Math 317 HW #11 Solutions
1. Exercise 5.3.1. Recall from Exercise 4.4.9 that a function f : A → R is “Lipschitz on A” if
there exists an M > 0 such that
f (x) − f (y) ≤M
x−y
for all x, y ∈ A. Show that if f is differentiable on a closed interval [a, b] and if f 0 is continuous
on [a, b], then f is Lipschitz on [a, b].
Proof. Suppose f is differentiable and f 0 is continuous on [a, b]. Then, since [a, b] is compact,
f 0 is bounded. In other words, there exists M > 0 such that |f 0 (x)| ≤ M for all x ∈ [a, b].
Then, for any x, y ∈ [a, b], the Mean Value Theorem guarantees that there exists c between
x and y such that
f (x) − f (y) = |f 0 (c)| ≤ M.
x−y
Since the choice of x and y was arbitrary, we see that the inequality
f (x) − f (y) ≤M
x−y
holds for all x, y ∈ [a, b], so f is Lipschitz on [a, b].
2. Exercise 5.3.6. Let g : [0, 1] → R be twice-differentiable (i.e., both g and g 0 are differentiable
functions) with g 00 (x) > 0 for all x ∈ [0, 1]. If g(0) > 0 and g(1) = 1, show that g(d) = d for
some point d ∈ (0, 1) if and only if g 0 (1) > 1.
Proof. (⇒) First, suppose g(d) = d for some d ∈ (0, 1). Then, by the Mean Value Theorem,
there exists c1 ∈ (d, 1) such that
g 0 (c1 ) =
g(1) − g(d)
1−d
=
= 1.
1−d
1−d
Using the Mean Value Theorem again, there exists c2 ∈ (c1 , 1) such that
g 00 (c2 ) =
g 0 (1) − g 0 (c1 )
g 0 (1) − 1
=
.
1 − c1
1 − c1
We know that g 00 (c2 ) > 0; therefore, since the denominator on the right hand side is positive,
the numerator must be positive as well, meaning that g 0 (1) > 1.
(⇐) Now, I intend to prove the contrapositive of the reverse direction. Suppose g(x) 6= x for
all x ∈ (0, 1). Let h(x) = g(x) − x. Then g(x) 6= x is equivalent to saying h(x) 6= 0 for all
x ∈ (0, 1).
We know that h(0) = g(0)−0 = g(0) > 0. Since h(x) 6= 0 for all x ∈ (0, 1) and h is continuous,
we know, by the converse of the Intermediate Value Theorem, that h(x) 6≤ 0 for all x ∈ (0, 1).
Thus, we know that h(x) > 0 for all x ∈ (0, 1), so
h(1) − h(x)
0 − h(x)
=
<0
1−x
1−x
1
for all x ∈ (0, 1). Hence, if (xn ) is a sequence in (0, 1) converging to 1, then for each n there
exists cn ∈ (xn , 1) such that
h(1) − h(xn )
h0 (cn ) =
< 0,
1 − xn
so h0 (cn ) < 0 for all n. We know that h0 is differentiable and, hence, continuous, so Theorem
4.3.2 implies that limn→∞ h0 (cn ) = h0 (1). Since h0 (cn ) < 0 for all n, this means that h0 (1) ≤ 0.
But now, since
h0 (x) = g 0 (x) − 1,
this implies that g 0 (1) ≤ 1, as desired.
3. Exercise 5.3.8. Assume g : (a, b) → R is differentiable at some point c ∈ (a, b). If g 0 (c) 6= 0,
show that there exists a δ-neighborhood Vδ (c) ⊆ (a, b) for which g(x) 6= g(c) for all x ∈ Vδ (c)
with x 6= c. Compare this result with Exercise 5.3.7.
Proof. Suppose
0 6= g 0 (c) = lim
x→c
g(x) − g(c)
.
x−c
Let = |g 0 (c)| > 0. Then, by definition of the functional limit, there exists δ > 0 such that
0 < |x − c| < δ implies
g(x) − g(c)
0
− g (c) < = |g 0 (c)|.
x−c
This means both terms on the left have the same sign and hence, in particular, that
g(x) − g(c)
6= 0.
x−c
Therefore, for all x ∈ Vδ (c) with x 6= c, we have that g(x) 6= g(c).
Exercise 5.3.7 gives a function g such that g(0) = 0 and g 0 (0) > 0, but g is not increasing
on any interval containing zero. The simplest way for this to happen would seem to be for
g to take on negative values on any interval containing zero. However, by the Intermediate
Value Theorem, this would imply that g(x) = 0 for some x 6= 0 in any interval containing
zero, which would contradict the statement we just proved.
It turns out that for all x > 0, g(x) > 0, and for all x < 0, g(x) < 0. However, in any interval
containing zero there will be some y > x > 0 (in fact, infinitely many such x and y) such that
g(x) > g(y). The below figure gives some sense of this:
2
0.2
0.16
0.12
0.08
0.04
-0.28
-0.24
-0.2
-0.16
-0.12
-0.08
-0.04
0
0.04
0.08
0.12
0.16
0.2
0.24
0.28
-0.04
-0.08
-0.12
-0.16
-0.2
4. Exercise 5.3.13. Prove the following version of L’Hôpital’s Rule:
Theorem 1. Assume f and g are differentiable on an interval containing a, except possibly
at a itself. If limx→a f (x) = 0 and limx→a g(x) = 0, then
f 0 (x)
=L
x→a g 0 (x)
lim
implies
lim
x→a
f (x)
= L.
g(x)
Proof. Since f and g are differentiable except possibly at a, f and g must be continuous
except possibly at a. Since limx→a f (x) = 0 and limx→0 g(x) = 0, we can define continuous
functions fe and ge by
(
f (x) if x 6= a
fe(x) :=
0
if x = a
and
(
g(x)
ge(x) :=
0
if x 6= a
if x = a.
Since fe(a) = 0 and ge(a) = 0, Theorem 5.3.6 applies to fe and ge, so we know that
fe0 (x)
= L implies
x→a g
e0 (x)
lim
Now, since
fe0 (x)
g
e0 (x)
=
f 0 (x)
g 0 (x)
and
fe(x)
g
e(x)
=
f (x)
g(x)
fe(x)
= L.
x→a g
e(x)
lim
for all x 6= a, Lemma 2 below implies that
f 0 (x)
fe0 (x)
= lim 0
0
x→a g (x)
x→a g
e (x)
lim
and
f (x)
fe(x)
= lim
.
x→a g(x)
x→a g
e(x)
lim
Combining this with (1), then, we can conclude that
f 0 (x)
= L implies
x→a g 0 (x)
lim
as desired.
3
lim
x→a
f (x)
= L,
g(x)
(1)
Lemma 2. Suppose two functions h1 and h2 are defined on a set A, that a ∈ A is a limit
point of A, and that h1 (x) = h2 (x) for all x =
6 a. Then
lim h1 (x) = lim h2 (x)
x→a
x→a
unless neither limit exists.
Proof. Suppose one of the limits exists; without loss of generality, assume limx→a h1 (x) = L.
Let > 0. Then, by definition of the functional limit, there exists δ > 0 such that 0 <
|x − a| < δ implies that |h1 (x) − L| < .
Then, if 0 < |x − a| < δ, we have that
|h2 (x) − L| = |h1 (x) − L| < ,
where the equality follows from the fact that h1 (x) = h2 (x) for all x 6= a. Since the choice of
> 0 was arbitrary, we see that for all > 0 there exists such a δ, so limx→a h2 (x) = L as
well.
5. Exercise 5.4.6.
(a) Without working too hard, explain why the partial sum gm = h0 + h1 + · · · + hm is
differentiable at x. Now, prove that, for every value of m, we have
0
0
|gm+1
(x) − gm
(x)| = 1.
Proof. Note that none of h0 , h1 , . . . , hm has a corner between xm and ym , so each of these
functions is linear on (xm , ym ). Since the sum of linear functions is linear, we know that
gm is linear and, hence, differentiable on (xm , ym ). In fact, since the derivative of a linear
0 is constant on (x , y ). We will use this fact in
function is constant, we know that gm
m m
part (b), but notice that this is more than enough to conclude that gm is differentiable
at x for all m.
Now,
gm+1 (x) − gm (x) = (h0 (x) + h1 (x) + · · · + hm (x) + hm+1 (x)) − (h0 (x) + h1 (x) + · · · + hm (x))
= hm+1 (x).
Since each hk is a line of slope ±1 away from its corners, we know that h0m+1 (x) = ±1.
Therefore,
0
0
|gm+1
(x) − gm
(x)| = |h0m+1 (x)| = 1.
(b) Prove the two inequalities
g(ym ) − g(x)
g(xm ) − g(x)
0
< gm
(x) <
.
ym − x
xm − x
4
Proof. Note that, for all n ≥ m,
1
p+1
1
= n h 2n−m (p + 1) = 0
hn (ym ) = n h 2n · m
2
2
2
since h is zero on every even integer. Hence,
g(ym ) =
∞
X
hn (ym ) =
n=0
m
X
hn (ym ) = gm (ym ).
n=0
Now, this implies that
g(ym ) − g(x) = gm (ym ) − g(x) = gm (ym ) − gm (x) + gm (x) − g(x).
Notice that
gm (x) − g(x) = −
∞
X
(2)
hn (x) < 0
n=m+1
since hn (x) > 0 for all n. This, along with (2), implies that
g(ym ) − g(x) < gm (ym ) − gm (x),
and, hence, that
g(ym ) − g(x)
gm (ym ) − gm (x)
<
.
ym − x
ym − x
(3)
On the other hand, by the Mean Value Theorem, there exists c ∈ (x, ym ) such that
0
gm
(c) =
gm (ym ) − gm (x)
.
ym − x
Since, as we argued in (a), gm is linear on (xm , ym ), g 0 (c) = g 0 (x) so combining the above
with (3) yields
g(ym ) − g(x)
0
< gm
(x).
ym − x
A completely analoguous argument gives the other inequality.
(c) Use parts (a) and (b) to show that g 0 (x) does not exist.
Proof. If g 0 (x) = limz→x
lim
m→∞
g(z)−g(x)
z−x
exists, then Theorem 4.2.3 implies that
g(xm ) − g(x)
g(ym ) − g(x)
= g 0 (x) = lim
.
m→∞
xm − x
ym − x
In other words,
lim
n→∞
g(xm ) − g(x) g(ym ) − g(x)
−
= 0.
xm − x
ym − x
However, (b) implies that
0
0 < gm
(x) −
g(ym ) − g(x)
g(xm ) − g(x) g(ym ) − g(x)
<
−
.
ym − x
xm − x
ym − x
5
Then, by the Squeeze Theorem,
g(ym ) − g(x)
0
,
0 = lim gm (x) −
m→∞
ym − x
so, adding limm→∞
implies
g(ym )−g(x)
ym −x
to both sides and using the Algebraic Limit Theorem
g(ym ) − g(x)
g(ym ) − g(x)
g(ym ) − g(x)
0
lim
+ lim
= lim gm (x) −
m→∞
m→∞
m→∞
ym − x
ym − x
ym − x
g(ym ) − g(x) g(ym ) − g(x)
0
= lim gm (x) −
+
m→∞
ym − x
ym − x
0
= lim gm (x).
m→∞
0 (x)) is not Cauchy, so it cannot converge.
But part (a) implies that the sequence (gm
From this contradiction, then, we can conclude that g 0 (x) does not exist.
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