HOMEWORK ASSIGNMENT 6
DUE 15 MARCH, 2016
1) Suppose f, g : A → R are uniformly continuous on A. Show that f + g is uniformly
continuous on A.
Solution First we note: In order to show that f + g is uniformly continuous on A, we must
show that for every ε > 0, there exists δ > 0 such that for any x, y ∈ A with |x − y| < δ,
|(f + g)(y) − (f + g)(x)| < ε.
Fix ε > 0. Since f is uniformly continuous, there exists δf > 0 such that for any x, y ∈ A
with |x − y| < δf , |f (x) − f (y)| < ε/2. Since g is uniformly continuous, there exists δg > 0
such that for any x, y ∈ A with |x − y| < δg , |g(x) − g(y)| < ε/2. Let δ = min{δf , δg }.
Then if x, y ∈ A are such that |x − y| < δ, then |x − y| < δf , so |f (x) − f (y)| < ε/2, and
|x − y| < δg , so |g(x) − g(y)| < ε/2. Therefore
|(f + g)(y) − (f + g)(x)| = |f (x) − f (y) + g(x) − g(y)|
6 |f (x) − f (y)| + |g(x) − g(y)|
< ε/2 + ε/2 = ε.
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DUE 15 MARCH, 2016
2) Let f : [0, 1] → R be a continuous function. For each 0 6 x 6 1, let
F (x) = max f (t).
06t6x
How do we know F is well-defined (that is, how do we know this maximum exists)? Show
that F is continuous.
Solution We know the value F (x) = max06t6x f (t) exists, since f is continuous on the
compact set [0, t], and it must therefore have a maximum on that set.
We show that F is continuous. We observe that F is non-decreasing on [0, 1]. To see this,
suppose that 0 6 x < y 6 1. Then [0, x] ⊂ [0, y], so the maximum of f on the set [0, y] is
at least as large as the maximum of f on the subset [0, x]. This means F (x) 6 F (y) when
0 6 x < y 6 1.
Fix ε > 0. Since f is continuous on the compact set [0, 1], f is uniformly continuous, and
there exists δ > 0 such that if x, y ∈ [0, 1] and |x−y| < δ, |f (x)−f (y)| < ε. We will show that
if x, y ∈ [0, 1] and |x−y| < δ, |F (x)−F (y)| < ε. To that end, fix x, y ∈ [0, 1] with |x−y| < δ.
Assume without loss of generality that x 6 y. Since F is non-decreasing, F (x) 6 F (y). This
means |F (y) − F (x)| = F (y) − F (x). We need to show that F (y) − F (x) < ε. Fix some
s ∈ [0, y] such that F (y) = f (s) (that is, since F (y) is the maximum value of f on [0, y],
we can pick an s ∈ [0, y] where this maximum is attained). We consider two cases. Either
s ∈ [0, x], in which case
F (x) = max f (t) > f (s) = F (y).
06t6x
In this case, F (x) 6 F (y) 6 F (x), and F (x) = F (y). In the second case, s ∈ (x, y]. This
means that
|s − x| = s − x 6 y − x < δ.
By our choice of δ, |f (s) − f (x)| < ε. Since F (x) = max06t6x f (t) > f (x), we deduce that
F (y) = f (s) < f (x) + ε 6 F (x) + ε,
so that F (y) − F (x) < ε. In either case, F (y) − F (x) < ε, as desired.
HOMEWORK ASSIGNMENT 6
3
3) If f : A → B is uniformly continuous and g : B → R is uniformly continuous, then
g ◦ f : A → R is uniformly continuous.
Solution We want to show that for every ε > 0, there exists δ > 0 such that for any x, y ∈ A
with |x − y| < δ, |g(f (x)) − g(f (y))| < ε.
Fix ε > 0. Since g is uniformly continuous on B, there exists η > 0 such that for any
w, z ∈ B with |w − z| < η, |g(w) − g(z)| < ε. Since f is uniformly continuous on A, there
exists δ > 0 such that for any x, y ∈ A with |x − y| < δ, |f (x) − f (y)| < η. We claim that
this δ is the one we want. To that end, fix x, y ∈ A with |x − y| < δ. Let w = f (x) and
z = f (y). Then by our choice of δ,
|w − z| = |f (x) − f (y)| < η.
By our choice of η,
|g(f (x)) − g(f (y))| = |g(w) − g(z)| < ε.
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DUE 15 MARCH, 2016
4) Suppose that I, J are disjoint, closed intervals, and f is a function which is defined on
I ∪ J, uniformly continuous on I, and uniformly continuous on J. Show that f is uniformly
continuous on I ∪ J. Give an example to show that this conclusion may not hold if I, J are
not assumed to be closed.
Solution If either I = ∅ or J = ∅, the result is trivial. We may therefore assume that
neither interval is empty. We want to show that for any ε > 0, there exists δ > 0 such that
for any x, y ∈ I ∪ J with |x − y| < δ, |f (x) − f (y)| < ε.
The first step of the proof is to establish the following: There exists a positive number δ0
such that for any x ∈ I and y ∈ J, |x − y| > δ0 . We note that since the sets are disjoint
intervals, it must be that either x < y for every x ∈ I and y ∈ J or x < y for every x ∈ J and
y ∈ I. That is, one of the intervals lies entirely to the left of the other. By switching I and J
if necessary, we may assume that I lies entirely to the left of J. This means that any member
of J is an upper bound for I, and sup I is finite. Since I is closed and the supremum of a set
lies in the closure of that set whenever the supremum is finite, we deduce that sup I ∈ I = I.
Similarly, inf J ∈ J = J. Therefore sup I < inf J. Let δ0 = inf J − sup I > 0. Then for any
x ∈ I and y ∈ J, since x < y, |x − y| = y − x. Moreover, since x 6 sup I and inf J 6 y,
y − x > inf J − sup I = δ0 .
This proves the claim at the beginning of the paragraph.
Next, fix ε > 0. Since f is uniformly continuous on I, there exists δI > 0 such that for
any x, y ∈ I with |x − y| < δI , |f (x) − f (y)| < ε. Since f is uniformly continuous on J, there
exists δJ > 0 such that for any x, y ∈ J with |x − y| < δJ , |f (x) − f (y)| < ε. Let
δ = min{δ0 , δI , δJ }.
Fix x, y ∈ I ∪ J. Since |x − y| < δ 6 δ0 , either x, y ∈ I or x, y ∈ J. That is, it is impossible
to have one of x, y in I and the other in J. This is because if x ∈ I and y ∈ J, |x − y| > δ0 ,
while we know that |x − y| < δ0 . If x, y ∈ I, then since |x − y| < δ 6 δI , |f (x) − f (y)| < ε.
If x, y ∈ J, then since |x − y| < δJ , |f (x) − f (y)| < ε. In either case, |f (x) − f (y)| < ε.
For the counterexample, we recall the first fact used in the previous proof: That there is a
positive distance δ0 between the sets I and J. For the counterexample, we wish to use some
I, J which are not separated by a positive distance.
For our first counterexample, let I = [−1, 0] and J = (0, 1]. Let f (x) = 0 if x ∈ I and
f (x) = 1 if x ∈ J. Then f is uniformly continuous on I, since it is constant on I. Also,
f is uniformly continuous on J, since it is constant on J. But the function f is not even
continuous on I ∪ J = [−1, 1], since there is a jump discontinuity from 0 to 1 at x = 0.
For our second counterexample, we will provide a function f which is uniformly continuous
on I and J, and continuous on I ∪ J, but not uniformly continuous. Let I = (−1, 0) and
J = (0, 1). Let f (x) = 0 for all x ∈ I and f (x) = 1 for all x ∈ J. Then I ∪J = (−1, 0)∪(0, 1).
We claim that f is continuous on I ∪ J, but not uniformly continuous. We first show
continuity. Fix x ∈ (−1, 0) ∪ (0, 1). Fix ε > 0 and let δ = |x|. Then if y ∈ I ∪ J and
HOMEWORK ASSIGNMENT 6
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|y − x| < |x|, we consider two cases. If x < 0, |x| = −x, and y − x < −x implies y < 0.
Therefore y ∈ (−1, 0), and
|f (y) − f (x)| = |0 − 0| = 0 < ε.
In the second case, x > 0, |x| = x, and |y − x| = |x − y| < x implies x − y < x, which in
turn implies that y > 0. Therefore y ∈ (0, 1) and
|f (y) − f (x)| = |1 − 1| = 0 < ε.
In either case, if |y − x| < δ = |x|, |f (y) − f (x)| < ε. This shows that f is continuous.
We next show that f is not uniformly continuous on I ∪ J. To obtain a contradiction,
assume f is uniformly continuous on I ∪ J. Then there exists δ > 0 such that for any
x, y ∈ I ∪ J with |x − y| < δ, |f (x) − f (y)| < 1. Let x = δ/3 and y = −δ/3. Then
x, y ∈ I ∪ J and |x − y| = δ/3 − (−δ/3) = 2δ/3 < δ. However
|f (x) − f (y)| = |1 − 0| = 1,
which is not less than 1.
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DUE 15 MARCH, 2016
5) Suppose f : R → R is a function such that (1) limx→0 f (x) = 1 and (2) f (x + y) =
f (x)f (y) for any x, y ∈ R. Complete the following steps.
(i) Show that f is continuous on R. Hint: Use limy→x f (y) = limy→0 f (x + y) to check
continuity.
(ii) Show that f (x) > 0 for all x. Hint: We know by continuity that f (0) = 1 > 0. First
prove that f (x) > 0 for every x > 0 by contradiction. Assume there exists a positive
number x such that f (x) 6 0. Then let s = inf{x > 0 : f (x) 6 0}. Show that s > 0
and then consider the relationship between f (s) and f ( 2s ). Then use property (2) to
show that f (−x) = f (x)−1 for every x ∈ R, and then use this to deduce that f (x) > 0
for every negative x.
(iii) Show that for every real number r, for every real number x, f (rx) = [f (x)]r . Hint:
First prove this when r is a natural number using property (2). Then prove this when r
is an integer. Extend this to the case when r is a rational number. Last, use continuity
to prove the result for all real numbers.
(iv) Show that if f (1) = 1, then f is constant. If f (1) > 1, then f is increasing. If
0 < f (1) < 1, then f is decreasing.
Solution (i) Fix x ∈ R. Then since limy→x f (y) = limy→0 f (x + y),
lim f (y) = lim f (x + y) = lim f (x)f (y) = f (x) lim f (y) = f (x).
y→x
y→0
y→0
y→0
This shows that f is continuous at x. Since x was arbitrary, f is continuous.
(ii) First suppose that there exists x0 > 0 such that f (x0 ) 6 0. Let A = {x > 0 :
f (x) 6 0}. Since A 6= ∅, s = inf A is finite. We know s > 0. We first claim that
s > 0. Indeed, if it were not so, then 0 = inf A. Then for any n ∈ N, 1/n is not a lower
bound for A, and there exists xn ∈ [0, 1/n) such that f (xn ) 6 0. This yields a sequence
(xn ). Since limn xn = 0 and since f is continuous at 0, 1 = f (0) = limn f (xn ). But
since f (xn ) ∈ (−∞, 0] for each n, limn f (xn ) ∈ (−∞, 0], which means 1 = f (0) 6 0, a
contradiction. Therefore s > 0. This means that s/2 < s. But since s is a lower bound
for A, s/2 ∈
/ A, and f (s/2) > 0. Then
s s
f (s) = f ( + ) = f (s/2)2 > 0,
2 2
a contradiction. This gives that f (x) > 0 for all x > 0.
Next, note that for any x ∈ R,
1 = f (0) = f (x + (−x)) = f (x)f (−x).
This means that neither f (x) nor f (−x) can be zero, and we can divide both sides by
f (x) to obtain f (−x) = f (x)−1 ). If x > 0, then this fact together with the previous
paragraph yields that f (−x) = f (x)−1 > 0.
(iii) We first note that f (0 · x) = f (0) = 1 and f (x)0 = 1, so the statement holds if
r = 0. Next, we prove by induction on r that f (rx) = f (x)r for all r ∈ N. The base
HOMEWORK ASSIGNMENT 6
7
case r = 1 is true since f (1 · x) = f (x) = f (x)1 . Next, assume f (rx) = f (x)r for some
r ∈ N. Then
f ((r + 1)x) = f (rx + x) = f (rx)f (x) = f (x)r f (x) = f (x)r+1 .
This finishes the induction and shows that f (rx) = f (x)r for all r ∈ N.
Next, suppose r is a negative integer. Then −r ∈ N. Applying the previous paragraph together with the fact from (ii) that f (−x) = f (x)−1 for any x, we deduce
that
f (rx) = f (−(−r)x) = f ((−r)x)−1 = [f (x)−r ]−1 = f (x)r .
Next, fix q ∈ Z, q 6= 0. Then by the previous paragraph,
1 1 q
f (x) = f q x = f x ,
q
q
1
and taking q th roots on both sides gives f 1q x = f (x) q . Let r be a rational number.
Then we may write r = pq for some p, q ∈ Z, q 6= 0. Then using the previous paragraph
and the beginning of this paragraph,
1 p
1 p p
1
f (rx) = f x = f p x = f x = [f (x) q ]p = f (x) q = f (x)r .
q
q
q
Next, let r be any real number. Fix a sequence (qn ) of rational numbers such that
limn qn = r. Then limn qn x = rx. Since f is continuous, limn f (qn x) = f (rx). But since
qn is rational for each n, f (qn ) = f (x)qn . By continuity of the function g(y) = f (x)y ,
and since limn qn = r, we deduce that limn f (x)qn = f (x)r . Therefore
f (rx) = lim f (qn x) = lim f (x)qn = f (x)r .
n
n
(iv) Let b = f (1). We know b > 0. For any x ∈ R, by (iii), f (x) = f (x · 1) = f (1)x =
bx . Thus f is just the exponential function bx . If b = 1, this is constant. If b > 1, this
is increasing. If 0 < b < 1, this is increasing.