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Math40002: Analysis I
We will pick up where MATH40002 left off in the Autumn term, rigorously devel-
oping the basic tools of calculus: continuity, differentiation, and integration.
We will continue to have infinite fun.
Syllabus
Continuity: Review of continuity. Sequential continuity. Uniform continuity. In-
termediate and extreme value theorems. Inverse function theorem for mono-
tonic functions.
Differentiation: Definitions, examples, and properties. Mean value theorem. Higher
derivatives and convexity. Differentiation of series.
Integration: Definition, examples, and properties of Riemann–Darboux integral.
Fundamental theorem of calculus. Techniques: integration by parts, substitu-
tion. Indefinite integration.
Books
Martin Liebeck, A Concise Introduction to Pure Mathematics.
Mary Hart, Guide to Analysis.
KG Binmore, Mathematical Analysis, A Straightforward Approach.
David Brannan, A first course in mathematical analysis.
Steven Lay, Analysis: with an introduction to proof.
Stephen Abbott, Understanding analysis.
I will post gappy notes on Blackboard, and a complete version at the end of term.
We will continue to use edStem for online discussion of the lecture material.
Assessment
20% Autumn term assessments, including the January test
3% Mini Blackboard quizzes every few weeks (due 1 Feb, 1 Mar, 22 Mar)
5% Midterm test (1 hour) – week of 14-18 February
1% Formal write-up of a problem sheet problem, due 4 February
1% Formal write-up of a midterm problem, due 4 March
70% May exam
Contents Math40002: Analysis I
Contents
1 Continuity 3
1.1 The intermediate value theorem . . . . . . . . . . . . . . . . . . . . . 5
1.2 The extreme value theorem . . . . . . . . . . . . . . . . . . . . . . . 9
1.3 Open, closed, and compact sets . . . . . . . . . . . . . . . . . . . . . 17
1.4 Uniform continuity and convergence . . . . . . . . . . . . . . . . . . . 21
2 Differentiation 31
2.1 Basic properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
2.2 The mean value theorem . . . . . . . . . . . . . . . . . . . . . . . . . 38
2.3 L’Hoˆpital’s rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
2.4 Higher derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
2.5 Second derivatives and convexity . . . . . . . . . . . . . . . . . . . . 51
2.6 Limits of differentiable functions . . . . . . . . . . . . . . . . . . . . . 55
3 Integration 63
3.1 Darboux sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
3.2 The Darboux integral . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
3.3 Basic properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
3.4 The fundamental theorem of calculus . . . . . . . . . . . . . . . . . . 83
3.5 More properties of integrals . . . . . . . . . . . . . . . . . . . . . . . 87
3.6 Limits of integrable functions . . . . . . . . . . . . . . . . . . . . . . 92
3.7 Improper integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
3.8 Lebesgue’s criterion for integrability . . . . . . . . . . . . . . . . . . . 103
3.9 The irrationality of pi . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
2
1 Continuity Math40002: Analysis I
1 Continuity
We begin by recalling what it means for a function to be continuous.
Definition. Given a function f : R→ R, we say that f is continuous at a ∈ R
if and only if
∀ > 0 ∃δ > 0 such that |x− a| < δ =⇒ |f(x)− f(a)| <
We say that f is continuous on R (or just “continuous”) if it is continuous at
all a ∈ R.
We also saw the equivalent notion of sequential continuity, which is sometimes more
convenient.
Theorem 1.1
f : R→ R is continuous at a ∈ R ⇐⇒ f(xn)→ f(a) ∀ sequences xn → a.
Let’s warm up by proving some basic properties of continuity.
Proposition 1.2. Let f, g : R → R be functions which are both continuous at
a ∈ R. Then the functions f + g, f − g, and f · g are all continuous at a, and if
g(a) 6= 0 then fg is also continuous at a.
Proof. We’ll make use of sequential continuity here. For any sequence xn → a, we
have f(xn) → f(a) and g(xn) → g(a) by hypothesis. The algebra of limits tells us
that
lim
n→∞
(
f(xn) + g(xn)
)
= lim
n→∞ f(xn) + limn→∞ g(xn) = f(a) + g(a).
Since this works for every sequence xn → a, we conclude that f + g is continuous.
The same argument works for f − g and f · g, and also for fg if g(a) 6= 0.
Now we can start to assemble a big collection of continuous functions almost for free
– no need for epsilons and deltas! We know that linear functions f(x) = mx+ c are
continuous, so we get
• any function f(x) = xn is continuous, where n ≥ 0 is an integer;
• any polynomial p(x) = anxn+an−1xn−1+ · · ·+a1x+a0 is continuous, where
the ai are real numbers;
• any rational function p(x)
q(x)
, where p and q are polynomials, is continuous at
all a ∈ R where q(a) 6= 0.
3
1 Continuity Math40002: Analysis I
The first two can be proved by induction on n (try it!), and the third follows from
the second.
Similarly, we saw last term that E : C → C defined by E(z) = ∑∞n=0 znn! is contin-
uous on C. Since x 7→ sin(x) and x 7→ cos(x) can be defined in terms of E (how?),
this tells us where the trigonometric functions are continuous:
• sin(x) and cos(x) are continuous on R;
• tan(x) = sin(x)
cos(x)
is continuous wherever cos(x) 6= 0, meaning at all x except
x =
(2k+1)pi
2 , k ∈ Z;
• likewise, sec(x), csc(x), cot(x) are continuous at any x where cos(x) 6= 0,
sin(x) 6= 0, and sin(x) 6= 0 respectively.
Proposition 1.3. If f : R → R is continuous at a, and g : R → R is continuous
at f(a), then the composition g ◦ f defined by x 7→ g(f(x)) is continuous at a.
Proof. Let xn be any sequence such that xn → a. Since f is continuous at a, we
have f(xn) → f(a). Now we can plug the convergent sequence yn = f(xn), with
limit y = f(a), into g: since g is continuous at y, we have g(yn)) → g(y). But this
is just another way of saying that g(f(xn)) → g(f(a)). Again, this works for any
sequence xn → a, so g(f(x)) must be continuous at a.
Example 1.4. We know that E : C → C defined by E(z) = ∑∞n=0 znn! is
continuous on C, and we saw that this implies that sin(x) is continuous on R.
So now
f(z) = sin
(
z2 + 1
z − 2
)
is continuous at all a ∈ R where z2+1z−2 is continuous, meaning at all a 6= 2.
Example 1.5. Consider the function f : R→ R defined by
f(x) =
{
x sin(1/x), x 6= 0
0, x = 0.
Since 1x is a rational function, it is continuous at all x 6= 0. Thus
• sin( 1x) is continuous at all x 6= 0, since x 7→ sin(x) is continuous;
4
1 Continuity Math40002: Analysis I
• x sin( 1x) is continuous at all x 6= 0, since both x and sin( 1x) are;
and finally we saw last term that f is continuous at x = 0 as well, so f is
continuous on all of R.
Question 1. Let f, g : R→ R be functions, and let h(x) = g(f(x)). Which of
the following is always true?
1. If f and g are not continuous, then h is not continuous.
2. If f is continuous and g is not, then h is not.
3. If f is not continuous but g is, then h is.
4. More than one of choices 1, 2, 3.
5. None of choices 1, 2, 3. X
The easiest thing to try would be functions that are constant, or maybe piecewise
constant, to see what happens. For item 1, we might try something like
f(x) =
{
0, x < 0
1, x ≥ 0 and g(x) =
{
0, −2 ≤ x ≤ 2
1, |x| > 2
so that g(f(x)) = 0 is constant. For 2, we take f = 0 and any discontinuous g, and
then h = f(0) is constant. For 3, we take g(x) = x and any discontinuous f , and
then h = f is discontinuous.
1.1 The intermediate value theorem
We now come to the first major application of continuity.
Theorem 1.6: Intermediate value theorem
Let f : [a, b] → R be a continuous function, and pick a value c between f(a)
and f(b). Then there is some x ∈ [a, b] such that f(x) = c.
Here’s a picture of what this might look like:
5
1 Continuity Math40002: Analysis I
a
b
f(a)
f(b)
x
c
Note that there might be several possible values of x; in this picture there are three.
This theorem certainly sounds obvious – a continuous function can’t jump, so it
can’t skip any values – but that doesn’t mean it’s easy to prove! We wouldn’t
even know where to begin if we hadn’t already built up the notions of limits and
continuity from scratch.
Proof. We can assume without loss of generality that f(a) < c < f(b). Indeed:
• if f(a) = c we take x = a, and if f(b) = c then we take x = b;
• if f(a) = f(b) then c = f(a) and we are already done; and
• if f(a) > f(b) then we can take g(x) = −f(x), so that g(a) < g(b), and ask
for x with g(x) = −c instead.
So now we consider the set
Sc = {y ∈ [a, b] | f(y) < c}.
This set is nonempty, because a ∈ Sc, and it is bounded above by b. Thus if we let
x = supSc then we have a ≤ x ≤ b.
Claim 1: f(x) ≥ c.
Let’s suppose this claim is false, so f(x) < c, and take = c− f(x) > 0. Note that
if f(x) < c then x 6= b, so we must have x < b here. Since f is continuous at x, we
have
∃δ > 0 such that |f(y)− f(x)| < ∀y ∈ (x− δ, x+ δ) ∩ [a, b].
In particular, this means that
f(y) < f(x) + = c ∀y ∈ (x− δ, x+ δ) ∩ [a, b].
So all of these y belong to Sc, and if we choose y ∈ (x, x+ δ) ∩ [a, b], say
y = x+
1
2
min(δ, b− x)
