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Homework 7, Advanced Calculus 1
1. Rudin Chapter 4 Exercise 20a.
Solution:
Suppose that ρE(x) = 0 = infz∈Ed(x, z). For all > 0, by definition of infimum, there is z∈ E so that 0 ≤ d(x, z) < and therefore x ∈ ¯E. Conversely, x ∈ ¯E implies that the set of {d(x, z)}z∈E consist of nonnegative numbers that are arbitrarily small, and therefore the infimum must be 0.
2. Rudin Chapter 4 Exercise 20b.
Solution:
Given x, y ∈ X, for every z ∈ E, we have ρE(x) ≤ d(x, z) ≤ d(x, y) + d(y, z). That is ρE(x) − d(x, y) ≤ d(y, z) for all z ∈ E, or ρE(x) − d(x, y) is a lower bound for {d(y, z)}z∈E. Therefore, ρE(x) − d(x, y) ≤ inf{d(y, z)}z∈E= ρE(y), or ρE(x) − ρE(y) ≤ d(x, y). Exchanging x and y, we get ρE(y) − ρE(x) ≤ d(y, x) = d(x, y). Therefore, |ρE(x) − ρE(y)| ≤ d(x, y) and result follows.
3. Rudin Chapter 4 Exercise 21.
Solution: By Exercise 20, ρK: K → F is a continuous function on X and therefore on the compact set K. Therefore ρF attains a minimum on K:
ρF(x0) = inf
x∈KρF(x).
It then suffices to show that ρF(x) > 0 ∀x ∈ K, which follows easily from Exercise 20. Indeed, if ρF(x) = 0 for some x ∈ K, then x ∈ ¯F . But since F is closed, we have ¯F = F ⇒ x ∈ K ∩ F , which is a contradiction.
4. Rudin Chapter 4 Exercise 22.
Solution:
Since ρA and ρB are both continuous function, f is continuous except at point p where ρA(p) + ρB(p) = 0. But since both functions are nonnegative, it only happens when ρA(p) = ρB(p) = 0, or p ∈ ¯A ∩ ¯B. However, since both sets are closed, we have x ∈ A ∩ B, contradicting the fact A ∩ B = ∅.
f (p) = 0 precisely when ρA(p) = 0 and ρA(p) + ρB(p) > 0, which are true iff p ∈ ¯A = A. f (p) = 1 ⇔ ρA(p) = ρA(p) + ρB(p) ⇔ ρB(p) = 0 ⇔ p ∈ ¯B = B.
It is clear that A = f−1(0) ⊂ V = f−1([0,12)) and similarly B ⊂ W . The openness of V and W follow from the fact that 0 ≤ f (X) ≤ 1 and therefore V = f−1([0,12)) = f−1((−12,12)) and W = f−1((−12, 1]) = f−1((12, 2)) plus the continuity of f .
The Cantor Function is defined by f : [0, 1] → [0, 1] with the following rules. Recall that every x ∈ [0, 1]
can be written in tertiary expression x =P
jaj3−j, with aj = 0, 1, 2. The expression is unique except that
N −1
X
j=1
aj3−j+ aN3−N+
∞
X
j=N +1
2 · 3−j =
N −1
X
j=1
aj3−j+ (aN + 1)3−N.
We pick the first expression to ensure uniqueness. The Cantor set C ⊂ [0, 1] is defined by those real numbers with aj6= 1 ∀j. We define f separately on C and Cc. For x =P
jaj3−j ∈ C, we define f (x) =X
j
aj
2 2−j. For x =P
jaj3−j ∈ Cc, we define
f (x) =
Jx−1
X
j=1
aj
2 2−j+ 2−Jx, where Jx is the first digit of x with aj= 1.
Problems 5,6 concern the Cantor function and related topics.
5. Prove that the Cantor function f is uniformly continuous on [0, 1] and differentiable on Cc.
Solution: Continuity follows clearly from Problem 6 and 7 below. It is differentiable on Cc since every x ∈ Cc is contained in an open interval on which f is constant. It is therefore differentiable at x with f0(x) = 0.
6. Prove that there exist constants K, α > 0 so that
|f (x) − f (y)| ≤ K|x − y|α ∀x, y ∈ [0, 1]. (1)
Solution: First we check that f is monotonic. Given y = P
jaj3−j < x = P
jbj3−j, there is N ∈ N so that aN < bN (ie. (0, 1), (1, 2), (0, 2)) and aj = bj ∀j < N . By definition of f , the first N − 1 digits of f (x) and f (y) are still the same and the Nth digit of f (y) is less or equal to that of f (x). The digits after the Nthare of course irrelevant to the order. Therefore f (y) ≤ f (x) and the function is monotonically increasing.
The inequality above is certainly true if x = y. Take y < x, with the expression in the first paragraph.
There are four possibilities:
• x, y ∈ C.
• x ∈ C, y ∈ Cc.
• x ∈ Cc, y ∈ C.
• x, y ∈ Cc.
We first obverse that it suffices to prove the first case. Indeed, for every t =P
jaj3−j ∈ Cc, let aN
be the first digit with aN = 1. Then there are t1< t < t2 with t1, t2∈ C and f (t1) = f (t) = f (t2).
t1 and t2 are the two endpoints of the open interval t belongs to. Explicitly, t1 =PN −1
j=1 aj3−j+ P∞
N +12 · 3−j and t2 =PN −1
j=1 aj3−j+ 2 · 3−N. With these observations, for every y < x in [0, 1],
there exist then y0 ≤ x0, both in C, so that f (x) = f (x0), f (y) = f (y0), and |x0− y0| ≤ |x − y|.
Therefore, if
|f (x0) − f (y0)| ≤ C|x0− y0|α, we certainly have
|f (x) − f (y)| ≤ C|x − y|α. We now prove the first case. For y =P
jaj3−j < x =P
jbj3−j, since x, y ∈ C, we aN = 0 < bN = 2.
Then,
|f (x) − f (y)| = 2−N+
∞
X
j=N +1
bj− aj
2 2−j
≤ 2−N+
∞
X
j=N +1
2−j
= 2−N +1. On the other hand
|x − y| = 2 · 3−N+
∞
X
j=N +1
(bj− aj)3−j ≥ 2 · 3−N−
∞
X
j=N +1
2 · 3−j = 3−N.
Therefore, take α = log32, we have |x − y|α≥ 2−N and
|f (x) − f (y)|
|x − y|α ≤ 2, which is the desired inequality.
7. Functions satisfying the condition in Problem 6 on its domain is said to be H¨older continuous with exponent α. Prove that
{H¨older continuous function} ( {Uniformly Continuous Functions}.
Solution: The inclusion is easy. Let f satisfies (1) in Problem 6. For all , take δ = Kα then uniform continuity follows.
For the properness of this inclusion, consider
f (x) = ( 1
log x ; x ∈ (0,12]
0 ; x = 0. (2)
The function is continuous on [0,12], and therefore uniformly continuous since [0,12] is compact.
However, for any α > 0, by L’Hospital’s rule, we have
x→0lim
1 log x
xα = ∞ and therefore (1) is impossible with y = 0.
8. Rudin Chapter 4 Exercise 25.
Solution:
(a) Note that the distance we use here is the Euclidean distance d(a, b) = ka − bk.
Following the hint, we show that (K + C)cis open. Let z ∈ (K + C)cand consider F = z − C :=
{z − c | c ∈ C}. It is clear that F ∩ K = ∅. Since if x ∈ F ∩ K, x = z − y for some y ∈ C, then z = y + x ∈ C + K, contradicting z ∈ (K + C)c. It is also clear that F is closed, since if x ∈ F0, z − x ∈ C0⊂ C as C is closed. Therefore x ∈ F .
We now have a closed set F and a compact set K disjoint from each other. By Exercise 21, there is δ > 0 so that d(p, q) > δ for all p ∈ K, q ∈ F . We finally show that
Bδ(z) ∩ (K + C) = ∅
and the openness of (K + C)c, or the closedness of K + C follows. To the contrary, suppose that we have a + b ∈ Bδ(z) ∩ (K + C), where a ∈ K, b ∈ C. Since a + b ∈ Bδ(z), we have
ka + b − zk = d(a + b, z) < δ.
On the other hand, since b ∈ C, we have z − b ∈ F . With a ∈ K, we have
ka + b − zk = ka − (z − b)k = d(a, z − b) > δ.
The two inequalities contradict each other and we have Bδ(z) ∩ (K + C) = ∅.
(b) C1= Z and C2= {mα | m ∈ Z} for some α /∈ Q. Both sets are closed since both of them have no limit point. We show that C = C1+ C2 is not closed. The result follows if we show that C is dense in R. This is true because C is clearly countable. If it is dense, then it can not be closed. Otherwise, we have
R = C = C
but the right hand side is countable while the left hand side is not.
We now show that C is dense. For any x ∈ R, define (x) = x − [x] to be the fractional part of x. Here, [x] is the largest integer ≤ x. Since α /∈ Q, we have (mα) 6= (m0α) if m, m0 are two distinct integers. Indeed, if (mα) = (m0α) while m 6= m0, we have mα − [mα] = m0α − [m0α].
But then α = [mα]−[mm−m00α] ∈ Q, a contradiction.
For each n, divide [0, 1] into n subintervals of length n1. By the discussion above, {(jα)}nj=1is a set of n distinct values in [0, 1]. By pigeonhole principle, two of these values, say (mα) < (m0α) belong to one subinterval. Their distance is then less than n1, and we have
0 < (m0α) − (mα) = m0α − [m0α] − mα + [mα] = (m0− m)α + ([m0α] − [mα]) < 1 n.
Let αn = (m0 − m)α + ([m0α] − [mα]) ∈ (0,n1). For any x ∈ [0, 1], it is now clear that x ∈ (rαn, (r + 1)αn) for some integer r, and the length of the interval is less than n1. Finally, for x ∈ R, we have x = [x] + (x). [x] ∈ Z and (x) ∈ [0, 1]. For all n ∈ N, there is αn ∈ C so that |(x) − αn| < 1n. Then [x] + αn ∈ C and |[x] + αn− x| < n1.
9. Rudin Chapter 4 Exercise 26.
Solution: Since g : Y → Z is one-to-one, g : Y → g(Y ) is surjective and we have g−1: g(Y ) → Y . By Theorem 4.17, g−1 is continuous. The continuity of f follows easily by the observation that f = g−1◦ h and both functions on the right are continuous.
In addition, since g is continuous and Y is compact, g(Y ) is compact and therefore g−1: g(Y ) → Y is uniformly continuous. Therefore, if h is uniformly continuous, so is f .
If Y is no longer compact but X and Z are, the statement fails by considering the following coun- terexample. X = [0, 1], Y = [0,π2), and Z be the unit circle in R2. Let f : X → Y be defined by
f (x) =
(0 ; x = 0
tan−1(− log x) ; x ∈ (0, 1). (3)
g(t) = (cos(4t), sin(4t)). Then h = f ◦ g : [0, 1] → S1 is continuous, and therefore uniformly continuous. g is continuous and one-to-one, however, f is not even continuous.
