maths 332: real analysis - university of auckland · math 332: real analysis ... 6.5 intermediate...

MATH 332: Real Analysis

Lecture Notes 2010

Lectured by Rod Gover and Warren Moors

April 23, 2010

Contents

1 Real Numbers 21.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.1.1 Prove√

5 /∈ Q . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 R is a Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 R is Ordered . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.3.1 If x ∈ R and x ≤ 0 then −x ≥ 0 . . . . . . . . . . . . . . . 21.3.2 x ≥ 0 =⇒ x2 ≥ 0 . . . . . . . . . . . . . . . . . . . . . . 21.3.3 x ≤ 0 =⇒ (−x)2 ≥ 0 . . . . . . . . . . . . . . . . . . . . 31.3.4 1 > 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3.5 If x > 0 then 1

x > 0 . . . . . . . . . . . . . . . . . . . . . . 31.4 Bernoulli Inequality . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.4.1 Corollary 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 31.4.2 Lemma 1: If a ∈ [0, 1] and n ∈ N then (1 + a)n ≤ 1 + 3na. 41.4.3 Corollary 2 . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.5 Archmedian Property . . . . . . . . . . . . . . . . . . . . . . . . 41.6 Maximums and Minimums . . . . . . . . . . . . . . . . . . . . . . 5

1.6.1 Lemma 2: X has at most 1 maximum. . . . . . . . . . . . 51.6.2 In [2, 4) no max exists. . . . . . . . . . . . . . . . . . . . . 5

1.7 Upper and Lower Bounds . . . . . . . . . . . . . . . . . . . . . . 51.8 Supremum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.8.1 Lemma 3: If a supremum exists, it is unique. . . . . . . . 51.8.2 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.8.3 Lemma 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.9 Theorem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.9.1 Lemma 5: Let x, y ∈ Q > 0 then x < y ⇐⇒ x2 < y2. . . 61.9.2 Example: When no supremum exists in Q . . . . . . . . . 6

1.10 Modulus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.11 Triangle Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . 71.12 Reverse Triangle Inequality . . . . . . . . . . . . . . . . . . . . . 71.13 Neighbourhoods . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1

2 Sequences 82.1 Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.1.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.1.2 Let xn =

√n2 + 6− n. Claim {xn} → 0 as n→∞ . . . . 8

2.1.3 Lemma 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.1.4 Corollary . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.1.5 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.1.6 Lemma 6 rewritten . . . . . . . . . . . . . . . . . . . . . . 92.1.7 If limn→∞ xn = l and limn→∞ yn = m and xn < yn ∀n ∈

N is l < m ? . . . . . . . . . . . . . . . . . . . . . . . . . . 92.2 Subsequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2.1 Lemma 7: A sequence (xn) converges to a limit l ⇐⇒every subsequence converges to the same limit l. . . . . . 9

2.2.2 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.3 Boundedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.3.1 Lemma 8: A convergent subsequence is bounded. . . . . . 102.4 Theorem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.4.1 Example: Find limn→∞ an . . . . . . . . . . . . . . . . . 112.5 Sandwich / Squeeze Theorem . . . . . . . . . . . . . . . . . . . . 11

2.5.1 Example: Let bn = (−1)n sin(n2)n . . . . . . . . . . . . . . . 11

2.5.2 Example: lim |xn| = 0 then limxn = 0 . . . . . . . . . . . 112.5.3 Example: Calculate lim n

√n . . . . . . . . . . . . . . . . . 12

2.5.4 Example: lim n√a, a > 0 . . . . . . . . . . . . . . . . . . . 12

2.6 Definition: Increasing or Decreasing Sequences . . . . . . . . . . 122.7 Monotone Convergence Theorem . . . . . . . . . . . . . . . . . . 12

2.7.1 Example: If (xn) is decreasing and bounded below then(xn) converges to the infimum. ie. limxn = inf{xn} . . . 12

2.7.2 Example: Calculate limit of (an) given by a1 = 1, an+1 =√an + 1, n ≥ 1. . . . . . . . . . . . . . . . . . . . . . . . . 13

2.7.3 Example: Find a sequence of rationals converging to√a, a >

0, a ∈ Q . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.8 Cauchy Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.8.1 Lemma 9: Every convergent sequence is a Cauchy sequence 142.8.2 Theorem 3: Every sequence in R has a monotone subse-

quence. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.8.3 Corollary: (Bolzano / Weierstraß) Theorem . . . . . . . . 152.8.4 Lemma 10: Every Cauchy sequence which has a conver-

gent subsequence is convergent. . . . . . . . . . . . . . . . 152.8.5 Lemma 11: Every Cauchy sequence is bounded. . . . . . . 152.8.6 Theorem 4 (Cauchy): Every Cauchy sequence is conver-

gent in R . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.9 Definition: Contractive Sequences . . . . . . . . . . . . . . . . . . 16

2.9.1 Lemma 12: Every contractive sequence is a Cauchy se-quence. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.9.2 Corollary: Every contractive sequence is convergent . . . 172.9.3 An application example in numerical analysis . . . . . . . 17

2.10 R is dense . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2

3 Series 183.1 Lemma 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3.1.1 Example: Geometric Series . . . . . . . . . . . . . . . . . 193.1.2 Example: Telescopic Series . . . . . . . . . . . . . . . . . 19

3.2 Lemma 14: N th Term Test . . . . . . . . . . . . . . . . . . . . . 203.2.1 Corollary . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3.3 Theorem 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.3.1 Interlude: Consider

∑1n . . . . . . . . . . . . . . . . . . . 20

3.4 Theorem 6: Comparison Test . . . . . . . . . . . . . . . . . . . . 213.4.1 Example:

∑1k2 . . . . . . . . . . . . . . . . . . . . . . . . 21

3.4.2 Lemma 15:∑xk is convergent ⇐⇒

∑cxk is convergent

for some constant c 6= 0 . . . . . . . . . . . . . . . . . . . 213.5 Theorem 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.5.1 Corollary 1: Limit Comparison Test . . . . . . . . . . . . 213.5.2 Corollary 2: Also Limit Comparison Test . . . . . . . . . 223.5.3 Example:

∑n+5

n3+2n+1 , xn = n+5n3+2n+1 . . . . . . . . . . . . 22

3.6 Definition: Absolute Convergence . . . . . . . . . . . . . . . . . . 223.7 Theorem 8: Every absolutely convergent series is convergent. . . 223.8 Ratio Test 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.9 Ratio Test 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.9.1 Example: For ρ = 1 . . . . . . . . . . . . . . . . . . . . . 233.9.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

4 Power Series 234.1 Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234.2 Theorem 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

4.2.1 Corollary 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 244.2.2 Corollary 2 . . . . . . . . . . . . . . . . . . . . . . . . . . 244.2.3 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244.2.4 More Examples . . . . . . . . . . . . . . . . . . . . . . . . 244.2.5 Some common functions as power series . . . . . . . . . . 254.2.6 Another example . . . . . . . . . . . . . . . . . . . . . . . 25

4.3 Theorem 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.3.1 If the series

∑xk is convergent then the set S is bounded

above and supS ≤∑∞n=1 xn . . . . . . . . . . . . . . . . . 25

4.3.2 If the set S is bounded above, then the series∑xn is

convergent and∑∞n=1 xn ≤ supS. . . . . . . . . . . . . . 25

4.3.3 If the series∑xn is convergent, then

∑∞n=1 xn = supS. . 25

4.3.4 Corollary (Rearrangement Theorem) . . . . . . . . . . . . 25

5 Limits of Functions 265.1 Open Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

5.1.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 265.2 Cluster Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

5.2.1 Theorem 11 . . . . . . . . . . . . . . . . . . . . . . . . . . 265.2.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

5.3 Definition of limits of a function . . . . . . . . . . . . . . . . . . 275.4 Limits of functions are unique . . . . . . . . . . . . . . . . . . . . 27

5.4.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 275.4.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

3

5.5 Theorem 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285.5.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 295.5.2 Example 2: Show limx→1

x+1x2+3 = 1

2 . . . . . . . . . . . . . 295.5.3 Example 3: Consider limx→0+

1x , where f(x) = 1

x ,∀x > 0 295.6 Theorem 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295.7 Squeeze Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

5.7.1 Example 1: limx→0+ x · sin( 1x ) = 0 . . . . . . . . . . . . . 30

5.7.2 Example 2: f(x) = sin( 1x ) has no limit at x = 0 . . . . . . 30

5.8 Definition: Limits to Infinity . . . . . . . . . . . . . . . . . . . . 305.8.1 Remark . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305.8.2 Example 1: Show limx→∞

1√x

= 0 . . . . . . . . . . . . . 305.8.3 Example 2: limx→−∞

xx+1 . . . . . . . . . . . . . . . . . . 31

5.8.4 Calculate limx→∞2x3−3x+6x3−5x+1 . . . . . . . . . . . . . . . . . 31

6 Continuity 316.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

6.1.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 316.1.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 316.1.3 Example 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

6.2 Theorem 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326.2.1 Corollary 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 336.2.2 Corollary 2 . . . . . . . . . . . . . . . . . . . . . . . . . . 336.2.3 Corollary 3 . . . . . . . . . . . . . . . . . . . . . . . . . . 336.2.4 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 336.2.5 Corollary 4 . . . . . . . . . . . . . . . . . . . . . . . . . . 336.2.6 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 346.2.7 Example 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

6.3 Boundedness Theorem . . . . . . . . . . . . . . . . . . . . . . . . 346.4 Global Maximum and Minimums . . . . . . . . . . . . . . . . . . 35

6.4.1 Min-Max Theorem (Weierstraß) . . . . . . . . . . . . . . 356.5 Intermediate Value Theorem (IVT I) . . . . . . . . . . . . . . . . 356.6 Bolzano Intermediate Value Theorem (IVT II) . . . . . . . . . . 35

6.6.1 Corollary 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 366.6.2 Corollary 2 . . . . . . . . . . . . . . . . . . . . . . . . . . 36

6.7 Monotone Functions . . . . . . . . . . . . . . . . . . . . . . . . . 366.7.1 Lemma 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . 366.7.2 Corollary 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 376.7.3 Corollary 2: Continuity Inverse Theorem . . . . . . . . . 376.7.4 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

6.8 Uniform Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . 386.8.1 Lemma 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . 386.8.2 Converse Counter Example 1 . . . . . . . . . . . . . . . . 386.8.3 Lemma 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . 386.8.4 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 396.8.5 Theorem 15 . . . . . . . . . . . . . . . . . . . . . . . . . . 396.8.6 Lipschitz Functions . . . . . . . . . . . . . . . . . . . . . . 396.8.7 Lemma 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . 406.8.8 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406.8.9 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

4

7 Differentiation 407.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407.2 Theorem 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

7.2.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417.3 Theorem 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417.4 Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

7.4.1 Lemma 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . 427.4.2 Remark . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437.4.3 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

7.5 Inverse Function Theorem . . . . . . . . . . . . . . . . . . . . . . 447.5.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

7.6 Relative Maximum . . . . . . . . . . . . . . . . . . . . . . . . . . 447.6.1 Interior Extremum Theorem . . . . . . . . . . . . . . . . 457.6.2 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 457.6.3 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

7.7 Rolle’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 457.8 Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 457.9 Corollary 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

7.9.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 467.10 Corollary 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 467.11 Corollary 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

7.11.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 477.12 Darboux’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 477.13 Cauchy’s Mean Value Theorem . . . . . . . . . . . . . . . . . . . 477.14 L’Hopital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

7.14.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 487.15 Interpolation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

7.15.1 Taylor (Mean Value Form) Theorem . . . . . . . . . . . . 497.15.2 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

7.16 Classification of Extrema . . . . . . . . . . . . . . . . . . . . . . 50

8 Riemann Integration 518.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

8.1.1 Lemma 21 . . . . . . . . . . . . . . . . . . . . . . . . . . . 518.1.2 Corollary . . . . . . . . . . . . . . . . . . . . . . . . . . . 528.1.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 528.1.4 Lower and Upper Integrals . . . . . . . . . . . . . . . . . 528.1.5 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

8.2 Riemann Integrable . . . . . . . . . . . . . . . . . . . . . . . . . . 538.3 Riemann Condition . . . . . . . . . . . . . . . . . . . . . . . . . . 548.4 Theorem 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

8.4.1 Lemma 22 . . . . . . . . . . . . . . . . . . . . . . . . . . . 578.5 Lemma 23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 598.6 Lemma 24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 598.7 Theorem 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 598.8 First Fundamental Theorem of Calculus (FTC 1) . . . . . . . . . 60

8.8.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . 608.9 Second Fundamental Theorem of Calculus (FTC 2) . . . . . . . . 618.10 Substitution Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

5

9 Pointwise and Uniform Convergence 639.1 Pointwise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

9.1.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 639.1.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

9.2 Uniform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 659.2.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 659.2.2 Lemma 25 . . . . . . . . . . . . . . . . . . . . . . . . . . . 659.2.3 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

9.3 Theorem 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 669.3.1 Corollary . . . . . . . . . . . . . . . . . . . . . . . . . . . 669.3.2 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

9.4 Theorem 21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 679.4.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

9.5 Theorem 22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

6

1 Real Numbers

1.1 Introduction

There are gaps in the rationals that we need to accommodate for. Hence theneed for the reals. Example below.

1.1.1 Prove√

5 /∈ Q

Proof. Suppose p ∈ Z, q ∈ N such that x = pq ≥ 0. Then p ≥ 0 so p ∈ N and(

pq

)2

= 5. Then p2 = 5q2. We know every natural number has a unique primefactorisation. Let k be number of factors 5 in the prime factorisation of p. Letl be likewise for q. Then p2 has 2k factors of 5. Similarly q2 has 2l factors of 5,and 5q2 has 2l + 1 factors of 5.

But p2 = 5q2 and by the uniqueness of the prime factors, there must be2k = 2l + 1 factors. A contradiction since 2k is even, and 2l + 1 is odd.

1.2 R is a Field

There exists operations · and + st:(A1) x+ (y + z) = (x+ y) + z(A2) ∃0 ∈ R : 0 + x = x(A3) ∀x ∈ R,∃y ∈ R : y + x = 0(A4) x+ y = y + x∀x, y ∈ R(A5) (x · y)z = x · (y · z)(A6) (x+ y)z = (xz) + (yz)(A7) ∃1 ∈ R : 1x = x, 1 6= 0(A8) xy = yx(A9) ∃y,∀x : xy = 1 if x 6= 0. ie. y = x−1 = 1

y

1.3 R is Ordered

Let B be a total ordering on R. Then, there is a relation ≤ on R st:(B1) x ≤ x, ∀x ∈ R(B2) if (x ≤ y and y ≤ x) then x = y(B3) if (x ≤ y and y ≤ z) then x ≤ z(B4) ∀x, y ∈ R then x ≤ y or y ≤ x(B5) if x ≤ y then x+ z ≤ y + z,∀x, y, z ∈ R(B6) if x ≤ y and u ≥ 0 then xu ≤ yu

1.3.1 If x ∈ R and x ≤ 0 then −x ≥ 0

Proof. Take z = −x and apply (B5). Then 0 = (−x) + x = x + (−x) ≤0 + (−x) = −x.

1.3.2 x ≥ 0 =⇒ x2 ≥ 0

Proof.0 = 0 · x ≤ x · x = x2

7

1.3.3 x ≤ 0 =⇒ (−x)2 ≥ 0

Proof. Let x ≥ 0. Then −x < 0 and (−x) · (−x) = −(−x ·x). Since −(−x ·x) +(−x · x) = 0 we know (−x)2 = x2. It follows that x2 > 0.

1.3.4 1 > 0

Proof. 1 = 12 >= 0. Since 1 6= 0 then 1 > 0.

1.3.5 If x > 0 then 1x > 0

Proof. Suppose not. Then 1x < 0 or 1

x = 0.(i) 1

x = 0 ⇐⇒ 1 = 0 · x = 0 (ii) 1

x < 0 =⇒ 1 = 1x · x ≤ 0 · x = 0.

1.4 Bernoulli Inequality

Let a ∈ R, a > −1, n ∈ N. Then (1 + a)n ≥ 1 + na.

Proof. Let P (n) be the statement that (1 + a)n ≥ 1 + na is true.P(1): (1 + a)n = 1 + a ≥ 1 + 1a = 1 + na, n = 1 is true.Let n ∈ N and suppose P (n) true. ie. (1 + a)n ≥ 1 + na true.

(1 + a)n+1 = (1 + a)n(1 + a) ≥ (1 + na)(1 + a) = 1 + a+ na+ na2

= 1 + (n+ 1)a+ na2

≥ 1 + (n+ 1)a

Therefore P (n) is true ∀n ∈ N by the principle of induction.

1.4.1 Corollary 1

Let c, q ∈ Q > 0 and n ∈ N, ε ∈ Q. Suppose qn > c. Then ∃ε st. (q − ε)n > c.

Motivation:(q − ε)n =

(q(

1− εq

))n= qn

(1− ε

q

)n. We know if a = − ε

q , we need a > −1

(for Bernoulli inequality to hold) so −εq > −1 =⇒ 1 > εq =⇒ q > ε.

Then, qn(

1− εq

)n≥ qn

(1 + n

(−εq

))= qn

(1− nε

q

).

In order for qn − nεqn−1 > c ⇐⇒ qn − c > nεqn−1, we choose:

ε <qn − cnqn−1

Proof. Take ε = 12 min(q, q

n−cnqn−1 , 1) then ε ∈ Q, ε > 0, and ε ≤ 1

2 · 1 < 1.

Moreover, because −εq ≥−12 q

q = − 12 > −1,

(q − ε)n =(q

(1− ε

q

))n= qn

(1− ε

q

)n≥ qn

(1− nε

q

)= qn − nεqnn−1

8

But ε ≤ 12 and qn−c

2nqn−1 <qn−cnqn−1 , so nqn−1ε < qn − c and qn − nqn−1ε > c.

Hence, (q − ε)n ≥ qn − nεnqn−1 > c as required.

1.4.2 Lemma 1: If a ∈ [0, 1] and n ∈ N then (1 + a)n ≤ 1 + 3na.

Proof. Let P (n) be our lemma statement.P (1) : for n = 1, (1 + a)n = (1 + a) ≤ 1 + 3a = 1 + 3na is true.Suppose P (n) is true. If a ∈ [0, 1] and n ∈ N then (1 + a)n ≤ 1 + 3na. Thus,

(1 + a)n+1 = (1 + a)n(1 + a) ≤ (1 + 3na)(1 + a)= 1 + a+ 3na+ 3na2 = 1 + a(1 + 3n + 3na)≤ 1 + a(3 · 3n) = 1 + 3n+1a

1.4.3 Corollary 2

Let c, p ∈ Q > 0 and n ∈ N. Suppose pn < c then ∃ε ∈ Q > 0 such that(p+ ε)n < c.

Motivation:

pn + 3nεpn+1 < c

3nεpn−1 < c− pn

ε <c− pn

3npn−1

Proof. Take ε = min(p, c−pn2·3npn−1 )

Since 0 ≤ εp ≤ 1 we can use Lemma 1 so that

(p+ ε)n =(p

(1 +

e

p

))n= pn

(1 +

ε

p

)n≤ pn

(1 + 3n

ε

p

)= pn + 3nεpn−1

< pn + 3n · c− pn

3npn−1· pn−1 = c

1.5 Archmedian Property

∀x, y ∈ R > 0, ∃n ∈ N|nx > y.

Proof. Let A = {nx : x ∈ R, n ∈ N}. Suppose the hypothesis is false, then Ais bounded above. Let ∃a = supA, then a − x < a and a − x < mx for somem ∈ N. But then a < mx− x = (m+ 1)x [since a is an upper bound].

9

1.6 Maximums and Minimums

Definition: Let X ⊂ R, X 6= ∅ and M ∈ R. Then M is a maximum for X iff:

1. M ∈ X

2. ∀x ∈ X, x ≤M

Similarly for minimums.

Notation: If M ∈ R is a maximum for X, we write M = maxX. Similarly forminX.

1.6.1 Lemma 2: X has at most 1 maximum.

Proof. Let M1,M2 be maximums for X. Since M1,M2 are maximum, M1,M2 ∈X. Note that M1 ≤ M2 and M2 ≤ M1. Therefore M1 = M2 via the antisym-metric property and hence the maximum is unique.

1.6.2 In [2, 4) no max exists.

Proof. Suppose M is max. But X = M+42 ∈ [2, 4) and M < X.

1.7 Upper and Lower Bounds

Definition: Let X ⊂ R and M ∈ R. Then M is an upper bound for X if:∀x ∈ X,x ≤ M . If X has an upper bound, it is bounded above. Similarly forlower bounds.

A set is bounded if it is bounded above and below. Consider [2, 4) thenX = {M ∈ R : M ≥ 4} is the set of upper bounds for [2, 4).

1.8 Supremum

Definition: Let X ⊂ R and s ∈ R. Then s is called a supremum for X if:

1. s is an upper bound for X.

2. s ≤M , ∀M ∈ {y : y ≥ x, ∀x ∈ X}.

ie. The supremum is the least upper bound. Similarly an infimum is the greatestlower bound.

1.8.1 Lemma 3: If a supremum exists, it is unique.

Proof. Let s1, s2 ∈ R and suppose both are supremum. Then s1, s2 are upperbounds. s1 ≤ s2 and s2 ≤ s1 hence s1 = s2.

eg. If X = [2, 4] =⇒ supX = 4, inf X = 2. If X = (2, 4) =⇒ supX =4, inf X = 2.

If X = ∅ then all x ∈ R are upper and lower bounds (vacuously true). HenceX does not have a supremum or infinum.

10

1.8.2 Example

inf N = 1. But, N has no upper bound.

Proof. Suppose M ∈ R and M is an upper bound for N. Then 1 ≤ M because1 ∈ N therefore M > 0 and 1

M > 0. Using the archimedian property we know∃n ∈ N st. n · 1

M > 1. So n > M and M is not an upper bound. . ThereforeN is not bounded above.

1.8.3 Lemma 4

Let X ⊂ R and s ∈ R. Then the following are equivalent.

1. s = supX

2. (a) s is an upper bound for X.(b) ∀ε > 0, ∃x ∈ X st. s− ε < x.

Proof. The “1 =⇒ 2” case: Suppose 1. is true. 2(a) is obviously true. Letε > 0 so that s− ε < s. Then s− ε is not an upper bound for X. Hence ∃x ∈ Xst. x > s− ε.

The “2 =⇒ 1” case: Suppose 2 is true. Then 2(a) is consistent with 1.

Let M ∈ R and suppose M is an upper bound for X. Suppose M < s. Setε = s−M > 0 then M = s− ε and by 2(b) there is an x ∈ X st s− ε < x. ThusM < x and M is not an upper bound for X. Therefore M ≥ s.

1.9 Theorem 1

Every non empty X ⊂ R that’s bounded above has a supremum. Taken as anaxiom in this course.

With this theorem, field properties, orderedness, archimedian property, wehave a unique system called the real number system.

1.9.1 Lemma 5: Let x, y ∈ Q > 0 then x < y ⇐⇒ x2 < y2.

Proof. “⇒” case: If x < y then x2 = x · x < x · y < y · y = y2

“⇐” case: We know if x < y then x2 < y2 and if y < x then y2 < x2 and ifx = y then x2 = y2.

Suppose x2 < y2. If x > y then x2 > y2 . If x = y then x2 = y2 .Therefore x < y.

1.9.2 Example: When no supremum exists in Q

In Q consider X = {x ∈ Q : 0 < x <√

5} = {x ∈ Q : x > 0 and x2 < 5}. ThenX 6= ∅ and is bounded above by 3. X has no supremum in Q since

√5 doesn’t

exist in Q (proved earlier).Claim: X is bounded above and has no supremum in Q.

Proof. ∀x ∈ X, x2 < 5 < 32 = 9 =⇒ X bounded above and x < 3 (by lemma5). Suppose X has a supremum in Q. Let q = supX. We have to consider q2

11

when q2 > 5, q2 < 5, q2 = 5. Already, q2 = 5 is impossible (proved earlier).

Suppose q2 < 5We know if c, q ∈ Q > 0, n ∈ N and qn < c then ∃ε ∈ Q > 0, st. (q + ε)n < c sothat (q + ε)n < 5. This implies q + ε ∈ X. But q is an upper bound for X, soq ≥ q + ε .

Suppose q2 > 5We know if c, q ∈ Q > 0, n ∈ N, and qn > c then ∃ε ∈ Q > 0 and 0 < ε < 1st. (q − ε)n > c. Then q2 > 5 > 1 and q > 1 by lemma 5. We have ε ∈ Q st.0 < ε < 1 and (q − ε)2 > 5 then q − ε > 1− 1 > 0.∀x ∈ X, one has (q − ε)2 > 5 > x2 so q − ε > x by lemma so q − ε is an upperbound for X. But q = supX ≤ q − ε. .

1.10 Modulus

Definition: If x ∈ R define |x| =

{x if x ≥ 0,−x if x < 0.

Properties:

• |x| ≥ 0

• |x| = 0 ⇐⇒ x = 0

• −|x| ≤ x ≤ |x|

• |xy| = |x||y|

1.11 Triangle Inequality

|x+ y| ≤ |x|+ |y| ∀x, y ∈ R

Proof. −|x| ≤ x ≤ |x| and −|y| ≤ y ≤ |y|=⇒ −(|x|+ |y|) ≤ −|x|+ y ≤ x+ y ≤ |x|+ y ≤ |x|+ |y|therefore |x+ y| ≤ |x|+ |y|

Note |x+ y| is the distance(metric) from x to y. Also:

|x− z| = |(x− y) + (y − z)| ≤ |x− y|+ |y − z|

1.12 Reverse Triangle Inequality

|x− y| ≥ ||x| − |y||∀x, y ∈ R

Proof. |x| = |(x− y) + y| ≤ |x− y|+ |y| =⇒ |x| − |y| ≤ |x− y|Similarly, |y| − |x| ≤ |y − x| = | − (x− y)| = |x− y|

So ||x| − |y|| =

{|x| − |y| ≤ |x− y| if |x| − |y| ≥ 0−(|x| − |y|) ≤ |x− y| if |x| − |y| < 0

Therefore ||x| − |y|| ≤ |x− y|

12

1.13 Neighbourhoods

Definition: Let a ∈ R and U ⊂ R. U is called a neighbourhood of a if∃ε > 0 st. (a− ε, a+ ε) ⊂ U . ie. U is a neighbourhood for a ⇐⇒∃ε > 0 st. ∀x ∈ R with |x− a| < ε one has x ∈ U .Sometimes we denote this by Bε(a).

2 Sequences

Definition: A sequence of real numbers is a function X : N→ R.Write xn = x(n), n ∈ N. Other notation includes x = (x1, x2, . . .), (xn),(xn)n∈N, {xn}∞n=1, xn.

2.1 Convergence

Definition: Let x1, x2, . . . ∈ R and L ∈ R. The sequence x1, x2, . . . converges toL iff:∀ε > 0, ∃N ∈ R, st. ∀n ∈ N∀ n ≥ N :

|xn − L| < ε ⇐⇒ L− ε < xn < L+ ε

2.1.1 Example

xn = 1n converges to 0.

Proof. xn = 1n . Let ε > 0. Take N = 2

ε . Then ∀n ∈ N with n ≥ N ,|xn − 0| = | 1n − 0| = 1

n ≤1N = ε

2 < ε

2.1.2 Let xn =√n2 + 6− n. Claim {xn} → 0 as n→∞

Let ε > 0. Take N = 7ε . Then ∀n ∈ N with n ≥ N one has that:

|xn − 0| = |√n2 + 6− n| =

√n2 + 6− n =

√n2 + 6−

√n2

= (√n2 + 6−

√n2) ·

(√n2 + 6 +

√n2

√n2 + 6 +

√n2

)=

n2 + 6− n2

√n2 + 6 +

√n2

=6

√n2 + 6 +

√n2≤ 6√

n2=

6n≤ 6N

=67ε < ε

Thus the sequence converges to 0.

2.1.3 Lemma 6

Let x1, x2, . . . and y1, y2, . . . be 2 sequences and let l,m ∈ R such that xn → land yn → m for n→∞. Suppose ∀n ∈ N that xn ≤ yn then l ≤ m.

Proof. Suppose to the contrary that l > m. Take ε = l−m2 . Since the sequence

x1, x2, . . . converges to l then ∃N1 st. ∀n ∈ N with n ≥ N1 we have thatl − ε < xn < l + ε. Since {yn} converges to m, ∃N2 st. m− ε < yn < m+ ε.

So, ∃n ∈ N with n ≥ max(N1, N2) st. both l − ε < xn < l + ε and m− ε <yn < m+ ε hold.

Thus we have that yn < m+ ε = l − ε < xn. A contradiction since xn ≤ yn. Therefore l ≤ m.

13

2.1.4 Corollary

Let a1, a2, . . . be a sequence and l,m ∈ R. Suppose that a1, a2, . . . converges tol and also to m. Then l = m.

Proof. Take xn = an and yn = an. Then ∀n : xn ≤ yn. By lemma: l ≤ m andsimilarly m ≤ l which implies l = m.

2.1.5 Limits

The sequence x1, x2, . . . is called convergent if there is an l ∈ R st. x1, x2, . . .converges to l. If the sequence is not convergent then it is divergent.

CAVEAT : Don’t assume that a sequence is convergent if you have to showthat: (limn→∞ xn = l).

Thus limn→∞ xn = l means that x1, x2, . . . converges to l.

2.1.6 Lemma 6 rewritten

If limn→∞ xn = l and limn→∞ yn = m and xn ≤ yn ∀n ∈ N then l ≤ m.

2.1.7 If limn→∞ xn = l and limn→∞ yn = m and xn < yn ∀n ∈ N is l < m?

NO. Counterexample: Take xn = 0 ∀n ∈ N and yn = 1n ∀n ∈ N.

NB: If 1n−37 = xn then limn→∞

1n−37 = 0. We ignore first 37 terms.

It also suffices to say that xn ≤ yn for all large n. ie. ∃M st ∀n ∈ N, n ≥Mone has that xn ≤ yn.

2.2 Subsequences

Definition: A subsequence of a sequence (xn)∞n=1 is a sequence of the form(xnk)∞n=1 such that 1 ≤ n1 < n2 < n3 < . . ..

2.2.1 Lemma 7: A sequence (xn) converges to a limit l ⇐⇒ everysubsequence converges to the same limit l.

Proof. =⇒ : Suppose (xn) converges to l and (xnk) is a subsequence. Sincexn → l, ∀ε > 0, ∃N : |xn− l| < ε, ∀n > N . But nk ≥ k so |xnk − l| < ε, ∀k > Nhence xnk → l as k →∞.

⇐= : Conversely if all subsequences converge to l, in particular, (xn) con-verges to l.

2.2.2 Example

((−1)n) = (−1, 1,−1, 1, . . .) does not converge. ie it diverges. We can takesubsequences (−1,−1,−1, . . .) and (1, 1, 1, . . .) which converge to −1 and 1 re-spectively. Because of uniqueness of limits in R and lemma 7, we know it isdivergent.

14

2.3 Boundedness

(xn) is bounded above by M if ∃M : xn ≤M , ∀n ∈ N. Similarly, bounded belowif ∃M > 0: −M ≤ xn, ∀n ∈ N. If it is bounded above and below ie. ∃M > 0:|xn| ≤M , ∀n ∈ N then it is bounded.eg. ((−1)n) is bounded below by −1 and above by 1 so is bounded. Butn = (1, 2, 3, . . .) is bounded below by 1 but not bounded above.

2.3.1 Lemma 8: A convergent subsequence is bounded.

Proof. Suppose limn→∞ = l. ie. ∀ε > 0, ∃N : |xn− l| < ε, ∀n > N ⇐⇒ l− ε <xn < l+ε. So xn ≤ max(x1, . . . , xN , l+ε), ∀n and x ≥ min(x1, . . . , xN , l−ε).

Note that boundedness does NOT imply convergence.

2.4 Theorem 2

Let (an), (bn) be convergent sequences. Then:

(i) limn→∞

(an + bn) = limn→∞

an + limn→∞

bn

(ii) limn→∞

(can) = c limn→∞

an, c ∈ R

(iii) limn→∞

(anbn) = ( limn→∞

an) · ( limn→∞

bn)

(iv) If b = limn→∞

bn 6= 0, then (∃N : bn 6= 0, ∀n > N) and limn→∞

(anbn

)=

lim anlim bn

(v) limn→∞

|an| =∣∣∣ limn→∞

an

∣∣∣Proof. : Let a := limn→∞ an and b := limn→∞ bn.(i) We want to make |(an + bn) − (a + b)| small by making |xn − a|, |xn − b|small.

|(an + bn)− (a+ b)| ≤ |an − a|+ |bn − b|

Since an → a, bn → b, ∃N1, N2 : ∀n > N1, |an − a| < ε2 , and ∀n > N2,

|bn − a| < ε2 :

|(an + bn)− (a+ b)| ≤ |an − a|+ |bn − b| <ε

2+ε

2= ε, ∀n > max(N1, N2)

(ii), (iii) Note if we take bn = c in (iii) then we have (ii). Since bn → b, b isbounded so that |bn| ≤M , ∀n ∈ N.

|anbn−ab| = |anbn−abn+abn−ab| since an → a, ∃N1 : |an−a| < 12εM , ∀n > N1

.Suppose a 6= 0, then since bn → b then ∃N2 : |bn − b| < 1

2ε|a| , ∀n > N2 so

that:

|a||bn − b| <12ε,∀n > N2 holds for all a

15

Hence for all n > max(N1, N2):

|anbn − ab| ≤ |bn||an − a|+ |a||bn − b| < M · 12ε

M+ |a| 1

2|a|ε = ε

(iv) In view of (iii) we need only to show that 1bn→ 1

b as n→∞. Since bn → b

and |b| > 0, then ∃N : |b− bn| ≤ |b|2 ,∀n > N . So by triangle inequality:

|b| − |bn| ≤ |b− bn| ≤|b|2,∀n > N

0 <|b|2≤ |bn| ⇐⇒

1|bn|≤ 2|b|,∀n > N

Since bn → b,∃N1 > N : |b− bn| < ε|b|22 ,∀n > N1. Hence:∣∣∣∣ 1

bn− 1b

∣∣∣∣ = |b− bn| ·1|bn|· 1|b|≤ ε|b2|

2· 2|b|· 1|b|

= ε

(v) Since an → a then ∃N : |an − a| < ε,∀n > N . By triangle inequality wehave that:

||an| − |a|| ≤ |an − a| < ε,∀n > N

2.4.1 Example: Find limn→∞ an

an =3n2 + 2n+ 1

2n2 + 1+

3n

=3 + 2

n + 1n2

2 + 1n2

+3n

Using all the rules:

limn→∞

an =3 + 2 limn→∞

1n +

(limn→∞

1n

)22 + limn→∞( 1

n )2+ 3 lim

n→∞

1n

=32

2.5 Sandwich / Squeeze Theorem

Let (an), (bn), (cn) be sequences with an ≤ bn ≤ cn ∀n ∈ N and lim an =lim cn = l. Then (bn) converges to l.

Proof. Let ε > 0. Since an → l, cn → l then ∃N1, N2 :l − ε < an < l + ε, n > N1 and l − ε < cn < l + ε, n > N2.For n > max(N1, N2) we have l−ε ≤ an ≤ bn ≤ cn ≤ l+ε hence |bn− l| < ε.

2.5.1 Example: Let bn = (−1)n sin(n2)n

Then −1n ≤ bn ≤

1n . So by sandwich theorem, lim bn exists and is 0, as lim 1

n = 0.

2.5.2 Example: lim |xn| = 0 then limxn = 0

Because −|xn| ≤ xn ≤ |xn| and lim |xn| = lim(−|xn|) = 0, then limxn = 0.

16

2.5.3 Example: Calculate lim n√n

Recall Bernoulli inequality. Let a = 1√n

. Then:

(1 +

1√n

)n≥√n ⇐⇒ n ≤

(1 +

1√n

)2n

⇐⇒ n√n ≤

(1 +

1√n

)2

⇐⇒ 1 ≤ n√n ≤

(1 +

1√n

)2

Assuming lim 1√n

= 0, lim(1 + 1√n

)2 = (1 + 0)2 = 1:

limn→∞

n√n = 1

2.5.4 Example: lim n√a, a > 0

Suppose a ≥ 1:√

1 ≤ n√a ≤ n

√n ie. lim n

√a = 1 for a ≥ 1.

If a < 1, lim n√a = lim 1

n√

1a

= 1 and 1a > 1 by the quotient rule of limits.

2.6 Definition: Increasing or Decreasing Sequences

We say (xn) is increasing if xn+1 ≥ xn and decreasing if xn+1 ≤ xn for alln ∈ N. We say (xn) is monotone if it is increasing or decreasing.

2.7 Monotone Convergence Theorem

Suppose (xn) is increasing and bounded above, then limxn = supxn = sup{xn :n ∈ N}.

Proof. Let ε > 0, l = supxn. Then l − ε is not an upper bound of {xn}∞n=1. ie.∃N : xN > l − ε. Since (xn) is increasing, xn ≥ xN , ∀n > N . ie.

l − ε < xN ≤ xn ≤ l < l + ε

Since l = supxn then |xn − l| < ε, ∀n ≥ N .

2.7.1 Example: If (xn) is decreasing and bounded below then (xn)converges to the infimum. ie. limxn = inf{xn}

Let (yn) : yn = −xn. Since yn is increasing and bounded above by -(lowerbound for xn) then:

limn→∞

yn = sup{yn} ⇐⇒ limn→∞

(−xn) = sup{−xn}

⇐⇒ − limn→∞

xn = − inf{xn}

⇐⇒ limn→∞

xn = inf{xn}

17

2.7.2 Example: Calculate limit of (an) given by a1 = 1, an+1 =√an + 1, n ≥ 1.

Claim: (an) is bounded above by 2, increasing, and hence has a limit.increasing : Using Induction. n = 1, a2,=

√a1 + 1 =

√1 + 1 =

√2, 1 = a1 <

a2 =√

2. Now suppose that an−1 ≤ an then an =√

1 + an−1 ≤√

1 + an =an+1 thus it’s increasing.

bounded above by 2 : Use induction again. a1 = 1 ≤ 2. Suppose an−1 ≤ 2.Then an =

√1 + an−1 ≤

√1 + 2 =

√3 <

√4 = 2. Hence an ≤ 2,∀n ∈ N by

induction.

Therefore an is increasing, bounded above by 2 and therefore there is a limitlim an = a by the monotone convergence theorem.

Now an+1 =√

1 + an =⇒ (an+1)2 = 1 + an,∀n ≥ 1, and (an+1)2 → a2,1 + an → 1 + a by the sum and product rules of limits. So then,

a2 = 1 + a ⇐⇒ a2 − a− 1 = 0 ⇐⇒ a = 1±√

5

Now an ≥ 1,∀n so lim an = a ≥ 1 and therefore:

a = lim an =1 +√

52

2.7.3 Example: Find a sequence of rationals converging to√a, a > 0,

a ∈ Q

x =√a ⇐⇒ x2 = a ⇐⇒ x2 − a = 0

Visualise a quadratic with the roots being√a,−√a. Let x1 > 0 and choose

xn+1 using Newton’s iterative method(assume for time being we’ve done differ-entiation). Then the slope of the tangent line is 2xn. Then the tangent lineequation can be constructed:

y − (x2n − a)

xn+1 − xn= 2xn

We know it intersects the x-axis when y = 0 so,

−(x2n − a) = 2xn(xn+1 − xn)−x2

n + a = 2xnxn+1 − 2x2n

x2n − 2xnxn+1 + a = 0

Solving for xn+1:

xn+1 =a+ x2

n

2xn=

12

(a

xn+ xn

)and xn > 0, ∀n ∈ N.

18

Claim: xn is decreasing for n ≥ 2.

xn+1 − xn =12

(a

xn+ xn

)− xn

= −12xn +

12a

xn

= −12

(xn −

a

xn

)< 0(since xn, a > 0)

ie. xn+1 < xn,∀n ≥ 1 ⇐⇒ (xn) decreasing.xn− a

xnis increasing because if we consider the quadratic, we get complex roots

if the discriminant less than 0, and we get repeated roots if the discriminant is0. But since we have distinct roots here, we have that the discriminant xn− a

xnis greater than 0.

Claim: x2n − a > 0,∀n ≥ 2

Now, x2n − 2xnxn+1 + a = 0 is a quadratic with root xn > 0 (and real). Since

this has real coefficients, its distinct roots are real if:discriminant = (−2xn+1)2 − 4a > 0=⇒ x2

n+1 − a > 0 =⇒ decreasing with xn ≥√a. Thus by the monotone

convergence theorem, it has a limit x and:

x2n−2xnxn+1+a = 0→ x2−2x2+a = 0 =⇒ x2 = a =⇒ x = ±

√a =⇒ x =

√a

Since xn > 0 =⇒ limxn ≥ 0 we infer x =√a. It can’t be both positive and

negative, due to uniqueness of limits.

2.8 Cauchy Sequences

Definition: A sequence x1, x2, . . . is called a Cauchy sequence if: ∀ε > 0, ∃N ∈ Rst. ∀n,m ∈ N and n,m ≥ N

|xn − xm| < ε

2.8.1 Lemma 9: Every convergent sequence is a Cauchy sequence

Proof. Let x1, x2, . . . be a convergent sequence. Let ε > 0 and l = limn→∞ xnexists (since the sequence is convergent). Since ε

2 > 0, ∃N : ∀n ∈ N, n > N wehave |xn − l| < ε

2 .Then ∀n,m ∈ N, with n,m ≥ N one has that |xn− l| < ε

2 and |xm− l| < ε2 .

So that:

|xn − xm| = |(xn − l)− (xm − l)| ≤ |xn − l|+ |xm − l| <ε

2+ε

2= ε

A Cauchy sequence in R is not necessarily convergent in Q. For example:∃(xn), xi ∈ Q, i ∈ N st limn→∞ xn =

√5 (in R). Then (xn) is a Cauchy sequence

in R and hence in Q. But the sequence is not convergent in Q since√

5 /∈ Q.

19

2.8.2 Theorem 3: Every sequence in R has a monotone subsequence.

Proof. Let x1, x2, . . . ∈ R. An element n ∈ N is called a lighted index if∀k > n, xk ≤ xn.

Case 1 : There are infinite many lighted indices.Call these n1 < n2 < n3 < . . ..Let i ∈ N. Since n is a lighted index and ni+1 < ni, taking k = ni+1 and n = nigives xni+1 ≤ xni . So xn1 , xn2 , . . . is a decreasing sequence.

Case 2 : There are finite many lighted indices.There exists an N ∈ N st. ∀n ≥ N the index n is not lighted. ie. ∀n ≥ N ,∃k > n: xk > xn. Now choose n1 = N . Then ∃k ∈ N, k > n st. xk > xn1 .Choose n2 = k. Then ∃n3 > n2 st. xn3 > xn2 and so on. Then xn1 , xn2 , . . . isa strictly increasing subsequence.

2.8.3 Corollary: (Bolzano / Weierstraß) Theorem

Every bounded sequence in R has a convergent subsequence.

Proof. Let (xn) be a bounded sequence in R. By theorem 3, (xn) has a mono-tone subsequence xn1 , xn2 , . . .. Then (xni), i ∈ N is also bounded. Since everybounded monotone sequence is convergent, then our subsequence (xni) is con-vergent.

2.8.4 Lemma 10: Every Cauchy sequence which has a convergentsubsequence is convergent.

Proof. Let (xn) be a Cauchy sequence and assume that xn1 , xn2 , . . . is a conver-gent subsequence. Let l = limk→∞ xnk exist.Let ε > 0 and since limxnk = l implies ∃N : ∀k ≥ N1, |xnk − l| < ε

2 . Since (xn)is a Cauchy sequence ∃N2 ∈ N st. ∀n,m ≥ N2 one has |xn − xm| < ε

2 . Now setN = N2.Then ∀n ∈ N with n ≥ N one has with k = max(N1, N2) that nk ≥ k ≥ N2 andN ≥ N2 so that |xn − xnk | < ε

2 . Moreover, k ≥ N1 so |xnk − l| < ε2 . Hence we

have:

|xn − l| = |xn − xnk + xnk − l| ≤ |xn − xnk |+ |xnk − l| <ε

2+ε

2= ε

2.8.5 Lemma 11: Every Cauchy sequence is bounded.

Proof. Let (xn) be our sequence. Let ε = 37. By our assumption, there existsan N ∈ N st. ∀n,m ∈ N with n,m ≥ N one has |xn − xm| < 37. Take m = N.Then ∀n ≥ N :

|xn| = |(xn − xN ) + xN | ≤ |xn − xN |+ |xN | < 37 + |xN |

Thus ∀n ∈ N , |xn| ≤ max(|x1|, . . . , |xN |, 37+ |xN |). Hence (xn) is bounded.

20

So ∀n,m ∈ N with n,m ≥ N + 1 one has:

Case 1 : If n > m then:

|xn − xm| ≤1

1− C· Cm−1 · |x2 − x1|

≤ 11− C

· CN |x2 − x1|

<1

1− C· (1− C)ε|x2 − x1|

· |x2 − x1| = ε

Case 2 : If n < m then similarly, |xn − xm| < ε.Case 3 : If n = m then |xn − xm| = 0 < ε.

2.9.2 Corollary: Every contractive sequence is convergent

Because every contractive sequence is Cauchy and every Cauchy sequence isconvergent.

2.9.3 An application example in numerical analysis

Consider the equation x3 − 7x+ 2 = 0 for x ∈ (0, 1). We have that 7x = x3 + 2and x = 1

7 (x3 + 2).Define x1 = 1

2 . Now, ∀n ∈ N, xn+1 = 17 (x3

n + 2).

Claim: xn ∈ (0, 1), ∀n ∈ N. Use induction. n = 1: x1 = 12 ∈ (0, 1) is true.

Suppose xn ∈ (0, 1). Then xn+1 = 17 (x3

n + 2) ≤ 17 (1 + 2) = 3

7 which is also in(0, 1). Therefore xn ∈ (0, 1) for all n ∈ N.

Claim: Sequence is contractive.Take n ∈ N. Then |xn+2 − xn+1| = | 17 (x3

n+1 + 2)− 17 (x3

n + 2)| = 17 |x

3n+1 − x3

n|.Recall that a2− b2 = (a− b)(a+ b); a3− b3 = (a− b)(a2 + ab+ b2); a4− b4 =

(a− b)(a3 + a2b+ ab2 + b3). So then:

17|x3n+1 − x3

n| =17|xn+1 − xn||x2

n+1 + xn+1xn + x2n|

≤ 17|xn+1 − xn|(|x2

n+1|+ |xn+1xn|+ |x2n|)

≤ 37|xn+1 − xn|

So the sequence is contractive with C = 37 < 1. If limxn = l then |xn−xm| ≤

11−C · C

m−1|x2 − x1|, ∀n > m. Now fix m and take the limit. Then

|l − xm| ≤Cm−1

1− C|x2 − x1| , ∀m ∈ N

Now all you have to do is calculate m using a computer.

22

2.10 R is dense

Proof. Let x ∈ R. Let xn = 10−nb10nxc ∈ Q. We claim:

limn→∞

xn = x

Now,

10−n(10nx− 1) ≤ 10−nb10nxc ≤ 10−n10nx

x− 10−n ≤ xn ≤ x

3 Series

Partial Sums: Given a sequence x1, x2, . . . ∈ R. The sequence x1, x2, . . . isconvergent if limxn exists. Now we define:

s1 = x1

s2 = x1 + x2

s3 = x1 + x2 + x3

sn = x1 + . . .+ xn =n∑k=1

xk nth partial sum

Definition: The series∑xk is the sequence s1, s2, . . . of partial sums. The

series∑xk is called convergent if lim sn exists. Otherwise it’s called divergent.

If the series∑xk is convergent then lim sn is called the sum.

Notation:∞∑k=1

xk := limn→∞

sn ∈ R

In other literature, some people use that same notation for the series, so it canbe a little confusing where each iteration gives you another partial sum in thesequence. Note that xk is called the kth term in the series

∑xk.

Sometimes∑∞k=5 xk is used to denote the limit of the sequence s5 + s6 + . . ..

3.1 Lemma 13

Let x1, x2, . . . and y1, y2, . . . be real sequences. Suppose the series∑xk and∑

yk are both convergent. Then∑

(xk + yk) is convergent.

Proof.

Define s(x)n =n∑k=1

xk , s(y)n =n∑k=1

yk , s(x+y)n =n∑k=1

(xk + yk)

Then s(x+y)n = s

(x)n + s

(y)n using a result in linear algebra because we know real

numbers are commutative.

limn→∞

s(x+y)n = limn→∞

(s(x)n + s(y)n ) = limn→∞

s(x)n + limn→∞

s(y)n

23

Therefore s(x+y)n is convergent and∑

(xk + yk) is convergent. Hence

∞∑k=1

(xk + yk) =∞∑k=1

xk +∞∑k=1

yk

3.1.1 Example: Geometric Series

Fix r ∈ R and set xk = rk, ∀k ∈ {0, 1, 2, . . .}. Set:

sn =n∑k=0

xk =n∑k=0

rk = 1 + r + r2 + . . .+ rn =

{1−rn+1

1−r if r 6= 1n+ 1 if r = 1

Recall that lim( 11+a )n = lim 1

(1+a)n = 0 or if r ∈ (0, 1) with a = 1r − 1 > 0

then lim rn = 0.Note if 0 < r < 1 then lim rn+1 = 0. If r = 0 then lim rn+1 = 0. If

−1 < r < 0 then lim |rn+1| = lim |r|n+1 = 0. Hence lim rn+1 = 0.Now, if r = 1 then lim rn+1 = 1.

If r > 1 then the sequence rn+1 is divergent. We can see that using theBernoulli inequality if we pick

rn+1 ≥ 1 + (n+ 1)(r − 1)

So that it is increasing for all n ∈ N and r > 1.If r < −1 then it is also divergent as |rn+1| is divergent. If r = 1, then

rn+1 = (−1)n+1 which we also know is divergent.

Therefore, the sequence (sn) is

{convergent if −1 < r < 1divergent if r ≥ 1, r ≤ 1

So the series∑rk is convergent ⇐⇒ |r| < 1 ⇐⇒ −1 < r < 1. Moreover,

∞∑k=0

rk =1

1− rif |r| < 1

3.1.2 Example: Telescopic Series

Let xk = 1k(k+1) , k ∈ N. Note that 1

k(k+1) = 1k −

1k+1 , ∀k ∈ N. Then one has

that ∀n ∈ N,

sn = x1 + x2 + . . .+ xn−1 + xn = (11− 1

2) + (

12− 1

3) + . . .+ (

1n− 1n+ 1

)

= 1− 1n+ 1

So then lim sn = lim(1 − 1n+1 ) = 1 exists. Thus the series

∑1

k(k+1) is con-vergent and

∑1

k(k+1) = 1.

The motivation for the week is to see if∑

1n2 converges.

24

3.2 Lemma 14: N th Term Test

If the series∞∑k=1

xk is convergent then limn→∞

xn = 0

Proof. Suppose the series∑xk is convergent, s = lim sn ∈ R exists. Note that,

sn − sn−1 = (x1 + . . .+ xn)− (x1 + . . .+ xn−1) = xn

Then limn→∞

xn = limn→∞

(sn − sn−1) = ( limn→∞

sn)− ( limn→∞

sn−1) = s− s = 0

3.2.1 Corollary

If we do not have limxn = 0 then∑xk is divergent. This is just the con-

trapositive statement of the lemma. Note however this does not mean that iflimxn = 0 then the series is convergent. A counter example is the harmonicseries

∑1n which diverges.

NB:If yn = xn+5 then

y1 + . . .+ yn = x6 + . . .+ xn+5 = x1 + . . .+ xn −5∑k=1

xk

∞∑k=1

yk =∞∑k=1

xk −5∑k=1

xk

3.3 Theorem 5

Suppose x1, x2, . . . are in [0,∞). Then the series∑xk is convergent ⇐⇒ the

sequence s1, s2, . . . of partial sums is bounded above.

Proof. Since x1, x2, . . . ≥ 0 the sequence s1, s2, . . . is increasing. So∑xk is

convergent if s1, s2, . . . is convergent. If this monotone sequence is boundedabove, then by the monotone convergence theorem, it is convergent.

3.3.1 Interlude: Consider∑

1n

s1 = 1

s2 = 1 +12

=32

s3 = 1 +12

+13

+14≥ 3

2+

12

=42

s8 = 1 + . . .+14

+15

+16

+17

+18≥ 4

2+

12

=52

s16 =52

+12

=62

s2n =n+ 2

2Where sn is the sequence of partial sums, so sn is not bounded above. In

particular, it is a divergent (Theorem 5). Note that lim 1k = 0 but

∑1k is

divergent. Also that lim( 12 )k = 0 and

∑( 12 )k is convergent.

25

3.4 Theorem 6: Comparison Test

Suppose that (xn) and (yn) are sequences. Suppose 0 ≤ xn ≤ yn ∀n ∈ N.

I : Suppose∑yn is convergent. Then the series

∑xk is convergent.

Proof. Since∑yk is convergent, the sequence

∑nk=1 yk is bounded above, so

∃M ∈ R: ∀n ∈ N∑nk=1 yk ≤ M . But then ∀n ∈ N:

∑nk=1 xk ≤

∑nk=1 yk ≤ M .

hence∑nk=1 xk is also bounded above. Thus by Theorem 5,

∑xk is convergent.

II : Suppose∑xk is divergent. Then

∑k yk is divergent. This is just the

contrapositive of the above statement.

3.4.1 Example:∑

1k2

Recall that∑

1k(k+1) is convergent.

1n(n+ 1)

≥ 1n(n+ n)

=1

n · 2n=

12· 1n2

So that 0 ≤ 12 ·

1n2 ≤ 1

n(n+1) . Then by the comparison test, the series∑

( 12 ·

1k2 )

is convergent. So that∑

1k2 is convergent (using lemma 15 below).

3.4.2 Lemma 15:∑xk is convergent ⇐⇒

∑cxk is convergent for

some constant c 6= 0

Proof. Let s1n =∑nk=1 xk and s2n =

∑nk=1 cxk for some c ∈ R. Note that:

s2n = cx1 + cx2 + . . .+ cxn = c(x1 + . . .+ xn) = cs1n

“⇒”: If lim s1n = S1 then lim s2n = lim(cs1n) = c lim(s1n) = c ∗ S1 and henceconvergent.“⇐”: If lim s2n = lim(cs1n) = c lim(s1n) = S2 then in particular, lim(s1n) is alsoconvergent and lim s1n = S2

c .

3.5 Theorem 7

Suppose (xn) : xn > 0 and (yn) : yn ≥ 0, ∀n ∈ N. Suppose M ∈ [0,∞) and∀n ∈ N : yn

xn≤ M . If the series

∑xk is convergent, then the series

∑yk is

convergent.

Proof. ∀n ∈ N one has 0 ≤ yn ≤Mxn. The series∑xk is convergent hence also

the series∑Mxk is convergent. So by comparison,

∑yk is also convergent.

3.5.1 Corollary 1: Limit Comparison Test

Suppose xn > 0 and yn > 0 ∀n ∈ N. Suppose also that lim ynxn

exists. If theseries

∑xk is convergent, then the series

∑yk is convergent.

Proof. Since lim ynxn

exists, we know it is bounded. So in particular, ∃M : ∀n ∈ Nynxn≤M . Now apply Theorem 7.

26

3.5.2 Corollary 2: Also Limit Comparison Test

Suppose xn > 0 and yn > 0 ∀n ∈ N. Suppose L := lim ynxn

exists.I : If L > 0 then the series

∑xk is convergent ⇐⇒

∑yk convergent.

Proof. “⇒”: Use Corollary 1. “⇐”: lim xnyn

= 1L exists. So use Corollary 1.

II : If L = 0 then if xn is convergent, =⇒∑yn convergent.

Proof. Use Corollary 1.

3.5.3 Example:∑

n+5n3+2n+1 , xn = n+5

n3+2n+1

Set yn = 1n2 then xn > 0 and yn > 0 ∀n ∈ N. Moreover,

limn→∞

ynxn

= limn→∞

(1n2· n

3 + 2n+ 1n+ 5

)=

1 + 2n2 + 1

n3

1 + 5n

= 1 > 0

But∑

1k2 =

∑yk is convergent. Hence by the comparison test the series

∑xk

is also convergent.

3.6 Definition: Absolute Convergence

A series∑xk is called absolutely convergent if the series

∑|xk| is convergent.

3.7 Theorem 8: Every absolutely convergent series is con-vergent.

Proof. Let∑xk be an absolutely convergent series. Then the series

∑|xk| is

convergent. Now, define yk := |xk| and zk := |xk| − xk. Then xk = yk − zk∀k ∈ N. obviously the series

∑yk is convergent. Note however that:

0 ≤ zk = |xk| − xk ≤ |xk|+ |xk| = 2|xk|

Thus the series∑

2|xk| is convergent, and by comparison the series∑zk is

convergent. Therefore∑xk =

∑(yk − zk) is also convergent.

3.8 Ratio Test 1

Suppose x1, x2, . . . ∈ R, xn 6= 0 ∀n ∈ N.1a: If ∃r < 1 (ie. fixed r) st. ∀n ∈ N

∣∣∣xn+1xn

∣∣∣ ≤ r. Then the series∑xk is

absolutely convergent. Note that only the tail is important as per usual. ie. forall large n.

Proof. ∀n ∈ N one has |xn+1| ≤ r|xn| so by induction, ∀n ∈ N one has 0 ≤|xn| ≤ rn−1|x1| < 1 therefore

∑ |x1|r rn is convergent.

By comparison, the series∑|xk| is also convergent. So the series

∑xk is

absolutely convergent and hence convergent.

1b: Suppose ∀n ∈ N :∣∣∣xn+1xn

∣∣∣ ≥ 1. Then the series is divergent.

Proof. ∀n ∈ N one has that |xn+1| ≥ |xn| so by induction |xn| ≥ |x1| ∀n ∈ N.Since |x1| > 0 one does not have limxn = 0. So the series

∑xk is divergent by

the n-th term test.

27

3.9 Ratio Test 2

Suppose x1, x2, . . . ∈ R, xn 6= 0 ∀n ∈ N. Suppose ρ := lim∣∣∣xn+1xn

∣∣∣ exists.Case 1 If ρ < 1 then the series

∑xk is absolutely convergent.

Case 2 If ρ > 1 then the series is divergent.Case 3 If ρ = 1 then it is inconclusive.

Proof. If ρ < 1, set r = 1+ρ2 so that ρ < r < 1. Choose ε = r− ρ > 0 then there

is an N ∈ N : ∀n ≥ N ρ− ε <∣∣∣xn+1xn

∣∣∣ < ρ+ ε. In particular,∣∣∣xn+1xn

∣∣∣ < ρ+ ε = r

∀n ≥ N . Then using 1a. the series is absolutely convergent.If ρ > 1 then similarly,

∣∣∣xn+1xn

∣∣∣ ≥ 1 ∀large n so that the series∑xk is

divergent by 1b.

3.9.1 Example: For ρ = 1∑1nρ with ρ ∈ N is fixed. xn = 1

nρ .

limn→∞

∣∣∣∣xn+1

xn

∣∣∣∣ = limn→∞

nρ

(n+ 1)ρ= limn→∞

(n

n+ 1

)ρ= 1ρ = 1

NB: If ρ = 1 then the series is divergent and if ρ > 1 the series is conver-gent(integral test).

3.9.2 Example 2

Let xn = nn−1 Then |xn+1

xn| =

n+1nnn−1

= n+1n ·

n−1n = n2−1

n2 < 1 ∀n ∈ N. However,∑xn is divergent because limxn = 1 6= 0 (n-th term test).

4 Power Series

Definition: Let the coefficients of the power series c ∈ R and c0, c1, c2, . . . ∈ Rbe fixed. Then the series

∑∞k=0 ck(x− c)k where x ∈ R is called a power series

about c.Note that we only consider terms for k ≥ 0 so no negative ones.

4.1 Problem

Given a power series, determine for which x ∈ R this series converges. Clearlyit is the case for x = c. However for the x = c case, we have interesting issues todeal with. First k cannot be negative or we will have division by zero. Secondly,remember the case where 00 = 1.

We may assume c = 0 otherwise translate and use y = x− c.

4.2 Theorem 9

Let∑∞k=0 ckx

k be a power series. Let x0 ∈ R. If∑ck · xk0 is convergent, then

∀y ∈ R with |y| < |x0| then the series∑ck · yk is also convergent.

28

Proof. We may assume that |x0| > 0. We know that the series∑ckx

k0 is

convergent. Therefore lim cnxn0 = 0 so the sequence (cnxn0 ) is bounded. That

means ∃M ∈ R : ∀n ∈ N |cnxn0 | ≤M . Then ∀n ∈ N:

|ckyk| = |ckxk0 |(|y||x0|

)k≤M

(|y||x0|

)kThe geometric series

∑M(|y||x0|

)kis convergent. Then, by comparison, the

series∑|ckyk| is also convergent so therefore the series

∑cky

k is absolutelyconvergent and hence convergent.

4.2.1 Corollary 1

If∑ckx

k0 is divergent, then

∑cky

k is divergent for all y ∈ R with |y| > |x0|.This is just the contrapositive.

4.2.2 Corollary 2

Let∑∞k=0 ck(x − c)k be a power series. Then there is a unique R called the

convergence radius, 0 ≤ R ≤ ∞ st. ∀x ∈ R : |x − c| < R the series∑ckx

k

is convergent, and that if |x − c| > R then it is divergent. It is called theconvergence radius because it looks like a circle in C or R2.

Proof. We may assume c = 0. If∑ckx

k is convergent, for all x ∈ R then setR =∞. Define otherwise R = sup{|x| : x ∈ R and

∑ckx

k is convergent }. If itis convergent only at x = 0 then set R = 0.

Remark : If x = R or x = −R then we are unsure if it converges or diverges.

4.2.3 Example

:∞∑k=0

xk is convergent ⇐⇒ |x| < 1. If R = 1 then it is divergent at both R and −R

∞∑k=1

xk

k2is convergent ⇐⇒ |x| ≤ 1. If R = 1, convergent at both R and −R.

∞∑k=1

xk

kis convergent ⇐⇒ −1 ≤ x < 1. If R = 1 Convergent at −R and divergent at R.

4.2.4 More Examples∑∞n=0

xn

n! . Use the ratio test. If x 6= 0 then

limn→∞

∣∣∣∣∣∣xn+1

(n+1)!xn

n!

∣∣∣∣∣∣ = limn→∞

∣∣∣∣xn+1

xn· n!

(n+ 1)!

∣∣∣∣ = limn→∞

|x| 1n+ 1

= 0 < 1

This implies ∀x ∈ R the power series is convergent. So R =∞.

29

4.2.5 Some common functions as power series

• exp:R→ R exp(x) =∑∞n=0

xn

n! = ex.

• sin(x) =∑∞n=0(−1)n x2n+1

(2n+1)! , R =∞.

• cos(x) =∑∞n=0(−1)n x2n

(2n)! , R =∞.

4.2.6 Another example

Find R, given∑

n3nx

n. Let x ∈ R and x 6= 0, then

limn→∞

∣∣∣∣∣(n+1)xn+1

3n+1

nxn

3n

∣∣∣∣∣ = limn→∞

∣∣∣∣n+ 1n· 3n

3n+1· x

n+1

xn

∣∣∣∣ = limn→∞

((1 +

1n

) · 13· |x|

)=|x|3

Let ρ = |x|3 here. Then the series is convergent if ρ < 1. ie. if |x| < 3. Similarly

it is divergent if ρ > 1 ie .|x| > 3. After testing at R and −R we conclude it isdivergent at both.

4.3 Theorem 10

Let x1, x2, . . . ≥ 0.

4.3.1 If the series∑xk is convergent then the set S is bounded above

and supS ≤∑∞n=1 xn

Proof. Let J ⊂ N, J 6= ∅, J finite. Say, J = {i1, . . . , ik}. If N = max(i1, . . . , ik)then ∑

n∈Jxn = xi1 + . . .+ xik ≤ x1 + . . .+ xN = sn ≤

∞∑n=1

xn

So∑∞n=1 xn is an upper bound for S. Moreover, supS ≤

∑∞n=1 xn.

4.3.2 If the set S is bounded above, then the series∑xn is conver-

gent and∑∞n=1 xn ≤ supS.

Proof. For all n ∈ N one has

sn = x1 + . . .+ xn =∑

k∈{1,...,n}

xk ≤ supS

So the partial sums are bounded above. Hence the series∑xn is convergent

and∑∞n=1 xn ≤ supS.

4.3.3 If the series∑xn is convergent, then

∑∞n=1 xn = supS.

4.3.4 Corollary (Rearrangement Theorem)

Let π : N → N be a bijection. Suppose x1, . . . ≥ 0 then∑xn is convergent

⇐⇒ the series∑xπ(n) :=

∑yn is convergent. Moreover,

∞∑n=1

xn =∞∑n=1

xπ(n)

30

5 Limits of Functions

Definition: An interval is a subset S of a totally ordered set T with the propertythat whenever x, y ∈ S and x < z < y then z ∈ S.

5.1 Open Sets

Let A ⊂ R. The set A is open if ∀x0 ∈ A: ∃δ > 0, (x0 − δ, x0 + δ) ⊂ A. ie.There exists a neighbourhood of x0 contained in A.

5.1.1 Examples

A union of open intervals is open. Suppose A is such a union, and x0 ∈ A.Then x0 ⊂ (a, b) ⊂ A. WLOG, suppose (a, b) is finite (and if it isn’t we canreplace it by (x0 − 1, x0 + 1) ∩ (a, b) ). Let δ = min{x0 − a, b − x0}. Then(x0 − δ, x0 + δ) ⊂ (a, b) ⊂ A. in particular, an open interval is open.

5.2 Cluster Points

A cluster point is sometimes called a limit point or an accumulation point. Wesay x0 is a cluster point of A ⊂ R if ∀δ > 0: ∃a ∈ A

0 < |x0 − a| < δ

ie. ∀δ > 0, (x0 − δ, x0 + δ) ∩ A\{x0} 6= ∅. The 0 < |x0 − a| guarantees thata 6= x0.

5.2.1 Theorem 11

x0 is a cluster point of A ⊂ R ⇐⇒ there is a sequence (an) of distinct pointsin A\{x0} which converges to x0.

Proof. “⇒”: Suppose x0 is a cluster point of A.Claim: We can choose a sequence (an) with 0 < |x0 − an| < 1

n and so by sand-wich theorem, an → x0 as n→∞.Suppose first n terms have been chosen. Let δ = min{ 1

n+1 , |a1 − x0|, |a2 −x0|, . . . , |an − x0|} and choose an+1 ∈ A with 0 < |x0 − an+1| < δ so that|an+1 − x0| < δ ≤ 1

n+1 .

For j = 1, . . . , n |aj −x0| > |an+1−x0| so that an+1 /∈ {a1, . . . , an}. ie. thatthey are distinct.

“⇐”: Suppose we have such a sequence in A\{x0}. Then for δ > 0, ∃N sothat 0 < |an − x0| < δ, ∀n > N hence x0 is a cluster point of A.

5.2.2 Examples

1. If A is an interval, then every point in A is a cluster point, as is everyfinite end point.

2. A finite set has no limit points

3. The infinite set N has no limit points.

31

4. The set A = { 1n : n ∈ N} has only one limit point 0 /∈ A.

5. (0, 1) ∩Q has cluster points [0, 1]

5.3 Definition of limits of a function

Let x0 be a cluster point of A ⊂ R and if f : A→ R and L ∈ R then we say Lis the limit of f at x0 if:∀ε > 0, ∃δ > 0

|f(x)− L| < ε , ∀x ∈ A, 0 < |x− x0| < δ

If A is an interval with endpoints a, b respectively, then we often write:

limx→a

f(x) = limx→a+

f(x)

limx→b

f(x) = limx→b−

f(x)

We write limx→x0

f(x0) = L or f(x)→ L as x→ x0

5.4 Limits of functions are unique

Proof. Suppose f has limits L1, L2 at x0. ie.∀e > 0, ∃δ1 > 0, δ2 > 0

|f(x)− L1| <ε

2∀x ∈ A, 0 < |x− x0| < δ1

|f(x)− L2| <ε

2∀x ∈ A, 0 < |x− x0| < δ2

Let δ = min{δ1, δ2}. Then ∀x ∈ A, 0 < |x− x0| < δ

|L1−L2| = |L1− f(x) + f(x)−L2| ≤ |f(x)−L1|+ |f(x)−L2| <ε

2+ε

2= ε

So then ∀ε > 0, 0 ≤ |L1 − L2| < ε =⇒ |L1 − L2| = 0 =⇒ L1 = L2.

5.4.1 Example 1

Suppose f is the linear polynomial given by f(x) = α+β, α, β ∈ R and x0 ∈ R.Then we claim that:

limx→x0

f(x) = αx0 + β

|f(x)− (αx0 + β)| = |αx+ β − (αx0 + β)|= |αx− αx0|= |α||x− x0|

Let ε > 0, then if α = 0, |f(x) − (αx0 + β)| = 0 < ε =⇒ limit is B. If α 6= 0then choose δ = ε

|α| > 0. So for 0 < |x− x0| < δ we have

|f(x)− (αx0 + β)| = |α||x− x0| < |α|δ = |α| ε|α|

= ε

Therefore, limx→x0

f(x) = αx0 + β

32

5.4.2 Example 2

Claim: limx→x0

x2 = x20

Consider: |x2 − x20| = |(x− x0)(x+ x0)|. Suppose |x− x0| < 1. Then

|x+ x0| = |x− x0 + 2x0| ≤ |x− x0|+ 2|x0| < 1 + 2|x0|

Let ε > 0 and choose δ = min{ ε2|x0|+1 , 1}. Then if 0 < |x − x0| < δ we have

|x+ x0| ≤ 2|x0|+ 1. So,

|f(x)−x20| = |x2−x2

0| = |x−x0||x+x0| ≤ |x−x0|(2|x0|+1) <ε

2|x0|+ 1(2|x0|+1) = ε

Therefore limx→x0

x2 = x20

5.5 Theorem 12

Let f : A→ R, A ⊂ R, x0 is a cluster point of A, and L ∈ R. Then the followingare equivalent:

1. limx→x0 f(x) = L

2. Given any ε-neighbourhood (L − ε, L + ε) of the point L, ∃δ > 0 withf((x0 − δ, x0 + δ) ∩A\{x0}) ⊂ (L− ε, L+ ε) = Bε(L).

3. For every sequence (an) of the points in A\{x0} which converges to x0,limn→∞ f(an) = L.

Proof. (1)→(2): Suppose limx→x0 f(x) = L. ie ∀ε > 0, ∃δ > 0, 0 < |x−x0| < δ,

|f(x)− L| < ε

⇐⇒ x ∈ (x0 − δ, x0 + δ) ∩A\{x0} =⇒ f(x) ∈ (L− ε, L+ ε)

Where |x−x0| < δ ⇐⇒ (x0− δ, x0 + δ) and A\{x0} ⇐⇒ x ∈ A, 0 < |x−x0|.

(2) → (3): Suppose (2) holds. Let (an) be a sequence in A\{x0} convergingto x0. For ε > 0 choose δ > 0 given by (2). Since an → x0, ∃N : ∀n > N :

|an − x0| < δ

So for n > N |an − x0| < δ =⇒ |f(an)− L| < ε. ie.

limn→∞

f(an) = L

(3) → (1): We will prove this via a contrapositive statement. Suppose (i) isfalse. Then We prove (iii) is false.(i) false ⇐⇒ ∃ε > 0 for which there is no δ > 0 such that |f(x) − L| < ε. ie.∀δ > 0,∃x ∈ A, 0 < |x−x0| < δ, |f(x)−L| ≥ ε. Take δ = 1

n then choose pointsan ∈ A, 0 < |an − x0| < 1

n and |f(an) − L| ≥ ε. Since |an − x0| > 0, (an) is asequence in A\{x0}, and also since |an − x0| < 1

n → 0 as n→∞.Hence by squeeze theorem, an → x0 and |f(an)−L| ≥ ε, ∀n ∈ N . Therefore

f(an) does not have a limit L.

33

5.5.1 Example 1

The limit limx→∞ f(x) is unique and is given by limn→∞ f(an) where (an) isany sequence in A\{x0} which converges to x0.

5.5.2 Example 2: Show limx→1x+1x2+3 = 1

2

Suppose xn → 1 as n→∞. Then limn→∞xn+1x2n+3 = lim xn+1

(lim xn)2+3 = 1+11+3 = 1

2 .So by (iii) → (i) we have that limx→1

x+1x2+3 = 1

2 .

5.5.3 Example 3: Consider limx→0+1x , where f(x) = 1

x ,∀x > 0

Let xn = 1n and xn → 0 as n → ∞ and also that (xn > 0 ∀n ∈ N) ie. ∀x > 0.

Then the sequence 1xn

= n does not converge (as it is unbounded). So then 1x

has no limit as x→ 0+.

5.6 Theorem 13

Suppose x0 is a cluster point of A and f, g : A→ R have limits at x0. Then:

(i) limx→x0

(f + g)(x) = limx→x0

f(x) + limx→x0

g(x)

(ii) limx→x0

(cf(x)) = limx→x0

c · limx→x0

f(x) = c · limx→x0

f(x) , c ∈ R

(iii) limx→x0

(fg)(x) = ( limx→x0

f(x))( limx→x0

g(x))

(iv) limx→x0

(f

g

)(x) =

limx→x0 f(x)limx→x0 g(x)

if limx→x0

g(x) 6= 0

NB: If limx→x0 g(x) 6= 0 then, ∀ε > 0, ∃δ > 0

|g(x)− L| < ε ⇐⇒ L− ε < g(x) < L+ ε

For x close to x0, g(x) 6= 0 so limx→x0

(fg

)is “well defined”.

Proof. (i) Suppose an → x0 where (an) is in A\{x0}. Then (f + g)(x) =f(an) + g(an). By the sequential convergence criterion for the limit of f and gto exist, f(an)→ f(x) and g(an)→ g(x).

So by the sequential criterion, f+g has the limit limx→x0 f(x)+limx→x0 g(x).Similarly, for (ii), (iii), (iv).

5.7 Squeeze Theorem

Suppose x0 is a cluster point of A ⊂ R, f, g, h : A→ R and f(x) ≤ g(x) ≤ h(x)∀x ∈ A (or just neighbourhood of x0) and limx→x0 f(x) = limx→x0 h(x) = L.Then:

limx→x0

g(x) = L

The proof is via sequential criterion again.

34

5.7.1 Example 1: limx→0+ x · sin( 1x ) = 0

Since −1 ≤ sin( 1x ) ≤ 1, for x 6= 0 and −x ≤ x · sin( 1

x ) ≤ x, ∀x > 0 andlimx→0+ |x| = 0 then by the squeeze theorem we have that:

limx→0+

x · sin(1x

) = 0

5.7.2 Example 2: f(x) = sin( 1x ) has no limit at x = 0

Since we take xn = 1π2 +nπ , then f( 1

xn) = sin(π2 + nπ) = (−1)n+1 diverges. So f

has no limit as x→ 0.

5.8 Definition: Limits to Infinity

Suppose A ⊂ R and (a,∞) ⊂ A for some a ∈ R. Then we say f : A→ R has alimit L ∈ R if ∀ε > 0, ∃M ∈ R:

|f(x)− L| < ε , ∀x > M

and we write

L = limx→∞

f(x)

Similarly, if (−∞, b) ⊂ A for some b ∈ R we say

limx→−∞

f(x) = L

if ∀ε > 0, ∃M

|f(x)− L| < ε , ∀x < M

5.8.1 Remark

Similar to limits of sequences of functions, the limits as n→ ±∞ are unique ifthey exist and they also satisfy the usual rules.

5.8.2 Example 1: Show limx→∞1√x

= 0

Let ε > 0, and x > 0.

| 1√x− 0| = | 1√

x| = 1√

x< ε ⇐⇒ x >

1ε2

Let ε > 0, and choose M = 1ε2 then ∀x > M = 1

ε2

| 1√x− 0| = 1√

x< ε

Therefore

limx→∞

1√x

= 0

35

5.8.3 Example 2: limx→−∞xx+1

| xx+1 − 1| = 1|x+1| and for x < −1 we have 1

|x+1| = 1−x−1 so 1

−x−1 < ε ⇐⇒ x <

−1− 1ε .

Let ε > 0 then choose M = −1− 1ε

| x

x+ 1− 1| = 1

|x+ 1|< ε , ∀x < M = −1− 1

ε

5.8.4 Calculate limx→∞2x3−3x+6x3−5x+1

Note by product rule, limx→∞1x = (limx→∞

1√x

)(limx→∞1√x

) = 0 · 0 = 0. Soby the product, sum and quotient rules

limx→∞

2x3 − 3x+ 6x3 − 5x+ 1

= limx→∞

2− 3( 1x )2 + 6( 1

x )3

1− 5( 1x )2 + ( 1

x )3=

2− 3 · 0 + 6 · 01− 5 · 0 + 0

= 2

6 Continuity

f is continuous at x0 if f(x) comes close to f(x0) as x comes close to x0.

6.1 Definition

Let A ∈ R and f : A→ R and x0 be a point of A. Then we say f is continuousat x0 if ∀ε > 0, ∃δ > 0

|f(x)− f(x0)| < ε , ∀x ∈ A, |x− x0| < δ

We say f is continuous on B ⊂ A if f is continuous at every point in B. “fis continuous” if it is continuous on A.

6.1.1 Example 1

If f is continuous on B then f is continuous on any subset C ⊂ B.

6.1.2 Example 2

Let f : N→ R be any function. Then f is continuous. Since if ε > 0, take δ ≤ 1and x0 ∈ N

|x−x0| < δ ≤ 1 =⇒ x = x0 =⇒ |f(x)− f(x0)| = |f(x0)− f(x0)| = 0 < ε

6.1.3 Example 3

Let f : R → R and f(x) =

{1 x ∈ Q0 x /∈ Q

Then f is continuous at NO points.

This is because, if x0 is fixed, x0 ∈ R and δ > 0 then we can choose q ∈ Q andy ∈ R\Q, |x0 − q|, |x0 − y| < δ and

|f(x0)− f(q)|, |f(x0)− f(y)| = {0, 1}

and so we cannot have |f(x0)− f(x)| < 1 for any x ∈ (x0 − δ, x0 + δ), ∀δ > 0.

36

6.2.1 Corollary 1

If f, g : A → R x0 ∈ A, f, g both continuous at x0. Then the following arecontinuous at x0:

1. (f + g)(x)(from III).

2. (f · g)(x) = (limn→∞ f(xn))(limn→∞ g(xn)) = f(x0) · g(x0)

3. fg (x) = (limn→∞ f(xn))

(limn→∞ g(xn)) = f(x0)g(x0)

, if g(x0) 6= 0.

Proof. If x1, x2, . . . ∈ A with limn→∞ xn = x0 then limn→∞ f(xn) = f(x0) andlimn→∞ g(xn) = g(x0) So limn→∞(f+g)(xn) = limn→∞ f(xn)+limn→∞ g(xn) =f(x0) + g(x0). Similarly, for the other proofs.

6.2.2 Corollary 2

Every polynomial and rational function is continuous. Rational function is onewhere you divide a polynomial with another.

6.2.3 Corollary 3

Let f : A→ R, g : B → R. Suppose f(A) ⊂ B. Suppose x0 ∈ A, f is continuousat x0 and g continuous at f(x0). Then gof(x) is continuous at x0.

Proof. Let x1, x2, . . . ∈ A and suppose limn→∞ xn = x0. Since f is continuous atx0, one has limn→∞ f(xn) = f(x0) hence limn→∞ g(f(xn)) = g(f(x0)). Becauseg is continuous at f(x0), and g(f(x0)) ⇐⇒ gof(x0) then gof(x) is continuousat x0.

6.2.4 Example 1

f : R→ R, f(x) =

{x sin 1

x if x 6= 00 if x = 0

Claim: f is continuous. We only need to show it is continuous at 0. Letε > 0 and choose δ = ε. Then ∀x ∈ R with |x− 0| < δ one has:case 1 : When x = 0, |f(x)− f(0)| = 0 < ε.case 2 : When x 6= 0, |f(x) − f(0)| = |x sin 1

x | − 0| = |x sin 1x | = |x|| sin 1

x | ≤|x| < δ = ε.

So f is continuous at 0.

6.2.5 Corollary 4

Define g : R→ R. g(x) =

{x2 sin 1

x if x 6= 00 if x = 0

then g is also continuous.

Proof. g(x) = x · f(x) is a product of continuous functions and therefore con-tinuous.

38

6.2.6 Example 2

Define f : R → R. f(x) =

{1 if x ∈ Q0 if x 6= Q

Let x0 ∈ R then f is not continuous

at any point x0.

Proof. ∀n ∈ N let xn = 10−nb10nx0c ∈ Q. Then 10−n(10nx0−1) ≤ 10−nb10nx0c ≤10−n10nx0.

=⇒ x0 − 10n ≤ xn ≤ x0

So by the squeeze theorem, limn→∞ xn = x0.If f is continuous at x0 then f(x0) = limn→∞ f(xn) = limn→∞ 1 = 1.

Claim: There is a sequence of irrational numbers tending to 0. ie. ∃y1, y2, . . . ∈R\Q st. limn→∞ yn = x0. If that is so, then it follows that f(x0) = limn→∞ f(yn) =limn→∞ 0 = 0. .

Proof of claim:case 1: x0 6= 0, yn = 1

10n√

2b10n√

2x0c. Then we reason similarly to aboveproof.case 2: x0 = 0, then yn = 10−n

√2.

6.2.7 Example 3

f : R→ R. f(x) = |x| then f is continuous.

Proof. Let xn ∈ R. Let ε > 0 Choose δ = ε. Then ∀x ∈ R with |x− x0| < δ onehas (with reverse triangle inequality)

|f(x)− f(x0)| = ||x| − |x0|| ≤ |x− x0| < δ = ε

6.3 Boundedness Theorem

Let f : [a, b]→ R be continuous. Then f is bounded. Recall a function f : A→R is bounded ⇐⇒ ∃M ∈ R: ∀x ∈ A, [|f(x)| ≤M ].

Proof. Suppose it is false. ie. ∀M , ∃x ∈ A [|f(x)| > M ]. Then ∀n ∈ N (withM = n), ∃xn ∈ [a, b] st, |f(xn)| > n. Then the sequence x1, x2, . . . is boundedand so by Bolzano-Weierstraß, has a convergent subsequence xn1 , xn2 , . . . ofx1, x2, . . . which converges to some x0 ∈ R.

But xnk ≥ a ∀k ∈ N so x0 = limk→∞ xnk ≥ a. Similarly, x0 ≤ b, so thatx ∈ [a, b]. But f is continuous at x0 so f(x0) = limk→∞ f(xnk). Hence thereexists an N ∈ N so that ∀k ≥ N

|f(xnk)− f(x0)| < 1

Then for all k ≥ N , k ≤ nk < |f(xnk)| ≤ |f(x0)|+ 1. This is a contradiction ifk > max(N, |f(x0)|+ 1).

39

6.4 Global Maximum and Minimums

Definition: Let f : A → R, M ∈ R and a ∈ A. Then f has a global maximumM at a if f(a) = M and ∀x ∈ A, f(x) ≤M . Similarly for global minimums.

6.4.1 Min-Max Theorem (Weierstraß)

Let f : [a, b] → R be continuous. Then f has a global maximum and globalminimum.

Proof. Since f is bounded, set M = sup{f(x) : x ∈ [a, b]}. We have to show∃x0 ∈ [a, b] st. f(x0) = M and ∀x ∈ [a, b], f(x0) ≤ M . Suppose not. Thenf(x) < M for all x ∈ [a, b].

Define g : [a, b] → R, g(x) = 1M−f(x) . Then g is continuous. By the bound-

edness theorem, g is bounded above. So there is an N ∈ R such that for allx ∈ [a, b] : g(x) ≤ N and N > 0. Then ∀x ∈ [a, b] one has 1

M−f(x) = g(x) ≤ N .So that M − f(x) ≥ 1

N ⇐⇒ M − 1N ≥ f(x). Since M − 1

N is an upper boundfor {f(x) : x ∈ [a, b]} hence M − 1

N ≥M as M is the supremum. .

The Min-Max theorem and Boundedness theorem combine to give the ex-treme value theorem which says that given a function f : [a, b] → R is con-tinuous, then the function is bounded and there exists x0 ∈ [a, b] such thatf(x0) = M is the maximum, and similarly for the minimum.

6.5 Intermediate Value Theorem (IVT I)

Let f : [a, b] → R be a continuous function. Suppose f(a) < 0 and f(b) > 0then there must exist a c ∈ (a, b) so that f(c) = 0.

Proof. Note that {x ∈ [a, b] : f(x) < 0} is non empty since it contains a andbounded above by b. Set s = sup{x ∈ [a, b] : f(x) < 0}. We claim that f(s) = 0.First, since f(a) < 0, ∃δ > 0 st. ∀x ∈ [a, a + δ) one has |f(x) − f(a)| < −f(a)

2 .Then for all x ∈ [a, a + δ), 3

2f(a) < f(x) < 12f(a) < 0. Therefore, [a, a + δ) ⊂

{x ∈ [a, b] : f(x) < 0}. Hence a+ δ ≤ s and s 6= a.Case 1 : Suppose f(s) < 0, then s 6= b. Arguing as above, ∃δ > 0 st for all

x ∈ [s, s+ δ) one has f(x) < 0. Then s+ δ ≤ s .Case 2 : Suppose f(s) > 0, then s 6= a. Arguing as above, δ > 0, such that

∀x ∈ (s − δ, s] one has f(x) > 0. So s − δ is an upper bound for {x ∈ [a, b] :f(x) < 0}. Thus, s ≤ s− δ .

Therefore, f(s) = 0.

6.6 Bolzano Intermediate Value Theorem (IVT II)

Let f : [a, b] → R be continuous and k ∈ R with either f(a) < k < f(b) orf(b) < k < f(a). Then ∃c ∈ (a, b) st. f(c) = k.

Proof. Suppose f(a) ≤ k ≤ f(b). If f(a) = b, then take c = a. If f(b) = k, thentake c = b. If f(a) < k < f(b) apply IVT(I) to g : [a, b] → R, g(x) = f(x)− k.Then ∃c ∈ (a, b) with g(c) = 0 so that f(c) = k.

Suppose f(a) ≥ k ≥ f(b). Then consider g : [a, b] → R and g(x) = −f(x).Then g(a) ≤ −k ≤ g(b) so that ∃c ∈ [a, b]: g(c) = −k, so that f(c) = k.

40

6.6.1 Corollary 1

Let f : [a, b] → R be continuous. Then ∃m,M ∈ R with m ≤ M such thatf([a, b]) = [m,M ]. Recall that {37} = [37, 37].

Proof. Since f is bounded (boundedness theorem), ∃m,M : m ≤ f(x) ≤ M ,∀x ∈ [a, b]. Namely, ∃m,M with m ≤ M so that f([a, b]) ⊂ [m,M ]. UsingBolzano IVT, ∀k ∈ [m,M ], ∃c ∈ [a, b]: f(c) = k such that f([a, b]) = [m,M ].

6.6.2 Corollary 2

Let I ⊂ R, be an interval with f : I → R be a continuous function. Then f(I)is also an interval.

Proof. Apply corollary 1.

6.7 Monotone Functions

Increasing Function Definition: Let A ⊂ R and f : A → R. Then f is calledincreasing iff. ∀x, y ∈ A, and x ≤ y we have f(x) ≤ f(y). “Strictly increasing”means x < y =⇒ f(x) < f(y). Similarly for decreasing.

6.7.1 Lemma 16

Let I ⊂ R be an interval and f : I → J be increasing. Let x0 ∈ I be aninterior point of I. ie. ∃ε > 0: (x − ε, x + ε) ⊂ I. Then sup{f(x) : x ∈ I, x <x0} = limx↑x0 f(x) = limx→x−0

f(x). Note that limx↑x0 f(x) := limx→x−0f(x)

(different notation only).In particular, the supremum and the limit exist and are equal.

Proof. {f(x) : x ∈ I, x < x0} 6= ∅ and bounded by f(x0) so sup{f(x) : x ∈I, x < x0} exists. We shall prove that limx↑x0 f(x) = s. Let ε > 0. Sets = sup{f(x) : x ∈ I, x < x0}. Then s − ε is not an upper bound for the set{f(x) : x ∈ I, x < x0}. So, ∃y ∈ I with y < x0: f(y) > s−ε. Set δ = x0−y > 0.Then ∀x ∈ I with x < x0 and |x − x0| < δ [equivalently, x ∈ (x0 − δ, x0) soy < x].

Since x ∈ I, x < x0, and f(x) is increasing, one has s − ε < f(y) ≤ f(x) ≤s ≤ s+ ε. Thus,

limx→x−0

f(x) = s

WLOG, same applies for

limx→x+

0

f(x) = i = inf{f(x) : x ∈ I, x > x0}

41

6.7.2 Corollary 1

Let I ⊂ R be an interval and f : I → J be an increasing function whereJ = f(I). Then f is continuous iff J is an interval.

Proof. “⇒”: Already done. A continuous function maps intervals into intervals.

“⇐”: Let x0 ∈ I. we show that f is continuous at x0.Case 1 : Suppose x is an interior point of I. We claim that limx→x−0

f(x) =f(x0).

If sup{f(x) : x ∈ I, x < x0} < f(x0) then f(I) is not an interval becausethere is no x ∈ I with f(x) = f(x0)+s

2 where s = sup{f(x) : x ∈ I, x < x0}.This is because if k = f(x0)+s

2 then ∀x < x0 we have f(x) ≤ s < k and x ≥ x0

gives f(x) ≥ f(x0) > k, placing k between f(x0) and f(s) causing a gap wherethe interval will not cover. Hence it would not be an interval.

So that means, since f is increasing, one has sup{f(x) : x ∈ I, x < x0} =f(x0). Then by lemma,

limx→x−0

f(x) = sup{f(x) : x ∈ I, x < x0} = f(x0)

Similarly the same argument follows for limx→x+0f(x) = f(x0).

Hence limx→x0 f(x) = f(x0) and f is continuous at x0. See Warren’s notesto prove that a limit exists iff limit from above and below are equal. Idea istrivial - just restrict cluster point to be taken from one side of the delta neigh-bourhood for the one sided limits.

Case 2 : If x0 is a boundary point, a similar argument follows.

6.7.3 Corollary 2: Continuity Inverse Theorem

Let I ⊂ R be an interval, f : I → J be a strictly increasing bijection that iscontinuous. Let g : J → I be the inverse of f . Then g is strictly increasing andcontinuous.

Proof. We know g is trivially strictly increasing and J is an interval. (see 6.6.2above, Corollary 2). Moreover, g(J) = I is an interval. By corollary 1, g iscontinuous.

6.7.4 Example

Fix n ∈ N. Define f : [0,∞)→ [0,∞) by f(x) = xn. Clearly, f(x) is continuousand strictly increasing. So f([0,∞)) [notation for range] is an interval. Thenf([0,∞)) ⊂ [0,∞). Moreover by Bernoulli inequality,

f(k + 1) = (1 + k)n ≥ 1 + nk ≥ k, ∀k ∈ (−1,∞)

Hence k ∈ f([0,∞)) for all k ∈ (0,∞) by the intermediate value theorem. Thisis possible because we can take the restriction on [0, k+ 1] and then apply IVTso that ∃c ∈ (0, k+ 1): f(c) = k, then since k ∈ f([0, k+ 1]) ⊂ f([0,∞)) so thatit is surjective. Injective part follows because it is strictly increasing. Thereforef([0,∞)) = [0,∞) so that f : [0,∞)→ [0,∞) is a bijection. By corollary 2, theinverse of f is continuous.

We denote this: n√x = g(x), ∀x ∈ [0,∞).

42

6.8 Uniform Continuity

Recall: f : A → R is a function. Then f is continuous iff. f is continuous atevery point a ∈ A.

⇐⇒ (∀a ∈ A)(∀ε > 0)(∃δ > 0)(∀x ∈ A)[|x− a| < δ =⇒ |f(x)− f(a)| < ε]

So that δ is allowed to depend on a ∈ A and ε > 0. In general, it depends onboth.Definition Let f : A→ R be a function. Then f is called uniformly continuousiff

(∀ε > 0)(∃δ > 0)(∀a ∈ A)(∀x ∈ A)[|x− a| < δ =⇒ |f(x)− f(a)| < ε]

Where δ can only depend on ε > 0 and NOT on a ∈ A.

6.8.1 Lemma 17

Every uniformly continuous function is continuous. This is trivial. The converseis not true.

6.8.2 Converse Counter Example 1

We will show a continuous function which is not uniformly continuous. Definef : (0, 1]→ R, f(x) = 1

x . We claim f is not uniformly continuous. Now, supposefor a contradiction it is.

Proof. Choose ε = 1, then ∃δ > 0 st. ∀x, ε ∈ A with |x− a| < δ

|f(x)− f(a)| < ε = 1

But take a = 12δ and x = 1

n where n ∈ N, and n > 1δ . Then when |x− a| < δ,

1 > |f(x)− f(a)| = |n− 2δ|, ∀n > 1

δ

So that n < 1 + 2δ for all n > 1

δ . . Because natural numbers are not boundedabove.

6.8.3 Lemma 18

Let f : A→ R. The following are equivalent:

1. f is not uniformly continuous.

2. ∃ε > 0, ∀δ > 0, ∃a ∈ A, ∃x ∈ A, [|x− a| < δ =⇒ |f(x)− f(a)| ≥ ε]

3. ∃ε > 0, ∃(xn) ∈ A, ∃(yn) ∈ A, [limn→∞ |xn − yn| = 0 and ∀n ∈ N|f(xn)− f(yn)| ≥ ε]

Proof. I ⇐⇒ II : Follows by definition.

43

II =⇒ III : Suppose (II) valid. Let ε > 0, ∀δ > 0, ∃x, y ∈ A [|x − a| <δ =⇒ |f(x)− f(y)| ≥ ε]. For all n ∈ N choose δ = 1

n then there are xn, yn ∈ Ast. |xn − yn| < δ = 1

n and |f(xn)− f(yn)| ≥ ε. Thus, we have

limn→∞

|xn − yn| = 0

III =⇒ II : Assume (III). Then ∃ε > 0 and x1, x2, . . . , y1, y2, . . . ∈ A st.limn→∞ |xn − yn| = 0 and ∀n ∈ N[|f(xn)− f(yn)| ≥ ε].

Let δ > 0. Since limn→∞ |xn − yn| = 0 ∃n ∈ N st. |xn − yn| < δ. Thenchoose x = xn and y = yn then II is satisfied.

6.8.4 Example

Let f : (0, 1] → R and f(x) = 1x . Choose ε = 1, xn = 1

n , yn = 1n+2 . Then

limn→∞ |xn − yn| = 0 but ∀n ∈ N one has

|f(xn)− f(yn)| = |n− (n+ 2)| = 2 ≥ ε

Hence f is not uniformly continuously.

6.8.5 Theorem 15

Let f : [a, b]→ R be a continuous function. Then f is uniformly continuous.

Proof. Suppose not. ie. f is not uniformly continuous. Then there are ε >0, (yn), (xn) ∈ [a, b] such that limn→∞ |xn − yn| = 0 and for all n ∈ N: |f(xn)−f(yn)| ≥ ε. By Bolzano-Weierstraß, the bounded sequence (xn) has a convergentsubsequence (xnk) with limit L = limk→∞ xnk .

Since [a, b] is closed, one has L ∈ [a, b]. Now limk→∞(xnk − ynk) = 0 sincelimk→∞ |xn − yn| = 0. Hence

limk→∞

ynk = limk→∞

xnk = L exists.

Since f is continuous at L then we have

limk→∞

f(xnk) = limk→∞

f(ynk) = f(L)

Hence

limk→∞

(f(xnk)− f(ynk)) = 0

Therefore, there is a k ∈ N st.

|f(xnk)− f(ynk)| < ε

6.8.6 Lipschitz Functions

A function f : A → R is called “Lipschitz”, Lipschitz-function, Lipschitz-continuous or satisfies a Lipschitz condition iff.

(∃L ≥ 0)(∀x, y ∈ A)[|f(x)− f(y)| ≤ L|x− y|]

44

6.8.7 Lemma 19

Every Lipschitz funciton is uniformly continuous.

Proof. Let ε > 0. Choose δ = εL+1 . Then, ∀a, x ∈ A with |x− a| < δ, one has

|f(x)− f(a)| ≤ L|x− a| ≤ (L+ 1)|x− a| < (L+ 1)δ = ε

6.8.8 Example

Define f : [1,∞). f(x) =√x. We claim that f is Lipschitz.

Proof. Let x, y ∈ [1,∞) then

|f(x)−f(a)| =∣∣∣∣(√x−√y)

(√x+√y

√x+√y

)∣∣∣∣ =∣∣∣∣ x− y√x+√y

∣∣∣∣ =|x− y|√x+√y≤ |x− y|

2

So then one can choose L = 12 .

Therefore f is uniformly continuous, on [1,∞) by the Lipschitz proposition.

6.8.9 Example

Define f : [0,∞)→ R, f(x) =√x. Claim: f is uniformly continuous.

Proof. Let ε > 0. Since f : [0, 2] → R is continuous, closed and bounded,∃δ1 > 0, ∀x, y ∈ [0, 2], with |x− y| < δ1: |f(x)− f(y)| < ε.

Moreover, ∃δ2 > 0 st. ∀x, y ∈ [1,∞): [|x − y| < δ2 =⇒ |f(x) − f(y)| < ε].Now take, δ = min(δ1, δ2, 1) > 0. Let x, y ∈ [0,∞) and suppose |x − y| < δ.Then either:

1. x, y ∈ [0, 2] which implies |f(x)− f(y)| < ε; or

2. x, y ∈ [1,∞) which implies |f(x)− f(y)| < ε

Therefore, f is uniformly continuous.

7 Differentiation

Definition: Let I ⊂ R and c a cluster point of I. The function f : I → R is saidto be differentiable at c if and only if:

limx→c

f(x)− f(c)x− c

exists.

We denote the limit by f ′(c).

7.1 Notationdn

dxn(f(x)) = f ′(x) (Leibnitz)

Dnf(x) = f ′(x)

f (n)(x) = f ′(x)

f(x) = f ′(x) (Newton)

45

7.2 Theorem 16

If f is differentiable at c then f is continuous at c.

Proof. Claim: If c is a cluster point of I, f is continuous at c. Sequentialcontinuity:

⇐⇒ limx→c

f(x) = f(c) ⇐⇒ limx→c

(f(x)− f(c)) = 0

Since:

f(x)− f(c) =f(x)− f(c)

x− c· (x− c)→ f ′(c) · (0− 0) = 0 as x→ c

then f is continuous at c.

7.2.1 Example

Let f(x) = |x|, x ∈ R. For c = 0,

f(x)− f(0)x− 0

=|x|x

= sign(x) :=

1 x > 00 x = 0−1 x < 0

Since limx→0±f(x)−f(0)

x−0 = ±1 the limit of f(x)−f(0)x−0 does not exist. So

f(x) = |x| is not differentiable at x = 0.

7.3 Theorem 17

If f, g : I → R, c a cluster point of I and f, g differentiable at c, then so are:

1. αf(x), α ∈ R with (αf)′(c) = αf ′(c)

2. (f + g)(x) with (f + g)′(c) = f ′(c) + g′(c)

3. (fg)(x) with (fg)′(c) = f(c)g′(c) + f ′(c)g(c) [Product Rule]

4. fg (x) when g′(c) 6= 0 with g(c)f ′(c)−g′(c)f(c)

(g′(c))2 [Quotient Rule]

Proof of first two left as exercises.

Proof of (c):

Proof.

(fg)(x)− (fg)(c)x− c

=f(x)g(x)− f(c)g(c)

x− c

=f(x)g(x)− f(x)g(c) + f(x)g(c)− f(c)g(c)

x− c

= f(x)g(x)− g(c)x− c

+ g(c)f(x)− f(c)

x− c

46

Since f is differentiable at c, f is continuous at c and f(x)→ f(c) as x→ c.So using rules for taking limits,

limx→c

(fg)(x)− (fg)(c)x− c

= (limx→c

f(x))( limx→c

g(x)− g(c)x− c

) +

(limx→c

g(c))( limx→c

f(x)− f(c)x− c

)

= f(c)g′(c) + g(c)f ′(c)

Proof of (d):

Proof.

fg (x)− f

g (c)

x− c=

(f(x)g(c)−f(c)g(x)

g(x)g(c)

)x− c

=f(x)g(c)−g(c)f(c)

x−c + g(c)f(c)−f(c)g(x)x−c

g(x)g(c)

=1

g(x)g(c)

(g(c)

f(x)− f(c)x− c

+ f(c)−(g(x)− g(c))

x− c

)→ 1

[g(c)]2(g(c)f ′(c) + f(c)(−g′(c)))

7.4 Chain Rule

Definition: Let g : I → R and f : J → R, f(J) ⊂ I. Suppose f is differentiableat c, g differentiable at f(c) so that g◦f is differentiable at c. Then, (g◦f)′(c) =g′(f(c)) · f ′(c).

7.4.1 Lemma 20

If f is differentiable at c then there exists a function φ such that f(x)− f(c) =φf,c(x)(x− c) where φf,c is continuous at c with

limx→c

φf,c(x) = f ′(c)

Proof. Let φ(x) :=

{f(x)−f(c)

x−c x 6= c

f ′(c) x = c

Therefore φ(x) satisfies the conditions of the lemma.

Back to the chain rule proof:

Proof. Since g is differentiable at f(c) then by the lemma above, g(f(x)) −g(f(c)) = φg,f(c)(f(x))(f(x)− f(c)) where φg,f(c) is continuous at f(c) with

limy→f(c)

φg,f(c)(y) = g′(f(c))

47

Since f is differentiable at c, f(x) − f(c) = φf,c(x)(x − c) where φf,c iscontinuous at c and

limx→c

φf,c(x) = f ′(c)

Hence

(g ◦ f)(x)− (g ◦ f)(c)x− c

=g(f(x))− g(f(c))f(x)− f(c)

·f(x)− f(c)x− c

= φg,f(c)(f(x))·φf,c(x)

Since f is differentiable at c, f is continuous at c. so (φg,f(c) ◦ f)(x) is a compo-sition of continuous functions at f(c) and is thus continus at c. In particular,

limx→c

(φg,f(c) ◦ f)(x) = φg,f(c)(f(c)) = g′(f(c))

Hence the Newton quotient is a product of functions which have limits asx→ c so by the product rule of limits,

limx→c

(g ◦ f)(x)− (g ◦ f)(c)x− c

= g′(f(c)) · f ′(c)

7.4.2 Remark

One can show(but we won’t) that sin′(x) = cos(x) and cos′(x) = − sin(x).

7.4.3 Example

Let f : R→ R with f(x) =

{x2 sin(−1

x ) x 6= 00 x = 0

.

Claim: f is differentiable at every point in R.

Proof. For x 6= 0 we differentiate using the usual rules to obtain:

f ′(x) = x2 cos(−1x

) · (−1x2

) + 2x sin(−1x

)

= 2x sin(−1x

)− cos(−1x

)

f ′(x) does not have a limit as x→ 0, however, f is differentiable at 0 since,

limx→0

f(x)− f(0)x− 0

= limx→0

x sin(1x

) = 0

So f is differentiable at 0 with f ′(0) = 0.Note that even though f(x) is continuous, f ′(x) is not continuous at 0.

48

7.5 Inverse Function Theorem

Let I be an interval and f : I → R and f strictly monotone on I. Let g : J → R,J = f(I) be its strictly monotone inverse. If f is differentiable at c and f ′(c) 6= 0,then g is differentiable at f(c) and

g′(f(c)) =1

f ′(c)

This is an equivalent formulation to (f−1)′) = 1f ′◦f−1 (like in Warren’s notes)

except here we’ve taken the composition into account from the start.

Proof. Since f is differentiable at c, ∃φ:

f(x)− f(c) = φ(x)(x− c)

with φ continuous at c and φ(c) = f ′(c). Let y = f(x) and d = f(c). Sinceg = f−1 then g(y) = x and g(d) = c. So that

y − d = φ(g(y))(g(y)− g(d))

Since φ(g(d)) = φ(c) = f ′(c) 6= 0 and φ is continuous, we can choose a neigh-bourhood of c, (c− δ, c+ δ) ∩ I on which φ is non zero.

Let y ∈ f((c−δ, c+δ)∩I) and y = f(x). Then φ(g(y)) = φ(g(f(x))) = φ(x).So,

g(y)− g(d)y − d

=1

φ(g(y))

Then (φ ◦ g) is continuous at d, and g is continuous at d, with

limy→d

(φ ◦ g)(y) = (φ ◦ g)(d) = φ(c) = f ′(c)

So

limy→d

f(y)− f(d)y − d

=1

f ′(c)

7.5.1 Example

Let f(x) = x3 then it has an inverse g such that g(f(x)) = x. Suppose gis differentiable at f(0) = 0. We expect the value of derivative to be 1 sinceg(f(x)) = x. But by the chain rule 1 = g′(f ′(0)) = g′(f(0))f ′(0) = 0..

7.6 Relative Maximum

Let f : I → R. Then f has a relative maximum at c ∈ I if there is a neighbour-hood of c, so that

v = (c− δ, c+ δ) ∩ I

with f(x) ≤ f(c), ∀x ∈ V . Similarly, for minimums.

49

7.6.1 Interior Extremum Theorem

Let I be an interval and f : I → R. If f has a relative extremum at an interiorpoint c ∈ I and if f ′(c) exists, then f ′(c) = 0.

Proof. WLOG, suppose f has a local maximum at c. If not, consider −f .Suppose f ′(c) 6= 0.

Further suppose f ′(c) > 0. Since

limx→c

f(x)− f(c)x− c

= f ′(c) > 0

there is a neighbourhood of c with f(x)−f(c)x−c > 0, ∀x ∈ (c− δ, c+ δ) ∩ I.

So for x > c, we have f(x) − f(c) > 0(x − c) = 0 ⇐⇒ f(x) > f(c).Therefore c is not a local maxima.

Similarly, if f ′(c) < 0 we have a similar contradiction.

7.6.2 Example 1

Consider the Weierstraß function(with the crazy zigzagging fractals). Thenthere are relative minimums and maximums but they are not differentiable atthose points.

7.6.3 Example 2

f(x) = x2, g(x) = |x|. They have relative extrema at 0. f ′(c) = 0 exists butg′(c) does not exist at 0.

7.7 Rolle’s Theorem

If f : [a, b]→ R is continuous and differentiable on (a, b), and f(a) = f(b) = 0,then ∃c ∈ (a, b) with f ′(c) = 0.

Proof. Either f = 0 =⇒ f ′ = 0, or f 6= 0 on [a, b] in which case (WLOG), wecan assume f takes a positive value(otherwise consider −f).

Since f is continuous, then by the maximum-minimum theorem, f attainsits supremum. ie. ∃c ∈ (a, b): f(c) = sup{f(x) : x ∈ [a, b]} > 0.

So c is a global and hence relative maximum of f . Then by the relativeextrema theorem, we must have f ′(c) = 0.

7.8 Mean Value Theorem

If f : [a, b] → R is a continuous function on [a, b] and differentiable on (a, b)then ∃c ∈ (a, b) for which f ′(c) = f(b)−f(a)

b−a .

Proof. Let φ be f minus the line from (a, f(a)) to (b, f(b)).

φ(x) = f(x)− f(a)− f(b)− f(a)b− a

(x− a)

Then φ satisfies the conditions of Rolles’ theorem so that ∃c ∈ (a, b):

50

φ(c) = 0 = f ′(c)− f(b)− f(a)b− a

⇐⇒ f ′(c) =f(b)− f(a)

b− a

7.9 Corollary 1

If f : I → R is continuous, I an interval, and f ′ = 0 on the interval of I, thenf is a constant.

Proof. Suppose [a, x] ⊂ I then the restriction of f to [a, x] ⊂ I satisfies theconditions of the mean value theorem. Then, ∃c ∈ (a, x):

f(x)− f(a) = f ′(c)(x− a) = 0 =⇒ f(x) = f(a)

Similarly if x < a then f(x) = f(a). Therefore f is constant.

7.9.1 Example

f(x) =

{1 if 0 ≤ x ≤ 1,0 if 2 ≤ x ≤ 3.

Then f ′ = 0 on [0, 1] ∪ [2, 3] but f is not constant.

7.10 Corollary 2

Let f : I → R, f differentiable and I an interval. Then f is increasing onI ⇐⇒ f ′ ≥ 0.

Proof. “⇐”: Suppose f ′ ≥ 0. Let x1 < x2 and x1, x2 ∈ I. Then [x1, x2] ⊂ I.We can apply MVT to the restriction of f to [x1, x2]. ∃c ∈ (x1, x2):

f(x2)− f(x1) = f ′(c)(x2 − x1) ≥ 0

Therefore f is increasing.“⇒”: Suppose f is increasing on I. Since f is increasing,

f(x)− f(c)x− c

≥ 0, x 6= c (1)

Then if x > c, f(x) ≥ f(c) =⇒ f(x)− f(c) ≥ 0 =⇒ f(x)−f(c)x−c ≥ 0.

If x < c, f(c) ≥ f(x) =⇒ f(c)−f(x)c−x ≥ 0.

Since f ′(c) exists, take limit of (1) above.

f ′(c) = limx→c

f(x)− f(c)x− c

≥ 0

(The idea is that f(x) ≤ g(x) and if the limits exist then lim f ≤ lim g.)Therefore f ′ ≥ 0. Similarly for decreasing functions.

51

7.11 Corollary 3

Let I be an interval and f : I → R. If f ′ > 0 then f is strictly increasing.

Proof. By MVT, x1 < x2 =⇒ f(x2)−f(x1) = f ′(c)(x2−x1) > 0 =⇒ f(x2) >f(x1) and f is strictly increasing.

Converse is not true. eg. f(x) = x3 is strictly increasing but at 0, f ′ = 0and not > 0.

7.11.1 Example

If f ′ is bounded on an interval I, then f (original function) is Lipschitz contin-uous.

Proof. Suppose |f ′(x)| ≤M , ∀x ∈ I. By Cauchy MVT, f(x)−f(y) = f ′(c)(x−y). [c is the interior of the interval with end points x, y]

This implies |f(x) − f(y)| = |f ′(c)||x − y| ≤ M |x − y|. Converse is almostbut not quite true(almost everywhere differentiable).

7.12 Darboux’s Theorem

If f is differentiable on [a, b] then f ′ need not be continuous, but f ′ satisfies theintermediate value property. ie. If f ′(a) < f ′(b) and f ′(a) < k < f ′(b), then∃c ∈ (a, b) with f ′(c) = k.

Proof. Let g(x) = kx−f(x). We want to show g′(c) = 0 for some c ∈ (a, b) =⇒f ′(c) = k. Let c ∈ [a, b] be a point where the continuous function g attains itsmaximum. If c ∈ (a, b) then we must have g′(c) = k − f ′(c) = 0.

Suppose c = a. g′(a) = k − f ′(a) > 0. Since g′(a) > 0, ∃δ > 0: g(x) > g(a),for all a ≤ x < a+ δ.

This contradicts g(a) being maximum so we can’t have c = a and similarlyc cannot be b either.

7.13 Cauchy’s Mean Value Theorem

Let f, g : [a, b] → R be continuous on [a, b] and differentiable on (a, b). If∀x ∈ (a, b), g′(c) 6= 0 then ∃c ∈ (a, b): (f(b)− f(a)) · g′(c) = f ′(c)(g(b)− g(a)).

Proof. Let φ(x) :=

∣∣∣∣∣∣f(x) g(x) 1f(a) g(a) 1f(b) g(b) 1

∣∣∣∣∣∣Then

φ(x) = f(x)(g(a)− g(b))− g(x)(f(a)− f(b)) +∣∣∣∣f(a) g(a)f(b) g(b)

∣∣∣∣If x = b or x = a then the matrix above is linearly dependent so that the

determination is 0. ie.

φ(a) = φ(b) = 0

52

φ is continuous on [a, b], and differentiable on (a, b). By Rolle’s theorem, ∃c ∈(a, b):

φ′(c) = 0 = f ′(c)(g(a)− g(b))− g′(c)(f(a)− f(b))

7.14 L’Hopital’s Rule

Suppose f, g : (a, b)→ R which are differentiable on (a, b) with

limx→a+

f(x) = limx→a+

g(x) = 0

and ∃δ > 0: g′(x) 6= 0 for all x ∈ (a, a+ δ). If

limx→a+

f ′(x)g′(x)

= L ∈ R then limx→a+

f(x)g(x)

= L

Proof. Let ε > 0. Suppose limx→a+f ′(x)g′(x) = L. Then ∃δ > 0: ∀a < u < a+ δ

L− ε < f ′(a)g′(a)

< L+ ε

where g′(a) 6= 0, ∀a < u < a+ δ. By the Cauchy MVT, for a < α < β < a+ δ,∃u ∈ (α, β):

f ′(u)g′(u)

=f(β)− f(α)g(β)− g(α)

Since a < u < a+ δ, ∀a < α < β < a+ δ

L− ε < f(β)− f(α)g(β)− g(α)

< L+ ε

Take the limit of the above inequality as α→ a+ then by the squeeze theorem,∀a < β < a+ δ

L− ε ≤ f(β)− limα→a+ f(α)g(β)− limα→a+ g(α)

=f(β)g(β)

≤ L+ ε

ie. limβ→a+

f(β)g(β)

= L

The x→ a− argument similarly follows. Since all ∞∞ cases can be manipulatedinto 0

0 cases, we’re done.

7.14.1 Example

Suppose f : [a, b)→ R and f ′(x) exists on [a, b). Either:

• limx→a+ f ′(x) does not exist; or

• limx→a+ f ′(x) = f ′(a).

53

(NB: limx→a+ f ′(x) cannot exist and be different to f ′(a).)

Proof. Suppose limx→a+ f(x) exists, then

f ′(a) = limx→a+

f(x)− f(a)x− a

is indeterminate of the type 00 . So by L’Hopital’s rule,

f ′(a) = limx→a+

f(x)− f(a)x− a

= limx→a+

f ′(x)

7.15 Interpolation

One can show that if x0, . . . , xn ∈ R with each element of the sequence notnecessarily distinct, then there exists a unique polynomial pn(fundamental the-orem of algebra) of degree n interpolating the values of a function f at x0, . . . , xnwhen we count multiplicities (which means that when we interpolate repeatedlyon the same point, we match up the values of f(x0), . . . , f (n)(x0) to it. ie theTaylor situation).

Special cases:

1. If all points are distinct, then the interpolation conditions are pn(xj) =f(xj), j = 0, . . . , n. pn is called the Lagrange interpolant.

2. When all points are equal. ie. x0 = x1 = . . . = xn. The interpolationconditions are p(j)

n (x0) = f (j)(x0), j = 0, . . . , n and pn is called the Taylorinterpolant to f at x0.

The Taylor interpolant has the form:

pn(f, x) := pn(x) = f(x0) + f ′(x0)(x− x0) + . . .+f (n)(x0)

n!(x− x0)n

7.15.1 Taylor (Mean Value Form) Theorem

Suppose f : [a, b]→ R, f (n) exists and is continuous on [a, b], and f (n+1) existson (a, b). ie. (fn is differentiable on (a, b).)

Then the error in Taylor interpolation of degree n at x0 ∈ [a, b] is given:

f(x)− pn(x) =f (n+1)(c)(n+ 1)!

(x− x0)n+1

where c is between x and x0.

Proof. If x = x0 then there is nothing to prove. WLOG, suppose x > x0.Let φ(x) = f(x) − pn(x) − B(x−x0)

n+1

(n+1)! . Choose B, so that φ(x) = 0. ie.

B = (f(x)−pn(x)) (n+1)!(x−x0)n+1 ). We want to show B = f (n+1)(c) ie. φ(n+1)(c) = 0

and f (n+1)(c) − B = 0. Now φ(x0) = φ(1)(x0) = . . . = φ(n)(x0) = 0 andφ(x) = 0. (NB: φ(n) is continuous on [a, b] and φ(n+1) exists on (a, b).)

54

Apply Rolle’s theorem to φ on [x0, x] to get c1 ∈ (x0, x): φ(c1) = 0. Sinceφ(1)(x0) = φ(1)(c1) = 0, apply Rolle’s theorem again to get c2 on [x0, c1] withx0 < c2 < c1 < x and φ(2)(c2) = 0. Continuing, we find that with c1 > c2 >. . . > cn+1 > x0 : φ(n+1)(cn+1) = 0 which leads us to c = cn+1 with x0 < c < ximplying φ(n+1)(c) = 0.

7.15.2 Example

cos(x) ≥ 1 − 12x

2, ∀x. The Taylor approximation of order 2 at x0 = 0 isp2(x) = 1− 1

2x2. So by the remainder theorem (Taylor MV):

cos(x)− (1− 12x2) =

cos(3)(c)3!

x3 =sin(c)

6x3

where c is between 0 and x so that the inequality holds if error ≥ 0.For x ∈ [0, π], c ∈ [0, π], sin(c) ≥ 0, sin(c)

6 x3 ≥ 0.If x ∈ [−π, 0], c ∈ (−π, 0), x < 0 and x3 < 0, sin(c) < 0, and sin(c)

6 x3 ≥ 0.

If |x| > π, cos(x)− (1− 12x

2) ≥ −1− 1 + 12 |x|

2 > −2 + 12π

2 > 0

7.16 Classification of Extrema

Suppose f (n) exists, and is continuous in the neighbourhood of x0. Furthersuppose f ′(x0) = . . . = f (n−1)(x0) = 0 for n ≥ 2 and f (n)(x0) 6= 0.

Then:

1. If n is even, f (n)(x0) > 0, then f has a local minima at x0.

2. If n is even, f (n)(x0) < 0, then f has a local maxima at x0.

3. If n is odd, then f has neither local maxima or local minima at x0.

Proof. First case will be considered. The other ones follow using similar argu-ments.

Let pn−1 be the Taylor approximation at x0 so that pn−1(x) = f(x0). Thenthe error is:

f(x)− f(x0) =f (n)(c)n!

(x− x0)n

We have used the fact that the f(k)(c)k! (x−x0)k terms disappear for 1 ≤ k ≤ n−1

due to f ′(x0) = . . . = f (n−1)(x0) = 0. Because f (n) is continuous at x0, we canchoose a neighbourhood of x0 for which f (n)(x) > 0, x ∈ (x0 − δ, x+ δ). Sincec is between x and x0 then f (n)(c) > 0. Thus,

f(x) = f(x0) +f (n)(c)n!

(x− x0)n

Because f(n)(c)n! > 0 and (x−x0)n > 0 (as x 6= x0) implies f(x) > f(x0) we have

a local minima.

55

8 Riemann Integration

8.1 Introduction

Basic idea for Riemann integration is to first define lower and upper sums ofpartitioned intervals over the domain or restriction. We define rectangles thatare the width of the interval, sprouting from the x axis and let the area of a rect-angle reaching the supremum and infimum (respectively) represent the heightof those rectangles and sum across the intervals. By reducing the width of thepartitioned rectangles into arbitrarily small values, we compare the upper andlower sums. If the sums are equal, we say the function is Riemann integrable.

Definition: A partition(P ) of [a, b] is a set {x0, . . . , xn} with

a = x0 < x1 < . . . < xn = b

Corresponding to P , we define the lower sum:

s(f, P ) :=n∑j=1

( inf(xj−1,xj)

f)(xj − xj−1)

where

inf(xj−1,xj)

f := infx∈(xj−1,xj)

f(x)

We are ignoring the end points because we can. The finite number of end-points will have an area of 0 effectively, as they can be arbitrarily bounded byany ε > 0.

Now define the upper sum:

S(f, P ) :=n∑j=1

( sup(xj−1,xj)

f)(xj − xj−1)

Define Q to be a “refinement” of P if P ⊂ Q. Sometimes this is denotedP ≤ Q.

8.1.1 Lemma 21

If Q is a refinement of P then:

s(f, P ) ≤ s(f,Q) ≤ S(f,Q) ≤ S(f, P )

Proof. Suppose Q is obtained from P by adding one point, say x∗, where P ={x0, . . . , xn} and xk−1 < x∗ < xk.

56

Then,

s(f, P ) =∑j 6=k

( inf(xj−1,xj)

f)(xj − xj−1) + ( inf(xk−1,xk)

f)(xk − xk−1)

=∑j 6=k

( inf(xj−1,xj)

f)(xj − xj−1) + ( inf(xk−1,xk)

f)((xk − x∗) + (x∗ − xk−1))

≤∑j 6=k

( inf(xj−1,xj)

f)(xj − xj−1) + ( inf(xk−1,x∗)

f)(x∗ − xk−1) + ( inf(x∗,xk)

f)(xk − x∗)

= s(f,Q)≤ S(f,Q) [since ( inf

(xj−1,xj)f) ≤ ( sup

(xj−1,xj)

f)]

≤ S(f, P ) (by similar argument)

In general, we can obtain Q from P by adding one point at a time.

P = P1 ⊂ P2 ⊂ . . . ⊂ Pn = Q

where Pj is obtained from Pj−1 by adding one point. Therefore,

s(f, P ) = s(f, P1) ≤ s(f, P2) ≤ . . . ≤ s(f,Q)

and

s(f,Q) ≤ S(f,Q) = S(f, Pn) ≤ S(f, Pn−1) ≤ . . . ≤ S(f, P1) = S(f, P )

Definition: The “common refinement” of partitions P, P ′ is P, P ′ ⊂ P ∪P ′ = Q.

8.1.2 Corollary

If P, P ′ are partitions, then s(f, P ) ≤ S(f, P ′).

Proof. Let Q = P ∪ P ′ be the common partition.

s(f, P ) ≤ s(f,Q) ≤ S(f,Q) ≤ S(f, P ′)

8.1.3 Examples

s(f, P ) ≤ S(f, {a, b}) and S(f, P ) ≤ S(f, {a, b}).

8.1.4 Lower and Upper Integrals

NB: P is any partition so that sup and inf are taken over all partitions.Lower Integral∫ b

a

f = supP a partition

s(f, P )

57

Upper Integral∫ b

a

f = infPS(f, P )

Since s(f, P ) ≤ S(f, P ) ∀P, P ′,

supPs(f, P ) =

∫ b

a

f ≤ S(f, P ′)

=⇒∫ b

a

f ≤ infPS(f, P ′) =

∫ b

a

f

8.1.5 Example

The upper and lower sums are not necessarily the same so the Riemann inte-gration condition cannot solely depend on one. Suppose it did. Then considerf =: [0, 1]→ {0, 1} and

f(x) =

{1 x ∈ Q0 x /∈ Q

s(f, P ) =∑j

( inf(xj−1,xj)

f)(xj − xj−1)

=∑j

0(xj − xj−1) = 0

Similarly,

S(f, P ) =∑j

sup f(xj − xj−1) =∑

1(xj − xj−1) = 1

Thus,∫ 1

0f = 0, and

∫ 1

0f = 1, a contradiction.

8.2 Riemann Integrable

We say f is Riemann integrable if∫ b

a

f =∫ b

a

f

and the Riemann integral is∫ b

a

f =∫ b

a

f(x)dx =∫ b

a

f =∫ b

a

f

58

8.3 Riemann Condition

f is Riemann integrable ⇐⇒ ∀ε > 0, ∃Pε partition (where P depends on ε):

S(f, P )− s(f, P ) < ε

Proof. “⇒”: Suppose f is Riemann integrable. Let ε > 0. Choose partitionsP1, P2, and Q = P1 ∪ P2 be the common refinement such that:

S(f,Q)−∫ b

a

f ≤ S(f, P1)−∫ b

a

f <ε

2∫ b

a

f − s(f,Q) ≤∫ b

a

f − s(f, P2) <ε

2

Then S(f,Q) ≤ S(f, P1) Similarly, −s(f,Q) ≤ −s(f, P2).

=⇒ S(f,Q)− s(f,Q) = S(f,Q)−∫ b

a

f +∫ b

a

f − s(f,Q) <ε

2+ε

2= ε

“⇐”: Suppose Cauchy condition holds. ∀ε > 0, ∃Pε:

S(f, P )− s(f, P ) < ε

Let Q be a refinement of P , then

S(f,Q) ≤ S(f, P ) < s(f, P ) + ε

Take the infimum over Q. Then,∫ b

a

f ≤ infQS(f, P ) ≤ s(f, P ) + ε

Take another refinement Q, then

s(f,Q) ≥ s(f, P ) ≥∫ b

a

f − ε

so that∫ b

a

f ≥ supQs(f, P ) ≥

∫ b

a

f − ε

This means

0 ≤∫ b

a

f −∫ b

a

f ≤ ε

Since ε is arbitrary, we must have∫ b

a

f −∫ b

a

f = 0

therefore, it is Riemann integrable.

59

8.4 Theorem 18

Let f, g be Riemann Integrable and c ∈ R. Then:

(I): cf is R-integrable and∫ b

a

cf = c

∫ b

a

f

(II): f + g is R-integrable and∫ b

a

(f + g) =∫ b

a

f +∫ b

a

g

(III): If f ≥ 0 then∫ b

a

f ≥ 0

(IV): If f ≤ g then∫ b

a

f ≤∫ b

a

g (monotonicity)

(V): |f | is R-integrable and

∣∣∣∣∣∫ b

a

f

∣∣∣∣∣ ≤∫ b

a

|f |

(VI): f2 is R-integrable.

(VII): fg is R-integrable.

(I):

Proof. Suppose c ≥ 0 (if c = 0, then∫cf =

∫0 = 0 = 0

∫f).

S(cf, P ) =n∑j=1

( sup(xj−1,xj)

cf)(xj−xj−1) = c

n∑j=1

( sup(xj−1,xj)

f)(xj−xj−1) = cS(f, P )

So, ∫ b

a

f = infPS(cf, P ) = inf

PcS(f, P ) = c inf

PS(f, P ) = c

∫ b

a

f

Then the argument proceeds similarly.Since f is Riemann integrable,∫ b

a

cf = c

∫ b

a

f = c

∫ b

a

f =∫ b

a

cf

Therefore∫ b

a

cf = c

∫ b

a

f

Now consider when c < 0. Since inf = − sup

S(cf, P ) = s(cf, P )

So by argument above,

s(cf, P ) = cs(f, P )

and similarly for the other case.

60

(II):

Proof. S(f+g, P ) ≤ S(f, P )+S(g, P ). Similarly, s(f, P )+s(g, P ) ≤ s(f+g, P ).In particular,

S(f + g, P )− s(f + g, P ) ≤ S(f, P )− s(f, P ) + S(g, P )− s(g, P )

Since f and g are Riemann integrable, for ε > 0, ∃P1, P2

S(f, P1)− s(f, P1) <ε

2and S(g, P2)− s(g, P2) <

ε

2

Let Q = P1 ∪ P2.

S(f + g,Q)− s(f + g,Q) ≤ S(f,Q)− s(f,Q) + S(g,Q)− s(g,Q)≤ S(f, P1)− s(f, P1) + S(g, P2)− s(g, P2)

<ε

2+ε

2= ε

Therefore f + g is Riemann integrable via the Cauchy criterion.For second part of proof, let ε > 0. Since f, g is Riemann integrable, there

exists a common partition P with

S(f, P )−∫ b

a

f <ε

2and S(g, P )−

∫ b

a

g <ε

2

so that

S(f + g, P )− (∫ b

a

f +∫ b

a

g) ≤ S(f, P ) + S(g, P )− (∫ b

a

f +∫ b

a

g) < ε

Let Q be a refinement of P.

S(f + g,Q) ≤ S(f + g, P ) ≤ ε+ (∫ b

a

f +∫ b

a

g)

Take the infimum over Q, then∫ b

a

(f + g) ≤ ε+ (∫ b

a

f +∫ b

a

g)

By similar argument,

(∫ b

a

f +∫ b

a

g)− ε ≤∫ b

a

(f + g)

Putting the two sides together, and using the fact that f + g is Riemann inte-grable,

(∫ b

a

f +∫ b

a

g)− ε ≤∫ b

a

(f + g) =∫ b

a

(f + g) ≤ (∫ b

a

f +∫ b

a

g) + ε

Since ε is arbitrary, it must be that∫ b

a

(f + g) =∫ b

a

f +∫ b

a

g

61

III:

Proof. f ≥ 0 =⇒ S(f, P ) ≥ 0. (Converse is not true.) Thus,∫ b

a

f = infPS(f, P ) ≥ 0

IV:

Proof.

f ≤ g =⇒∫ b

a

f ≤∫ b

a

g

f ≤ g =⇒ g − f ≥ 0 as g − f is Riemann integrable by I,II.∫ b

a

(g − f) =∫ b

a

g −∫ b

a

f by III

=⇒∫ b

a

f ≤∫ b

a

g

Before proving V, we first need a lemma.

8.4.1 Lemma 22

If f : A→ R and f bounded, then

supA|f | − inf

B|f | ≤ sup

Af − inf

Af

Proof. Let a, b ∈ A. Consider |f(a)− f(b)| = f(a)− f(b) or f(b)− f(b). By thereverse triangle inequality,

|f(a)| − |f(b)| ≤ |f(a)− f(b)| ≤ sup{f(a)− f(b) : a, b ∈ A}≤ sup

a∈Af(a) + sup

b∈A(−f(b))

= supAf − inf

Af

Furthermore,

|f(a)| ≤ |f(b)|+ supAf − inf

Af =⇒ sup

A|f | ≤ |f(b)|+ sup

Af − inf

Af

and

|f(b)| ≥ supA|f | − sup

Af + inf

Af =⇒ inf

A|f | ≥ sup

A|f | − sup

Af + inf

Af

Rearranging,

supA|f | − inf

A|f | ≤ sup

Af − inf

Af

62

V:

Proof. By lemma 22 above,

sup(xj−1,xj)

|f | − inf(xj−1,xj)

|f | ≤ sup(xj−1,xj)

f − inf(xj−1,xj)

f

=⇒ S(|f |, P )− s(|f |, P ) ≤ S(f, P )− s(f, P )

By the Cauchy criterion, ∃P : S(f, P ) − s(f, P ) < ε hence |f | is Riemann in-tegrable. Further note that −|f | ≤ f ≤ |f |. Since |f |, −|f |, f are Riemannintegrable, by IV, we get

−∫ b

a

|f | =∫ b

a

−|f | ≤∫ b

a

f ≤∫ b

a

|f |

Therefore∣∣∣∣∣∫ b

a

f

∣∣∣∣∣ ≤∫ b

a

|f | (since∫ b

a

|f | ≥ 0)

as required.

VI:

Proof. Observe (|f |2) = f2. WLOG, assume f ≥ 0.

S(f2, P )− s(f2, P ) =n∑j=1

[( sup(xj−1,xj)

f2 − inf(xj−1,xj)

f2)(xj − xj−1)]

=n∑j=1

[(( sup(xj−1,xj)

f)2 − ( inf(xj−1,xj)

f)2)(xj − xj−1)]

=n∑j=1

[(sup f − inf f)(sup f + inf f)(xj − xj−1)]

Since −M ≤ f ≤M =⇒ inf f ≤ sup f ≤M =⇒ inf f + sup f ≤ 2M ,

n∑j=1

[(sup f − inf f)(sup f + inf f)(xj − xj−1)] ≤ 2M(S(f, P )− s(f, P ))

Let ε > 0. Since f is Riemann integrable, choose P : S(f, P )− s(f, P ) < ε2M

giving us

S(f2, P )− s(f2, P ) < 2M · ε

2M= ε

Therefore f2 is Riemann integrable.

VII:

Proof. Observe fg = ((f + g)2 − (f − g)2) (polarisation). By I,II,VI, we havethat fg is Riemann integrable.

63

8.5 Lemma 23

If f is monotone, f is Riemann integrable. Proof provided in Warren’s notes,and is also a tutorial question.

8.6 Lemma 24

Let f : [a, b]→ R be continuous, then f is Riemann integrable.

Proof. Since f is continuous, it is uniformly continuous. ∀ε > 0 ∃δ ∀|x− y| < δ

|f(x)− f(y)| < ε

b− a

Let P be a partition with the largest subinterval of length less than δ. Then

sup(xj−1,xj)

f − inf(xj−1,xj)

f <ε

b− a

This implies

S(f, P )− s(f, P ) ≤n∑j=1

(ε

b− a)(xj − xj−1) = ε

Hence by the Cauchy criterion, f is Riemann integrable.

8.7 Theorem 19

Let a < c < b. Then f is Riemann integrable on [a,b] ⇐⇒ f is Riemannintegrable on [a, b] and [c, b] in which case∫ b

a

f =∫ c

a

f +∫ b

c

f

Proof. Define χ[a,c](x) =

{1 x ∈ [a, c]0 otherwise

to be the characteristic function for [a, c]. Let P be a partition of [a, b] whichcontains c.

S(f · χ[a,c], P ) = S(f · χ[a,c], P ∩ [a, c])

Taking the infimum over P ,∫ b

a

f · χ[a,c] =∫ c

a

f · χ[a,c] =∫ c

a

f

Similarly, for lower integral.

Case: “⇒”: WLOG consider [a, c]. Suppose f is Riemann integrable over[a, b]. Since P is a partition of [a, b] containing c,

S(χ[a,c], P ) =

xn∑j=1

1(xj − xj−1)

= c− a where xk = c.

64

Then f · χ[a,b] and χ[a,c] are Riemann integrable by VII. ie. f is Riemannintegrable over [a, c] and [c, b]. Then∫ c

a

f =∫ b

a

f · χ[a,c] =∫ b

a

fχ[a,c] =∫ c

a

f

Case: “⇐”: Suppose f is Riemann integrable over [a, c] and [c, b]. ie. f ·χ[a,c]

and f ·χ[c,b] are integrable over [a, b]. Since f = f ·χ[a,c]+f ·χ(c,b], f is integrableby II.

Finally in both cases the linearity of the integral gives∫ b

a

f =∫ b

a

(f · χ[a,c] + f · χ(c,b]) =∫ b

a

χ[a,c] · f +∫ b

a

χ(c,b]f =∫ c

a

f +∫ b

c

f

8.8 First Fundamental Theorem of Calculus (FTC 1)

If f : [a, b]→ R is Riemann integrable, and let F : [a, b]→ R such that

F (x) :=∫ x

a

f

Then,

1. F is continuous (assignment question)

2. If f is continuous at x, then F is differentiable at x with F ′(x) = f(x).

8.8.1 Definition

Let ∫ a

a

f := 0 ,∫ a

b

f = −∫ b

a

f

with this we have the proposition:∫ b

a

f =∫ c

a

f +∫ b

c

f ∀a, b, c ∈ R

Now we can prove the FTC.

Proof. Let x, y ∈ [a, b], x 6= y. Since∫ y

a

f −∫ x

a

f =∫ y

a

f +∫ a

x

f =∫ y

x

f

and that as y draws near x, f(y) draws near f(x), we can infer

F (y)− F (x)y − x

=1

y − x(∫ y

a

f −∫ x

a

f) =1

y − x

∫ y

x

f

As f is continuous at x, ∀ε > 0, ∃δ > 0 ∀t : |x− t| < δ

f(x)− ε < f(t) < f(x) + ε

65

If |y − x| < δ, then for all t in the interval with endpoints x, y

f(x)− ε < f(t) < f(x) + ε

So integrating over the interval with endpoints x, y∫ y

x

(f(x)− ε) ≤∫ y

x

f ≤∫ y

x

(f(x) + ε)

If x < y then

(y − x)(f(x)− ε) ≤∫ y

x

f ≤ (y − x)(f(x) + ε)

Divide by y − x and we get

f(x)− ε ≤ 1y − x

∫ y

x

f ≤ f(x) + ε

Therefore

limy→x

F (y)− F (x)y − x

= f(x)

If x > y then a similar argument follows for the left hand limit.

8.9 Second Fundamental Theorem of Calculus (FTC 2)

Suppose f : [a, b] → R is differentiable on [a, b] and suppose f ′ is Riemannintegrable. Then

f(b)− f(a) =∫ b

a

f ′

Proof. Let P be a partition of [a, b].

f(b)− f(a) =n∑j=1

(f(xj)− f(xj−1))

Since f is differentiable, then by the mean value theorem,

f(xj)− f(xj−1) = f ′(x∗j )(xj − xj−1) for some x∗j ∈ (xj−1, xj)

So

f(b)− f(a) =n∑j=1

f ′(x∗j )(xj − xj−1)

Now

inf(xj−1,xj)

f ′ ≤ f ′(x∗j ) ≤ sup(xj−1,xj)

f ′

66

and multiply by (xj − xj−1) and sum over j to get

s(f ′, P ) =n∑j=1

( inf(xj−1,xj)

f ′)(xj − xj−1)

≤n∑j=1

f ′(x∗j )(xj − xj−1) = f(b)− f(a)

≤n∑j=1

( sup(xj−1,xj)

f ′)(xj − xj−1)

= S(f ′, P )∫ b

a

f ′ ≤ f(b)− f(a) ≤∫ b

a

f ′

Since f ′ is Riemann integrable,∫ b

a

f ′ =∫ b

a

f ′

Therefore,

f(b)− f(a) =∫ b

a

f ′

8.10 Substitution Rule

Suppose f : [a, b] → R is continuous. Suppose also that φ : [α, β] → R with φdifferentiable and φ′ continuous, and φ([α, β]) = [a, b]. Then∫ β

α

f(φ(t))φ′(t)dt =∫ φ(β)

φ(α)

f(x)dx

Proof. Let F : [a, b]→ R and

F (u) :=∫ u

φ(α)

f(x)dx

By FTC 1, since f is continuous on [a, b] and F is differentiable on [a, b] we haveF ′(u) = f(u) ∀u ∈ [a, b]. Then by FTC 2,∫ φ(β)

φ(α)

f(x)dx =∫ φ(β)

φ(α)

F ′(x)dx = F (φ(β))− F (φ(α))

By FTC 1 again,

F (φ(β))− F (φ(α)) =∫ β

α

(F ◦ φ)′

Since F is differentiable and φ is differentiable, then by the chain rule (F ◦φ)′ = F ′(φ) · φ′ hence∫ β

α

(F ◦ φ)′ =∫ β

α

f(φ(t))φ′(t)dt

67

9 Pointwise and Uniform Convergence

9.1 Pointwise

Definition: Let A ⊂ R and ∀n ∈ N let fn : A→ R be a function. Let f : A→ R.Then

limn→∞

fn = f pointwise ⇐⇒ ∀a ∈ A[ limn→∞

fn(a) = f(a)]

9.1.1 Example

A = [0, 1], fn(x) = xn. Then ∀x ∈ [0, 1]

limn→∞

fn(x) =

{0 x < 11 x = 1

So define f : A→ R

f(x) =

{0 x ∈ [0, 1)1 x = 1

Then

limn→∞

fn = f pointwise

9.1.2 Problems

Is it the case that

1. If ∀n, fn is continuous =⇒ f is continuous?

2. If fn differentiable =⇒ f differentiable?

3. If fn and f differentiable, limn→∞

f ′n(a) = f ′(a) ?

4. If fn is integrable =⇒ f integrable and limn→∞

∫ b

a

fn =∫ b

a

f

Unfortunately pointwise convergence is not strong enough for these proper-ties to hold in general. Consider the counter examples in Warren’s notes, aswell as these:

1. Let fn : [0, 1] → R be defined by fn(x) := xn. Then we have a family ofcurves where the gradient curves out from the y = x line more and more as nincreases. Let

f(x) :=

{0 x ∈ [0, 1)1 x = 1

Then

limn→∞

fn = f pointwise

68

but obviously not continuous at x = 1.

Another example is to define a sequence of functions that are 0 everywhereexcept at 1

2 −1n where it linearly ascends towards 1 and reaches it at 1

2 and thendescends symmetrically back down to 0 at 1

2 + 1n . Then the pointwise function

is 0 everywhere except at x = 0, then there’s discontinuity as the value is 1.

2. Define fn : R→ R and fn(x) = arctan(nx). Then

limn→∞

f(x) =

π2 x > 00 x = 0−π2 x < 0

Then fn is differentiable for all n ∈ N but f is not continuous or differentiable(in general).

3. f : R→ R, fn(x) = 1n · arctan(nx). Then ∀x ∈ R

limn→∞

fn(x) = 0

so f(x) = 0 ∀x ∈ R. But f ′n(x) = 1n ·

n1+(nx)2 = 1

1+(nx)2 . Then

f ′n(0) = 1 6= 0 = f ′(0) ∀n ∈ N

4. Define fn : [0, 1] → R so that the function linearly ascends to a peak of2n before symmetrically falling down to 0 again at 1

n where it remains for therest of the function. Then

limn→∞

fn = 0 pointwise (except maybe at 0, where it ascends towards infinity)

But ∫ 1

0

fn = 1 ∀n ∈ N so limn→∞

∫ 1

0

fn = 1 6= 0 =∫ 1

0

f

Consider also [0, 1] ∩ Q = {q1, q2, . . .} with qi 6= qj if i 6= j. Then it iscountable (as Q is countable). For all n ∈ N define fn : [0, 1] → R withfn = χ{q1,...,qn}. Then

fn(x) =

{1 x ∈ {q1, . . . , qn}0 otherwise

Define f = χQ∩[0,1] then

limn→∞

fn = f pointwise

Clearly fn is Riemann integrable ∀n ∈ N since it’s a finite number of disconti-nuities, but f is not Riemann integrable.

69

9.2 Uniform

Definition:

limn→∞

fn = f uniformly ⇐⇒ ∀ε > 0 ∃N ∈ N ∀a ∈ A ∀n ≥ N

|fn(a)− f(a)| < ε

where N is allowed to depend on ε and not on a.

9.2.1 Example 1

fn : [0, 1]→ R , fn(x) = (1− x)xn. Clearly, ∀x ∈ [0, 1]

limn→∞

fn(x) = 0

It is also clear that uniform convergence implies pointwise convergence.Take f : [0, 1]→ R f(x) = 0 ∀x ∈ [0, 1]. Claim: limn→∞ fn = f uniformly.

fn(x) = xn − xn+1 =⇒ f ′n(x) = nxn−1 − (n + 1)xn = xn−1(n − (n + 1)x)since fn(0) = fn(1) = 0 and fn ≥ 0 it follows that fn has a maximum atx = n

n+1 . ie. maximum is fn( nn+1 ) = 1

n+1 · (nn+1 )n.

Let ε > 0 Take N = 1ε . ∀x ∈ [0, 1] and ∀n ≥ N

|fn(x)− f(x)| = |fn(x)| ≤ 1n+ 1

· ( n

n+ 1)n <

1n+ 1

<1N

= ε

9.2.2 Lemma 25

Let∑an be an absolutely convergent series. Then∣∣∣∣∣∞∑n=1

an

∣∣∣∣∣ ≤∞∑n=1

|an|

Proof. Let N ∈ N. Then∣∣∣∣∣N∑n=1

an

∣∣∣∣∣ ≤N∑n=1

|an|

Since x 7→ |x| is continuous∣∣∣∣∣∞∑n=1

an

∣∣∣∣∣ =

∣∣∣∣∣ limN→∞

N∑n=1

an

∣∣∣∣∣ = limN→∞

∣∣∣∣∣N∑n=1

an

∣∣∣∣∣ ≤ limN→∞

N∑n=1

|an| =∞∑n=1

|an|

9.2.3 Example 2

Power series. Let c0, c1, . . . ∈ R and consider the power series∑cnx

n. LetR ∈ [0,∞] be the convergence radius. Suppose R > 0. Let r > 0, r ∈ R :r < R. For all n ∈ N define Sn : [−r, r]→ R

Sn(x) =n∑k=0

ckxk

70

Further define S : [−r, r]→ R

S(x) =∞∑k=0

ckxk

This is convergent since |x| < R. Clearly limn→∞ Sn = S pointwise.Claim: limn→∞ Sn = S uniformly.

Proof. Let ρ ∈ R : r < ρ < R. Then the power series(∑ckρ

k) is convergent.Then by N th term test,

limk→∞

ckρk = 0

Therefore (ckρk) is bounded. Then there exists M ∈ R: ∀k ∈ N |ckρk| ≤ M .Let ε > 0. Take N = [fix]. Then for all n ∈ N with n ≥ N and all x ∈ [−r, r]

|Sn(x)− S(x)| =

∣∣∣∣∣n∑k=0

ckxk −

∞∑k=0

ckxk

∣∣∣∣∣ =

∣∣∣∣∣∞∑

k=n+1

ckxk

∣∣∣∣∣ ≤∞∑

k=n+1

|ckxk|

=∞∑

k=n+1

|ck||x|k ≤∞∑

k=n+1

M

(|x|ρ

)k≤∞∑k=N

M

(r

ρ

)k

= M

(r

ρ

)N ( 11− r

ρ

)< ε

For a chosen N large enough.Therefore limn→∞ Sn = S uniformly.

9.3 Theorem 20

If fn is continuous for all n ∈ N and limn→∞ fn = f uniformly then f iscontinuous.

Proof. Let a ∈ A. We shall prove that f is continuous at a. Let ε > 0. Sincelimn→∞ fn = f uniformly, then ∃N : ∀x ∈ A ∀n ≥ N |fn(x)− f(a)| < ε

3 . Nowchoose n = N . fn continuous ⇐⇒ ∃δ > 0: |x−a| < δ =⇒ |fn(x)−fn(a)| < ε

3 .Then ∀x ∈ A |x− a| < δ one has

|f(x)− f(a)| = |f(x)− fn(x) + fn(x)− fn(a) + fn(a)− f(a)|≤ |f(x)− fn(x)|+ |fn(x)− fn(a)|+ |fn(a)− f(a)|

<ε

3+ε

3+ε

3= ε

9.3.1 Corollary

x 7→∞∑k=0

ckxk is continuous on (−R,R)

71

Proof. Let a ∈ (−R,R). Then there exists an r ∈ R: |a| < r < R. Now defineSn, S : [−r, r]→ R as before then S is continuous on [−r, r].

Since a ∈ (−r, r) it follows that x 7→∑∞k=0 ckx

k is continuous at a.ie. If f : (−R,R), f(x) =

∑∞k=0 ckx

k then the restriction of f to [−r, r] iscontinuous at r but f is not necessarily continuous over all (−R,R).

9.3.2 Example

If fn : (−1, 1) → R, fn(x) =∑nk=0 x

k , f(x) = 11−x . Then limn→∞ fn = f

pointwise but not uniformly on (−1, 1).

9.4 Theorem 21

Suppose a < b, fn, f : [a, b] → R. Then ∀n ∈ N: Suppose fn is Riemannintegrable and limn→∞ fn = f uniformly. Then f is Riemann integrable and∫f = lim

∫fn. ie.∫ b

a

limn→∞

fn = limn→∞

∫ b

a

fn

Proof. Let ε > 0. Since limn→∞ fn = f uniformly, ∃N ∈ N: ∀x ∈ [a, b] one has

|fn(x)− f(x)| < ε

Since fn is Riemann integrable there exists a partition P of [a, b]:

S(fn, P )− s(fn, P ) < ε

Let P = {a1, . . . , am}.

S(f, P ) =

(m∑i=1

sup(ai−1,ai)

f(x)

)(ai − ai−1)

≤

(m∑i=1

sup(ai−1,ai)

fn(x) + ε

)(ai − ai−1)

= S(fn, P ) +m∑i=1

ε(ai − ai−1) = S(fn, P ) + ε(b− a)

Similarly, s(f, P ) ≥ s(fn, P )− ε(b− a). Hence

S(f, P )− s(f, P ) ≤ S(fn, P ) + ε(b− a)− s(fn, P ) + ε(b− a)< ε+ 2ε(b− a) = ε(1 + 2(b− a))

Since ε is arbitrary, effectively we can adjust the inequalities to make it lessthan any ε we choose. Therefore, f is Riemann integrable.

Secondly, let ε > 0. By uniform convergence, ∃N ∈ N: ∀n ≥ N ∀x ∈ [a, b]

|fn(x)− f(x)| < ε

72

Then ∀n ≥ N∣∣∣∣∫ f −∫fn

∣∣∣∣ =∣∣∣∣∫ (f − fn)

∣∣∣∣ ≤ ∫ |f − fn| = ∫ b

a

|f(x)− fn(x)|dx

≤∫ b

a

ε dx = ε(b− a)

Therefore

limn→∞

∫fn =

∫f

9.4.1 Example

Let 0 < r < 1. Sn, S : [0, r]→ R, Sn(x) =∑nk=0(−1)kx2k. S(x) =

∑∞k=0(−1)kx2k.

Recall that∞∑k=0

xk =1

1− x, (radius 1) =⇒

∞∑k=0

(−x2)k =1

1 + x2, (radius 1)

where “radius” refers to the convergence radius. Then limn→∞ Sn = S uni-formly.

∫ r

0

11 + x2

dx = arctan(r) =∫ r

0

S(x)dx = limn→∞

∫ r

0

Sn(x)dx

= limn→∞

∫ r

0

n∑k=0

(−1)kx2kdx = limn→∞

n∑k=0

(−1)k1

2k + 1r2k+1

=∞∑k=0

(−1)k

2k + 1r2k+1

Similarly for [−r, 0].

9.5 Theorem 22

Suppose f1, f2, . . . : [a, b] → R are differentiable(and hence fn are continuous).Let f : [a, b] → R. Suppose limn→∞ fn(x) = f(x) for all x ∈ [a, b] (ie. point-wise). Suppose also that φ : [a, b] → R and limn→∞ f ′n = φ uniformly. Then fis differentiable and f ′(x) = φ(x) ∀x ∈ [a, b]. ie.

( limn→∞

fn)′ = limn→∞

(f ′n)

Proof. For all x ∈ [a, b] one has ∀n ∈ N

fn(x)− fn(a) =∫ x

a

f ′n

Take limn→∞ then fn(x)→ f(x), and fn(a)→ f(a). So that by FTC 2,

f(x)− f(a) =∫ x

a

φ

73

f ′n is continuous ∀n so φ is continuous by theorem 20. Since x →∫ xaφ is

differentiable(by FTC 1), the derivative is φ(FTC 2). So φ = (f − f(a))′ andthus f is differentiable and φ(x) = f ′(x) ∀x ∈ [a, b].

74

maths 332: real analysis - university of auckland · math 332: real analysis ... 6.5 intermediate...

Documents