MAS221 Analysis, Semester 1,

MAS221 Analysis, Semester 1, 2018-19 Sarah Whitehouse Contents About these notes 2 1 Numbers, inequalities, bounds and completeness 2 1.1 What is analysis?.......................... 2 1.2 Irrational numbers......................... 3 1.3 Inequalities............................. 5 1.4 Bounds and completeness..................... 6 1.5 The existence of 2......................... 10 2 Sequences 10 2.1 Convergence of sequences...................... 10 2.2 The algebra of limits........................ 13 2.3 Subsequences............................ 15 2.4 Cauchy sequences.......................... 16 3 Limits of functions 17 3.1 Functions on the real line..................... 17 3.2 The limit of a function....................... 19 3.3 Divergence.............................. 21 4 Continuous Functions 24 4.1 Definition of continuity, basic properties and examples..... 24 4.2 Discontinuity............................ 26 4.3 Continuity on Intervals....................... 27 4.3.1 The intermediate value theorem.............. 28 4.3.2 The boundedness theorem................. 28 4.3.3 Inverses........................... 29 1

About these notes These are outline notes for the whole of semester 1. This version does not include the proofs or solutions to examples. This material will be presented on blackboards in the lectures. I plan to make a full version of the notes covering this material available towards the end of the semester. Note, however, that I expect to draw lots of graphs and diagrams on the blackboards in the lectures and many of these will not be the notes. You are strongly encouraged to attend lectures and take your own notes of the blackboard material, especially the graphs and diagrams. 1 Numbers, inequalities, bounds and completeness 1.1 What is analysis? Broadly speaking, analysis is the theory of the limit. It serves as a foundation for the calculus, and also enables us to give a mathematically rigorous description of the real number line. Like algebra, geometry and topology, it is one of the central themes of pure mathematics and is still being actively developed by research mathematicians. It is also a vital tool in many applications, for example in quantum theory, probability theory, finance and economics,... In this course, we will study real analysis, so our focus is on real numbers and real-valued functions. In later years, you may choose to study complex analysis (MAS332), functional analysis (MAS436) where analysis meets linear algebra, or other parts of analysis that build on the work we ll do here, such as metric spaces (MAS331), and measure theory (MAS350/451). The modern theory of the calculus, which is a mathematical theory of change and motion, was to a large extent, discovered independently by two great scientists: Isaac Newton (1642 1727) and Gottfried Leibniz (1646 1716), but despite its amazing success in applications, it was not a logically coherent theory at the time that it first burst on the scene, more a collection of useful mathematical methods. You all know the definition of the derivative at a point x of a sufficiently nice function f: df dx = lim x 0 f(x + x) f(x) ; x but how we know that df/dx exists? What does lim x 0 really mean? What 2

is the infinitesimal quantity x, and how can it tend to zero? In the eighteenth century, the calculus was fiercely criticised by the British philosopher George Berkeley, who called infinitesimals like x: the ghost of departed quantities. Many mathematicians rose to Berkeley s challenge and tried to give a satisfactory foundation to the calculus, but it wasn t until the nineteenth century that modern analysis was born, and a logically rigorous meaning was given to the concept of the limit. This was mainly the work of three great mathematicians: Augustus Louis Cauchy (1789 1857), Bernhard Bolzano (1781 1848), and Karl Weierstrass (1815 97). One of the key aims of this course is to understand the definition of the limit, discover its properties, learn how to use these, and also give a more thorough foundation to the real number line, and to the calculus. In Semester 1, we will study sets of real numbers, sequences and their limits, limits of functions and what it means for a function to be continuous. Along the way, we will prove a lot of new theorems, many of which are also very useful in applications. In Semester 2, we will study differentiation as a limit, the convergence of infinite series, integration, and we will extend our work on sequences and series of numbers to sequences and series of functions. 1.2 Irrational numbers We use the following notation for sets of numbers: N = {1, 2, 3, 4,...}, N 0 = N {0}, Z = {..., 3, 2, 1, 0, 1, 2, 3,...}, Q = {p/q; p Z, q N}, R the natural numbers, the non-negative integers, the integers, the rational numbers, the real numbers. Our intuitive understanding of R is the set of all points on an infinite straight line of zero width. We write R = Q Q c, where Q c is the set of all irrational numbers, i.e. real numbers that cannot be written as ratios of whole numbers. You are aware that 2, e, π are irrational numbers. In MAS114, you showed that the sum of a rational and an irrational number is irrational. It isn t hard 3

to prove that the product of a non-zero rational and an irrational number is irrational. To increase our supply of irrational numbers, we will prove that p is irrational for every prime number p, so 2, 3, 5, 7, etc are irrational. Since there are infinitely many prime numbers, this immediately tells us that there are infinitely many irrational numbers. We will need the prime factorisation theorem which was proved in MAS114. This states that every natural number is uniquely expressible as a product of primes. So if n N, we may write n = 2 m 1 3 m 2 5 m 3 7 m4 p m N, where p is the largest prime that we need, and m 1, m 2,..., m N N 0, e.g. 7920 = 2 4.3 2.5.11, so in this case, p = 11, m 1 = 4, m 2 = 2, m 3 = 1, m 4 = 0, m 5 = 1. The next result presupposes that square roots exist, and we will revisit this later. Theorem 1.2.1 If p is a prime number then p is irrational. The result just proved has a very interesting consequence. Theorem 1.2.2 Given any two rational numbers a and b, with a < b, we can find infinitely many irrational numbers q such that a < q < b. The next result goes a little further than Theorem 1.2.1. Recall that a square number (sometimes called a perfect square) is a natural number of the form n 2, for a natural number n. Theorem 1.2.3 If N is a natural number then either it is a square number or N is irrational. From this, we see that 2, 3, 5, 6, 7, 8, 10, 11, 12,... are irrational. In MAS114, it was shown that Q is countable, (i.e. it can be put into one to one correspondence with N), but that R is uncountable. 1 Since the union of two countable sets is countable, it follows that the set Q c of irrationals is also uncountable. 1918). 1 The proof of this uses the important diagonal argument, due to Georg Cantor (1845 4

There are many examples of real numbers for which it is still an open question as to whether they are rational, or irrational, for example, n 1 e + n 2 π, for n 1, n 2 Z \ {0}. 1.3 Inequalities Inequalities are vital tools in analysis. You ve already met some important examples of these in MAS111. Many of these involve the modulus a of a real number a, which measures its size, or absolute value: { a if a 0, a = a if a < 0. The following alternative description can be quite useful: a = max{a, a}. We ll often use the (easily verified) fact that ab = a. b for all a, b R. One of the most important inequalities you ll meet is the triangle inequality: a + b a + b, for all a, b R. If a, b R, the number a b is important, as it measures the distance between a and b, e.g. 7 1 = 8, 7 ( 1) = 6. But beware that a b a b is NOT true (e.g. try a = 1, b = 2). The following result is sometimes called the corollary to the triangle inequality : Theorem 1.3.1 For all a, b R, a b a b. Another very famous inequality, which was proved in MAS111, is the theorem of the means: if a 1, a 2,..., a n 0, then n a1 a 2 a n a 1 + a 2 + + a n. n In the problems, you can establish two other very useful inequalities: Bernoulli s inequality: for all n N, x > 1. (1 + x) n 1 + nx, 5

Cauchy s inequality: if a 1, a 2,..., a n and b 1, b 2,..., b n are real numbers then ( n n ) 1 ( ) 1 2 2 n a i b i a 2 i b 2 i. i=1 i=1 1.4 Bounds and completeness Let A R. We will use the convenient notation A = { a a A}, so, for example, if A = { 3, 2, 5}, then A = {3, 2, 5} = { 5, 2, 3}. Let A R be finite and non-empty. We use max(a) and min(a) to denote the largest (smallest) numbers in A, respectively. For finite (non-empty) sets, it is clear that such minimal and maximal elements exist. You might want to try to prove that min(a) = max( A). We ll establish a more general result later on in this section. We are also interested in infinite sets of real numbers. Let a, b R, a < b. Important roles in analysis are played by the closed intervals: and the open intervals: We also use the following notation: Half-open intervals such as i=1 [a, b] = {x R a x b}, (a, b) = {x R a < x < b}. (, a) = {x R x < a}, (a, ) = {x R x > a}. (a, b] = {x R a < x b}, [a, b) = {x R a x < b} are also useful. Observe that [a, b] has a maximum element b, and a minimum element a. However, infinite subsets of R need not have minimum or maximum elements. 6

Proposition 1.4.1 The open interval (a, b) has no maximum or minimum element. We are now going to develop generalisations of the idea of maximum and minimum elements that are extremely useful tools in analysis. First we need to think about bounds. Definition 1.4.2 Let A R. (1) We say that A is bounded above if there exists M R such that x M for all x A. (2) We say that A is bounded below if there exists L R such that x L for all x A. (3) We say that A is bounded if it is both bounded above and below. If A is bounded above, a number M as in part 1) is called an upper bound for A, and if A is bounded below, then a number L as in part 2) is called a lower bound for A. For example, (, 1) is bounded above, but not below; ( 47, π) is bounded. Notice (see Problem?) that A being bounded above and below is equivalent to saying that there exists K R such that x K for all x A. For a, b R with a < b, the set (a, b) has many upper bounds, b + 1/2, b + 1, b + 2, b + 3 etc. but it is clear that b is special as it is the smallest. If A is an arbitrary non-empty subset of R that is bounded above, is it obvious that a least (i.e. smallest) upper bound for A exists? In fact this is true and a very important property of the real numbers. First let s make a precise definition. Definition 1.4.3 If A is non-empty and bounded above, the least upper bound of A is a real number α such that α is an upper bound for the set A, i.e. a α for all a A and if β is any upper bound for A then β α. The least upper bound of A is usually called the supremum, and written sup(a). The alternative notation lub(a) is sometimes used. 7

Completeness property for R. Every non-empty set of real numbers that is bounded above has a least upper bound. We will assume the completeness property. On the course website you can find a proof that the completeness property follows from the construction of the real numbers as limits of Cauchy sequences of rational numbers. 2 It follows from the definition that the supremum of A is unique, which is why we wrote the least upper bound, not a least upper bound. Important examples are sup((a, b)) = b and if the set A has a maximum element, you can show (exercise) that sup(a) = max(a). The following property of the supremum can be rather useful. Here the Greek letter ε (epsilon) is used to denote a real number that can be chosen to be as small as you like ; this will become a familiar pattern when we study limits. Proposition 1.4.4 If A R is non-empty and bounded above, then given any ε > 0, there exists a A such that a > sup(a) ε. Now suppose that A is non-empty and bounded below. In the next theorem, we will show that it has a greatest lower bound. Again let s have a precise definition first. Definition 1.4.5 If A is non-empty and bounded below, the greatest lower bound of A is a real number α such that α is a lower bound for the set A, i.e. α a for all a A and if β is any lower bound for A then β α. The least upper bound of A is usually called the infimum, and written inf(a). The alternative notation glb(a) is sometimes used. Theorem 1.4.6 A set A R is bounded below if and only if A is bounded above. Furthermore if A is bounded below, then it has a greatest lower bound and inf(a) = sup( A). 2 Characterising completeness of the real number line ; I suggest reading this after Chapter 2. 8

An important example is inf((a, b)) = a. The completeness property does not hold in Q. For example, consider the set A = {q Q q 2 < 2}. It is non-empty, and bounded above by, for example, 3/2, which is rational. When regarded as a subset of the real numbers its supremum is 2, which is irrational. It has no supremum in the set Q. We finish this chapter by deducing some useful consequences of the completeness property of the real numbers. The next result is often called the Archimedean property of the real numbers, in honour of Archimedes of Syracuse (287 BCE 212 BCE). Theorem 1.4.7 (The Archimedean Property) Let x and y be positive real numbers. Then there exists n N such that nx > y. For applications of Theorem 1.4.7 to convergence of sequences, we often take x = 1 and y to be a large number. In fact, suppose that ε > 0 is a very small real number. Then 1/ε is very large, and Theorem 1.4.7 tells us that there exists n N so that n > 1/ε. But then 1/n < ε. This last fact is very useful. Some of the founders of calculus believed in the existence of infinitesimals positive numbers dx that can be taken to be arbitrarily small. What does this mean? If such a dx exists, then 0 < dx < q for any positive rational number q = m/n, where m, n N. But then ndx < m for all m, n N, and this contradicts the Archimedean property of R, so we have shown that infinitesimals don t exist on the real number line. Instead of asserting the existence of an aritrarily small number, the correct way to deal with this is to write arguments that work for all positive real numbers ε, no matter how small, as we will see soon. We ll finish this section with a delightful and intriguing property of the real numbers. This is sometimes referred to as the density of the rational numbers in the real numbers. Theorem 1.4.8 Given any two real numbers x and y with x < y there exists a rational number q such that x < q < y. 9

If we fix x R, and take y = x+ε, where ε > 0 is very small, then Theorem 1.4.8 tells us that there exists q Q so that x q < ε, i.e. we can approximate an arbitrary real number as closely as we like by a rational number. If x and y are both irrational numbers then Theorem 1.4.8 tells us that there exists a rational number q such that x < q < y, i.e. there is a rational number between every pair of irrationals. On the other hand, in Theorem 1.2.2, we showed that there are infinitely many irrational numbers between every pair of rationals. This gives us some insight into the complicated structure of the real number line. 1.5 The existence of 2 We happily use square roots. But how do we know they really exist? In this section, we ll prove that 2 exists in R as a consequence of the completeness property. To do this we ll need the following inequality: for a, b > 0 with a b: ( ) 2 a + b > ab. (1.5.1) 2 This is easily deduced by adding 4ab to both sides of (a b) 2 > 0. Theorem 1.5.1 There exists α R, with α > 0, such that α 2 = 2. 2 Sequences 2.1 Convergence of sequences A sequence of real numbers is a function f : N R. We will often write f(1) = a 1, f(2) = a 2, and in general f(n) = a n. We usually identify the sequence with the range of f, which is written as (a 1, a 2, a 3,... ) or (a n, n N), or (a n ) n 1 or just simply (a n ). It is very important to distinguish the sequence (a n ) from its nth term a n. Sequences are important for a number of reasons. They are the simplest context in which we can study the notion of convergence, and they will play a vital role later when we study limits of functions, continuity, and infinite series. 10

You ve already met the definition of convergence of a sequence in MAS114, and investigated some of its consequences. We ll revise these ideas, and then go further. Consider the sequence (1/n) = (1, 1/2, 1/3, 1/4, 1/5,...). It is clear that as n gets larger and larger, the terms get closer and closer to zero, but no value of n will give 1/n = 0. Of course, zero is the limit, but how do we make this precise? More generally, suppose that we have a sequence (a n ), and we have good reason to believe that it converges and its limit is the real number l. How do we prove that l really is the limit? We need to understand more deeply what a limit really is. Although the terms of the sequence may never reach l, we can require the distance between the terms and l to become arbitrarily small. Recall that for ε > 0, a n l < ε if and only if l ε < a n < l + ε. We want to say that as n gets very large, a n l gets very small. How small? As tiny a positive number as we like! The smaller we require this distance to be, the further along the sequence we expect to have to go. Given a positive real number ε, we want there to be a stage of the sequence beyond which all terms satisfy a n l < ε. How small do we take ε? How large do we take N? Suppose that I want to convince you that (a n ) converges to l and you are sceptical. You might say what if I take ε = 0.001, or 0.00001, or 3.18 10 700? To convince you, I have to show you that no matter how small an ε you give me, I can always find a stage N so that for all n > N, a n l < ε. This leads to the following key definition. Definition 2.1.1 A sequence (a n ) is said to converge to a limit l R if given any ε > 0, there exists N N so that for all n > N, we have a n l < ε. This definition is central to this module. You need to learn it, and understand it, and be able to work with it, both in examples and in proofs. Notation. If (a n ) converges to l, we write lim n a n = l, or a n l as n, or a n n l. 11

Example 2.1.2 Show lim n 1/n = 0. Note that we needed the completeness property of R to prove Theorem 1.4.4 which we have just used. Also, although we ve shown that (1/n) converges to zero, how do we know that it is the only limit? We ll prove that limits are unique as our next theorem, but first let s try another example. Example ( 2.1.3 Guess the limit as n of the sequence whose nth term is 3 1 ) 7 4 2 n, and then carefully prove that your guess is correct. 3 Theorem 2.1.4 If a sequence converges to a limit, then that limit is unique. More precisely if (a n ) is a sequence such that lim n a n = l, and also lim n a n = l then l = l. Example 2.1.5 Prove that lim n r n = 0, whenever 0 r < 1. Definition 2.1.6 A sequence which does not converge is said to diverge. Divergent sequences may display different types of behaviour. For example, a sequence (a n ) is said to diverge to (respectively, diverge to ), if given any K > 0, there exists N N so that for all n > N, a n > K (respectively, a n < K). In this case, we may write lim a n =, (respectively, lim a n = ). n n A divergent sequence may also oscillate between different values, for example, the sequence (( 1) n ) takes only two values +1 and 1. Definition 2.1.7 A sequence (a n ) is said to be bounded above (respectively, bounded below, bounded) if the set {a n n N} is bounded above (respectively, bounded below, bounded). For example (1/n) and (( 1) n ) are both bounded. Theorem 2.1.8 If a sequence (a n ) is convergent, then it is bounded. The converse to Theorem 2.1.8 is not true; for a counter example, consider the sequence whose nth term is ( 1) n ; it is bounded, but not convergent. 12

2.2 The algebra of limits Consider the sequence whose nth term is a n = 2n+3n2. If you do some numerical n 2 +4 experiments, you might conjecture that the limit is 3. How would we prove this? Divide every term in the numerator and denominator of the fraction by the highest power of n that occurs. This is n 2 and we get: a n = 2 + 3 n 1 + 4. n 2 Now we know that lim n 1 n = 0, lim n 3 = 3, lim n 1 = 1 and lim n 2 = 2. We could indeed argue that lim n a n = 3 if we could justify writing lim a n = n 1 2 lim n + 3 n 1 1 + 4 lim n lim n n 1. n It turns out that this sort of reasoning is indeed justified, and the general result that we need is given in the next theorem which is often known as the algebra of limits. Theorem 2.2.1 (The algebra of limits) Suppose that (a n ) and (b n ) are convergent sequences with lim n a n = l and lim n b n = m. Then: (1) The sequence whose nth term is a n + b n converges to l + m. (2) The sequence whose nth term is a n b n converges to lm. (3) If c is any real number then the sequence whose nth term is ca n converges to cl. (4) If b n 0 for all n and also m 0 then the sequence whose nth term is a n b n converges to l m. Notice that we could write the statement of part 1) of the theorem as lim a n + b n = lim a n + lim b n, n n n with the slogan the limit of a sum is the sum of the limits, as long as we are careful to remember that all of the limits have to exist for this to make sense. Similar remarks apply to the other parts. Now that we ve proved Theorem 2.2.1 you should go back to the sequence we considered at the beginning of the section, and convince yourself that every step can be justified to prove that the limit is 3. 13

2n 7n 3 Example 2.2.2 Find lim n 6n 2 + 11n. 3 Corollary 2.2.3 (1) If the sequence (a n ) is monotonic increasing, then either it converges or it diverges to +. (2) If the sequence (a n ) is monotonic decreasing, then either it converges or it diverges to. Proof. We only prove (2) as (1) is so similar. Suppose that (a n ) is monotonic decreasing. Then either it is bounded below or it isn t. If it is bounded below then it converges by Theorem?? (2). If it isn t bounded below then given any K < 0 we can find N N such that a N < K, for otherwise K would be a lower bound. But then since the sequence is monotonic decreasing we have a n < K for all n N and so the sequence diverges to. Example 2.2.4 (Exam style question) Consider the sequence (a n ) given by a 1 = 0, a n+1 = 3a n + 1 a n + 3 (a) Use induction to show that 0 a n 1 for all n N. (b) Show that (a n ) is monotonic increasing. (c) Explain why lim n a n exists, and find its value. for all n > 1. (2.2.1) Example 2.2.5 (The golden ratio as a limit) Consider the sequence (a n ) given by a 1 = 1, a n+1 = 1 + a n for n 1. (2.2.2) Show that this sequence is bounded above and monotonic increasing. Show that it converges to the golden ratio, 1 + 5. 2 Example 2.2.6 (e as a limit) Show that the sequence (a n ), where ( a n = 1 + n) 1 n, converges. Also show that the limit is between 2 and 3. The limit of this sequence is one of the most natural definitions of the number e, which is, among other things, the base of natural logarithms. 14

In Semester 2, we will relate this definition of e to other expressions. 1 Through our work on series we ll see that e =, and more generally, n! n=0 for all x R, e x x n = n!. n=0 2.3 Subsequences By considering some, but (generally) not all, of the terms in a sequence, we get a subsequence. Definition 2.3.1 A sequence (y n ) is a subsequence of a sequence (x n ) if there is a strictly increasing function σ : N N so that y n = x σ(n) for all n N. Equivalently a subsequence of (x n, n N) is a sequence of the form (x nr, r N), i.e. (x n1, x n2, x n3,...), where n 1 < n 2 < n 3 <. For example, (1/2n), (1/3n), (1/(5n 3)) are all subsequences of (1/n). In fact (1/an + b) is a subsequence of (1/n) for all a N, b Z with a + b > 0. Proposition 2.3.2 If the sequence (a n ) converges to a limit l, then every subsequence of (a n ) also converges to l. It can be more interesting to seek convergent subsequences of divergent sequences. For example, the sequence whose nth term is ( 1) n, has two obvious convergent subsequences, obtained by taking odd and even terms, respectively. Another way in which subsequences can be important, is to provide tools to prove that a sequence converges; sometimes it turns out that the best way to do this is to first find a subsequence converging to l, say, and then show that the whole sequence converges to l. We ll see an example of this in the next section (Theorem 2.4.2), where we will also use the important Bolzano Weierstrass theorem, which we will prove after the next result, which is a key step on the way. Theorem 2.3.3 Every sequence has a monotone subsequence. We will see in the next section how important the next result is. Theorem 2.3.4 [Bolzano-Weierstrass] Every bounded sequence has a convergent subsequence. 15

2.4 Cauchy sequences In this last section of the chapter, we return to one of the themes of Chapter 1: the completeness of the real numbers. In fact, you met Cauchy sequences before, at the end of MAS114 semester 1. Definition 2.4.1 A sequence (a n ) is said to be a Cauchy sequence if given any ε > 0, there exists N N such that for all m, n > N we have a m a n < ε. Notice that the definition says nothing about convergence; but it does tell us that terms of the sequence get arbitrarily close to each other if we move far enough along it. The next two results are left for you to prove for yourself in the exercises (Problems 41 and 42). Every convergent sequence is Cauchy. Every Cauchy sequence is bounded. The next result is very important. It is equivalent to the completeness property of the real numbers. A proof of that fact can be found on the course website. 3 Theorem 2.4.2 If (a n ) is a Cauchy sequence (of real numbers), then it converges to a limit in R. Definition 2.4.3 A non empty subset A R is said to be complete if every Cauchy sequence taking values in A converges to a limit in A. Theorem 2.4.2 tells us that R is complete. But Q is not, for example, consider the sequence (1, 1.4.1.41, 1.414, 1.4142,...) of rational approximations to 2. It is Cauchy (as it converges in the real numbers to 2), and each term of the sequence lies in Q, but the limit does not. One way of constructing the real number line is to take the union of the rational numbers with all limits of Cauchy sequences of rational numbers. Of course this requires quite a lot of work to do carefully in full detail and to check it gives what we want. 3 Look for the file characterising completeness of the real number line. 16

3 Limits of functions 3.1 Functions on the real line In this section, we revise some aspects of functions and look at some examples. This is mostly revision of material from MAS110. We will be interested in functions f : A B, where A and B are some subsets of the real numbers R (perhaps the whole of R). Recall that f being a function from A to B means that for each element a A, there is specified exactly one element f(a) B, the value of the function f on the element a. The domain of f is A and the codomain is B. The range or image of the function f is the set {b B b = f(a) for some a A}; that is, it is the set of values of the function. It is important in this course that a function has a specified domain and codomain. In particular, giving a formula, like 1/x, for example, is not giving a function, although it may be part of the information that specifies a function. We could give a function by, for example, saying f : R \ {0} R, given by f(x) = 1/x. And this is a different function from f : (0, 1) R, also given by f(x) = 1/x. Notice that we have used the difference of sets notation here: recall that if X and Y are arbitrary sets, then X \ Y = X Y c = {x X x / Y }. If we have some formula in x that we would like to use for the value f(x) of a function, it is often useful to think about the largest domain f could have, that is, the set of all x in R for which f(x) makes sense as a real number. This will not include any x R for which we might be tempted to write f(x) = 0/0 or f(x) = ±. Example 3.1.1 A polynomial is a function f : A R of the form f(x) = a 0 + a 1 x + + a n 1 x n 1 + a n x n, where the coefficients a 0, a 1,..., a n 1, a n are in R, and the degree n is a natural number. Here, A can be any subset of R. Example 3.1.2 A rational function is a function f : A R of the form f(x) = p(x)/q(x), where p and q are polynomials. The largest subset of R we can take as its domain A is {x R q(x) 0}. 17

For example, if f(x) = R \ { 1, 3}. x 2 + 5, the largest possible domain of f is (x + 1)(x 3) Other important functions are the functions f : R R, given by f(x) = e kx, f(x) = sin(kx) and f(x) = cos(kx), where k R. In analysis, they are best defined as convergent power series, and that topic will be picked up in Semester 2. Example 3.1.3 The sign function f : R R is defined by 1 if x < 0, sign(x) = 0 if x = 0, 1 if x > 0. The range of this function is { 1, 0, 1}. Example 3.1.4 The indicator function of the closed interval [a, b] is the function 1 [a,b] : R R given by { 1 if x [a, b], 1 [a,b] (x) = 0 if x / [a, b]. Similarly, we can define an indicator function of an open interval, or an arbitrary subset of R. The indicator function of [0, ) is called the Heaviside function by engineers. We can add functions, multiply them by scalars, multiply them together and divide them, but we need to be careful with domains. So for example, the function f + g can be given by (f + g)(x) = f(x) + g(x), whenever both f and g are defined at x. So if f : A R and g : B R, where A and B are subsets of R, then we can define the sum f +g as a function f + g : A B R. Similarly, for α R, we have αf : A R, fg : A B R and f/g : (A B) \ {x R g(x) = 0} R, where (αf)(x) = αf(x), (fg)(x) = f(x)g(x), ( ) f (x) = f(x) g g(x). 18

Remark. Let F(R) denote the set of all functions from R to R. Then we can add and scalar multiply and it turns out that F(R) is a vector space over R (but it is not finite-dimensional). We can also multiply and all this structure gives what is called an algebra over R. These structural properties are important in higher analysis. The composition g f of two functions f and g is defined only when the codomain of f is the same as the domain of g. Note that we are using the usual convention that g f means f first. So, if f : A B and g : B C, then we have the composition g f : A C, given by (g f)(x) = g(f(x)). Of course, the order of composition matters: we typically do not have f g = g f, even at points where both are defined. For example, if f : R R is given by f(x) = x+2 and g : R R is given by g(x) = 3x then g f : R R given by (g f)(x) = 3x+6, whereas f g : R R given by (f g)(x) = 3x+2. 3.2 The limit of a function We want to make sense of the notion of lim x a f(x) for a function f : A R, where A is a subset of R and a is some real number (which may be an element of the domain A, but doesn t have to be). You have already studied this in MAS110 and you should have good intuition for this situation. Here we will make the idea precise, by using sequences, building on the previous chapter. So we consider sequences in the domain approaching a and we study the values of the function on the terms of such sequences. Definition 3.2.1 Let f : A R, where A is a subset of R, and let a R be such that there is some sequence (x n ) satisfying (1) for all n N, x n A and x n a, (2) lim n x n = a. Then we say that f has limit l at the point a, if lim n f(x n ) = l for every sequence (x n ) which satisfies the conditions above. In this case, we write Notes. lim f(x) = l. x a 19

(1) We require some such sequence (x n ) to exist for it to make sense to consider what happens to the values of the function as we approach a. (2) We must have convergence of (f(x n )) to l for every such sequence (x n ). (3) The real number a may or may not be in the domain A. Example 3.2.2 Consider f : R \ { 5, 3, 5} R, given by f(x) = x 5 (x 2 25)(x 3). Investigate lim x a f(x) for each a R. (You will need to think about points in the domain and each of the three special points 5, 3, 5 separately.) Theorem 3.2.3 [Algebra of limits] Suppose that f : A R, g : B R, and a R is such that lim x a f(x) = l and lim x a g(x) = m, then (1) lim x a (f + g)(x) = l + m, (2) lim x a (fg)(x) = lm, (3) lim x a (αf)(x) = αl, for all α R, ( ) f (4) lim x a (x) = l, if m 0. g m From a geometric perspective, the idea of a limit is that as x gets closer and closer to a, so f(x) should get closer and closer to f(a). More insight to this is given by the following theorem, which establishes the important (ɛ δ) criterion for existence of limits. Theorem 3.2.4 Let f : A R, where A is a subset of R, and let a R be such that there is some sequence (x n ) satisfying for all n N, x n A and x n a, with lim n x n = a. Then the following are equivalent: (1) lim x a f(x) = l. (2) the (ɛ δ) criterion: given any ε > 0 there exists δ > 0 such that whenever x A with 0 < x a < δ, then f(x) l < ɛ. 20

Theorem 3.2.5 (Sandwich rule for functions) Suppose that f : A R, g : B R and h : C R and suppose that there exists an interval (a, b) A B C such that for all x (a, b) f(x) g(x) h(x). If there exists c (a, b) such that lim x c f(x) = lim x c h(x) = l, then lim x c g(x) exists and equals l. 3.3 Divergence In the same setting as Definition 3.2.1, we say that a function f : A R diverges at x = a if lim x a f(x) does not exist. There are some interesting types of behaviour involved here. For example, we say that the function f diverges to infinity at x = a and we write lim x a f(x) = if for every sequence (x n ), with x n A, x n a, which satisfies lim n x n = a, we have lim n f(x n ) =, i.e. the sequence (f(x n )) diverges to infinity in the sense given in Chapter 2. The notion of divergence to minus infinity is defined similarly, and in that case we write lim x a f(x) =. For an example of divergence to infinity, consider e.g. lim x 0 1/x 2. The details are left to you. To investigate divergence further, suppose that the following occurs: whenever (x n ) is a sequence in A converging to a, with x n < a for all n N, then there exists l 1 R, so that lim n f(x n ) = l 1. In this case, we call l 1 the left limit of f at a, and we write lim x a f(x) = l 1. Similarly, it can be the case that whenever (x n ) is a sequence in A converging to a, with x n > a for all n N, then there exists l 2 R, so that lim n f(x n ) = l 2. In that case, we call l 2 the right limit of f at a, and we write lim x a f(x) = l 2. 4 We will show below that a function f for which both left and right limits exist at a is divergent there if and only if l 1 l 2. For example, consider Heaviside s indicator function H(x) = 1 [0, ), as defined in Example 3.4. It is not difficult to verify that lim x 0 H(x) = 0 and lim x 0 H(x) = 1. This is an example of a discontinuity; we will have more to say about continuous functions, and discontinuities, in Chapter 4. There is an (ɛ δ) criterion for left and right limits. Theorem 3.3.1 Let f : A R, where A is a subset of R. 4 Alternative notation that you may meet is to write lim x a instead of lim x a, and lim x a+ instead of lim x a. 21

(1) Let a R be such that there is some sequence (x n ) in A satisfying x n < a for all n N and lim n x n = a. Then lim x a f(x) = l 1 if and only if given any ε > 0 there exists δ > 0 such that whenever x A with 0 < a x < δ, then f(x) l 1 < ɛ. (2) Let a R be such that there is some sequence (x n ) in A satisfying a < x n a for all n N and lim n x n = a. Then lim x a f(x) = l 2 if and only if given any ε > 0 there exists δ > 0 such that whenever x A with 0 < x a < δ, then f(x) l 2 < ɛ. The proof is very similar to that of Theorem 3.2.4, and is left to you to do as one of the problems. Note that 0 < a x < δ if and only if a δ < x < a and 0 < x a < δ if and only if a < x < a + δ. Theorem 3.3.2 Let f : A R, where A is a subset of R. Let a R and suppose that (b, a) (a, c) A, for some b, c R. Then f has limit l at a if and only if both left and right limits exist at a and they are both equal to l. The condition (b, a) (a, c) A ensures that there are sequences in A converging to a from below and above, so that both the left and right limits are defined. We also meet functions that diverge in different ways when approached from the left and from the right of a given point. For example, consider f : R \ {0} R given by f(x) = 1/x. Then it is easily verified that lim x 0 f(x) =, and lim x 0 f(x) =. 22

Figure 1: Graph of f(x) = 1/x 23

4 Continuous Functions 4.1 Definition of continuity, basic properties and examples Intuitively we think of functions as being continuous if we can draw their graphs without ever taking our pen/pencil off the paper; so the graph contains no gaps/jumps/discontinuities. So if we think of a typical point a in the domain of the function f, then as x gets closer and closer to a, we expect that f(x) will get closer and closer to f(a). But from the work of chapter 3, recall that as x gets closer and closer to a, then f(x) gets closer and closer to its limit at a (if this exists). This leads to the following definition. Definition 4.1.1 Let f : A R where A is a subset of R and let a A. We say that f is continuous at the point a if lim x a f(x) exists and equals f(a). We say that f is continuous on a set S A if it is continuous at every point of S. Theorem 4.1.2 Let f : A R where A is a subset of R and let a A be such that there is some sequence (x n ) in A \ {a} with lim n x n = a. Then the following statements are equivalent. (1) f is continuous at a. (2) Given any sequence (x n ) with x n A for all n N, such that lim n x n = a, we have lim n f(x n ) = f(a). (3) Given any ɛ > 0, there exists δ > 0 such that whenever x A with x a < δ, we have f(x) f(a) < ɛ. Remark. We can rewrite Theorem 4.1.2(3) as Given any ɛ > 0, there exists δ > 0 such that whenever x A with x (a δ, a + δ), we have f(x) (f(a) ɛ, f(a) + ɛ). An open interval of the form (a δ, a + δ), where δ > 0 is called an open neighbourhood of a R. This reformulation of the notion of continuity in terms of open neighbourhoods is important in advanced analysis and can be viewed as the starting point for topology. Example 4.1.3 Fix c R, then the constant function f : R R given by f(x) = c is continuous on R. To see this, let a R be arbitrary and let (x n ) be any sequence converging to a. Then f(x n ) = c for all n N and 24

so the sequence (f(x n )) clearly converges to c = f(a). Thus the function is continuous at a. Example 4.1.4 The linear function f : R R given by f(x) = x is also continuous on R. Again given any sequence (x n ) converging to a, f(x n ) = x n for all n N, and so the sequence (f(x n )) clearly converges to f(a) = a. To build more interesting examples, we need the following. Theorem 4.1.5 [Algebra of limits revisited] Suppose that f : A R and g : B R are both continuous at a A B. The following functions are also continuous at a. (a) f + g, (b) fg, (c) αf, for all α R, (d) f, provided g(a) 0. g Using algebra of limits and Examples 4.1.3 and 4.1.4, we can expand our supply of continuous functions. Using Theorem 4.1.5(b) repeatedly, we can show that f(x) = x n is continuous on R for all n N. Then using Theorem 4.1.5(a) and (c), it follows that every polynomial p(x) = a 0 + a 1 x + a n x n is continuous on R. Finally using Theorem 4.1.5(d) we conclude that rational functions r(x) = p(x)/q(x) are continuous at every point where q(x) 0. The continuity on R of functions such as f(x) = e kx, g(x) = sin(kx), h(x) = cos(kx), for k R, will be established in semester 2, using power series arguments. We also have a result for composition. Theorem 4.1.6 Let f : A B and g : B R, so we have the composition g f : A R, given by (g f)(x) = g(f(x)). If f is continuous at a and g is continuous at f(a), then g f is continuous at a. 25

With Theorem 4.1.6 and results from semester 2, we can immediately deduce continuity on R of functions such as f : R R given by f(x) = sin ( 2x 1 x 2 +1). Example 4.1.7 Consider the function g : R \ {0} R given by g(x) = x sin(1/x). Using Theorems 4.1.5 and 4.1.6, it is not hard to see that g is continuous at every point of its domain. In Problem 51, we showed that lim x 0 g(x) exists and is 0. Now define a new function g : R R by { g(x) if x R \ {0}, g(x) = 0 if x = 0. Then g is continuous on the whole of R. Definition 4.1.8 Given two functions f 1 : A R and f 2 : B R, we say that f 2 is an extension of f 1, (and that f 1 is a restriction of f 2 ) if A B and f 1 (x) = f 2 (x) for all x A. If f 1 is continuous on A and f 2 is continuous on B, we say that f 2 is a continuous extension of f 1 (to B). So in Example 4.1.7, g is a continuous extension of g. On the other hand, consider the function h : R \ {0} R given by h(x) = sin(1/x). It has many extensions to the whole of R: we could, for example, define h(x) = { h(x) if x 0 0 if x = 0, but since, as was shown in Problem 50, lim x 0 h(x) does not exist, there are no continuous extensions of h to R. 4.2 Discontinuity A function f : A R is said to have a discontinuity at a A if it fails to be continuous there. In this case we say that f is discontinuous at a. For example, the indicator function 1 [a,b] : R R is discontinuous at a, and at b, but is continuous on R \ {a, b}. To show that a function is discontinuous at a, it is sufficient to find a single sequence (x n ) in A \ {a}, such that lim n x n = a, but lim n f(x n ) f(a). We can learn more about what happens at a discontinuity by using right and left limits. Definition 4.2.1 We say that f : A R is left continuous at a A if lim x a f(x) = f(a), and right continuous at a A if lim x a f(x) = f(a). 26

Example 4.2.2 The function 1 [a,b] : R R is left continuous at x = b, and right continuous at x = a. Theorem 4.2.3 A function f : A R is continuous at a A if and only if it is both right and left continuous there. If f is discontinuous at a but both lim x a f(x) and lim x a f(x) exist (i.e. they are real numbers) and are unequal, we say that f has a jump discontinuity at a. In this case the jump at a is defined to be J f (a) = lim x a f(x) lim x a f(x). For example, the function 1 [a,b] has jump discontinuities at a and b, with J f (a) = 1 and J f (b) = 1. Next we consider two fascinating examples of discontinuity, which are a little more complicated. Both of these are due to Pierre Lejeune Dirichlet (1805-59). Example 4.2.4 (Dirichlet s function). Show that the function 1 Q : R R is discontinuous at every point in R. Example 4.2.5 (Dirichlet s other function). Consider the function g : [0, 1) R defined by 1 if x = 0, 1/n if x = m/n Q, g(x) = where the fraction is written in its lowest terms, 0 if x R \ Q. Show that g is continuous at every irrational number. (It is also discontinuous at every rational number - this is left to you to do in Problem 64.) 4.3 Continuity on Intervals In this section we will study functions on closed intervals, f : [a, b] R, which are continuous at every point of [a, b]. Many well known functions have this property, including polynomials, sines, cosines and exponential functions. It turns out, as we will see, that there are some very important and powerful theorems that can be proved in this context. 27

4.3.1 The intermediate value theorem Theorem 4.3.1 (The intermediate value theorem) Let f : [a, b] R be continuous with f(a) > 0 and f(b) < 0, or f(a) < 0 and f(b) > 0. Then there exists c (a, b) such that f(c) = 0. Corollary 4.3.2 Let f : [a, b] R be continuous with f(a) < f(b). Then for each γ (f(a), f(b)), there exists c (a, b) with f(c) = γ. Proof. This is left for you to do as Problem 66. Note that Corollary 4.3.2 tells us that the image (or range) of the interval [a, b] under the continuous function f contains the interval [f(a), f(b)], i.e. [f(a), f(b)] f([a, b]). The next result gives a nice application of analysis to the theory of equations. Corollary 4.3.3 Every polynomial of odd degree p has at least one real root. Of course, there is no analogue of Corollary 4.3.3 when m is even, e.g. p(x) = x 2 + 1 has no real roots. 4.3.2 The boundedness theorem Definition 4.3.4 A function f : A R is bounded on a non-empty set S A if there exists K > 0, such that f(x) K for all x S. In this situation, the set {f(x) x S} is a non empty bounded set of real numbers and so, by completeness, it has a supremum, which we write as sup x S f(x), and an infimum, which we write as inf x S f(x). Definition 4.3.5 We say that f : A R attains its bounds on S A if there exist a, b S such that f(a) = sup f(x), x S f(b) = inf x S f(x). Examples of functions f : R R that are bounded on R are f(x) = sin(x) and f(x) = cos(x). Both functions attain their bounds, e.g. f(x) = sin(x) has supremum 1 and this is attained at all points of the form (4n + 1)π/2, where n Z; similarly the infimum 1 is attained at all points of the form (4n 1)π/2, where n Z. 28

We are interested in the case where S is an interval and f is continuous. The function f(x) = 1/x is continuous on the interval (0, 1), but it is not bounded as it diverges to infinity at 0. The function f(x) = x is clearly bounded on (0, 1), but it does not attain its bounds. When we restrict to closed intervals, we have a very nice result. Theorem 4.3.6 [The boundedness theorem] If f : [a, b] R is continuous, then it is bounded on [a, b] and it attains its bounds there. It s important to be clear what Theorem 4.3.6 is telling us. It says nothing about boundedness on R. Indeed any non-constant polynomial is unbounded on R, but its restriction to every closed interval [a, b] is bounded. Corollary 4.3.7 If f : A R is continuous and non constant on [a, b] A, then there exists m < M so that 4.3.3 Inverses f([a, b]) = [m, M]. Let A and B be arbitrary sets and f : A B be a function. Recall from MAS114 (semester 2) that f is surjective if the image of f is equal to B, i.e. for all y B there exists x A such that f(x) = y, injective if whenever f(x 1 ) = f(x 2 ) for some x 1, x 2 A, then x 1 = x 2. bijective if it is both surjective and injective. invertible if there exists a function f 1 : B A, called the inverse of f, for which f 1 (f(x)) = x, for all x A and f(f 1 (y)) = y, for all y B. In MAS114, you also proved the following. Proposition 4.3.8 The function f : A B is invertible if and only if it is bijective. Now we return to consider functions f : A R where A R. We extend to such functions some ideas that we considered for sequences in chapter 2. 29

Definition 4.3.9 Let f : A R where A R. We say that f is monotonic increasing if whenever x, y A with x < y, we have f(x) f(y), monotonic decreasing if whenever x, y A with x < y, we have f(x) f(y), monotone if it is either monotonic increasing or decreasing, and strictly increasing/decreasing when the or in the above definitions is replaced with < or > (respectively). Here are some simple examples; it is easy to see that f : R R given by f(x) = x is strictly increasing, and that f : R \ {0} R given by f(x) = 1/x is strictly decreasing. Our key theorem considers the invertibility of monotone functions that are continuous on closed intervals. Theorem 4.3.10 [The inverse function theorem] If f : [a, b] R is continuous and strictly increasing (respectively, strictly decreasing), then f is invertible and f 1 is strictly increasing on [f(a), f(b)] (respectively, strictly decreasing on [f(b), f(a)]) and continuous on [f(a), f(b)] (respectively, on [f(b), f(a)]). Example 4.3.11 (nth roots) Fix n N and let f : [0, ) [0, ) be given by f(x) = x n. Then you can prove in Problem 73 that f is strictly monotonic increasing on every [a, b] [0, ). We already know that f is continuous. Considering the restriction f : [a, b] [a n, b n ], we can apply Theorem 4.3.10, to see that we have an inverse f 1 : [a n, b n ] [a, b] which is is continuous and strictly monotonic increasing. Since we can do this for all intervals [a, b], this allows us to obtain the function [0, ) [0, ) that we write as f(x) = x 1/n. So in particular, Theorem 4.3.10 has given us a unified method for proving the existence of positive nth roots of any positive real number. Example 4.3.12 Next semester, we ll prove that f : R (0, ) given by f(x) = e x is monotonic increasing, as well as showing it is continuous. By a similar argument to that of Example 4.3.11, we can deduce that it has a continuous, monotonic increasing inverse f 1 : (0, ) R. Of course, in this case f 1 (x) = ln(x) (or log e (x), if you prefer) and so we are using Theorem 4.3.10 to deduce that every positive real number has a natural logarithm. 30