can estimate the above three integrals as follows: δ< x <π Let p k (x) = (p(x)) k, k. Let s ) + + f(x)p k (x)dx. η x δ x <η f(x)p k (x)dx < ( ɛ )k B,

10 x <η η< x <δ f(x)p k (x)dx ( + ɛ f(x)p k (x)dx, f() )k η, as k, where B is an upper bound of f(x) since f(x) is integrable on [, π], hence it is bounded. Thus Hence f(n) = for any n. f(x)p k (x)dx, as k. If f is a complex-valued function. Write f(x) = u(x) + iv(x). Then u(x) and v(x) are real-valued continuous functions at x and and We also observe that u(x) = v(x) = f(x) + f(x) f(x) f(x). i f(x) = f(x)e inx dx = f(x)e inx dx = f( n) =. Hence we obtain u(n) = = v(n) for any n. Applying the above argument, we have u(x ) = v(x ). Hence f(x) =. The following are two immediate consequences of the theorem. Corollary.4 Let f is a continuous and - periodic function. If f(n) = for any n, then f(x) = for any x. The following corollary answers the convergence problem formulated before. Corollary.5 Suppose that f is a continuous function on [, ] and that the Fourier series of f is convergent absolutely, i.e., f(n) <. Then the Fourier series converges uniformly to f, that is, n= S N (f)(x) f(x). Proof. Let s consider the function g(x) = n= f(n)e inx = lim N f(n)e inx.

11 By the condition that n= f(n) <, it follows that the partial sums lim N N f(n)e inx = lim N f converges uniformly. Hence its limit function g(x) is continuous by the continuity theorem we showed in last chapter. The Fourier coefficients of g are exactly f(n) since ĝ(n) = s g(x)e inx dx = = = f(n) m= N m= f(m)e imx e inx dx f(m)e imx e inx dx Here in the second last equality, we exchanged infinite sum and integral since the infinite sum is convergent absolutely. Now we apply the corollary to obtain f g = since both have the same Fourier series.

12 3. Convergence of Fourier Series In this section we discuss the convergence of Fourier series and present two results: global and local. The Global convergence is also called mean-square convergence theorem, which says that the norm of f S N (f) converges to as n. The local convergence is focused on the local behavior of f at a given point x. As a consequence of local convergence we obtain a surprising result: Riemman s localization principal, which states that the convergence of S N (f) only depends on the local behavior of f near x. This is a surprise since Fourier series are obtained by integrating the function f over the entire interval. 3. A Global Convergence. The goal of this subsection is to prove the following Mean-Square Convergence Theorem Suppose that f is an integrable periodic function on [, ], then f(x) S N (f)(x) dx, as N. Remark. Using norm, we observe that f S N (f) = ( f(x) S N (f)(x) dx). But the fact that the limit of f S N (f) approaches doesn t guarantee f S N (f) since our function f is only integrable. See exercise 5. Before we prove the theorem, we need some lemmas. Lemma 3.. (Best Approximation) If a function f is integrable on [, ] and its Fourier coefficients are a n. Let c n C be a sequence of complex numbers. Put T N (x) = c n e inx. Then f S N (f) f T N. Proof. We first notice that the vectors f S N (f) and S N (f) c n e inx are orthogonal.

13 3 We prove this by looking at the inner product (f S N (f), S N (f) (f S N (f), (a n c n )e inx ) = (f, = = =. (a n c n )e inx ) (S N (f), f(x)a n c n e inx dx a n c n f(n) N n= N a n c n f(n) Now using the above fact and Pythagorean theorem, we obtain since S N (f) f c n e inx = f S N (f) + S N (f) = f S N (f) + S N (f) c n e inx. c n e inx c n e inx f S N (f), c n e inx ). We have (a n c n )e inx ) We also need the following two lemmas to prove the Mean-Square convergence theorem. Unfortunately the proofs of these lemmas will be skipped due to the lack of time. You can read the proofs in the book mentioned above. Lemma 3.. Let f be a continuous function on [a, b]. polynomial P of degree M, such that Then there is a trigonometric S N (f)a n c n e inx dx f(x) P (x) < ɛ x [a, b]. Lemma 3..3 Suppose f is integrable and periodic function on [, ] and bounded by B. Then there exists a sequence {f k } of continuous functions on [, ] so that sup f k (x) B, k =,,. and f(x) f k (x) dx as k.

14 4 Now we are ready to prove the Mean-Square convergence theorem. Proof of Mean-Square Convergence Theorem. Let s first assume that f is continuous on [, ]. By lemma 3.., there is a trigonometric polynomial P of degree M such that f(x) P (x) < ɛ, x. Thus f P dx ɛ. Using the best approximation lemma, we obtain Here we used the fact that We can see that by looking at f S N (f) f S M (f) f P ɛ, N > M. f S N (f) f S M (f), N > M. f S M (f) = f S N (f) + S N (f) S M (f). Let v = f S N (f) and v = S N (f) S M (f). Then f S N (f)+s N (f) S M (f) = v +v = (v +v, v +v ) = v +(v, v )+(v, v )+ v. So it suffices to compute (v, v ). We have (v, v ) = (f S n (f), S N (f) S M (f)) = (f S N (f), N n >M f(n)e inx ) = N n >M = N n >M f(x) f(n)e inx dx (S N (f), f(n) f(n) N n >M N n >M f(n) =. f(n)e inx Hence the inequality. Summing up, we have f S N (f), as N. Now we assume that f is only integrable. By lemma 3..3, there exists a continuous function g on [, ], such that sup g(x) sup f(x) = B and f(x) g(x) dx < ɛ.

15 5 From these, we obtain f g = f g dx = B f g dx B ɛ. f g f g dx For this g, we apply the results we obtained in the above, so there is a trigonometric polynomial P such that g P < ɛ. Combining all the above, we obtain f P f g + g P < Cɛ. The rest of proof follows the case of continuous function exactly. Hence the statement. Using Mean-Square Convergence theorem, we have Theorem 3..4 (Parseval s Identity) If f is an integrable periodic function on [, ] with Fourier coefficients a n. Then a n = f. n= Proof. We observe that vectors f S N (f) and S N (f) are orthogonal since (f S N (f), S N (f)) = (f, S N (f)) S N (f) = f(n) f = Here we used the fact that S N (f) = we have f(n) by the orthogonality of the sequence {e inx }. Thus using Pythagorean theorem, f = f S N (f) + S N (f) = f S N (f) + f(n). By the Mean-Sqare convergence theorem, we see that f S N (f) as N. Hence we obtain f = a n. n= n= As an easy consequence of Parseval s identity, we have Theorem 3..5 (Riemman-Lebesgue Lemma) If f is integrable and periodic function on [, ], then f(n)

16 6 as n. obtain Proof. By Parseval s identity, we see that the series lim a n =, n n= a n converges. Hence we i.e., lim f(n) =. n 3. A Local Convergence of Fourier Series The following is a local convergence of Fourier Series. As an easy consequence of the local convergence, we show the Riemman s localization principal. Theorem 3.. Let f be an integrable and periodic function on [, π] which is differentiable at x. Then S N (f)(x ) f(x ) as N. Proof. Let D N (x) = e inx. Using the fact that f(x ) = f(x )D N (t)dt. Then we have S N (x ) f(x ) = = = f(x t) f(x ) t π D N (t)dt =, we can write f(x t)d N (t)dt f(x )D N (t)dt (f(x t) f(x ))D N (t)dt f(x t) f(x ) td N (t)dt t, if t ; Let F (t) = Since f is differentiable at x, then there is a f (x ), if t =. δ > such that F (x) is bounded on t < δ. Moreover on [, δ] [δ, π], F is integrable. So we may choose δ small enough so that the function F on t < δ is negligible. Hence F is integrable on [, π]. Now we use the following formula: D N (t) = sin(nt + t ) sin t = sin Nt cos t + cos Nt sin t sin t, to obtain S N (f)(x ) f(x ) = F (t)( t cos t sin Nt sin t + t cos Nt)dt.

17 7 By Riemann-Lebesgue lemma, f(n) as n, this is equivalent to and f(x) cos nxdx π f(x) sin nxdx as n. Applying the lemma to Riemann integrable functions F (t)t cos t obtain the result. and F (t)t, we Using the above local convergence, we have the following surprising localization principal of Riemann. Theorem. Suppose f and g are two integrable periodic functions defined on [, π], and for some x there exists an open interval I containing x such that f(x) = g(x) for any x I. Then S N (f)(x ) S N (g)(x ) as N. Proof. Let s consider f g. Then f g is differentiable at x since f(x) g(x) = for any x I. Now apply the localization to f g to finish the proof. Remark In the above theorem we see that the convergence of Fourier series at a point x only depends on a local behavior of f(x) near x. This is a surprise since Fourier coefficients f(n) are defined as integrals over [, ]. So these are global behavior of f(x) on [, ].

18 8 Exercise. Suppose f is -periodic function and integrable on any finite interval. Prove that if a, b R, then Also prove that b a f(x)dx = b+ f(x + a)dx = a+ f(x)dx = f(x)dx = b a +a +a f(x)dx. f(x)dx.. Consider the periodic odd function defined on [, π] by f(x) = x(π x). Compute the Fourier coefficients of f and show that f(x) = 8 π n odd 3. Let f be the function on [, π] by f(x) = x. sin kx k 3. (a) Compute the Fourier coefficients of f and show that (b) Taking x =, prove that π, if n = ; f(n) = + ( ) n πn, if n n odd n = π 8 and n= n = π Let f(x) = {, for x = ; ln x, for < x <. and define a sequence of functions on R([, ]) by {, for x n f n (x) = ; f(x), for n < x. Prove that {f n } is a Cauchy sequence in R([, ]). However f does not belong to R([, ]). (Hint: Show that b a ln xdx if < a < b and b, by using the fact that the derivative of x(ln x) x ln x + x is equal to ln x.)

19 9 5. Recall the vector space R([, ]) of integrable functions, with it inner product and norm f = ( f(x) dx). (a) Show that there exists a non-zero integrable functions f for which f =. (b) However, show that if f R([, ]) with f =, then f(x) = whenever f is continuous at x. (c) Conversely, show that if f R([, ]) vanishes at all of its points of continuity, then f =. 6. For any N N, define the Dirichlet kernel by D N (x) = (a) D N (x) is a continuous, -periodic function. (b) D N (x) is an even function. (c) e inx. π D N (x)dx = D N (x)dx =. π (d) (e) D N () = N +. (f) For all x, D N (x) N +. (g) For all < x < π, D N (x) π x. D N (x) = sin((n + )x) sin x. 7. Let f be the function defined on [, π] by f(x) = x. Use Parseval s identity to find the sums of the following two series: n= (n + ) 4 and n= n Consider the -periodic odd function defined on [, π] by f(x) = x(π x). Show that n= (n + ) 6 = π6 96 and n= n 6 = π Show that for α not an integer, the Fourier series of π sin πα ei(π x)α

20 on [, ] is given by Apply Parseval s identity to show that n= n= e inx n + α. (n + α) = π (sin πα).. Prove that sin x x dx = π. (Hint: Start with the fact that the integral of D N (x) equals, and note that the function sin x is continuous on [, π]. Apply the Riemann-Lebesgue lemma.) x. Prove that the Fourier series of a continuously differentiable function f on [, ] is absolutely convergent. (Hint: Use Cauchy-Schwarz inequality and Parseval s identity for f.). Suppose that f is periodic and of class C k. Show that (Hint: Use the Riemann-Lebesque lemma.) lim n n k f(n) =. 3. Let f be a -periodic and Riemann integrable on [, π]. Show that f(n) = f(x + π n )e inx dx Hence f(n) = f(x) f(x + π 4π n ) e inx dx.