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Modern Analysis
Prof. Chang-Lara
Dept. of Mathematics Columbia University Notes by Yiqiao Yin in LATEX
February 1, 2017
Abstract This is the lecture notes from Prof. Chang-Lara Modern Analysis, an upper level course offered in the school year of 2016-2017 (two semesters, I & II). Similar to Calculus, Real Analysis studies the behavior of functions. In this course we mainly consider real valued functions defined over a subset of the real line. The emphasis will be to establish rigorous proofs for some of the theorems and properties you already know from your Calculus class: Sequences and series, continuity, differentiability, Riemann integral, sequences and series of functions, uniform continuity, Arzela- Ascoli Theorem and Wierstrass approximation Theorem.
This note is dedicated to Hector Chang-Lara.
Contents
- 1 §The Real and Complex Number Systems§
- 1.1 Introduction
- 1.2 The Real Numbers: Sets, order and field axioms
- 1.3 The Real Numbers: Ordered Fields and Completeness
- 2 §Basic Topology§
- 2.1 Metric Spaces and Point Set Topology
- 2.2 Compact Sets
- 3 §Numerical Sequences and Series§
- 3.1 Sequences
- 3.2 Series
- 3.3 Rearrange
- 4 §Continuity§
- 4.1 Continuity
- 4.2 Weierstrass Theorem (Extreme Value Theorem)
- 4.3 Intermediate Value Theorem
- 4.4 Uniform Continuity
- 4.5 Lipschitz Function
- 4.6 H¨older Function
- 5 §Differentiation§
- 5.1 Derivative
- 5.2 Mean Value Theorem
- 5.3 L’Hˆopital’s Rule
- 5.4 Taylor’s Theorem
- 6 §Riemann-Stieltjes Integral§
- 6.1 Riemann Integral
- 6.2 Properties of the Integral
- 6.3 Fundamental Theorems of Calculus
- 7 §Sequences and Series of Functions§
- 7.1 Uniform Convergence
- 7.2 Uniform Convergence and Continuity
- 7.3 Uniform Convergence and Integration
- 7.4 Uniform Convergence and Differentiation
- 7.5 Stoke’s Theorem
- 7.6 The Arzela-Ascoli Theorem
- 7.7 Wierstrass Approximation Theorem
- 7.8 Convolution
- 7.9 The Gamma Function
- 8 §Some Special Functions§
- 8.1 Power Series
- 8.2 The Exponential and Logarithmic Functions
- 9 §Functions of Several Variables§
- 10 §Integration of Differential Forms§
Definition 1.3. (S, <) is an ordered set if
(1) Trichotomy: a < b, b > a, or a = b (2) Transitivity: ∀a, b, c, ∈ S, we have a < b and b < c ⇒ a < c. Examples: (N, <), (Z, <), and (Q, <). A field is a set where two (binary) operations are defined, namely addition and multiplication. We have the following axioms to define a field.
Definition 1.4. A field, (S, +, ×), is a set where two binary operations are defined. A field, S, satisfies the following axioms [10].
Axioms of summation: (S, +) (1) Closed: ∀a, b ∈ S, we have a + b ∈ S (2) Commutative: ∀a, b ∈ S, a + b = b + a (3) Associative: ∃a, b, c ∈ S, (a + b) + c = a + (b + c) (4) Neutral: ∃ 0 ∈ S, s.t. ∀a ∈ S, we have a + 0 = a (5) Opposite: ∀a ∈ S, ∃ − a ∈ S, s.t. a + (−a) = 0 Axioms of product: (S, ×) (1) Closed: ∀a, b ∈ S, we have a × b ∈ S (2) Commutative: ∀a, b ∈ S, a × b = b × a or ab = ba (3) Associative: ∃a, b, c ∈ S, (a × b) × c = a × (b × c) or (ab)c = a(bc) (4) Unit: ∃ 1 6 = 0 ∈ S, s.t. ∀a ∈ S, we have a × 1 = a (5) Inverse: ∀a ∈ S|{ 0 }, ∃ (^1) a ∈ S, s.t. a × ( (^1) a ) = 1
We have Distributivity stating that ∀a, b, c ∈ S, we have a × (b + c) = a × b + a × c. Some famous examples: (Q, +, ×), ({ 0 , 1 }, +, ×).
Lemma 1.5. 0 × a = 0
Proof: First, we have 0 × a = 0 × (0 + a) by neutral quality from summation. Then we have 0 × a = 0 × (0 + a) = 0 × 0 + 0 × a by distributivity. Given 0 × (0 + a) = 0 × 0 + 0 × a, we use opposite to this equation and obtain 0 × a + (− 0 × a) = 0 × 0 + 0 × a + (− 0 × a). We get 0 = 0 × 0.
Q.E.D. Some examples of field are Q and Z 2 = { 0 , 1 }. For the set of Z 2 = { 0 , 1 }, we have addition and multiplication defined to be the following:
0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1, 1 + 1 = 0
0 × 0 = 0, 0 × 1 = 0, 1 × 0 = 0, 1 × 1 = 1
We have the field of ({ 0 , 1 }, +, ×):
× 0 1
We looked at all the possibilities that we have to define the opera- tions in Z 2 and we conclude the operations listed above. One interesting deduction is that 0 × 0 = 0. Then we can move on to discuss some of the properties we see. First, we have cancellation, i.e. x + y = x + z ⇒ y = z. We also know for x 6 = 0, we have xy = xy ⇔ y = z. We can prove this easily by subtracting x on both sides. Secondly, we have uniqueness of inverses. That is, we have the follow- ing identity x + y = 0 ⇔ y = −x. For x 6 = 0, we have xy = 1 ⇔ y = (^1) x. Thirdly, We have inverse of inverse. That is, for x 6 = 0, we have (^11) x
= x. We also have opposite of opposite, −(−x) = x. Moreover, we
have multiplication by zero, ∅ × a = 0. We can also set product to zero. That is, ∀x, y ∈ R, xy = ∅ ⇒ x = 0 ∨ y = 0. We can also have different signs. We can have (−x)y = −xy = x(−y), and (−x)(−y) = xy. After the understanding of definitions and basic properties, we move on to discuss some more advanced properties [10].
Proposition 1.6. The axioms for addition imply
(a) If x + y = x + z, then y = z. (b) If x + y = x, then y = 0. (c) If x + y = 0, then y = −x. (d) If −(−x) = x.
Proposition 1.7. The axioms for multiplication imply
(a) If x 6 = 0 and xy = xz, then y = x. (b) If x 6 = 0 and xy = x, then y = 1. (c) If x 6 = 0 and xy = 1, then y = (^) x^1. (d) If x 6 = 0, then 1 /( (^1) x ) = x.
Proposition 1.8. The field axioms imply
(a) 0 x = 0. (b) If x 6 = 0 and y 6 = 0, then xy 6 = 0. (c) If (−x)y = −(xy) = x(−y). (d) If (−x)(−y) = xy.
Remark 1.9. We can define n ∈ F , while F is a field, and n = 1+1+...+1. Is it possible for 1 + 1 = 0? The answer is yes and we can observe such result in Z 2 = { 0 , 1 }.
We can define F to be a field of characteristic zero if 1+1+... = 1 6 = 0, ∀n ∈ N. The consequence is the following. If F has a characteristics zero, then we have N ⊆ F, Z ⊆ F, Q ∈ F. This is where Newton’s Binomial Identify comes in. We have (a + b)n^ and we want to see how to describe it. The idea is to write the term into a product of a series of terms, i.e. (a + b)n^ = (a + b)(a + b)...(a + b). For each of the n factor, for m terms, by either picking a or b. The next question we are interested is the amount of terms we have. The answer is 2N^ , since this is a combination of a
left with the case of 0 and the case of k + 1, = ak+1^ +
∑k 1
(k+ r
ar^ bk−r+1^ + bk+1, since
( (^) k r− 1
)(k r
(k+ r
∑k+ r=
(k+ r
ar^ bk+1−r^ , by taking the sum adding the case k + 1 back to the range of the summation. Thus we have
(a + b)n^ =
k∑+
r=
k + 1 r
ar^ bk+1−r
Q.E.D.
Definition 1.11. An ordered field is a field F which is also an ordered set, s.t.
(i) x + y < x + z if x, y, z ∈ F and y < z. (ii) xy > 0 if x ∈ F , y ∈ F , x > 0, and y > 0. If x > 0, we call x positive; if x < 0, x is negative. Example 1.12. For example, Q is an ordered field.
Proposition 1.13. The following statements are true in every ordered field [10]
(a) If x > 0 , then −x < 0 , and vice versa. (b) If x > 0 and y < z, then xy < xz. (c) If x < 0 and y < z, then xy > xz. (d) If x 6 = 0, then x^2 > 0. In particular, 1 > 0. (e) If 0 < x < y, then 0 < (^1) y < (^1) x. We shall discuss a little about the inequalities here. (1) We have x^2 ≥ 0 and the equivalence, =, holds if and only if x = 0. (2) Assume ∀x ≥ 0, ∃
x ∈ F , s.t. x + (^1) x ≥ 2, ∀x > 0. Then we have x − 2 + (^) x^1 ≥ 0. This is true from the first inequalities. We have a complete square (
x + √^1 x )^2 ≥ 0. (3) Cauchy-Schwarz Inequality, see the following theorem.
Theorem 1.14. Let (a 1 , a 2 , ..., an) and (b 1 , b 2 , ..., bn) be two sequences of real numbers, then ( (^) ∑n
i=
a^2 i
)( (^) ∑n
i=
b^2 i
( (^) ∑n
i=
aibi
Proof: The Cauchy-Schwarz inequality is an elementary inequality at the same time a powerful inequality. We consider the following complete square, which we know from the first inequality that it is larger or equal or zero. Then we open up the brackets and collect the identical terms.
∑^ n
i=
∑^ n
j=
(aibj − aj bi)^2
∑^ n
i=
a^2 i
∑n
j=
b^2 j +
∑^ n
i=
b^2 i
∑n
j=
a^2 j − 2
∑^ n
i=
aibi
∑n
j=
bj aj
( (^) ∑n
i=
a^2 i
)( (^) ∑n
i=
b^2 i
( (^) ∑n
i=
aibi
The left-hand side of the equation is a sum of the squares of real numbers, that is, greater or equal to zero, thus we have
( (^) ∑n
i=
a^2 i
)( (^) ∑n
i=
b^2 i
( (^) ∑n
i=
aibi
( (^) ∑n
i=
a^2 i
)( (^) ∑n
i=
b^2 i
( (^) ∑n
i=
aibi
Q.E.D.
An interesting question is, when does = hold? This leads to some other famous inequalities.
Theorem 1.15. AM-GM Theorem (Arithmetic Mean-Geometric Mean). The theorem states that for any set of nonnegative real numbers, the arith- metic mean of the set is greater than or equal to the geometric mean of the set. That is, for a set of nonnegative real numbers a 1 , a 2 , ..., an, the following holds a 1 + a 2 + ... + an n
≥ (a 1 a 2 ...an)
(^1) n
We can also write in the shorthand notation with summations and prod- ucts: (^) n ∑
i=
ai n
∏^ n
i=
ai^1 n
Proof: Note that the function x → ln(x) is strictly concave. Then, by Jensen’s Inequality (proof see Jensen’s Inequality), we have
ln
i
λiai ≥
i
λiln(ai) = ln
i
aλ ii
with equality if and only if all the ai are equal. Since x → ln(x) is a strictly increasing function, it then follows that ∑
i
λiai ≥
i
aλ ii
with equality if and only if all the ai are equal, as desired.
Q.E.D.
Theorem 1.16. GM-HM Theorem (Geometric Mean-Harmonic Mean). For a set of nonnegative real numbers a 1 , a 2 , ..., an, the following holds
(a 1 a 2 ...an) n^1 ≥
n 1 a 1 +^ ...^ +^
1 an
Proof: Proof is similar.
Theorem 1.17. Jensen’s Inequality. A real function φ(x) is convex on the interval [a, b] if ∀θ ∈ (0, 1) we have
φ(θa + (1 − θ)b) ≤ θφ(a) + (1 − θ)φ(b)
Proof: Since x and y are real numbers, and x 6 = 0, yx is a real number. By the Archimedean property, we can choose an n ∈ N such that n > yx. Then nx > y.
Q.E.D.
Theorem 1.24. The Density of Real Numbers. Let x, y ∈ R be any two real numbers where x < y. Then there exists a rational number r ∈ Q such that x < r < y.
Proof: Suppose that x > 0. Since x < y we have that y >0 and then we have y − x > 0. By the Archimedean properties, since y − x > 0, then there exists a natural number n ∈ N such that (^) n^1 < y − x. If we multiply n on both sides, we get 1 < ny − nx and rewrite as nx + 1 < ny. Now we know that since n > 0 and since x > 0, and by the Archimedean properties that since nx > 0 then there exists a natural number, call it A ∈ N such that A − 1 ≤ nx ≤ A or equivalently A ≤ nx + 1 ≤ A + 1. Therefore nx ≤ A ≤ nx + 1 ≤ ny and so nx < A < ny and so the rational number r = An works for x < r < y.
Q.E.D.
Definition 1.25. For E ⊆ S bounded above (below), supE has a smallest upper bound. Let supE = α if
(1) α ≥ x, ∀x ∈ E (2) assume ∃α′^ > x, ∀x ∈ E, ⇒ α′^ ≥ α
Remark 1.26. Let E ⊆ S bounded above (below) ⇒ 6 supE exists. For example we have S = Q and E = {r ∈ Q : r^2 < 2 }. Is this bounded? Yes, we can simply consider 10 ≥ r, ∀r ∈ E. Is supE exists in Q? No. If α =supE ∈ Q, then we have α^2 = 2. Thus we have ( mn )^2 = 2 ⇒ m^2 = 2n^2 ⇒ m, n are even which contradicts assumption.
Definition 1.27. Let E ⊆ S bounded below in F , E = β if
(1) β ≤ x, ∀x ∈ F (2) β′^ ≤ x, ∀x ∈ E ⇒ β′^ ≤ β
Theorem 1.28. The Completeness Axiom. Every non-empty subset E of R that is bounded from above has a least upper bound supE ∈ R. Every non-empty subset E of R that is bounded from below has a greatest lower bound infE ∈ R.
Proof: Let us prove the case that is bounded from below. Let T be the set {−s : s ∈ E}. Since E is bounded from below there is an m ∈ R such that m ≤ s for all s ∈ E. This implies that −s ≤ −m for all s ∈ S and so t ≤ −m for all t ∈ T. Then T is bounded from above, hence by axiom, supT exists. Let u = supT. We show −u =infE. That is, ∀s ∈ E, −u ≤ s and ∀s ∈ E, we have t ≤ s ⇒ u ≤ q. Set q = −t, we have −s ≤ −t, ∀s ∈ E ⇒ u ≤ −t. Hence, ∀s ∈ E, t ≤ s ⇒ t ≤ −u.
Q.E.D. Next, we discuss Dedekind cut. The idea is simple. A Dedekind cut in an ordered field is a partition of it, (A, B), such that A is nonempty and closed downwards, B is nonempty and closed upwards. This satisfies all the axioms we discussed above, and hence is unique up to isomorphisms.
Let us recall the density of Q. ∀a, b ∈ F , ∃r ∈ Q s.t. a < r < b. Consider the following question. How to define
2? Let E = {r ∈ Q : r^2 < 2 }, then
(1) E 6 = ∅ while 0 ∈ E (2) Bounded above (we can consider 10 > r ∀r ∈ E) Therefore, we have
2 = supE = α. We can check the result. In this case, we have only three possibilities, α^2 > 2, α^2 < 2, or α^2 = 2. We can see that it is neither of the first two cases. We construct
2 = sup{r ∈ R : r^2 < 2 }, then it is well defined when sup{r ∈ R : r^2 < 2 } = 2, excluding the case α^2 > 2 and α^2 < 2. The consequence is
2 6 ∈ R.
Let us take a closer look at this problem. If α^2 > 2, then the goal is to find r ∈ R. Since r < α, α is an upper bound for E. If α^2 < 2, then the goal is to find r ∈ E, s.t. α < 2. We can take r = α + (^) n^1 and choose n → ∞ and n ∈ Q.
Proposition 1.29. We have following properties for powers
(i) xr^ xs^ = xr+s (ii) (xr^ )s^ = xrs (iii) xr^ yr^ = (xy)r
Definition 1.30. Dedekind Cut. Let E ⊂ Q, s.t.
(1) α ∈ E and α′^ < α ⇒ α′^ ∈ E (2) α ∈ E, ∃β ∈ E, s.t. β > α. We can construct R and look at the set of all cuts. (1) Ordered. E 1 ⊆ E 2 if E 1 ( E 2 (2) Operations. E 1 < E 2 , if E 1 ⊂ E 2 , with addition (+) and multipli- cation (×) (3) Ordered field. We can use cancellation. (4) Completeness. We can check bounded or not. We take a cut and check whether it is upper bound. Then look at the largest union. That is, there exists a complete ordered field that satisfies three properties (see theorem).
Theorem 1.31. There exists a complete ordered field that
R = {E ⊆ Q : (1)E 6 = Q, (2)P ∈ E and q < p ⇒ q ∈ E, (3)p ∈ E → ∃q ∈ E, s.t.q > p}
With the concepts discussed in mind, we can ask questions like how big are these sets? Are all of the infinite sets countable? We can move on to the next section.
2 §Basic Topology§
Go back to Table of Contents. Please click TOC
Proof: Let every set En be arranged in a sequence {xnk , k = 1, 2 , 3 , ..., and consider the infinite array
Figure 1: Graphic illustration of array
in which the elements of En from the nth row. The array contains all elements of S. As indicated by the arrows, these elements can be arranged in a sequence x 11 , x 21 , x 12 , x 31 , x 22 , x 13 , ...
If any two of the sets En have elements in common, these will appear more than once in the series above. Hence there is a subset T of the set of all positive integers such that S ∼ T , which shows that S is at most countable. Since E 1 ⊂ S, and E 1 is infinite, S is infinite, and thus countable [10].
Corollary 2.5. Suppose A is at most countable, and for every α ∈ A, Bα is at most countable. Put
T =
α∈A
Bα.
Then T is at most countable. For T is equivalent to a subset of S such that S =
n=1 En. There is famous story here called “Hilbert’s hotel”. Suppose a hotel has countably many rooms, numbered 1, 2, 3, ... with guest gi occupying room i; so the hotel is fully booked. Now a new guest x arrives asking for a room, whereupon the hotel manager tells him: sorry, all rooms are taken. No problem, says the new arrival, just move guest g 1 to room 2, g 2 to room 3, g 3 to room 4, and so on, and I will then take room 1. To the manager’s surprise this works; he can still put up all guests plus the new arrival x! Now it is clear that he can also put up another guest y, and another one z, and so on. In particular, we note that, in contrast to finite sets, it may well happen that a proper subset of an infinite set M has the same size as M. In fact, this is a characterization of infinity: a set is infinite if and only if it has the same size as some proper subset [1].
Figure 2: Graphic illustration of Hilbert’s Hotel
Theorem 2.6. Let A be a countable set, and let Bn be the set of all n- tuples (a 1 , ..., an), where ak ∈ A (k = 1, ..., n), and the elements a 1 , ..., an need not be distinct. Then Bn is countable.
Proof: That B 1 is countable is evidence, since B 1 = A. Suppose Bn− 1 is countable (n = 2, 3 , 4 , ...). The elements of Bn are of the form
(b, a), (b ∈ Bn− 1 , a ∈ A).
For every fixed b, the set of pairs (b, a) is equivalent to A, and hence countable. Thus Bn is the union of a countable set of countable sets. Then Bn is countable, which follows by induction [10].
Q.E.D.
Corollary 2.7. The set of all rational numbers is countable.
Proof: We apply previous theorem, with n = 2, noting that every rational r is of the form b/a, where a and b are integers. The set of pairs (a, b), and therefore the set of fractions b/a, is countable [10].
Q.E.D.
Theorem 2.8. Let A be the set of all sequences whose elements are the digits 0 and 1. This set A is uncountable. The elements of A are sequences like 1,0,0,1,0,1,1,...
Proof: Let E be a countable subset of A, and let E consist of the sequences s 1 , s 2 , s 3 , ... We construct a sequence s as follows. If the nth digit in sn is 1, we let the nth digit of s be 0, and vice versa. Then the sequence s differs from every member of E in at least one place; hence s 6 ∈ E. But clearly s ∈ A, so that E is a proper subset of A. We have shown that every countable subset of A is a proper subset of A. It follows that A is uncountable (for otherwise A would be a proper subset of A, which is absurd) [10]. This leads us to the famous Cantor-Berstein theorem.
Theorem 2.9. If each of two sets M and N can be mapped injectively into the other, then there is a bijection from M to N , hat is, |M | = |N |.
Proof: We assume that M and N are disjoint - if not, then we just replace N be a new copy.
Figure 6: Case 3. The one-way infinite chains of distinct elements that start at
the elements m 0 ∈ M \ g(N ).
Figure 7: Case 4. The one-way infinite chains of distinct elements that start at
the elements n 0 ∈ N \ f (M ).
Q.E.D.
With this understanding, we can move on to discuss metric spaces. Definition 2.10. A set X, whose elements we shall call points, is said to be metric space if with any two points p and q of X there is associated a real number d(p, q), called the distance from p to q, such that (a) d(p, q) > 0 if p 6 = q; d(p, p) = 0 (b) d(p, q) = d(q, p); (c) d(p, q) ≤ d(p, r) + d(r, q), ∀r ∈ X. Any function with these three properties is called a distance function, or a metric. We can think of the following examples. For real set, R, the distance between two points d(x, y) = |x − y|, and the triangle inequality tells us |x + y| ≤ |x| + |y|. Moreover, for Rn, we have d(x, y) = ||x − y||, and ||x|| =
x ◦ x while x > y =
∑n i=1 xiyi. Here it worth to note the details of triangle inequality. Theorem 2.11. d(x, y) ≤ d(x, z) + d(z, y) Proof: Let d(x, y) = ||x − y||, d(x, z) = ||x − z||, and d(z, y) = ||z − y||. Taking the power of two on both sides and open up the brackets, we have ||a + b||^2 = ||a||^2 + ||b||^2 + 2ab. Then ||a||^2 + ||b||^2 + 2ab ≤ ||a||^2 + ||b||^2 + 2 ||a||||b||. ⇒ ab ≤ ||a||||b|| which is the Cauchy - Schwarz equation. Definition 2.12. Let X be a metric space. All points and sets mentioned below are understood to be elements and subsets of X. (a) A neighborhood of p is a set Nr (p) consisting of all q such that d(p, q) < r, for some r > 0. The number r is called the radius of Np(p). (b) A point p is a limit point of the set E if every neighborhood of p contains a point q 6 = p such that q ∈ E.
(c) If p ∈ E and p is not a limit point of E, then p is called an isolated point of E. (d) E is closed if every limit point of E is a point of E. (e) A point p is an interior point of E if there is a neighborhood N of p such that N ⊂ E. Or we can notate the definition with shorthand notations. The set of interior points is as the following: Eo^ = {x ∈ X : d(x, Ec) > 0 } =
Br(X)⊆E Br(x) (f) E is open if every point of E is an interior point of E. (g) The complement of E (denoted by Ec) is the set of all points p ∈ X such that p 6 ∈ E. (h) E is perfect is E is closed if every point of E is a limit point of E. (i) E is bounded if there is a real number M and a point q ∈ X such that d(p, q) < M for all p ∈ E. (j) E is dense in X if every point of X is a limit point of E, or a point of E (or both). From the definition of metric space, limit point, interior point, perfect, and closures, we can show the following proofs.
Theorem 2.13. Every neighborhood is an open set.
Proof: Consider a neighborhood E = Nr (p), and let q be any point of E. Then there is a positive real number h s.t.
d(p, q) = r − h.
For all points s s.t. d(q, s) < h, we have then d(p, s) ≤ d(p, q) < h, we have then d(p, s) ≤ d(p, q) + d(q, s) < r − h + h = r,
so s ∈ E. Thus q is an interior point of E.
Q.E.D.
We can take a look at the following examples to understand closed, open, perfect, and bounded properties. We can simply consider the following subsets of R^2.
(a) The set of all complex z s.t. |z| < 1. (b) The set of all complex z s.t. |z| ≤ 1. (c) A nonempty finite set. (d) The set of all integers. (e) The set consisting of the numbers (^1) n (n = 1, 2 , 3 , ...). Let us note that this set E has a limit point (namely, z = 0) but that no point of E is a limit point of E; we wish to emphasize the difference between having a limit point and containing one. (f) The set of all complex numbers (that is, R^2 ). (g) The segment (a, b).
Then we have the following checklist:
