MAT 636: Lectures on Fourier series

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1 MAT 636: Lectures on Fourier series Brad Rodgers [Lecture ] Introduction The subject of Fourier series concerns decomposing periodic functions into sums of somewhat more basic functions that have nice algebraic and analytic properties. In this course we will develop the basic theory of this subject. In order to be concrete, in what follows we will work almost exclusively with functions f, defined on the real line, which are of period. The same theory, however, can be developed with only minor and obvious changes for functions that are of period L, for any L. Likewise, the same theory that we will develop here can be applied to functions that are defined on the unit circle, or, if you know some group theory, to functions defined on T := R/Z with the group operation of addition. All such types of functions are, so to speak, different embodiments of the same mathematical object. For functions f of period, the more basic functions into which we will decompose f(x) are {e i2πnx } n Z. There is an equivalent theory in which these functions are replaced by {sin 2πnx, cos 2πnx} n N, but it is not as nice algebraically. Some knowledge of measure theory (from Analysis III at UZH) is assumed in these notes, but with a little work, one could probably follow most of these notes with only a superficial understanding of that subject. Please do not hesitate to me with any questions. We begin with a definition. Definition. For a function f of period on the real line and such that f L [, ), the Fourier coefficients of f are defined by ˆf(n) := f(x)e i2πnx dx. We note the following fundamental facts, easily verified by calculus Proposition 2. For j an integer e i2πjx dx = {, j = j.

2 Corollary 3. For j, k integers e i2πjx e i2πkx dx = δ j=k := {, j = k j k. Some texts at this point define the formal Fourier series ˆf(k)e i2πkx, and write f k= k= ˆf(k)e i2πkx. It is important to stress that this identification is formal, and we don t know at this point that this formalism has any literal meaning; indeed we don t even know whether the sum on the right is guaranteed to converge in the traditional sense. As we will later see, it sometimes doesn t. A central question in the theory of Fourier series is the extent to which f(x) actually does equal the sum on the right. 2 The Dirichlet kernel In order to be more rigorous, we make the following definition. Definition 4. S n (f, x) := n k= n ˆf(k)e i2πkx. By the formal Fourier series we ve written above, one might have the somewhat vague intuition that for large n, S n (f, x) should look basically like f(x). This is certainly the case when f is a trigonometric polynomial, that is when f(x) is a finite linear combination of the functions e i2πkx. Proposition 5. If f(x) = k= M c k e i2πkx, where M and N are finite numbers and c k are constants, then as n, S n (f, x) f(x). This is easy to verify. Indeed, in this case, as long as n N, M, we have that S n (f, x) = f(x). Despite this proposition, it is not always the case, even for continuous f, that S n (f, x) f(x) for all x (a fact due to Paul du Bois-Reymond). But we will see very soon that statements very close to this are true. For the moment we simply note some properties of these partial sums. 2

3 Proposition 6. For f of period and in L [, ), S n (f, x) = f(t)d n (x t) dt where D n (x) := k= n n e i2πkx. Moreover, D n (x) = ( (n ) ) sin + 2 2πx sin ( 2 2πx). Proof. The first of these claims is trivial to verify by using the definition of D n to evaluate the integral, and we did so in class. For the second, let ω = e i2πx. Then D n (x) = ω n ( + ω + + ω 2n ) = ω n ω2n+ ω = ω(n+/2) ω (n+/2) ω /2 ω /2. This may be seen to be equal to the formula in the proposition, using the fact that (e ix e ix )/2i = sin x. D n (x) is called the Dirichlet kernel, in honor of Lejeune Dirichlet. We are deferring a proof that sometimes S n (f, x) does not converge to f(x) until later. Nonetheless, if we take this claim on faith for the moment, it is natural to then ask the weaker question whether given a sequence of Fourier coefficients ˆf(k) we can even recover the original function f(x). This is what we will turn to in the next two sections. 3 Summability methods In this section, we introduce a very first example of a much larger subject known as summability methods. It allows us to expand what is meant by a sequence converging to a value, and this broadening of our understanding has many consequences. For us, the importance of summability methods is that they will allow us to define analogues of the partial sums S n (f, x) that converge to f(x) even when the partial sums do not. We begin with an example sequence: α n = ( ) n. Here α n oscillates around, taking the values of and, and plainly doesn t converge to anything. Indeed, several months ago there was a sort of controversy on the internet about the formula = /2. Obviously such a formula can t literally be true, but it can be understood using summability methods (albeit summability methods that lie deeper than we will make use of in this course), and once understood it is seen to be an important identity in the theory of numbers. 3

4 Nonetheless, the average of this sequence tends to : N + ( ) n = + ± N + n= = or N +. Moreover, whenever a sequence converges to some value, so to will the average of the sequence: Theorem 7 (Cesàro). If α n α, then α + + α N N + α. Proof. Define ɛ n by α n = α + ɛ n. Then for any ɛ > there exists some M such that ɛ n ɛ for all n M. Now α + + α N N + Moreover, for any ɛ > there is M as above, and = α + ɛ + + ɛ N. () N + as N. Hence ɛ + + ɛ N N + and as ɛ was arbitrary ɛ + + ɛ M + ɛ M+ + + ɛ N N + N + ɛ + + ɛ M ɛ(n M) + N + N + + ɛ, lim sup N lim N ɛ + + ɛ N N + ɛ + + ɛ N N + ɛ, =. This and () prove the theorem. When (α + + α N )/(N + ) α, we say that α is the Cesàro limit of α N. What we have shown is that anytime α is the limit of α N, it will also be the Cesàro limit, and moreover, from the example, a sequence can sometimes have a Cesàro limit when the sequence itself has no limit. This widened concept of a converging sequence would have been familiar to the young mathematician Lipót Fejér a little over a century ago, and it ended up being exactly what he needed to answer the question with which we ended section : can we recover f(x) from the sequence ˆf(k)? 4

5 4 The Fejér kernel Theorem 8 (Fejér). For f of period and continuous, uniformly. σ N (f, x) := S (f, x) + S (f, x) + + S N (f, x) N + Before proving this theorem, we prove a lemma. Lemma 9. For f of period and in L [, ), σ N (f, x) = Moreover, Proof. Note that where K N (x) = N + f(t)k N (x t) dt where K N (x) := N + = N + = N + ( K N (x) = ( sin N+ 2 2πx ) ) 2 ( ). N + sin 2 2πx σ N (f, x) = N + = D n (x) n= n= f(t)k N (x t) dt, f(t)d n (x t) dt f(x), (N+ j )e i2πjx. j= N ( + (ω + + ω) + + (ω N + ω (N ) + + ω N ) (N + j )ω j, j= N where as before ω = e i2πx. This verifies the first claim. Yet one may check that (N + j )ω j = ( ω N/2 ( + ω + + ω N ) ) 2 j= N ( ω (N+)/2 ω (N+)/2 = ω /2 ω ( /2 ( sin N+ 2 2πx ) ) 2 = ( ), sin 2 2πx ) 2 ) 5

6 which verifies the second claim. Here the first line follows from a counting argument, and the second by summing the geometric series and some simple algebra. After Fejér, the function K N is called the Fejér kernel. The reader should take the time to graph both the functions S N and K N for various values of N to get a feel for their behavior. Lemma. Both the following are true: i) K N (x) for all x ii) K N (x) dx = Proof. i) This is clear from Lemma 9, because the square of any real number is non-negative. ii) By Lemma 9 and then Proposition 2 K N (x) dx = N + (N + j )e i2πjx dx =. j= N From this we have the very important Proposition. All of the following are true: () K N(x) dx = (2) K N(x) dx is bounded as N (3) For δ >, uniformly for δ x /2. lim K N(x) = N Proof. () and (2) follow from Lemma. For (3), note that for x [δ, /2], as N. K N (x) N + (sin πδ) 2 6

7 We can now outline the idea behind Fejér s proof. We have that for small δ and large N, σ N (f, x) = = /2 /2 δ δ f(x) f(x) f(t)k N (x t) dt f(x τ)k N (τ) dτ f(x τ)k N (τ) dτ by (3) of Proposition δ δ /2 /2 K N (τ) dτ by the continuity of f and (2) of Prop. K N (τ) dτ by (3) of Proposition = f(x) by () of Proposition. It is not too difficult to make this argument rigorous once this outline is in place. [Lecture 2] Proof of Theorem 8. We define κ := sup N K N L [ /2,/2]. By (2) of Proposition, we know κ < +. Let ɛ > be an arbitrary positive number and define ɛ := ɛ/κ. We proceed backwards through the outlined steps above. Note that for any δ >, lim K N (τ) dτ =, N δ τ /2 by (3) of Proposition. Note also, by property () of this proposition, that δ /2 K N (τ) dτ = K N (τ) dτ K N (τ) dτ δ /2 = δ τ /2 K N (τ) dτ Hence, δ sup f(x) K N (τ) dτ f(x) = sup f(x) x x δ δ τ /2 δ τ /2 K N (τ) dτ (2) as N. Yet because f is continuous and periodic, f must be uniformly continuous 2. So there must exist some δ > such that f(x) f(x τ) ɛ 2 To see this, note that f is uniformly continuous on [, 2] as this interval is compact, and the periodicity of f implies that this uniform continuity may be extended to all of R 7

8 for all x R and all τ < δ. But then for all N and x, δ δ δ f(x τ)k N (τ) dτ f(x) δ Moreover, note that sup x δ δ sup f(x) x /2 f(x τ)k N (τ) dτ δ τ /2 δ K N (τ) dτ ɛ K N (τ) dτ /2 K N (τ) dτ δ ɛ. (3) f(x τ)k N (τ) dτ. (4) Finally, it is indeed clear that σ N (f, x) = /2 /2 by the periodicity of f and K N. Yet sup x σ N (f, x) f(x) = sup x sup x /2 /2 δ + δ f(x τ)k N (τ) dτ, δ f(x τ)k N (τ) dτ f(x τ)k N (τ) dτ δ f(x τ)k N (τ) dτ f(x) δ δ K N (τ) dτ δ + f(x) K N (τ) dτ f(x) δ /2 δ f(x τ)k N (τ) dτ f(x τ)k N (τ) dτ /2 δ δ δ f(x τ)k N (τ) dτ f(x) K N (τ) dτ δ δ δ f(x) K N (τ) dτ f(x), + sup x + sup x δ by the triangle inequality. Each of these terms as N is controlled by the equations (4), (3), and (2) respectively, so we see that ( lim sup sup σ N (f, x) f(x) ) + ɛ + = ɛ. N x As ɛ was arbitrary, this proves the claim. This proof may seem somewhat hard-going at first; if so, it may help to review the outline given above it once again. 8

9 5 Consequences of Fejér s theorem We note some consequences of this remarkable result. Corollary 2. If f and g are continuous functions of period, and ˆf(k) = ĝ(k) for all k Z, then f(x) = g(x) for all x. Proof. By Fejér s theorem, σ N (f, x) f(x) and σ N (g, x) g(x) uniformly. But if ˆf(k) = ĝ(k) for all k, then σ N (f, x) = σ N (g, x). We will later extend this corollary to functions that are only integrable, rather than continuous. Closely related to Corollary 2, we also have that trigonometric polynomials are dense in the space of continuous functions with supremum norm. Said in a different language, Theorem 3. Let f be a continuous function of period. For any ɛ >, there exists a trigonometric polynomial such that P (x) = k= M c k e i2πkx sup f(x) P (x) < ɛ. x Proof. We need only note that σ N (f, x) is a trigonometric polynomial, and σ N (f, x) f(x) uniformly. We can bootstrap this theorem to show that trigonometric polynomials are dense in L as well. In the more concrete language of the theorem above, Theorem 4. Let φ L [, ) also be of period on the real line. For any ɛ >, there exists a trigonometric polynomial P (t) such that φ P L [,) < ɛ. Proof. We recall (from Analysis III or, for instance, Theorem 3.4 of Rudin s Real and Complex Analysis ) that for any ɛ > there exists a continuous function f such that φ f L [,) < ɛ/2. But likewise from Theorem 3, there exists a trigonometric polynomial P such that f P L [,) f P L [,) < ɛ/2. Hence, φ P L [,) φ f L [,) + f P L [,) < ɛ. 9

10 There is a simple estimate that lies very near the start of the theory of Fourier series: Proposition 5. For f L [, ) of period, for all k. ˆf(k) f L [,) Hence Fourier coefficients of a given function remain bounded. The proof of this Proposition is a straightforward exercise of taking absolute values inside an integral (with the triangle inequality) and is left to the reader. In fact, we can now show rather more than this. Theorem 6 (Riemann-Lebesgue lemma). For f L [, ) of period lim ˆf(k) =. k Proof. Choose any ɛ >. There exists a trigonometric polynomial such that P (x) = k= M c k e i2πkx f P L [,) < ɛ by Theorem 4. Note that for k > N, M, ˆP (k) =. Yet, as f = (f P ) + P, we have by the triangle inequality, ˆf(k) (f(x) P (x))e i2πkx dx + f P L [,) + ˆP (k) ɛ P (x)e i2πkx dx when k is larger than N and M. As ɛ has been chosen arbitrarily, this is the definition of the statement, lim ˆf(k) =. k We will not prove so in this class, but not much more can be said about the rate at which the Fourier coefficients of arbitrary integrable functions f decay than that ˆf(k). In fact, for any positive sequence α k, there exists an integrable function f such that ˆf(k) α k for all k. (See Katznelson, Section

11 I.4, and exercise 4., if you re curious.) Later, we will see that the Fourier coefficients of functions in L 2, rather than L, can be described more exactly. We do now have, however, enough information to show that when a function s Fourier coefficients are absolutely summable, then their partial sums converge to the original function. Proposition 7. For f continuous and of period, if ˆf(k) < +, k Z then for all x. S n (f, x) f(x) Proof. A proof very similar was covered in homework 7. We know that by Fejér, lim σ n (f, x) = f(x) for all x. On the other hand as ˆf(k) is absolutely summable, lim S n (f, x) exists for all x. Cesàro s theorem thus implies that lim σ n (f, x) = lim S n (f, x) so that lim S n (f, x) = f(x) for all x. We note in passing: it is easy to see that in the situation of Proposition 7, f L [,) ˆf l (Z), as f(x) = ˆf(k)e i2πkx k for all x. Note that Proposition 5 can be phrased, k ˆf(k) ˆf l (Z) f L [,), which is an interesting relationship... Finally we turn to a claim mentioned at the start of this section. In fact, with the theory we have developed so far, the proof is not completely straightforward. I give a proof below for the sake of completeness, but you may find it a little difficult depending upon your background do not worry if so. Theorem 8. If f and g in L [, ) are of period and ˆf(k) = ĝ(k) for all k Z, then f(x) = g(x) almost everywhere.. In our proof we will require Lemma 9. For φ L [, ], φ dx = sup β βφ dx, where the supremum is taken over all β C[, ] with β(x) for all x.

12 This may be found in a number of analysis textbooks; I provide a sketch of a proof here. Proof. Since any integrable function may be approximated arbitrarily closely by a continuous function, it is sufficient to prove this claim for φ C[, ]. (You may want to fill in the details of the claim I ve just made, if you re not convinced.) But, for φ continuous, φ dx = sgn(φ) φ dx = lim n β n φ dx, where sgn(φ(x)) is the sign function equal to when φ(x) is positive, when φ(x) is negative, and when φ(x) is, and where β n (x) = sgn(φ(x)) min(, n dist(x, E)), where dist(x, E) is the distance of x to the set E of zeros of φ. (The above equality follows from the fact that β n (x) sgn(φ(x)). Why is this fact true for continuous φ?) β n is continuous for all n, so φ dx sup β βφ dx, But clearly so our proof is complete. φ dx sup β βφ dx, Proof of Theorem 8. From the lemma above, by approximating each β uniformly by trigonometric polynomials (which is possible by Theorem 3), one may see that φ dx = sup P P φ dx, where the supremum is taken over all trigonometric polynomials P with P (x) for all x. 3 Hence, f(x) g(x) dx = sup P P (f g) dx. But for any trigonometric polynomial P (x) = N M c ke i2πkx, P (f g) dx = k= M c k ( ˆf( k) ĝ( k)) =, 3 See a discussion on the piazza discussion board for an elaboration on why, as well an argument due to Roland Prohaska that takes a slightly different path. 2

13 as ˆf( k) ĝ( k) = for all k. This of course implies f(x) g(x) dx =, which implies f(x) = g(x) almost everywhere (by a well known result proved in Analysis III). Note: There is another proof of this theorem that depends upon a proof that for f L, we have f σ N (f) L [,). We will not take this approach in this class, but a proof may be found in Katznelson, section I.2 or Krantz section.4. [Lecture 3] 6 Summability kernels In this section we generalize Fejér s theorem to other methods of summability. Definition 2. A summability kernel is a sequence {k n } of continuous functions, all of period, such that () k n(x) dx = (2) k n(x) dx is bounded as n (3) For δ >, uniformly for δ x /2. lim k n(x) = n Remark: Sometimes in other texts the condition (3) is replaced with the weaker condition (3 ) δ x /2 k n(x) dx, depending on the application the author has in mind. Remark: The sequence {k n }, with n =, 2,... may be replaced by a continuous parameter family {k r } with r, for instance. We will see an example of this later with the Poisson kernel. We also introduce the notation of convolutions: Definition 2. For two functions of period, f L [, ) and g L [, ), f g(x) := g(x t)f(t) dt Proposition 22. f g(x) = g(τ)f(x τ) dτ. 3

14 This is a simple exercise in making the change of variable τ := x t. Another way to write the above proposition is f g = g f. Note that S n (f, x) = D n f(x), σ N (f, x) = K N f(x). Remark: The condition that g L can be weakened to g L, and f g will still be well defined almost everywhere, but we do not prove this fact in these notes. Summability kernels are of interest because of the following result. Theorem 23. If {k n } is a summability kernel and f is continuous and of period, then k n f(x) f(x) uniformly. Proof. Exactly as for the Fejér kernel. Aside from the Fejér kernel, one other example of a summability kernel that is of particular importance is the Poisson kernel. Definition 24. For r <, and f continuous and of period, ρ r (f, x) := k= r k ˆf(k)e i2πkx. Remark: The notation ρ r to denote this sum is not universal. Proposition 25. For r <, where ρ r (f, x) = P r f(x), P r (t) = = k= r k e i2πkt r 2 2r cos 2πt + r 2. Proof. We have ρ r (f, x) = k r k e i2πkx f(t)e i2πkt dt = f(t) k r k e i2πk(x t) dt where swapping the order of integration and summation (in the second equality) follows from dominated convergence. 4

15 Moreover, letting ω = e i2πx, k= r k e i2πkx = + k r k ω k + k r k ω k = + rω rω + rω rω r 2 = r(ω + ω ) + r 2. The Poisson kernel P r is particularly useful in complex analysis, where it corresponds to power series. It is left to the reader to verify from the definition that Theorem 26. P r is a summability kernel. This is done in much the same way as for the Fejér kernel. [Section 7 below is somewhat more difficult and optional] 7 The Dirichlet kernel revisited: a theorem of du Bois-Reymond Theorem 27 (du Bois-Reymond). There exists a continuous function f of period, such that the partial sums S n (f, x) diverge at some point x. Clearly, given Theorem 23 and Lemma 9, this can only be the case if the Dirichlet kernel D n is not a summability kernel. This is in fact the case. Condition (2) is not met. Lemma 28. D n L [,] 4 π 2 log n. Proof. Note that sin x x for all x, so that D n = /2 = 2π 4 π 2 /2 n+/2 /2 D n (x) dx 2 n k+ k= k sin πt dt t 2 π dτ τ = 4 log(n + ). π2 sin ( (n + /2) 2πx) ) dx πx n k sin πt dt = 4 n k k= k π 2 k k= 5

16 In fact, it is not hard to modify this proof to show that D n = 4 log n + O(), π2 though we do not need this fact in what follows. For us, the lemma in particular implies by Lemma 9 that for each n there exists a continous function β n such that β n, yet D n β n () 2 log n (5) π2 Clearly each function β n has a Fourier series that is badly behaved near the origin, at least for the n-th partial sum. Our approach will be to add together a large number of these functions to obtain a single function badly behaved for an infinite number of partial sums. This approach is very powerful, and foreshadows what is called the uniform boundedness principle in functional analysis. In fact, it is by an appeal to the uniform boundedness principle that Theorem 27 is usually proved in textbooks. (See Rudin, Real and Complex Analysis, ch. 5, or Katznelson section II.2, from which our proof is adapted.) The proof we give below, instead of appealing directly to this somewhat abstract theorem, borrows the technique of its proof to construct a concrete counterexample. The price we pay for being concrete is that our proof appears a little messier than what one would find in Rudin s textbook for instance. Proof of Theorem 27. We have constructed functions β n above in (5). We now construct trigonometric polynomials α n (x) := σ n 2(β n, x) = K n 2 β n (x). Because K n 2 = O(), we have that α n = O(), and S n (α n, x) S n (β n, x) k n = 2. k n ˆα n (k) ˆβ n (k) k n 2 + ˆβ n (k) In particular, We define λ n := 2 3n and S n (α n, ) 2 log n 2. (6) π2 f(x) := n= n 2 α λ n (λ n x). (7) 6

17 t. We will show that f is continuous, yet lim sup S n (f, ) =. The continuity follows because the sum (7) is absolutely convergent for all On the other hand, we have that α λj (λ j x) = ˆα λj (k)e i2πλjkx, k λ 2 j so that ( S λ 2 n (f, ) = S n ) λ 2 n j 2 α λ j (λ j x), + j=n+ n = j 2 α λ j () + n 2 S λ n (α λn, ) + 2 log λ n π 2 O(), n 2 with the inequality in the third line following from (6). [Resumption of lecture 3] j 2 ˆα λ j () j=n+ 8 Fourier coefficients and smoothness j 2 ˆα λ j () We briefly expand on an exercise in the homework, which showed that a twice continuously differentiable function has Fourier coefficients that decay quadratically. General principle: The smoother a function f is, the faster ˆf(k), and vice-versa. Likewise, in general, the smoother a function is, the more quickly the partial sums (or summability kernels) of its Fourier series will converge to the function itself. An instance of this principle, proved in the same manner as your homework is the following: Proposition 29. For f C l (R) and of period, ( ) ˆf(k) = O + k l. Because this is so similar to your homework we do not outline the proof in detail, but only say that it follows easily from the fact that where f (l) is the l-th derivative of f. (f (l) )ˆ(k) = (i2πk) l ˆf(k), 7

18 9 An application: uniform distribution As an application of the theory we have developed, we consider the distribution of the fractional parts of the sequence ϑ, 2ϑ, 3ϑ,..., for numbers ϑ. Definition 3. x = max{k Z : k x} and {x} = x x. x and {x} are called the floor and fractional part of x respectively. If ϑ = p/q, then {nϑ} must be one of, /q, 2/q,..., (q )/q so we see that Proposition 3. If ϑ is rational, then the sequence { {nϑ} } is discrete. On the other hand, for ϑ irrational, the behavior of this sequence is more interesting. Proposition 32. If ϑ is irrational, then the sequence { {nϑ} } is dense in [, ]. This theorem will be familiar to anyone who as a child ever tried to avoid the periodic cracks between sidewalk tiles while walking. By walking in the ordinary fashion (with each step the same length, ϑ), unless you carefully step in ratio to the length of tiles, you re certain sooner or later to step on one of the sidewalk s cracks no matter in what location you begin. We will prove more general theorem than Proposition 32 below, but for the moment let us at least outline a self-contained proof of this proposition that could be filled in by interested readers. Place the points {nϑ} where n N around the unit circle by marking points at 2πnϑ radians. If you suppose that this collection of points in not dense, then there must be some non-empty open interval on the unit circle in which none of these points occur. By a compactness argument, there exists some such interval I which is of the longest length. Consider I rotated by 2πϑ radians clockwise. By assumption, this interval too must contain none of the points that have been marked on the circle. Continue rotating I in this way, an infinite number of times. None of these rotated intervals can contain any points that have been marked on the circle. Because ϑ is not rational, none of these rotated intervals can exactly coincide. But on the other hand, none of the rotated intervals can overlap without exactly coinciding, since this would produce a larger interval that contains no points. We have thus produced an infinite collection of disjoint intervals, all of the same size and contained on the unit circle plainly a contradiction. Whether you have chosen to fill in the details of the above outline or not, is natural having thought about Proposition 32 to ask whether anything else can be said about the distribution of the points { {nϑ} } that that they are dense. In fact more can be said: Theorem 33. For (a, b) [, ), lim N N #{ {nϑ} (a, b) : n N } = b a. Note that this implies Proposition 32, as (a, b) may be centered around any number and of arbitrarily small length. We will prove Theorem 33 from 8

19 Theorem 34. For f continous and of period, lim N N f({nϑ}) = n N f(x) dx. Proof of Theorem 34. Step : It is trivial to verify that the theorem is true when f is constant. Step 2: We see that the theorem is true when f(x) = e i2πkx, for any fixed k. For, Yet, for α = e i2πkϑ, N e i2πkx dx =. e i2πk(nϑ) = N (α + α2 + + α N ) n= = α α N+ N α ( ) = O N. Here we have made critical use of the fact that ϑ is irrational, so α for any k. Step 3: For general f, for all ɛ >, there exists a trigonometric polynomial such that f(x) From this we have, f(x) dx c = M2 c k e i2πkx < ɛ/2, k= M f(x) dx x. P (x) dx = f(x) P (x) dx f(x) P (x) dx ɛ/2. Moreover, Yet N f({nϑ}) N n= lim N N P ({nϑ}) ɛ/2. n= P ({nϑ}) = c. n= 9

20 So implying lim sup N lim sup N N N n= f({nϑ}) c ɛ/2, n= f({nϑ}) As ɛ was arbitrary, this proves the theorem. f(x) dx ɛ, Proof of Theorem 33. It is easy to see that for all ɛ >, there exist continuous functions of period, f and f + such that f (x) (a,b) (x) f + (x), for all x, yet f + (x) (a,b) (x) dx ɛ (a,b) (x) f (x) dx ɛ. Note that (a,b) (x) dx = b a. Hence we have that, with and Finally, n N f ({nϑ}) (a,b) ({nϑ}) f + ({nϑ}), N N N N N n N f ({nϑ}) n N f + ({nϑ}) n N n N f (x) dx (b a) ɛ, f + (x) dx (b a) + ɛ. (a,b) ({nϑ}) = # { {nϑ} (a, b) : n N }. n N Again because ɛ is arbitrary, we have shown that the limit in the theorem is as claimed. 2

21 Another application: the Weierstrass approximation theorem We use Fejér s theorem to give a quick proof of another approximation theorem due originally, with a different proof, to Weierstrass. Theorem 35 (Weierstrass). For f C[, /2], for all ɛ > there is a polynomial Q(x) = a + a x + + a h x h such that f(x) Q(x) ɛ, x [, /2]. Here we have used the interval [, /2] out of convenience, but it is not hard to see that this same theorem may be extended with any arbitrary interval [a, b] in place of [, /2] (since re-scalings and translations of polynomials remain polynomials). Proof. Define g(x) := f( x ) for x [ /2, /2] and define g elsewhere by periodicity so that it is of period. Then g C(R), and can be approximated uniformly by trigonometric polynomials. But each term e i2πkx that would appear as a summand in a trigonometric polynomial can be uniformly approximated on [, /2] by a sum M l! (i2πkx)l, l= for some M, by a Taylor approximation. Since a finite linear combination of such sums remains a polynomial, we can approximate trigonometric polynomials arbitrarily closely by ordinary polynomials, and thus we may approximate the function we began with as well. [Lectures 4 and 5] Inner product spaces We have seen that the partial sums S n (f, x) may not converge pointwise to f(x), even for a continuous function f. However, in the next two sections we will show that for f L 2 [, ), we do have f(x) S n (f, x) 2 dx. In this section and the next we review some facts about inner product spaces and introduce/review the notion of Hilbert spaces, the latter of which came about historically in large part from the study of Fourier series. We will use these notions to prove the convergence result above. (Throughout this section H will denote an arbitrary inner product space.) Definition 36. A complex vector space H is an inner product space if there exists an inner product, : H 2 C such that for all x, y H, and α C, 2

22 (i) x, y = y, x (ii) x + y, z = x, z + y, z (iii) αx, y = α x, y (iv) x, x (v) x, x = if and only if x =. From these definitions it is trivial to see that Proposition 37. We have i., y = ii. x, αy = α x, y iii. x, y + z = x, y + x, z. One can define a real inner product space in the same way, with scalars always real instead of complex. The reader should check that that following are examples of inner product spaces: (a) C n with x, y = n x i y i. i= (Similarly R n is a real inner product space.) (b) L 2 [, ) with the inner product f, g = fg dx, as long as we make the identification that f = f 2 whenever f (x) = f 2 (x) almost everywhere. (c) (somewhat less canonically) the subspace of L 2 [, ) consisting of polynomial functions a n x n. Definition 38. x := x, x. x is always non-negative and is called the norm of x, and we will demonstrate shortly that it possesses the properties of a norm. We first prove a preliminary but very important result: Theorem 39 (Cauchy-Schwarz). For x, y H Before proving this, we first prove x, y x y. 22

23 Lemma 4. For u, v H u, v 2 u v 2. One may note the similarity of this inequality to the inequality 2ab a 2 +b 2 for a, b positive reals. This follows from the fact that (a b) 2. This strategy of proof we imitate in the proof below. Proof of Lemma. Let α = be such that α u, v = u, v. Note that it follows from the axioms of an inner product space that u, v = αu, v = v, αu as well. Then we have αu v, αu v = u 2 2 u, v + v 2. Proof of Cauchy-Schwarz. Clearly the claim is true if x = or y =. If x, y, let u = x/ x and v = y/ y. Then u = v =, and the lemma tells us that u, v. But, u, v = giving us the inequality we wanted. x, y x y, One consequence is that inner products are continuous in their arguments. Corollary 4. For any z H, x x, z is continuous. Proof. x, z x 2, z = x x 2, z x x 2 z. Another consequence of Cauchy-Schwarz: Theorem 42 (Triangle Inequality). For x, y H, Proof. x + y x + y. This is just the statement that H is a normed linear space. x + y 2 = x + y, x + y = x 2 + x, y + y, x + y 2 x x y + y 2 = ( x + y ) 2. Corollary 43. For all x, y, z H x z x y + y z. 23

24 This corollary of the triangle inequality is just a reminder that d(x, y) = x y defines a metric. Inner product spaces have slightly more structure that most normed linear spaces however. In particular, we have a notion of orthogonality: Definition 44. For x, y H, we say x and y are orthogonal and write x y if x, y =. Definition 45. For x H and S H, we say x is orthogonal to S and write x S if x, y = for all y S. Definition 46. A set E is an orthogonal set if x, y = for all x, y E such that x y. Definition 47. A set E H is an orthonormal system if it is an orthogonal set and x = for all x E. The reader should check that the following are examples of orthonormal systems: (a) the set of all basis vectors {e, e 2,..., e n } C n, (b) the set of functions {e i2πnx } n Z L 2 [, ) (c) the set {e, e 2, e 3 } C 4 (d) the set of functions {e i2πj2x } j N L 2 [, ). The following result is very important: Theorem 48 (Pythagorean Theorem). If x, y H are orthogonal, x + y 2 = x 2 + y 2. Proof. x + y, x + y = x 2 + x, y + y, x + y 2 = x 2 + y 2. By induction, this clearly implies Corollary 49. If {x,..., x n } H is an orthogonal set (x i x j for all i j), then x + x x n 2 = x 2 + x x n 2 Corollary 5. Let {φ, φ 2,..., φ n } be a finite orthonormal system in H, and α, α 2,..., α n C. Then n 2 α j φ j = A deeper consequence is as follows: n α j 2. 24

25 Corollary 5 (Best approximation). Let {φ,..., φ n } be an orthonormal system, and u H. Define a j := u, φ j. Then n u n a j φ j u b j φ j, for all b, b 2,..., b n C, with equality if and only if a j = b j for all j. For reasons that will become clear in the next section, the coefficients u, φ n are called Fourier coefficients of the vector u, and the best approximation corollary above says that the closest one can approximate u by linear combinations of the vectors {φ j } is a linear combination whose coefficients are exactly the Fourier coefficients. Proof. Note that u n a j φ j, φ k = a k a k =, for all k. That is to say, u a j φ j is orthogonal to the set {φ, φ 2,..., φ n }. Let δ j = a j b j, so that u b j φ j = u a j φ j + δ j φ j. Then the Pythagorean Theorem and its corollaries imply n u 2 n b j φ j = u 2 n a j φ j + δ j 2 n u 2 a j φ j Another consequence of the observation that u a j φ j is orthogonal to the set {φ, φ 2,..., φ n } is Bessel s inequality: Theorem 52 (Bessel s inequality). For a n := u, φ n, n a j 2 u 2. Proof. Applying the Pythagorean Theorem and its corollaries as before, n u 2 = u 2 n n a j φ j + a j 2 a j 2. 25

26 Remark: In fact we know that Bessel s inequality does not express the whole truth, in the sense that we know in C n or R n this inequality is in fact an equality: α e + α n e n Cn = α α n 2. On the other hand, if we began with the orthonormal system {e, e 2, e 3 } in C 4, we would indeed have a strict inequality; To take for example the vector u = e + e 2 + e 3 + e 4 = (,,, ), we have while u 2 = 4, u, e 2 + u, e u, e 3 2 = 3. In C 4, what distinguishes the orthonormal system {e, e 2, e 3 } where Bessel s inequality is a strict inequality from the orthonormal system {e, e 2, e 3, e 4 } where it would be an equality? The answer, of course is that the linear span of {e, e 2, e 3, e 4 } is C 4, while the linear span of {e, e 2, e 3 } is not. Can we reproduce this sort of reasoning in more complicated inner product spaces like L 2? We can, but we must work with a certain refined type of inner product space known as a Hilbert space. 2 Hilbert space 2. General theory Definition 53. If an inner product space H is complete (with respect to the metric induced by ), then H is called a Hilbert space. The reader should check that the following are examples of Hilbert spaces (a) C n with x, y = (Similarly R n is a real Hilbert space.) (b) L 2 [, ) with the inner product f, g = n x i y i. i= fg dx, as long as we make the identification that f = f 2 whenever f (x) = f 2 (x) almost everywhere. (c) l 2 (N) := {((a, a 2,...)) : a i 2 < + } with x, y = a i b i. i= 26

27 In verifying that (b) and (c) are Hilbert spaces, one must use the nontrivial result, proved in Analysis III and also in most real analysis textbooks (for instance Rudin s Real and Complex Analysis, pp ): Theorem 54. L 2 [, ) and l 2 (N) are both complete. (Recall that a complete metric space is one which contains the limit point of every Cauchy sequence.) The definitions of orthogonal vectors, sets, and orthonormal systems remains the same in Hilbert spaces as in inner product spaces. We can quickly prove the following result: Theorem 55. Let {φ, φ 2,...} be an orthonormal system in a Hilbert space H, and ((α, α 2,...)) l 2 (N). Then converges in H. Moreover, α j φ j (8) α j φ j, φ n = α n. (9) Proof. H is complete, so to see that (8) converges, we need only verify that S N := j N α jφ j is a Cauchy sequence: S N S M 2 = j=m+ from Corollary 5. But this is no more than n=m+ α n φ n 2 = α n 2 n=m+ α n 2, as M, because α l 2 (N). To verify (9), note that because the inner product is continuous (Corollary 4), N α j φ j, φ n = lim α j φ j, φ n = α n. N The following sorts of orthonormal systems are of particular importance: Definition 56. A complete orthonormal system in a Hilbert space H is an orthonormal system {φ j } such that span{φ j } = H. 27

28 The reader should verify that the following are examples of complete orthonormal systems: (a) {e, e 2,..., e n } in C n (b) {((,,,...)), ((,,,..)), ((,,,...)),...} in l 2 (N). The following is an abstract version of the theorem about Fourier series we set out to prove. Theorem 57. Let H be a Hilbert space, and {φ j } be a complete orthonormal system. For any u H, define n S n (u) = a j φ j, where a j = u, φ j as before. Then u S n (u). The idea of our proof is this: we can approximate u arbitrarily closely by linear combinations of elements φ j, because these elements compose a complete orthonormal system. But by the best approximation corollary, for any linear combination we use to approximate u, there will be a better approximation of the sort S n (u). Proof. Choose ɛ >. Because span{φ j } = H, there exists a finite set of scalars α, α 2,..., α N such that N u α j φ j < ɛ. But by Corollary 5 (best approximation), for n N, n u N a j φ j u α j φ j. As ɛ is arbitrary this is just the statement that u S n (u). We can also refine Bessel s inequality, Theorem 58 (Parseval). For all u H, with H a Hilbert space, and for a j as in Theorem 57, with {φ j } a complete orthonormal system, u 2 = Moreover, if v H and b j := v, φ j, u, v = a j 2. () a j b j. () 28

29 Proof. We begin by proving () as in the proof of Bessel s inequality. We have n u 2 = u 2 n a j φ j + a j 2 = a j 2, by letting n, since u S n (u). (3) is similar but requires a little more bookkeeping. It is easy to verify that n S n (u), S n (v) = a j b j (2) and as u = (u S n (u)) + S n (u) and v = (v S n (v)) + S n (v), we have u, v = u S n (u), v S n (v) + u, v S n (v) + u S n (u), v + S n (u), S n (v) n = a j b j + o(). (3) This last line follows from (2) and the relations n, u S n (u), v S n (v) u S n (u) v S n (v) = o(), u, v S n (v) u v S n (v) = o(), u S n (u), v u S n (u) v = o(). Taking the limit n in (3) completes the proof. 2.2 Application to Fourier series We begin this section by adding one more set to our list of complete orthonormal systems. Lemma 59. {e i2πjx } j Z is a complete orthonormal system in L 2 [, ). Proof. We need only show that for all f L 2 [, ) and all ɛ >, there exists a trigonometric polynomial such that P (x) = k= M f P 2 < ɛ. c k e i2πkx We define the function Trim A (f) : x sgn(f(x)) min(a, f(x) ). Note that by dominated convergence lim A (f Trim A (f)) 2 dx =, 29

30 so that there exists g L [, ) such that f g 2 ɛ/3. (Just take g = Trim A (f) for sufficiently large A, for instance.) Let B := g. We can find φ C[, ) such that and φ B. Hence, φ g 2 2 = φ g (ɛ/3)2 B, φ g φ g dx B (ɛ/3)2 }{{} B B implying φ g 2 ɛ/3. (To this point, what we have shown, in effect, is that continuous functions are dense in L 2 [, ).) This established, we can find a trigonometric polynomial P such that implying that Hence, as we wanted. φ(x) P (x) ɛ/3, x, φ P 2 φ P ɛ/3. f P 2 f g 2 + g φ 2 + φ P 2 ɛ, With this lemma proved, Theorem 57 immediately implies Theorem 6. For f L 2 [, ), Said another way, f(x) = lim n n k= n f S n (f) 2. ˆf(k)e i2πkx in the L 2 [, ) norm. Because this is such an important result, we review the proof of Theorem 57, now put in the concrete language of Theorem 6. The proof went as follows: choose any f L 2 [, ) and any ɛ >. From Lemma 59, we have that there is a trigonometric polynomial P (x) = N M c ke i2πkx such that f P 2 ɛ. But by best approximation (Corollary 5) for n N, M, f S n (f) 2 f P 2 ɛ. This is the same claim that the statement in the theorem makes. Other results we have proved in the abstract setting of Hilbert spaces now translate directly to results about Fourier series as well. Theorem 55 implies 3

31 Theorem 6. For any {c j } l 2 (Z), there exists a unique f L 2 [, ) such that c k = ˆf(k) for all k Z. Likewise Theorem 58 implies Theorem 62 (Parseval). For f L 2 [, ), we have f L 2 = ˆf l 2. That is, Moreover for f, g L 2 [, ), [Lecture 6 - optional] f(x) 2 dx = f(x)g(x) dx = k= k= ˆf(k) 2. ˆf(k)ĝ(k). 3 Riemann s localization principle In this section we prove a few last convergence results for Fourier series. Theorem 63. For f of period and in L [, ), suppose f is differentiable at x. Then lim n S n(f, x ) = f(x ). Proof. We define for t [ /2, /2), { f(x t) f(x ) F (t) := t if t, t [ /2, /2) f (x ) if t =. As f is differentiable at x, F (t) is bounded in a neighborhood of and therefore we see that F is integrable on [ /2, /2). Note that S n (f, x ) f(x ) = Yet td n (t) = = = /2 /2 /2 /2 /2 /2 f(x t)d n (t) dt f(x ) (f(x t) f(x ))D n (t) dt F (t) ( td n (t) ) dt. t sin πt sin ( 2π(n+/2)t ) = e t i2πnt eiπt 2i ( t ) e sin πt (as D n (t) dt = ) i2πnt e iπt 2i ( t ), sin πt and both eiπt 2i ( e iπt sin πt ) and sin πt ) remain bounded in t, so that the theorem is implied by the Riemann-Lebesgue lemma (Theorem 6). 2i ( t 3

32 Note that the differentiability of a function at a point is an entirely local condition, so that the convergence of S n (f, x) depends in the above theorem upon how f behaves only at x. This leads to the following localization principle of Riemann, which is far from obvious from the definition of S n (f, x ), which depends on the behavior of f everywhere, not just at x. Theorem 64 (Riemann). For f, g L [, ) and of period, if f(x) = g(x) for all x in some neighborhood U of x, then Proof. Note that S n (f, x ) S n (g, x ). S n (f, x ) S n (g, x ) = S n (f g, x ), but (f g)(x ) = and moreover (f g)(x) = for x U, so that certainly f g is differentiable at x. Theorem 63 then applies. These theorems only scratch the surface of what is known about the convergence of Fourier series for various sorts of functions. We have discussed some more advanced results in class, but any proofs of them exceed the scope of this class. The books by Katznelson or Krantz are an excellent place to start learning about them in more depth. 4 Flat polynomials We conclude these notes with a discussion of a problem due to Littlewood that, despite its elementary character, remains unsolved. The problem is as follows: Do there exist positive constants c and c 2 such that for all n there is some choice of signs ε = ±, ε 2 = ±,..., ε n = ± such that (Here {ε,..., ε n } = {ε (n) n c n ε k e i2πkx c 2 n, for all x? (4) k=,..., ε(n) n } may vary with n.) Polynomials of this sort, such that ε k z k are roughly of order n for all z = are known as flat polynomials. Why is n the right constant here? The reason is that for any choice {ε k }, n 2 ε k e i2πkx dx = k= n ε k 2 = n. In particular, for at least some x, the sum ε k e i2πkx must be at least n, and for some x the sum must be no more than n. (If you don t see why, assume for instance the sum is always less than n and derive a contradiction.) k= 32

33 This reasoning shows that if ε can be chosen so that the sum is nearly constant for all x, the value it must always be close to is n. I should make at this point two confessions about the above flat polynomial problem. The first is that there is no consensus among experts that the problem has an affirmative answer. Indeed, there does not seem to be any special heuristic evidence either for it or against it. However, if we allow the coefficients ε k to be any complex scalar of absolute value, rather than just ±, then it is known that for all n there do exist flat polynomials of the sort (4). Indeed, something even stronger is true, that for any δ > one can choose c = δ and c 2 = + δ. That is, for all sufficiently large n there exist complex coefficients ε depending on n such that ε k = for all k, and ( δ) n n ε ke i2πkx ( + δ) n, for all x. k= Results of this sort are due to Byrnes, Körner, Kahane, and Bombieri & Bourgain, among others. The restriction that ɛ k takes only the values or is more severe, but one may hope it is not so severe that a result of this sort is lost entirely. I leave the second confession to Tom Körner (who as listed above has worked on this problem and from whom I take much of the exposition below): Admittedly the problem has no deep mathematical significance, but I will happily send the cost of a bottle of good champagne to the first person to answer it. I mention the problem at the end of our course mainly to convey that even at a very elementary level, nice problems can be asked about trigonometric polynomials and Fourier series, to which a solution is still not known. Perhaps you might like to think up some of your own! What we will demonstrate in what follows answers one half of the question above. Theorem 65 (Golay-Shapiro-Rudin). For each n there exists ε (n), ε(n) 2,...ε(n) n {, } such that n ε (n) k ei2πkx A n, k= where A is the absolute constant /( 2 ). Proof. We define the polynomials P m and Q m of degree 2 m inductively by P (z) = z, Q (z) = z, and P m+ (z) = P m (z) + z 2m Q m (z), Q m+ (z) = P m (z) z 2m Q m (z). 33

34 It is easy to verify inductively that the coefficients of P m and Q m remain either or, and for z =, P m+ (z) 2 + Q m+ (z) 2 = 2 ( P m (z) 2 + Q m (z) 2). But this implies inductively that for z =, and so P m (z) 2 + Q m (z) 2 = 2 m( P (z) 2 + Q (z) 2), P m (z) 2 (m+)/2, for all z =. P m (z) is thus a polynomial of degree 2 m satisfying the hypothesis of the theorem for n = 2 m and A = 2. Yet for any integer n we may write n = l 2 m(k), k= by expanding n into its binary representation. It is not difficult to verify that P (z) :=z m()+m(2)+ +m(l ) P m(l) (z) + z m()+m(2)+ +m(l 2) P m(l ) (z) + + z m() P m(2) (z) + P m() (z), is equal to for some coefficients ε (n) k z =, n k= ε (n) k zk, that take the values or for all k. Moreover, for P (z) l k= 2 (m(k)+)/2 2 (m(l)+)/2( ) / /2 n, 2 as we wanted. 34