Chapter 5
Differentiation
In this chapter we add more structure to the functions we consider. Not surprisingly,
we also strengthen our results.
Definition 5.1. Let f : D → R with x0 a cluster point of D and x0 ∈ D. We say
f is differentiable at x0 if and only if
lim
x→x0
f (x) − f (x0 )
x − x0
exists, in which case we call the limit f ′ (x0 ).
Theorem 5.2. Suppose f : I → R with I an interval and x0 ∈ I. Then f is
differentiable at x0 if and only if, for all sequences {xn }n∈N ∈ I with xn 6= x0 for
all n ∈ N and converging to x0 , we have the limit
(5.1)
exists.
f (xn ) − f (x0 )
n→∞
xn − x0
lim
Proof. This is the sequential characterization of Limits, Theorem 4.21.
�
Example 5.3. Show the function f (x) = |x| is not differentiable at the origin.
Proof. Consider the sequences
1
{xn }n∈N =
n n∈N
{yn }n∈N =
−1
n
.
n∈N
Note that
f (xn ) − f (0)
f (yn ) − f (0)
= 1,
= −1,
xn − 0
yn − 0
for all n ∈ N. This implies the limit in (5.1) does not exist.
�
As expected differentiation is stronger than continuity.
51
52
5. Differentiation
Theorem 5.4. Let f : I → R be differentiable at x0 ∈ I. Then f is continuous at
x0 .
Proof. Set
f (x) − f (x0 )
.
x − x0
By assumption T has a limit at x0 . Note the function
f (x) − f (x0 )
(x − x0 ) + f (x0 )
f (x) =
x − x0
= T (x)(x − x0 ) + f (x0 )
T (x) =
also, by the algebra of limits, has a limit at x0 . Indeed, lim f (x) = f (x0 ). By
x→x0
Theorem 4.11 f is continuous at x0 .
�
Theorem 5.5. (Algebra of Derivatives) Suppose f, g : I → R are differentiable at
x0 ∈ I. Then
(i) (f + g)′ (x0 ) = f ′ (x0 ) + g ′ (x0 ),
(ii) (f g)′ (x0 ) = f (x0 )g ′ (x0 + f ′ (x0 )g(x0 ),
′
f ′ (x0 )g(x0 ) − f (x0 )g ′ (x0 )
f
(x0 ) =
(iii) If g(x0 ) 6= 0,
g
g 2 (x0 )
Proof. We use Theorem 5.1 and the algebra of limits. Let {xn }n∈N ⊂ I/{x0 } be
any sequence converging to x0 . By assumption the sequences
f (xn ) − f (x0 )
g(xn ) − g(x0 )
,
xn − x0
xn − x0
n∈N
n∈N
converge to f ′ (x0 ) and g ′ (x0 ) respectively. Then
(f (x0 ) + g(x0 ))′
f (xn ) + g(xn ) − (f (x0 ) + g(x0 ))
xn − x0
f (xn ) − f (x0 )
g(xn ) − g(x0 )
= lim
+ lim
n→∞
n→∞
xn − x0
xn − x0
= f ′ (x0 ) + g ′ (x0 ).
=
lim
n→∞
To show (ii) note
g(xn ) − g(x0 )
f (xn ) − f (x0 )
f (xn )g(xn ) − f (x0 )g(x0 )
= f (xn )
+ g(x0 )
.
xn − x0
xn − x0
xn − x0
Moreover, since f is differentiable at x0 it is continuous there. Thus lim f (xn ) =
(5.2)
n→∞
f (x0 ). Taking the limit of (5.2) and applying the algebra of limits, we find (ii).
Part (iii) is similar. Since g is continuous at x0 and g(x0 ) 6= 0, the sign
preserving property, Lemma 4.28, implies there is an interval, I ′ ⊆ I, containing x0
such that g 6= 0 on I ′ . We apply Theorem 4.21 to I ′ . Then
f (xn )
g(xn )
−
f (x0 )
g(x0 )
g(x0 ) f (xn ) − f (x0 )
f (x0 ) g(xn ) − g(x0 )
=
−
.
xn − x0
g(x0 )g(xn )
xn − x0
g(x0 )g(xn )
xn − x0
As before we may take the limit of both sides and obtain the result.
�
53
5. Differentiation
We must not forgot the chain rule. The proof is based on the one given in [1]
and more difficult than one might expect.
Theorem 5.6. (Chain Rule) Let I and J be intervals with f : I → J and g : J → R.
Let c ∈ I with f differentiable at c and g differentiable at f (c). Then the composite
function g ◦ f is differentiable at c and
(g ◦ f )′ (c) = g ′ (f (c))f ′ (c).
Proof. Let y0 = f (c) and define
(
h(y) =
g(y)−g(y0 )
y−y0
− g ′ (y0 ) y 6= y0
.
0
y = y0
Since g is differentiable at y0 , we see lim h(y) = 0 = h(y0 ) and so h is continuous
y→y0
at y0 . Moreover, since f is continuous at c, h ◦ f is continuous at c (Theorem 4.17).
Thus
lim (h ◦ f )(x) = (h ◦ f )(c) = h(y0 ) = 0.
x→c
We calculate
g(f (x)) − g(y0 )
− g ′ (y0 )
f (x) − y0
if f (x) 6= y0 and zero otherwise. Thus, for x ∈ I and x 6= c
f (x) − f (c) g(f (x)) − g(f (c)) ′
.
= (h ◦ f )(x) + g (y0 )
x−c
x−c
The result now follows from the algebra of limits. Indeed
g(f (x)) − g(f (c))
(g ◦ f )′ (c) = lim
x→c
x−c
f (x) − f (c)
= lim (h ◦ f )(x) + g ′ (y0 ) · lim
x→c
x→c
x−c
′
′
= (0 + g (y0 ))f (c)
(h ◦ f )(x) =
= g ′ (f (c))f ′ (c).
�
You might try to find a function which is differentiable but whose derivative is
not continuous. Here is one.
Example 5.7. Set
x2 sin x1 x 6= 0
.
0
x=0
Away from x = 0 the chain rule applies and f ′ (x) = 2x sin(1/x) − cos(1/x). At
x = 0 we must apply the definition. We find
f (x) − 0
= 0.
f ′ (0) = lim
x→0 x − 0
However, f ′ is not continuous at x = 0 since lim f ′ (x) does not exist.
f (x) =
x→0
54
5. Differentiation
5.1. Mean-Value Theorems
In this section we establish some familiar results from elementary calculus.
Definition 5.8. Let f : I → R with I an open interval. A point, x0 ∈ I, is a
relative maximum (relative minimum) of f if there exists a δ > 0 such that for all
x ∈ I and |x − x0 | < δ, f (x0 ) ≥ f (x) (f (x0 ) ≤ f (x)).
Theorem 5.9. Let f : [a, b] → R and suppose f has either a relative maximum or
a relative minimum at x0 ∈ (a, b). If f is differentiable at x0 , then f ′ (x0 ) = 0.
Proof. Suppose x0 is relative maximum. Then a δ > 0 exists so that, for any x
satisfying x0 − δ < x < x0 + δ, x ∈ [a, b] and f (x) ≤ f (x0 ). Since the derivative
exists at x0 , we may compute it by taking any sequence {xn }n∈N ⊂ [a, b]/{x0 }
converging to x0 . We choose a sequence such that x0 − δ < xn < x0 for all n ∈ N.
For this choice note that, since x0 is a relative maximum f (xn ) ≤ f (x0 ),
f (xn ) − f (x0 )
≥ 0.
xn − x0
This implies the limit is non negative and f ′ (x0 ) ≥ 0. Similarly we could choose
{yn }n∈N ⊂ [a, b]/{x0 } converging to x0 with x0 < yn < x0 + δ for all n ∈ N. Then
f (yn ) − f (x0 )
≤0
yn − x0
for n ∈ N. This implies the limit is non positive and f ′ (x0 ) ≤ 0. Thus f ′ (x0 ) = 0
as required. A similar argument holds for a relative minimum.
�
Theorem 5.10. (Rolle’s Theorem) Let f : [a, b] → R continuous on [a, b] and
differentiable on (a, b). If f (a) = f (b) = 0, there exists c ∈ (a, b) such that f ′ (c) = 0.
Proof. If f (x) = 0 for all x ∈ [a, b], f ′ (x) = 0 and any point c ∈ (a, b) will
suffice. Suppose f (x) 6= 0 for some x ∈ [a, b]. By the Extreme-Value Theorem f
assumes a minimum and maximum, xM , xm in [a, b]. Since f (a) = f (b) = 0, one of
�
xM , xm ∈ (a, b), say xM . By the previous theorem f ′ (xM ) = 0.
The next theorem is a generalization of Rolle’s Theorem.
Theorem 5.11. (Mean-Value Theorem). Let f : [a, b] → R be continuous on [a, b]
and differentiable on (a, b). Then there exists a c ∈ (a, b) such that
f ′ (c) =
f (b) − f (a)
.
b−a
Proof. Set
f (b) − f (a)
(x − a) − f (a).
b−a
Then g(a) = g(b) = 0, and by the algebra of continuous/differentiable functions g
is continuous on [a, b] and differentiable on (a, b). Rolle’s Theorem applies, and a
c ∈ (a, b) exists so that g ′ (c) = 0. But g ′ (c) = f ′ (c) − (f (b) − f (a))/(b − a).
�
g(x) = f (x) −
The next theorem generalizes it even more.
5.2. Applications of the Mean-Value Theorem
55
Theorem 5.12. (Cauchy Mean-Value Theorem). Let f, g : [a, b] → R continuous
on [a, b] and differentiable on (a, b). Then there exists a c ∈ (a, b) such that
(f (b) − f (a))g ′ (c) = (g(b) − g(a))f ′ (c).
Remark 5.13. If g(x) = x we recover the Mean-Value Theorem.
Proof. Set
h(x) = (f (b) − f (a))g(x) − (g(b) − g(a))f (x),
for x ∈ [a, b]. Then h(x) is continuous on [a, b] and differentiable on (a, b). Moreover,
h(a) = h(b). The Mean-Value Theorem applies. Differentiating h gives the result.
�
5.2. Applications of the Mean-Value Theorem
The usefulness of the Mean-Value Theorem cannot be overstated. Interestingly it
does not extend to higher spatial dimensions.
Theorem 5.14. Let f : [a, b] → R continuous on [a, b] and differentiable on (a, b).
(i) If f ′ (x) 6= 0 for all x ∈ (a, b), f is injective on [a, b].
(ii) If f ′ (x) = 0 for all x ∈ (a, b), f is constant on [a, b].
(iii) If f ′ (x) > 0 for all x ∈ (a, b), x, y ∈ [a, b], x < y implies f (x) < f (y).
(iv) If f ′ (x) < 0 for all x ∈ (a, b), x, y ∈ [a, b], x < y implies f (x) > f (y).
(v) If |f ′ (x)| ≤ M for all x ∈ (a, b), and some M , then f is uniformly continuous
on [a, b].
Proof. These follow immediately from the Mean-Value Theorem, and in particular
the expression
f (b) − f (a)
= f ′ (c).
b−a
We only check (v). We have for any x, y ∈ [a, b]
|f (x) − f (y)| ≤ |f ′ (cxy )||x − y| ≤ M |x − y|
for some cxy ∈ (a, b) and between x and y. A function satisfying the inequality
|f (x) − f (y)| ≤ M |x − y| is called Lipschitz. It implies f is uniformly continuous
on [a, b] if we choose δ = ǫ/M .
�
Theorem 5.15. Let f, g : [a, b] → R continuous on [a, b], differentiable and f ′ (x) =
g ′ (x) on (a, b). Then f (x) = g(x) + k for all x ∈ [a, b] and constant k.
Proof. Set h(x) = f (x) − g(x). Then h satisfies the conditions of Theorem 5.14,
(ii).
�
The next theorem is unexpected since the derivative of a function need not be
continuous.
Theorem 5.16. (Darboux’s Theorem) Suppose f : [a, b] → R is differentiable on
[a, b]. Then f ′ enjoys the intermediate-value property.
56
5. Differentiation
Proof. Suppose that, for some differentiable function h : [a, b] → R, h′ (x) 6= 0 on
(a, b). Then by Theorem 5.14, (i) h is injective. Moreover, by Exercise 4.24, h is
monotone. If h is increasing
h(x) − h(y)
≥ 0.
x−y
This implies h′ ≥ 0 on [a, b]. If h is decreasing, h′ ≤ 0 on [a, b]. Thus h′ (x) 6= 0 on
(a, b) implies either h′ (x) ≤ 0 or h′ (x) ≥ 0 on [a.b] - h′ cannot change signs.
To show f ′ has the intermediate-value property, let x, y ∈ [a, b], x < y and λ be
any point between f ′ (x) and f ′ (y). Set h(z) = f (z) − λz. Then h is differentiable
on [a, b] and h′ (z) = f ′ (z) − λ. One checks that under our assumptions about λ,
h′ (x) and h′ (y) have opposite signs. Thus h′ must be zero for some c between x
and y. That is, f ′ (c) = λ for some x < c < y.
�
Example 5.17. There is no function on the interval [−1, 1] whose derivative is
−1 x ≤ 0,
h(x) =
1 x > 0.
The next theorem is call L’Hospital’s Rule. It appeared in a calculus book
written by L’Hospital: Analysis of the Infinitely Small to Understand Curved Lines,
(1696). The result is thought to have been discovered by Johann Bernoulli.
Theorem 5.18. (L’Hospital’s Rule) Suppose f, g : [a, b] → R are continuous on
[a, b] and differentiable on (a, b). Let x0 ∈ [a, b], and
(i) g ′ (x) 6= 0 for all x ∈ (a, b),
(ii) f (x0 ) = g(x0 ) = 0,
(iii) f ′ /g ′ has a limit at x0 .
Then
lim
x→x0
f (x)
f ′ (x)
= lim ′
.
g(x) x→x0 g (x)
Proof. Let {xn }n∈N ⊂ (a, b)/{x0 } be any sequence converging to x0 . By the
Cauchy Mean-Value Theorem, Theorem 5.12, there exists a sequence {cn }n∈N with
cn between x and x0 for all n ∈ N such that
(f (xn ) − f (x0 ))g ′ (cn ) = (g(xn ) − g(x0 ))f ′ (cn ).
Since g ′ 6= 0 on (a, b), g is injective. Thus g(xn ) 6= g(x0 ) for all n ∈ N. Moreover,
since f (x0 ) = g(x0 ) = 0, we have
f (xn ) − f (x0 )
f ′ (cn )
f (xn )
=
= ′
.
g(xn )
g(xn ) − f (x0 )
g (cn )
By the Squeeze Theorem, Theorem 3.24, {cn }n∈N converges to x0 . The result
follows.
�
Note L’Hospital’s rule says the existence of limit limx→x0
limx→x0
f (x)
g(x)
exists. The converse is not true.
f ′ (x)
g′ (x)
implies the limit
57
5.2. Applications of the Mean-Value Theorem
Example 5.19. Set
x sin x1 x ≤ 0,
0
x = 0.
√
and g(x) = x. Then f (0) = g(0) = 0. Note
√
1
f (x)
= lim x sin = 0
lim
x→0
x→0 g(x)
x
while
√
2
1
1
f ′ (x)
,
= lim 2 x sin − √ sin
lim ′
x→0
x→0 g (x)
x
x
x
does not have a limit.
f (x) =
Example 5.20. Prove
lim x ln x = 0.
x→0+
ln x
= 0. The limit tends to −∞/∞
1/x
which is not in the form discussed in Theorem 5.18. Nevertheless, the proof still
works. Set f (x) = ln x and g(x) = 1/x. Notice −x = f ′ /g ′ → 0 as x → 0+ . Thus,
for any ǫ > 0, there exists a 0 < x1 < 1 such that
′
f (x) g ′ (x) < ǫ
Proof. We can rewrite the limit as lim
x→0+
for all 0 < x < x1 . Applying the Cauchy Mean-Value Theorem, we find
f (x) − f (x1 ) f ′ (x2 ) =
g(x) − g(x1 ) g ′ (x2 ) < ǫ
for some x2 between x and x1 . However,

f (x)  1 −
f (x) − f (x1 )
=
g(x) − g(x1 )
g(x) 1 −
f (x1 )
f (x)
g(x1 )
g(x)

 = f (x) h(x).
g(x)
We see h(x) → 1 as x → 0+ . Thus, h(x) > 1/2 for x small enough, and
′
f (x) 1 f (x) ≤
h(x) ≤ f (x2 ) < ǫ
g ′ (x2 ) g(x) 2 g(x) for x small. Thus given any ǫ > 0 there exists a δ > 0 such that
f (x) < 2ǫ,
|x ln x − 0| = g(x) and the limit is established.
�
(n)
Theorem 5.21. (Taylor’s Theorem) Let f : [a, b] → R. Suppose f (x) is continuous on [a, b] and f (n+1) (x) exists on (a, b) for some n ∈ N. For any x, x0 ∈ [a, b],
c between x and x0 exists such that
f ′′ (x0 )
(x − x0 )2 + · · ·
f (x) = f (x0 ) + f ′ (x0 )(x − x0 ) +
2!
f (n) (x0 )
f (n+1) (c)
+
(x − x0 )n +
(x − x0 )n+1
n!
(n + 1)!
= Pn (x) + Rn (x).
58
5. Differentiation
The last term in the sum, Rn (x) is called the Lagrangian form of the remainder.
Proof. Note that, if n = 0, Taylor’s theorem is the mean-value theorem. For any
n ∈ N we define the number E to satisfy
f (x)
f ′′ (x0 )
(x − x0 )2 + · · ·
2!
f (n) (x0 )
E
+
(x − x0 )n +
(x − x0 )n+1
n!
(n + 1)!
f (x0 ) + f ′ (x0 )(x − x0 ) +
=
For any x, t ∈ [a, b] set
ϕ(t) = f (x) − f (t) − f ′ (t)(x − t) − · · · −
f (n) (t)
E
(x − t)n −
(x − t)n+1 .
n!
(n + 1)!
By the algebra of continuous/differentiable functions ϕ(t) is continuous on [a, b]
and differentiable on (a, b). Moreover, ϕ(x) = 0 = ϕ(x0 ). By Rolle’s theorem there
is a c between x and x0 such that ϕ′ (c) = 0. We calculate
ϕ′ (t)
= −f ′ (t) + f ′ (t) − f ′′ (t)(x − t) + f ′′ (t)(x − t)
f ′′′ (t)
f ′′′ (t)
(x − t)2 +
(x − t)2
−
2!
2!
E
f (n+1) (t)
(x − t)n + (x − t)n .
+···−
n!
n!
Since ϕ′ (c) = 0, we find
E = f (n+1) (c).
�
Example 5.22. Show
ex = 1 + x +
x3
xn
x2
+
+ ···+
+ ···
2!
3!
n!
Proof. For the moment we assume f (x) = ex has infinitely many derivatives on
R and f ′ (x) = ex . The number e is defined in Equation (5.3). Taylor’s Theorem
applies and
x3
xn
ec
x2
+
+ ···+
+
xn+1
ex = 1 + x +
2!
3!
n!
(n + 1)!
for some c between 0 and x. That is, the remainder is
Rn (x) =
ec
xn+1 .
(n + 1)!
For a fixed x consider the sequence {an }n∈N = {|Rn |}n∈N . Problem 3.14 shows
an+1
= L with
that any sequence {an }n∈N satisfying an > 0 for all n ∈ N and lim
n→∞ an
L < 1 converges to zero. Here
an+1
|x|
=
.
an
n+2
Thus Rn goes to zero as n → ∞ for all x, c.
�
59
5.2. Applications of the Mean-Value Theorem
Example 5.23. In Example 1.3 we examined the Taylor expansion of
−1
e x2 x 6= 0,
f (x) =
0
x = 0.
In Exercise 5.19, we find f (n) (0) = 0 for all n ∈ N. Thus Pn (x) = 0 and all of the
information about the function is in the remainder Rn (x), and the Taylor expansion
of f does not converge to f .
A nice application of Taylor’s Theorem was produced by Charles Hermite
(1873). We define the number e to be the unique number such that
Z e
1
(5.3)
1=
dx.
x
1
The integral will be defined in the next chapter.
Theorem 5.24. The number e is irrational.
Proof. We know from Example 5.22 that e may be written
1
1
1
ec
e = 1 + 1 + + + ···+
+
.
2! 3!
n! (n + 1)!
Suppose e = p/q (reduced form) with p, q ∈ N. Fix n ≥ max{q, ec }. Then
ec
1
1
1
n! +
.
n!e = 1 + 1 + + + · · · + +
2! 3!
n!
n+1
Now n!e = n!p/q and n ≥ q, thus n!e ∈ N, and therefore, so is
n ≥ ec , and so n + 1 > ec or ec /(n + 1) < 1, a contradiction.
ec
n+1 .
However
�
60
5. Differentiation
Theorems so far. See if you remember the ideas of the proofs and the way the one
theorem was used to prove another.
The Completeness Axiom
❍
❍❍
❍❍
❄
❥
Monotone-Convergence Theorem ❍
Archimedean Principle
❄
Nested-Cell Theorem
❇
❇❇◆
Density of Q
❄
Bolzano-Weierstrass Theorem
❍❍
✟✟
❥
❍
✙
✟
❄
Cauchy Theorem
Uniformly Continuous Extreme-Value Thm
❄
Rolle’s Theorem
❄
Mean-Value Theorem
❄
Cauchy MVT
❄
L’Hôsiptal’s Rule
5.2. Applications of the Mean-Value Theorem
61
A Collection of Strange Functions
(a) The following function is differentiable for all x ∈ R; however, its derivative is
not continuous.
2
x sin(1/x) x 6= 0,
g(x) =
0
x = 0.
(b) The function, h(x), below is differentiable and continuous at exactly one point.
2
x x ∈ Q,
h(x) =
0
x 6∈ Q.
(c) The following function is differentiable everywhere. In addition, f (n) (0) = 0 for
all n ∈ N, and the function has no Taylor series expansion.
exp(−1/x2 ) x 6= 0,
f (x) =
0
x = 0.
(d) The next function is continuous on the irrationals but discontinuous on the
rationals (it is also Riemann integrable on [0, 1]).
1/q x = p/q ∈ Q, (reduced form)
g(x) =
0
x 6∈ Q.
(e) The next function is not continuous anywhere (it is not Riemann integrable as
we will see in the next chapter)
1 x ∈ Q,
f (x) =
0 x 6∈ Q.
(f ) Consider the function on the real line given by
x
x ∈ Qc ,
f (x) =
1 − x x ∈ Q.
The function is continuous at x = 1/2 and discontinuous everywhere else.
(g) Finally, the function below is continuous everywhere, but differentiable nowhere:
∞
X
1
cos (3n x) .
f (x) =
n
2
n=1
For a proof of (g) see [4], A Primer of Real Functions by Ralph P. Boas Jr., (1981),
QA331.5 B63x.
62
5. Differentiation
5.3. Homework
Exercise 5.1. Using the definition find the derivative of f (x) =
√
x for x > 0.
Exercise 5.2. Let f be differentiable on an interval I. If f is monotone increasing
on I, show f ′ (x) ≥ 0 on I.
Exercise 5.3. Let f, g, : [a, b] → R, and suppose f, g are differentiable at x0 ∈ [a, b].
In addition, f (x0 ) = g(x0 ) and f (x) ≤ g(x) for all x ∈ [a, b]. Does f ′ (x0 ) = g ′ (x0 )?
Exercise 5.4. Provide a more direct proof of Darboux’s Theorem (Theorem 5.16)
by considering the maximum of the function g(t) = f (t) − yt, where y satisfies
f ′ (a) > y > f ′ (b). Hint: first show that, if a function satisfies g ′ (a) > 0 and
g ′ (b) < 0, then the maximum must be in the interior.
Exercise 5.5. Let ǫ > 0. Show there does not exist a differentiable function on
[0, ∞) with f ′ (0) = 0 and f ′ (x) ≥ ǫ for x > 0.
Exercise 5.6. Suppose f is differentiable on R and that f has n distinct roots.
Show f ′ has at least n − 1 distinct real roots. Show by example that f ′ can have
more real roots than f .
Exercise 5.7. Let f : (0, 1] → R be differentiable with |f ′ (x)| ≤ 1 for x ∈ (0, 1].
For each n ∈ N, set an = f (1/n). Show that the sequence {an }n∈N converges.
Exercise 5.8. Let f be continuous on [0, ∞) and differentiable on (0, ∞), and
let f (0) = 0. If f ′ is monotone increasing on (0, ∞), show that g(x) = f (x)/x is
monotone increasing on (0, ∞).
Exercise 5.9. Let f be continuous on [a, b] and differentiable on (a, b). Suppose
lim f ′ (x) exists. Show f is differentiable at a and that f ′ (a) = lim f ′ (x).
x→a
x→a
Exercise 5.10. A function f : I → R is called uniformly Hölder continuous with
exponent θ if there exists a constant, c, such that for all x, y ∈ I, |f (x) − f (y)| ≤
c|x − y|θ with 0 < θ < 1. It is called uniformly Lipschitz if θ = 1.
(a) Prove that Hölder and Lipschitz continuous functions are in fact continuous.
(b) What can you say about the differentiability of such functions?
(c) Give a characterization of the functions that are Hölder continuous with exponent θ > 1.
Exercise 5.11. Recall a function f is said to have a fixed point if f (c) = c.
(i) Suppose f is differentiable on R and |f ′ (x)| < 1 for all x ∈ R. Show that f
can have at most one fixed point.
(ii) Let f (x) = x + (1 + ex )−1 . Show f satisfies Part (i), but has no fixed points.
Exercise 5.12. Let P be any polynomial of degree two and let [a, b] be any interval. Find the c guaranteed by the Mean-Value Theorem. Do you notice anything
interesting about this point? Repeat for f (x) = 1/x with a > 0.
g ′ (x)
f ′ (x)
=
on some interval I. Find, without using
f (x)
g(x)
logarithms, a relationship between f and g.
Exercise 5.13. Suppose
63
5.3. Homework
Exercise 5.14. Suppose f , g are differentiable on [a, b] an f ′ (x) ≥ g ′ (x) for all
x ∈ [a, b]. Show f (b) − f (a) ≥ g(b) − g(a).
√
Exercise 5.15. Prove 1 + x ≤ 1 + x/2 for all x > 0.
Exercise 5.16. Prove | sin x|/|x| ≤ 1 for all x ∈ R
Exercise 5.17. Show (x − 1)/x < ln x < x − 1 for x > 1, and hence, ln(1 + x) < x
for x > 0.
Exercise 5.18. Prove ex ≥ 1 + x for all x ≥ 0
Exercise 5.19. By using induction find a general form for the derivative of
−1
e x2 x 6= 0,
f (x) =
0
x = 0,
and hence, show f (n) (0) = 0 for all n ∈ N. Thus the Taylor expansion for f is zero.
Exercise 5.20. Show, for x > 0,
x2
x2
< ex < 1 + x + ex .
2
2
Exercise 5.21. Let C, S be two real-valued functions on the real line satisfying
the following properties:
1+x+
S(x + y) = S(x)C(y) + C(x)S(y), C(x + y) = C(x)C(y) − S(x)S(y);
S(0) = 0, S ′ (0) = 1, C(0) = 1, C ′ (0) = 0.
(a) Show S ′ (x) exists for all x ∈ R and is C(x).
(b) Show C ′ (x) exists for all x ∈ R and is −S(x).
(c) Use (a), (b), and Theorem 5.14 (ii) to show S 2 + C 2 = 1.
∞
X
x2n+1
(−1)n
(d) show S(x) =
(2n + 1)!
n=0
(e) Show that S(x) = sin(x).
2
Exercise 5.22. (a) Show that |x| < ex for all real x.
2
(b) Show that f (x) = e−x is uniformly continuous on R. (Hint: use the MVT
and Part (a)).