CHAPTER 4-无代写
时间:2023-03-31
CHAPTER 4
1. Functions
Dirichlet’s function:
f(x) =
{
1 if x ∈ Q
0 if x 6∈ Q.
Is this function continuous? Why or why not?
Roughly speaking, a function f is continuous at a number c if when x is close to c,
f(x) is close to f(c). If we let x = b10
6

2c
106
, then x is pretty close to

2, but f(x) = 1
because f is rational. However, f(

2) = 0.
Modified Dirichlet’s function:
f(x) =
{
x if x ∈ Q
0 if x 6∈ Q.
Is this function continuous? Why or why not?
If c 6= 0, then f isn’t continuous at c because there are rational and irrational
numbers that are both close to c.
As x→ 0, f(x)→ 0 regardless of whether x is rational or not.
1
2 4
Thomae’s function:
This is the function f(x) =
{
1
n
if x ∈ Q and x = m
n
with gcd(m,n) = 1 and n > 0
0 if x 6∈ Q .
Where is this function continuous?
If c is irrational, then rational numbers approaching x must have denominators
going to infinity and so f(x)→ 0 as x→ c.
However, there are irrational numbers close to every rational, and so if c ∈ Q,
then f(c) > 0 but there are lots of x close to c so that f(x) = 0.
This function is continuous at all irrational numbers, but discontinuous at every
x ∈ Q.
2. FUNCTIONAL LIMITS 3
2. Functional limits
Definition 4.2.1. Let f : R → R and x0 ∈ R. We say that lim
x→x0
f(x) = L if,
∀ > 0, ∃δ > 0 (δ depends on and x0) so that whenever 0 < |x− x0| < δ, it follows
that |f(x)− L| < .
Definition 4.2.2. A function f : R→ R is continuous at x0 ∈ R if lim
x→x0
f(x) =
f(x0).
Example 4.2.3. Show that f(x) = x2 + 1 is continuous at x = 3.
We have f(3) = 10 so we want to show that limx→3 f(x) = 10. Fix > 0. How
do we pick δ?
We want to show that there is some δ > 0 so that |x−3| < δ =⇒ |f(x)−10| < .
We have f(x) − 10 = x2 + 1 − 10 = x2 − 9 = (x + 3)(x − 3). We can pick δ,
and if δ ≤ 1, then x ∈ (2, 4) and so 5 < x + 3 < 7. Under this assumption,
|f(x)−10| = |x+3||x−3| < 7|x−3|. So if δ is also ≤ /7, then 7|x−3| < 7(/7) < .
So we set δ = min{1, /7}. If |x− 3| < δ, then x ∈ (2, 4) and so |x+ 3| < 7. Thus
|f(x)− 10| = |x+ 3||x− 3| < 7 · (/7) = .
Example 4.2.4. Show that f(x) = 1
x
is continuous on (0,∞).
We’ll show that if c > 0, then limx→c 1x =
1
c
. Fix > 0. We want to show that
∃δ > 0 so that |x− c| < δ =⇒ ∣∣ 1
x
− 1
c
∣∣ < .
We have 1
x
− 1
c
= c
cx
− x
cx
so
∣∣ 1
x
− 1
c
∣∣ = |c−x|
c|x| . To make sure this is not too large,
we need to make sure x is not too small. Assume that δ < c
2
so that x ∈ (c/2, 3c/2).
Then 1|x| <
2
c
is not too big.
Then |c−x|
c|x| <
2|c−x|
c2
. To make this less than , we need |c − x| < c2
2
. So we let
δ = min
{
c
2
, c
2
2
}
. (This explains how to choose δ. I’ll skip the formal proof that
|x− c| < δ =⇒ |1/x− 1/c| < .)
4 4
Example 4.2.5. Show that f(x) = x
x+1
is continuous on (0,∞).
Let c ∈ (0,∞) and fix > 0. We have
f(x)− f(c) = x
x+ 1
− c
c+ 1
=
x(c+ 1)
(x+ 1)(c+ 1)
− c(x+ 1)
(x+ 1)(c+ 1)
=
x− c
(x+ 1)(c+ 1)
.
If x ≥ 0, then 1
x+1
≤ 1 and since c > 0, we have 1
c+1
≤ 1. If we choose δ ≤ c we have
that x ≥ 0. So take δ = min(c, ). Then if |x− c| < δ, then
|f(x)− f(c)| = |x− c||x+ 1||c+ 1| ≤ |x− c| < .
Example 4.2.6. Show that f(x) = 3x3 + x is continuous at every point in its
domain.
The domain is all real numbers. So let c ∈ R. We have
f(x)− f(c) = (3x3 + x)− (3c3 + c) = 3(x3 − c3) + (x− c)
= 3(x− c)(x2 + xc+ c2) + (x− c)(1)
= (x− c)(3(x2 + xc+ c2) + 1).
Suppose that |x− c| ≤ |c|+ 1. Then
|3(x2 + xc+ c2) + 1| ≤ 3(|c|+ 1)2 + 3(|c|+ 1)|c|+ |c|2 + 1 ≤ 9|c|2 + 9|c|+ 4.
So let δ = min
(
|c|+ 1,
9|c|2+9|c|+4
)
.
Q: Why would it NOT work to instead assume |x−c| ≤ |c| and let δ = min
(
|c|,
9|c|2+1
)
?
3. CONTINUITY 5
3. Continuity
Theorem 4.3.2. Let f : R→ R and let c ∈ R. Then f is continuous at c if and
only if:
(1) ∀ > 0, ∃δ > 0 such that |x− c| < δ implies |f(x)− f(c)| < .
(2) ∀ > 0, ∃δ > 0 such that x ∈ Bδ(c) implies f(x) ∈ B(f(c)).
(3) ∀{xn}∞n=1 so that xn → c, f(xn)→ f(c).
(4) lim
x→c
f(x) = f(c).
Rephrasing of item 2: If U ⊆ R is open, then f−1(U) = {x ∈ R : f(x) ∈ U} is
also open.
Proof that (1) and (3) are equivalent: Suppose that f is continuous and xn → c.
Fix > 0. Then there is some δ > 0 so that |x − c| < δ =⇒ |f(x) − f(c)| < .
Because xn → c, there is some N so that n ≥ N implies |xn − c| < δ. Then the
continuity of f gives that |f(xn)− f(c)| < and so f(xn)→ f(c).
Suppose now that f is not continuous at c. Then there is an > 0 so that for all
δ > 0, there exists some x with |x − c| < δ and |f(x) − f(c)| ≥ . For each n ∈ N,
choose some xn so that |xn − c| < 1/n and |f(xn)− f(c)| ≥ . Then xn → c is true,
but f(xn)→ f(c) is false.
Corollary 4.3.3. Let f : R→ R and let c ∈ R. If ∃(xn) so that xn → c but
f(xn) 6→ f(c), then f is not continuous at c.
Example 4.3.4. Prove or disprove f(x) = sin(1/x) if x 6= 0, and f(0) = 0 is
continuous at x = 0.
Let xn =
1
2pin+pi/2
. Then xn → 0. However,
f(xn) = sin(1/xn) = sin(2pin+ pi/2) = sin(pi/2) = 1.
Thus, xn → 0, but f(xn) 6→ f(0) = 0.
6 4
Theorem 4.3.5 (Algebraic Continuity Theorem). If f, g are functions that are
continuous at c and k is a constant, then:
(1) kf is continuous at c,
(2) f ± g is continuous at c,
(3) fg is continuous at c, and
(4) f
g
is continuous at c so long as g(c) 6= 0.
Main idea: Use part (3) of Theorem 1.9 to convert to a problem about convergence
of sequences. Then apply the algebraic limit theorem.
Theorem 4.3.9. If f : R→ R is continuous at c ∈ R and g : R→ R is continuous
at f(c), then g ◦ f = g(f(x)) is continuous at c.
Proof: Homework.
Example 4.3.10. Prove or disprove f(x) = x sin(1/x) if x 6= 0, and f(0) = 0 is
continuous at x = 0.
This is true. Fix > 0 and let δ = . If |x− 0| < δ, then
|f(x)− f(0)| = |x sin(1/x)− 0| ≤ |x|| sin(1/x)| < · 1 = .
4. CONTINUOUS FUNCTIONS ON COMPACT SETS 7
4. Continuous functions on compact sets
Theorem 4.4.1. Let f(x) be continuous on a compact set K ⊆ R. Then f(K) is
also a compact set.
Suppose that (bn) is a sequence in f(K) and for each n ≥ 1, let an ∈ K so that
f(an) = bn. Since (an) is a sequence in K and K is compact, there is a subsequence
(ank) that converges to an element L of K. Since ank and L are in K, f is continuous
at L and so
lim
k→∞
bnk = lim
k→∞
f(ank) = f(L).
Theorem 4.4.2. Let f(x) be continuous on a compact set K ⊆ R. Then max{f(x) :
x ∈ K} and min{f(x) : x ∈ K} exist.
By Theorem 1.15, f(K) is compact. By Lemma 1.9 from Chapter 3, since f(K)
is compact, min{x ∈ f(K)} and max{x ∈ f(K)} exist.
8 4
Definition 4.4.4. We say that f : R → R is uniformly continuous on a set
A ⊂ R if, for every > 0, there is a δ > 0 (which may depend on ) so that
∀x, y ∈ A, |x− y| < δ =⇒ |f(x)− f(y)| < .
How is this different from continuous?
Pairs of inputs are used rather than one “fixed” input c and one “variable” input
x. Also, the δ chosen only depends on , and not on anything else.
Note: The choice of A can matter. If f is uniformly continuous on A ⊆ R, it
might not be uniformly continuous on a larger subset B ⊆ R.
Theorem 4.4.5. Suppose f : R → R and A ⊂ R. Then f is not uniformly
continuous on A if and only if there exist sequences {xn}∞n=1, {yn}∞n=1 ⊂ A and a
number 0 > 0 so that |xn − yn| → 0 but |f(xn)− f(yn)| ≥ 0 for all n ≥ 1.
Suppose that there exist some 0 > 0 and sequences (xn) and (yn) so that |xn −
yn| → 0 but |f(xn)− f(yn)| ≥ 0 for all n. We want to show that f is not uniformly
continuous.
If it were, then there would be some δ > 0 so that |x − y| < δ implies that
|f(x)− f(y)| < 0. But since |xn− yn| → 0, ∃N ∈ N so that |xN − yN | < δ. But then
uniform continuity implies that |f(xN)− f(yN)| < 0, and the assumption says that
|f(xN)− f(yN)| ≥ 0. These contradict, and so f cannot be uniformly continuous.
For the other direction, assume that f is not uniformly continuous. Then ∃0 > 0
so that ∀δ > 0, ∃xδ, yδ ∈ A so that |xδ − yδ| < δ and |f(xδ) − f(yδ)| ≥ 0. For each
n ∈ N, choose δ = 1/n. Then there exist xn, yn ∈ A so that |xn − yn| < 1/n and
|f(xn)− f(yn)| ≥ 0. We have |xn − yn| → 0 and |f(xn)− f(yn)| ≥ 0.
4. CONTINUOUS FUNCTIONS ON COMPACT SETS 9
Example 4.4.6. x2 on [0,∞)
This is not uniformly continuous. Let xn = n and yn = n+1/n. Then |xn−yn| →
0, but f(xn) = n
2 and f(yn) = n
2 + 2 + 1/n2. Thus, |f(xn)− f(yn)| ≥ 2 for all n.
Example 4.4.7. 1
x
on (0,∞)
This is not uniformly continuous. Let xn = 1/n and yn = 1/(n + 1). Then
|xn− yn| → 0 but f(xn) = n and f(yn) = n+ 1 and so |f(xn)− f(yn)| ≥ 1 for all n.
Example 4.4.8. x on (0,∞)
This is uniformly continuous. Fix > 0. If |x − y| < , then |f(x) − f(y)| =
|x− y| < .
10 4
Theorem 4.4.9. If K ⊆ R is compact and f : K → R is continuous, then f is
uniformly continuous.
Note: One of the reasons that the open cover has a finite subcover version of
compactness is used is because of the proof I’m about to show you.
Fix > 0. For each x ∈ K, there is some δx so that if |y − x| < δx, then
|f(y)− f(x)| < /2. We therefore have an open cover
K ⊆

x∈K
Bδx/2(x).
(Why do we divide the radius of each ball by 2? Stay tuned.)
This cover has a finite subcover
K ⊆ Bδx1/2(x1) ∪Bδx2/2(x2) ∪ · · · ∪Bδxn/2(xn).
Draw a picture! (Imagine that each xi is a “major city” and every point in K is in a
“suburb of a major city.”)
Let δ = min{1
2
δxi : 1 ≤ i ≤ n}. Suppose that x, y ∈ K with |x − y| < δ. Then,
there is some xi so that x ∈ B 1
2
δxi
(xi) and we have |x − xi| < 12δxi . (So x is in the
suburb of the major city xi.)
Claim: |xi − y| < δxi . (That is, y is in a suburb of the same major city as x.)
We have |xi − y| ≤ |xi − x|+ |x− y| < 12δxi + δ ≤ 12δxi + 12δxi = δxi .
So we have |x−xi| < 12δxi and |xi−y| < δxi . This implies that |f(x)−f(xi)| < /2
and |f(xi)− f(y)| < /2 and so
|f(x)− f(y)| ≤ |f(x)− f(xi)|+ |f(xi − f(y)| < .
QED!!!
5. THE INTERMEDIATE VALUE THEOREM 11
5. The Intermediate Value Theorem
Theorem 4.5.1. Let f : [a, b]→ R be continuous. If L is a real number satisfying
f(a) < L < f(b) or f(b) < L < f(a), then there exists a point c ∈ (a, b) where
f(c) = L.
This is kind of like the proof that there’s a real square root of 2 (Theorem 1.4.5).
Assume that f(a) < L < f(b). Let K = {x ∈ [a, b] : f(x) ≤ L}. Then K is
nonempty and K is bounded above by b. Thus, c = sup(K) exists. We’ll prove that
f(c) = L by contradiction.
Case I: f(c) < L.
Since f is continuous at c, ∃δ > 0 so that |x− c| < δ implies that |f(x)− f(c)| <
L − f(c). This implies that for x ∈ (c − δ, c + δ) that f(x) < L. This implies that
there are some values of x > c for which f(x) ≤ L. These values of x are in K, but
we assumed that c = sup(K), a contradiction.
Case II: f(c) > L.
Since f is continuous at c, ∃δ > 0 so that |x− c| < δ implies that |f(x)− f(c)| <
f(c)− L. This implies that for x ∈ (c− δ, c+ δ) that
f(c)− (f(c)− L) < f(x) < f(c) + (f(c)− L).
In particular, this implies that f(x) > L for all x with c− δ < x ≤ c. However, since
c = sup(K) there must be some x ∈ K with x > c − δ, and this is a contradiction
(since for this x, we have f(x) > L and also f(x) ≤ L).
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