Introduction to Differential Equations and Fourier Series: Math 110 Section Notes. Christopher Eur

Transcription

1 Introduction to Differential Equations and Fourier Series: Math 110 Section Notes Christopher Eur May 20, 2015

2 As the second time course assistant for this course, I have decided to go a bit further than making section notes for Math 110 Spring 2015 and expound my understanding of the course material into a lengthy document, rather than series of disparate collection of notes. I do not claim in anyway that the content of this document is correct nor elegant, but I have tried to bring a more modern flavor to the material when possible, as Holland s Applied Analysis by Hilbert Space Method does not do so. Please any errata you may find to ceur@college.harvard.edu. Note for students: This is intended to supplement and in no way substitute the lecture outlines for Math 110. The primary purpose is to review the material and occasionally provide different perspectives. My hope is that it will help provide a big picture to the material being covered in the lectures. Enjoy! 1

3 Contents 1 Preliminaries Linear Algebra Review Vector Spaces of Functions Definition of Differential Equations Definition of Linear ODE First Order Linear ODE Integrating Factor Variation of Parameters Green s Function Power Series Preliminaries for Second Order ODE The Wronskian and Abel General Facts on the Kernel of Second Order ODE Second Order Homogeneous ODE Constant a, b, c Case Laplace Transforms Green s function for 2nd order ODE Hilbert Spaces Metric Space Preliminaries On convergence of sequence of functions Interlude: Lebesgue Integral Basics L 2 space An aside: on convergence of sequence of functions Sturm-Liouville Theory Classical Fourier Series Isoperimetric inequality

4 CHAPTER 1 Preliminaries In this chapter, we review and introduce the linear algebraic language that we will be using through out this course. We finish with the definition of (linear) differential equations. 1.1 Linear Algebra Review We assume that the reader is familiar with basic linear algebra. The following section can be skipped for anyone with sufficient linear algebra background. A good reference is Axler s Linear Algebra Done Right Ch. 1,2,3,5 and Artin s Algebra Ch. 1,3,4. Definition A field is a set where one can add, subtract, multiply, and divide (except by 0) elements. Example The set of rational numbers Q, real numbers R, and the complex numbers C are all examples of a field. The set of integers Z is not a field because division is closed in Z. The set of natural numbers N is also not a field because it doesn t include negative numbers (i.e. 5 8 for example doesn t exists in N). In this class, a field will be either R or C unless stated otherwise. Definition A vector space V over a field F is a set V with maps V V + V and F V V (called vector addition and scalar multiplication respectively) such that: 1. v + w = w + v v, w V 2. element denoted 0 V such that 0 + v = v = v + 0 v V 3. v V w V such that v + w = w + v = v = v and (ab) v = a (b v) a, b F, v V 5. a(v + w) = av + aw and (a + b)v = av + bv a, b F, v, w V 3

5 Definition (Concise definition). A vector space V over a field F (i.e. R or C) is a set V with a binary operation V V + V and a field action F V V such that V, + is an Abelian group and distributivity holds. Example A field F is a vector space over itself. R n with the usual definition of addition and scalar multiplication is a vector space over R. P C := {polynomials with coefficients in C} is a vector space over C. Definition Let V be a vector space. A subset U of V that is also a vector space is a subspace of V. Given U V, in order to show that U is a subspace we need check: 1. closed under addition: u, v U u + v U 2. closed under scalar multiplication: a F and u U au U Example V := {(x, y, 0) R 3 x, y R} is a vector subspace of R 3. P C n, the set of polynomials of degree n, is a vector subspace of P C. Exercise 1.1.A (Non-examples). Give an example of a subset of R 2 that is not a vector subspace and: (i) satisfies condition 1. but not 2. (in Definition ); (ii) satisfies 2. but not 1. Solution. (i) Z 2 R 2 is not a subspace (satisfies 1. but not 2.). (ii) (x-axis) (y-axis) R 2 is not a subspace (satisfies 2. but not 1.) Definition Let V, W be vector spaces over F. A function T : V W is a linear map if T (v +v ) = T v +T v and T (av) = a(t v) for any v, v V and a F. Equivalently, T is linear if T (av + v ) = a(t v) + T v for any v, v V and a F. Definition Let T : V W be a linear map. The kernel of T is ker T := {v V T v = 0} and the image of T is Im(T ) := {w W w = T v for some v V }. Exercise 1.1.B. Prove that ker T V is a vector subspace, and Im(T ) W is a vector subspace. Solution) For ker T, note that if v, w ker T then T (v + w) = T v + T w = = 0 and T (av) = a(t v) = a0 = 0. For Im(T ), if w, w Im(T ) then there exists v, v V such that T v = w, T v = w, and thus w + w = T v + T v = T (v + v ) and aw = a(t v) = T (av). Definition A basis for a vector space V (over F) is an ordered list of vectors B = (v 1,..., v n ) such that every element v V can be uniquely written as v = a 1 v a n v n for some (a 1,..., a n ) F n. In other words, a basis for V establishes a bijective correspondence between F n and V via (a 1,..., a n ) a 1 v 1 + a n v n. Given a basis B for V, let s denote by M(v) the vector in F n that corresponds to v via the bijection in the definition given above. 4

6 Example Consider P R 2, the set of polynomials of degree 2 with real coefficients. One can see that (1, x, x 2 ) is a basis for P R 2, and with this basis 1 + x2 corresponds to the 1 vector 0. Moreover, if one could also use (1, 1 + x, 1 + x + x 2 ) as the basis, in which case x 2 corresponds to 1. 1 Proposition Let (v 1,..., v n ), (w 1,..., w m ) be the bases for V, W respectively, and let T : V W be a linear map such that T v j = a 1j w a mj w m for j = 1,..., n. Then M(T v) = [A]M(v) where [A] is the matrix [A] = w 1. w m v 1... v j... v n a a 1j... a 1n... a m1... a mj... a mn Example Let s consider the map d : P R 3 PR 2 given by differentiation (which is known to be linear). With bases (1, x, x 2 ) and (1, x) respectively, the matrix [A] in this [ ] 1 [ ] case is. Sanity check: [A] 0 0 = and (1 + x ) = 2x Vector Spaces of Functions Now we can develop a language to discuss the central objects of our study. One of the very useful thoughts in math was to view the space of real-valued functions on a given set as a vector space (an R-algebra, really), as described below: Proposition Let X be any set, and define R X := {f : X R}. Then R X is a vector space over R with appropriate definition of + and. Proof. If f, g R X and c R, define (f + g) to be a function x f(x) + g(x), and define cf to be a function x cf(x). Checking that these make R X into a vector space is not difficult: the zero vector is the zero function 0(x) = 0, additive inverse to f is f, distributivity follows from distributivity in R, etc. Exercise 1.2.C. What is R X isomorphic to when X = {1, 2,..., n}? Give a basis of R X. 5

7 Solution. R X R n above. R X is the set of all the functions from X to R, but giving a function from X = {1,..., n} to R is same thing as picking n-tuple of real numbers a 1,..., a n. That is, if the function f R X, then we can identify f with the n-tuple (f(1), f(2),..., f(n)). A natural basis to pick is (f 1,..., f n ), where f i is a function that has value 1 at i and 0 otherwise. Remark R R is the set of all single-variable real-valued functions Much of real analysis is concerned with nice subspaces of R R. Here are some important examples: Definition Define C(R) := {f : R R f is continuous}, C k (R) := {f : R R f (k) (the k-th derivative) is continuous}, and P n := {polynomials of degree n}. Moreover, define the set of smooth functions to be C (R) := {f : R R f (k) C(R) k N} (i.e. infinitely differentiable functions). Proposition C(R), C k (R), and P n are linear subspace of R R. In fact, they are also subspaces of each other in the order P n C k (R) C(R). Proof) From calculus we know that sums and scalar multiples of continuous functions are continuous. Moreover, differentiation is linear ((cf + g) = cf + g ), so sums and scalar multiples of C k (R) functions are also C k (R). Lastly, polynomials are closed under sum and scalar multiplication. So, P n, C k (R), C(R) are all vector subspaces of R R. Moreover, differentiable implies continuous, and polynomials are infinitely differentiable, and thus P n C k (R) C(R). Remark The set of smooth functions is strictly bigger than the set of real-analytic functions (i.e. functions that can be written faithfully as a Taylor series). An example of smooth function that is not analytic is one defined as :f(x) = 0 for x 0 and f(x) = e 1/x2 for x > Definition of Differential Equations Definition A differential equation is an equation containing one or more derivatives of the function in consideration. The order of the differential equation is the highest power of derivative that appears in it. To solve the differential equation means finding a function f that satisfies a given differential equation. Example Differential equations with appropriate functions under consideration: 1. f (x) = e x + xe x (solution: f(x) = xe x ) 2. f (x) f (x) = e x (solution: same as above) 6

8 3. ( f f x )( y ) f(x, y) = 0 (solution: f(x, y) = xy) Definition An ordinary differential equation (ODE) is differential equation where the function under consideration is a single-variable function f : R R. In more concrete terms, an ordinary differential equation of order k is a differential equation of the form: [an expression in f(x), f (x),..., f (k) (x)] = h(x), h(x) some function of x Note that 1. and 2. (but not 3.) in Example are ordinary differential equations. Example 3. is called a partial differential equation (PDE). It is assumed that you have worked with ODEs before, but in case you have not, the following is essentially all the ODE you need to know for the course: Exercise 1.3.D. (Conventionally, we write y in place of f(x).) Solve the following ordinary differential equations: 1. y = 1/x 2. y = ky (alt. y /y = k) 3. y /y = f(x) where F (x) = f(x) for some function F. Solution. 1. Rewrite as dy/dx = 1/x so that dy = 1 xdx, and integrate both sides to get y = ln x +C; most of the time we assume x > 0 so ln x is fine. 2. Again, rewriting y /y = (dy/dx)/y = (dy/y)/dx, we get 1 y dy = kdx, and Integrating, we get ln y = kx + C, so y = e kx+c, i.e. y = ±e C e kx, and absorbing ± into the constant term, we get y = Ce kx. 3. We solve exactly as we did for 2., and get y = Ce F (x). Definition Suppose we are given an ODE, say: [expression in f, f,..., f (k) (x)] = h(x) Then we can consider the map f [expression in f, f,..., f (k) ] as a map F : C k (R) R R. We call such F a differential expression. Remark So the study of ODE can be rephrased as: Given F : C k (R) R R and h R R, can we find f C k (R) such that F (f) = h? and how so? 7

9 1.4 Definition of Linear ODE We now define what is the central object of our course: Definition An ODE whose differential expression F is a linear map is called a linear ODE. Equivalently, a linear ODE of order k is an ODE of the form: a 0 (x)f(x) + a 1 (x)f (x) + + a k (x)f (k) (x) = h(x), (a k (x) 0) When ODE is linear, we call F differential operator and denote it as l. We will only study linear ODEs in this course. Our aim is to figure out how to exploit the fact that F : C k (R) R R is a linear map of two vector spaces, since vector spaces and linear maps are rather well-understood objects. From this point on, an ODE means linear ODE unless otherwise stated. Here are two immediate consequences where our linear algebraic view pays off: Proposition A linear ODE is called homogeneous if the h(x) in the equation is zero. If f, g are solutions to a homogeneous ODE, then so are any linear combinations of f, g. Proof) That f, g are solutions to the homogeneous ODE means that f, g ker(l), where l is the differential operator of the homogeneous ODE. But ker(l) is a subspace. Proposition Let l(y) = h(x) be a linear ODE, y p a particular solution. Then y p + ker(l) is the general solution. (In fact, previous proposition is just a special case of this). Exercise 1.4.E. Prove the above proposition Remark Suppose l(y) = h(x) where h(x) P n, and we want a polynomial solution (of degree < n). Then we can restrict l to l : P n P n, a linear map of finite dimensional vector spaces, which we understand very well, and so the problem becomes very easy with linear algebra. Now we are ready to study linear ODEs. 8

10 CHAPTER 2 First Order Linear ODE First order linear ODEs are of the form a(x)y (x) + b(x)y(x) = h(x) for some functions a, b, h. In other words, the differential operator is l(y) = ay + by. Luckily, for first order case the kernel is easy to find: Proposition The kernel of the differential operator l(y) = ay +by is a 1-dimensional space spanned by e b/a. Proof) Solving for ay + by = 0, we have y /y = b/a and hence Ce b/a. We have four main methods of solving first order linear ODEs: integrating factor, variation of parameters, Green s function, and power series. The first three are closely related. 2.1 Integrating Factor Let ay + by = h be given. Note that if the LHS is of the form (µy) for some µ(x), the differential equations is easy to solve. One way to make ay + by into such form is by multiplying by an appropriate function µ. That is, if we multiply µ to the expression, we get aµy + bµy = Definition Let ay + by = h be given; equivalently, y + b a y = h a. An integrating factor µ(x) is a function such that (µ(x)y) = µ(x)(y + b a y). Proposition If µ is the integrating factor for ay + by = c, then the solution to the differential equation is y(x) = 1 x h(t)µ(t) dt µ(x) x 0 a(t) And in fact, µ(x) = exp( b a dx) Proof) Lecture. 9

11 Example Crofton s method: Suppose we want to express some quantity y as function of x, and suppose it is reasonable to believe that y(x) is differentiable. Then translate the problem into a ODE problem and then solve for y(x). 2.2 Variation of Parameters Proposition The kernel of the map l : y ay + by is one dimensional. In fact, it is spanned by f(x) = exp( b a dx). Proof) Lecture Exercise 2.2.A. Note that that (integrating factor) 1 =(kernel of l). Why is it so? Use this to motivate the method of variation of parameters stated below. Proposition Let l(y) = ay + by and f ker(l). Then the variation of parameter method, which is setting y = gf, yields the solution to l(y) = h as: x h(t) y(x) = f(x) x 0 a(t)f(t) dt Proof) Lecture In sum, either through integrating factors or variation of parameters, we conclude: Theorem The solution to ay + by = h with initial condition y(x 0 ) = 0 is ( y(x) = exp b ) x a dx h(t) a(t) exp( b a dt)dt 2.3 Green s Function Here we develop a language to state Theorem in a slightly different way. We will see the benefit of doing this later in the course. Definition The Heaviside function H(x) is defined as: { 0 if x < 0 H(x) = 1 if x 0 and the Dirac delta function δ(x) is defined as H (x). Remark Strictly speaking, δ(x) is what we call a generalized function, or a distribution. For our purposes, we just treat δ(x) as H (x). x 0 10

12 We now list some properties of these two functions: Proposition Let H(x) and δ(x) be defined as above, and let a < b: 1. A function that is 1 on (a, b] and 0 elsewhere is H(b t) H(a t) 2. b a f(x)δ(x t)dx = f(t) if t (a, b) and 0 otherwise. 3. f(x)δ(x t) = f(t)δ(x t) Proof) Lecture Definition Let l(y) = ay + by with the condition y(x 0 ) = 0, and f ker(l). Then the Green s function G(x, t) for l is defined as: G(x, t) = f(x) a(t)f(t) [H(x t) H(x 0 t)] Theorem (cf. Theorem 2.2.3) Suppose G(x, t) is the Green s function for l(y) = ay + by with y(x 0 ) = 0. Then, 1. y(x) = G(x, t)h(t)dt is the solution to the differential equation. 2. G(x, t) regarded as a function of x with t fixed satisfies l(g(x, t)) = δ(x t). Remark Let a differential expression l(y) = ay + by and a condition y(x 0 ) = 0 be given, and let G(x, t) be the Green s function. Noting that V := {f C 1 (R) f(x 0 ) = 0} is a subspace of C 1 (R), we may restrict l to V ; i.e. we have l : V R R. Now define W = {h R R G(x, t)h(t)dt C1 (R)}. Then, a linear transformation l 1 : W V given by h(x) G(x, t)h(t)dt is the inverse of l. 2.4 Power Series Power series approach does not provide much interesting results in first order linear ODEs. It can be summed up as: write y = y 0 +y 1 x+y 2 x 2 + and using the differential equation solve for the coefficients y 0, y 1,.... We thus discuss this method in more depth in the next chapter. 11

13 CHAPTER 3 Preliminaries for Second Order ODE 3.1 The Wronskian and Abel The kernel of second (or higher) order differential operator is no longer guaranteed to be one-dimensional (nor is it as easy to compute as in the first order case). So, we first need to develop a tool to test whether a set of functions are linearly independent. Observation Recall: Let {f 1, f 2,..., f k } R R be linearly independent set of functions. If a 1 f 1 + a n f n = 0 for a i R (i = 1,..., n), then a 1 = a 2 = = a n = 0. Definition Let f 1, f 2,..., f n C n 1 (R). The Wronskian of f 1,..., f n, denoted W (f 1,..., f n ), is a function of x defined as: f 1 f n W (f 1,..., f n ) := det..... f (n 1) 1 f n (n 1) Theorem Let {f 1, f 2,..., f n } C n 1 (R) be a set of functions. If they are linearly dependent, then W (f 1,..., f n ) 0. Equivalently, If W (f 1,..., f n )(x) 0 for some x, then f 1,..., f n are linearly independent. Proof) Lecture Remark The converse of the theorem if not true: x 2, x x are linearly independent, but W (x 2, x x ) 0. Remark Given f 1, f 2 linearly independent, W (y, f 1, f 2 ) = 0 yields a second order linear ODE with solutions f 1 and f 2. Theorem (Abel s Formula). Let f 1, f 2 be any linearly independent solutions to l(y) = ay + by + cy = 0, and let W = W (f 1, f 2 ). Then aw + bw = 0, and hence W = C exp( b a dx) Proof) Lecture 12

14 3.2 General Facts on the Kernel of Second Order ODE Before we discuss the kernel of second order linear ODE, we need an important theorem: Theorem Let a linear ODE be given: a(x)y +b(x)y +c(x) = 0, with a(x), b(x), c(x) analytic around x 0 and a(x 0 ) 0. Then, for any y 0, y 1 R, there exists unique analytic solution y(x) around x 0 such that y(x 0 ) = y 0 and y (x 0 ) = y 1. Proof) Lecture Corollary Let l(y) = ay + by + c = 0 satisfy the conditions as in the previous theorem. Then, dim ker(l) = 2 Proof) Lecture The previous corollary shows that most normal-looking second order linear ODEs have 2-dimensional kernel. For most cases, we only see a, b, c be polynomials, trigonometric/exponential functions, etc. Hence, unless otherwise stated, for the rest of the section assume that all the second order linear ODEs we consider satisfy dim ker(l) = 2. With such assumption, we can do much much more. Proposition Let l(y) = 0 be a second order linear ODE. Then, for any f 1, f 2 ker(l) linearly independent, W (f 1, f 2 ) is the same up to a multiplicative constant. Hence, Wronskian is inherent to l, not particular bases of ker(l) Proof) Lecture Proposition Let l(y) = 0 be second order linear ODE, W (x) be the Wronskian, and suppose f 1 ker(l) is known, then the other linearly independent f 2 ker(l) equals: x W (t) f 2 (x) = f 1 (x) (f 1 (t)) 2 dt Proof) Lecture Before moving on, we briefly treat what to do when a(x 0 ) = 0 (i.e. is regular singular). Proposition (Method of Frobenius). Suppose we are given y + p x y + q x 2 y = 0, then we can try to solve for y of the form y = y 0 x k + y 1 x k+1 +. k satisfies the indicial equation k 2 + (p 0 1)k + q 0 = 0. 13

15 CHAPTER 4 Second Order Homogeneous ODE Here we give methods to find the kernel of a second order differential operator (i.e. how to solve second order homogeneous ODEs). 4.1 Constant a, b, c Case Proposition If given ay + by + c = 0 with a, b, c constants, the general solution is C 1 e αx + C 2 e βx if α, β are two distinct solutions of ar 2 + br + c = 0. If α = β, we have C 1 e αx + C 2 xe αx. Proof) Lecture 4.2 Laplace Transforms Definition The Laplace Transform is the linear operator L defined on the space of integrable functions that does not grow faster than any exponential, defined by: L(f) = f(s) = 0 f(x)e sx dx Remark Often the problem becomes easier by applying Laplace Transform on the differential equation a(x)y (x) + b(x)y (x) + c(x) = h(x) to find what ỹ(s) has to be, and then we do the inverse of the Laplace transform to find y(x). Proposition Let y ỹ be the Laplace transform. Then: 1. If y(x) = c (constant), then ỹ(s) = c s 2. ỹ (s) = sỹ(s) y(0) 3. ỹ (s) = s 2 ỹ(s) sy(0) y (0) 14

16 Proof) Lecture Corollary Suppose we are given ay + by + c = 0 (a, b, c constants). Then the solution y satisfies: ỹ(s) = ay(0)s + by(0) + ay (0) as 2 + bs + c Proof) Lecture Proposition We establish some Laplace Transforms: 1. If y(x) = e αx, then ỹ(s) = 1 s α. 2. If y(x) = x n e αx, then ỹ(s) = n! (s α) n If y(x) = e (α+iω)x, then ỹ(s) = 1 s α iω = 4. If y(x) = e αx cos ωx, then ỹ(s) = s α (s α) 2 +ω 2, if y(x) = e αx sin ωx, then ỹ(s) = ω (s α) 2 +ω 2. s α+iω (s α) 2 +ω Green s function for 2nd order ODE 15

17 CHAPTER 5 Hilbert Spaces We now come to introductory functional analysis. In this course, we will exclusively focus on Fourier series, both the classical sine/cosines and the more general Sturm-Liouville theory. In this chapter, we introduce the fundamental object underlying the study of Fourier series: the Hilbert space. We first begin with the brief treatment of metric space topology that will set the language throughout our discussion. We will then specialize this to the space of functions, which we are ultimately interested in. In the end, we arrive at L 2 space, which is a subspace of functions that is the playing field of our Fourier series analysis. 5.1 Metric Space Preliminaries We here give a brief treatment of metric space topology that will be useful and/or necessary for our forthcoming discussion of Hilbert spaces. Definition A metric space is a set X with a distance function, or a metric, d : X X [0, ) such that: 1. d(x, y) = 0 x = y 2. d(x, y) = d(y, x) 3. d(x, z) d(x, y) + d(y, z) Example One easy example is R with the absolute value of the difference as the distance, i.e. d(x, y) = x y. When we say R, we will always assume R to have this metric unless stated otherwise. One can generalize this to R n in many different ways. If x, y R n, then one can define d 1 (x, y) := (x 1 y 1 ) (x n y n ) 2 (this is the usual Euclidean distance), or d 2 (x, y) := max( x 1 y 1,..., x n y n ) (this is usually called the infinity norm ). While 16

18 the two metrics d 1, d 2 on R n are different, we will later see that they are equivalent in that if a sequence {a n } R n converges with respect to d 1 it also converges with respect of d 2, and vice versa. Once we have a metric space, we can sensibly talk about convergence of a sequence, continuity, Cauchy sequences, and completeness. Definition Let X be a metric space with a metric d, and let x 1, x 2,... be a sequence in X. For some x X, we say that the sequence x n converges to x, or lim x n = x, or n x n x as n, if ɛ > 0, N s.t. m N d(x m, x) < ɛ. Exercise 5.1.A (Easy). Show that the sequence { 1 n } n=1,2,... converges to 0 in R with its usual metric. Show that the sequence 1, 1, 1, 1,... does not converge in R. Exercise 5.1.B. Prove the claim that if a sequence {a n } R n converges in metric d 1, then it also converges in metric d 2 of R n, and vice versa. When we say R n, we will assume d 1, or equivalently d 2, to be the metric when dealing with convergent sequences. Definition Let (X, d X ), (Y, d Y ) be metric spaces. A function f : X Y is: continuous if for all x 0 X and ɛ > 0 there exists δ > 0 such that d X (x 0, x) < δ implies d Y (f(x 0 ), f(x)) < ɛ. uniformly continuous if for all ɛ > 0 there exists δ > 0 such that d X (x, x ) < δ implies d Y (f(x), f(x )) < ɛ. One should confirm that with X = Y = R, the above definition is exactly the ɛ-δ continuity definition that one usually sees in an introductory calculus class. Note that the two definitions, while looking similar, are actually different. Uniform continuity implies continuity, but not conversely. Exercise 5.1.C. Show that uniform continuity implies continuity. Show that sin x is uniformly continuous whereas 1/x on x (0, ) is continuous but not uniformly continuous. A great feature of metric spaces is that continuity can be detected by convergent sequences. More precisely, Proposition Let (X, d X ), (Y, d Y ) be metric spaces. Then f : X Y is continuous if and only if for every convergent sequence {x n } X converging to x X, the sequence {f(x n )} Y is a convergent sequence in Y converging to f(x). In other words, f is continuous for any sequence x n x in X, f(x n ) f(x) in Y Exercise 5.1.D. Prove the above proposition. 17

19 To test if a given sequence {x n } converges, one must produce x such that x n x, but producing such x just by looking at {x n } may sometimes be a difficult task. Is there a way to test convergence without explicitly knowing what the sequence converges to? This brings us to the topic of Cauchy sequences and completeness. If we can t really pick out x that x n may converge to, the next best thing would be to require that the sequence x n clusters together. More precisely, Definition Let X be a metric space with a metric d. A sequence {x n } n N is a Cauchy sequence if ɛ > 0 N s.t. m, n N d(x n, x m ) < ɛ. Definition A metric space X (with metric d) is complete if every Cauchy sequence is convergent. Not every metric space is complete. A typical example is Q with the usual absolute value metric d(q 1, q 2 ) = q 1 q 2. In this metric, one can check that the sequence 3, 3.1, 3.14, 3.141, ,... is Cauchy, but is not convergent in Q since if it converges it must converge to π / Q. Remark (Warning). Note that whether a Q is complete or not really depends on the metric it is endowed with. We just saw that Q with the usual metric is not closed. Now, consider metric d on Q defined as d (x, y) = 1 if x y and d (x, x) = 0. One can check that with respect to d, Q is in fact complete because the Cauchy sequences are exactly the sequences that are eventually constant. Proposition Here are some facts about Cauchy sequences: 1. A convergent sequence is a Cauchy sequence. 2. A Cauchy sequence is bounded. 3. If a Cauchy sequence {x n } admits a convergent subsequence {x nk } that converges to x, then {x n } converges to x. Exercise 5.1.E. Prove the above proposition. Theorem R is complete. Proof. Let {x n } be a Cauchy sequence in R. Then {x n } [ R, R] for some R > 0 by 2. in the previous proposition. Thus, {x n } is a sequence in a compact set, and hence has a subsequence that converges. By 3. of the previous proposition, we have that {x n } is convergent. Corollary R n is complete, and C is complete. 18

20 Proof. Endow R n with the d 2 metric and apply Theorem to each of the coordinates. C = {a + bi : a, b R} is complete because it has the same metric as R 2 = {(a, b) a, b R}. In the case when X is a vector space with a norm, we can define a particularly nice metric. A norm on a vector V is defined as follows: Definition A norm on a vector space V over R or C is a function : V [0, ) such that for all v, w V, λ R or C 1. v = 0 v = 0 2. λv = λ v 3. v + w v + w A vector space V with a norm is called a normed vector space, written (V, ). Proposition Show that d(v, w) := v w is a metric on V. In other words, a norm on V naturally induces a metric on on V. Exercise 5.1.F. Prove the above proposition. Example The absolute value on R is indeed a norm that induces the usual metric on R. The infinity norm on R n defined as x := max( x i ) is also a norm on R n that induces the metric d 2 we saw. The usual norm on R n is the Euclidean norm defined as x := x x2 n, which induces the usual Euclidean metric d 1 that we have seen. Definition A normed vector space V is a Banach Space if it is complete (with respect to the metric induced by the norm). The particularly nice feature of the metric induced by a norm on a vector space is that we have an equivalent criterion for completeness that is often easier to work with: Proposition A normed vector space V is Banach (complete) if and only if absolute summability implies summability. In other words, a normed vector space V is complete if and only if for any series v n, if v n converges in R then v n converges in (V, ). n=1 Proof. Hint: any sequence w 1, w 2,... can be transformed into a sequence of partial sums v 1, v 1 + v 2,... where v 1 = w 1 and v n = w n w n 1, and vice versa. We now come to the nicest normed vector spaces, which are normed vector spaces whose norms come from an inner product. We will see what we mean precisely by nicest. 19

21 Definition An inner product, on a vector space V (over R or C) is a map V V R or C such that: 1. linear in the first factor: av + v, w = a v, w + v, w 2. sesquilinear in the second factor: v, bw + w = b v, w + v, w 3. Hermitian symmetric: f, g = g, f 4. positive definite: f, f 0 and f, f = 0 f = 0 Exercise 5.1.G. Show defining v := v, v gives a norm on V ; in particular, v+w v + w. In other words, an inner product on a vector space naturally gives it a norm v := v, v. Proof. Suffices to v + w 2 ( v + w ) 2 since both sides are non-negative. Well, v + w 2 = v + w, v + w = v 2 + w Re( v, w ) v 2 + w v, w. Proposition (Cauchy-Schwartz). Let V be a vector space with inner product,. Then, v, w v w Proof. Lecture (Proof 2.1). Exercise 5.1.H. In this exercise, we clarify why a norm from an inner product is a very nice norm. Let V be a vector space with an inner product,. Let v V and let V 0 V be a 1-dimensional subspace of V. What is the vector w V with the smallest w such that v 0 + w = v for some v 0 V 0? If we did not have an inner product,, but just a norm on V, is it possible to find such w? (Is it even well-defined?) Definition A Hilbert space H is a Banach space whose norm is given by an inner product. In other words, a Hilbert space is a vector space with inner product that is complete with respect to the norm that the inner product induces. Example R n with the usual inner product is an Hilbert space. Example l 2 space (see lecture) 5.2 On convergence of sequence of functions The spaces we are interested in are the vector spaces of functions, such as C (R) or P n. In order to make these spaces into metric spaces, we need to define a metric d on the spaces. There are many ways one can attempt to measure the distance between two functions f and g. One way is to look at the maximum difference sup f(x) g(x) between f and g. In this subsection, we discuss the pointwise convergence of sequence of functions, and its relationship with the metric we have just discussed. 20 x R

22 Definition Let f 1, f 2,... be sequences of functions f i : X R. Then we say that the sequence {f i } pointwise converges to a function f, if for every x X, the sequence {f i (x)} converges to f(x). Example Let sequences of functions f k : R R be defined as f k (x) := 1 k x. Then f k 0 pointwise. Definition Let E R be compact. Then d(f, g) := sup f g x E is a metric on C(E), the space of continuous functions on E, or on B(E), the space of bounded functions on E. Definition A sequence of functions f i : X R converges uniformly to a function f if for every ɛ > 0, there exists N N such that n > N implies f n (x) f(x) < ɛ for all x X. Exercise 5.2.I. Let B(X) be the space of bounded functions on X with the metric given as d(f, g) = sup x X f g. Show that convergence of sequence of elements {f k } B(X) in the metric d is the same as uniform convergence. 5.3 Interlude: Lebesgue Integral Basics Measure motivation Measure definition Lebesgue Integral loose definition MCT DCT 5.4 L 2 space 5.5 An aside: on convergence of sequence of functions 21

23 CHAPTER 6 Sturm-Liouville Theory We study what are called the Sturm-Liouville problems in this chapter as an application of the Hilbert space machinery that we have built in the previous chapter. We begin with the most classical problem l(y) = y, which leads to the classical Fourier series of sines and cosines. Then we present the general Sturm-Liouville problem and some fundamental ideas around it. We finish with two particularly important Sturm-Liouville problems, which are Legrendre s operator and Hermite s operator. 6.1 Classical Fourier Series We start with the linear differential operator defined as: Eigenvalue problem l(y) = y 22

24 Bibliography [RF10] H. Royden, P. Fitzpatrick. Real Analysis (4th ed.). [MC14] C. McMullen. Math 114 Course Notes. [SS] Stein, Shakarchi. Functional Analysis [Hol90] S. Holland. Applied Analysis by the Hilbert Space Method. 6.2 Isoperimetric inequality In this exercise, we prove the following (famous) isoperimetric inequality: Theorem (Isoperimetric inequality). For C a closed curve in R 2, if A is the area and L is the length of the perimeter, then 4πA L 2 In particular, the area is maximized when C is a circle. WLOG let L = 2π, and identifying R 2 C, let γ : [ π, π] C be an arc-length parametrization of the curve C (in other words, γ = 1 constant). We are now in a complex L 2 ( π, π) space, where f, g = π π f(t)g(t)dt, and any periodic function f : R C with period 2π can be expanded as a series f = n Z c ne int using the usual Fourier coefficient: c n = f,eint. Moreover, since C is a closed curve, γ is periodic. e int,e int (a) Using Green s theorem, show that the area A enclosed by C can be computed as A = 1 2i π π γ (t)γ(t)dt (b) Using the fact that γ = n Z c n e int, and doing term by term differentiation, show that the above integral equals n Z πnc 2 n (c) By noting that nc 2 n n 2 c 2 n for all n Z, conclude the isoperimetric inequality. 23