Math 317 HW #10 Solutions
1. Exercise 4.5.7. Let f be a continuous function on the closed interval [0, 1] with range also
contained in [0, 1]. Prove that f must have a fixed point; that is, show f (x) = x for at least
one value of x ∈ [0, 1].
Proof. Define g(x) = f (x) − x. Notice that, since f is continuous, the Algebraic Continuity
Theorem implies that g is continuous.
Now
g(0) = f (0) − 0 = f (0) ≥ 0.
If g(0) = 0, then f (0) = 0 and 0 is the desired fixed point. Otherwise, g(0) > 0.
Also,
g(1) = f (1) − 1 ≤ 0.
If g(1) = 0, then f (1) = 1 and 1 is the desired fixed point. Otherwise, g(1) < 0.
If neither g(0) nor g(1) is equal to zero then, by the Intermediate Value Theorem, there exists
c ∈ (0, 1) such that g(c) = 0, meaning that f (c) = c, so c is the desired fixed point.
Whatever the case, we see that f has at least one fixed point.
2. Exercise 5.2.5. Let
(
xa sin(1/x)
ga (x) =
0
if x 6= 0
if x = 0.
Find a particular (potentially noninteger) value for a so that
(a) ga is differentiable on R but such that ga0 is unbounded on [0, 1].
Answer. Let a = 4/3 (in fact, a similar proof will work for any a ∈ (1, 2)). Notice that
away from zero we can just use the product and chain rules to differentiate ga :
0
g4/3
(x) = x4/3 cos(1/x) ·
for x 6= 0. If xn =
1
2nπ ,
−1 4 1/3
− cos(1/x) 4 √
+ x sin(1/x) =
+ 3 x sin(1/x)
2
x
3
3
x2/3
then
0
g4/3
(xn ) =
√
2
−1
3
=
−
2nπ
.
(1/2nπ)2/3
0 (x )) is unbounded, so g 0
Hence, the sequence (g4/3
n
4/3 is unbounded on (0, 1).
It remains only to show that g4/3 is differentiable at zero. To do so, notice that
g4/3 (x) − g4/3 (0)
g4/3 (x)
√
x4/3 sin(1/x)
= lim
= lim
= lim 3 x sin(1/x).
x→0
x→0
x→0
x→0
x−0
x
x
√
√
But now, since limx→0 − 3 x = 0 and limx→0 3 x = 0, the Squeeze Theorem implies that
0 (0) = 0.
the above limit is equal to 0, meaning that g4/3
lim
1
(b) ga is differentiable on R with ga0 continuous but not differentiable at zero.
Answer. Let a = 7/3 (again, a similar proof will work for any a ∈ (2, 3]). Then away
from zero we have that
√
−1 7
7
0
(x) = x7/3 cos(1/x) · 2 + x4/3 sin(1/x) = − 3 x cos(1/x) + x4/3 sin(1/x),
g7/3
x
3
3
0 (0) = 0. Notice
and a similar proof to the one given above in part (a) implies that g7/3
0
that the above expression goes to 0 as x → 0, so g7/3
is continuous at zero.
However, if xn =
1
2nπ ,
then
r
0
(xn )
g7/3
Hence,
0 (x ) − g 0 (0)
g7/3
n
7/3
xn − 0
=−
=
3
1
−1
.
·1+0= √
3
2nπ
2nπ
0 (x )
g7/3
n
xn
which is unbounded, meaning that limx→0
differentiable at zero.
√
√
−1/ 3 2nπ
2
3
=
= − 2nπ ,
1/2nπ
0
0
g7/3
(x)−g7/3
(0)
x−0
0
does not exist, so g7/3
is not
(c) ga is differentiable on R and ga0 is differentiable on R, but such that ga00 is not continuous
at zero.
Answer. Let a = 10/3 (or any number in (3, 4]). Then a similar proof to those already
given in (a) and (b) will show that g10/3 is twice-differentiable, but that the second
derivative is not continuous at zero.
3. Exercise 5.2.8. Decide whether each conjecture is true or false. Provide an argument for those
that are true and a counterexample for each one that is false.
(a) If a derivative function is not constant, then the derivative must take on some irrational
values.
Answer. False. Consider the function f (x) = |x|. Then
(
−1 if x < 0
0
f (x) =
1
if x > 0
(notice that f 0 is not defined at 0). The function f 0 is not constant (since, e.g., f 0 (−1) 6=
f 0 (1)), but it only takes on two values, both rational.
(b) If f 0 exists on an open interval, and there is some point c where f 0 (c) > 0, then there
exists a δ-neighborhood Vδ (c) around c in which f 0 (x) > 0 for all x ∈ Vδ (c).
Answer. False. Let
(
x2 sin(1/x) if x 6= 0
g(x) =
0
if x = 0
on the interval (−1, 1) and let f (x) = g(x) + x. I’ll focus on the point c = 0. As seen in
Section 5.1,
(
− cos(1/x) + 2x sin(1/x) if x 6= 0
g 0 (x) =
.
0
if x = 0
2
By Theorem 5.2.4(i), then,
(
− cos(1/x) + 2x sin(1/x) + 1
f 0 (x) = g 0 (x) + 1 =
1
Let δ > 0. Then, if n >
1
2πδ ,
we have that
f 0 (1/2nπ) = − cos(2nπ) +
1
2nπ
if x 6= 0
if x = 0.
∈ Vδ (0), but
2
sin(2nπ) + 1 = −1 + 0 + 1 = 0.
2nπ
Hence, since our choice of δ > 0 was arbitrary, there is no δ such that x ∈ Vδ (0) implies
that f 0 (x) > 0.
(c) If f is differentiable on an interval containing zero and if limx→0 f 0 (x) = L, then it must
be that L = f 0 (0).
Answer. True. The basic idea is that f 0 (0) 6= L violates Darboux’s Theorem.
More precisely, suppose f 0 (0) > L (a similar proof works when f 0 (0) < L) and let
= f 0 (0) − L. Since limx→0 f 0 (x) = L, there exists δ > 0 such that 0 < |x − 0| < δ
implies |f 0 (x) − L| < /2. In particular, this implies that there is no x ∈ (0, δ/2) such
that
3
f 0 (x) = L + .
4
But this is a violation of Darboux’s Theorem, since f (δ/2) ∈ (L − /2, L + /2), meaning
0
0
that L + 3
4 is strictly between f (0) and f (δ/2).
From this contradiction, then, we can see that f 0 (0) must equal L.
(d) Repeat conjecture (c) but drop the assumption that f 0 (0) necessarily exists. If f 0 (x)
exists for all x 6= 0 and if limx→0 f 0 (x) = L, then f 0 (0) exists and equals L.
Answer. False. Let
x2 + x
f (x) =
.
x
Since x2 + x = x(x + 1),
f (x) = x + 1
whenever x 6= 0.
Hence, for all x 6= 0,
f 0 (x) = (x + 1)0 = 1,
so limx→0 f 0 (x) = 1. However, f is not even well-defined at 0, so f 0 (0) certainly does
not exist.
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