**Unformatted text preview: **Real Analysis Homework VII solutions
November 8th, 2019 Exercise 3.4.3
Show that f : (c, ∞) → R for some c > 0 and defined by f (x) := 1
x is Lipschitz continuous. Proof. By Definition 3.4.7, we must show that there exists an L such that for all x, y > c,
|f (x) − f (y)| < L|x − y|. We compute that for x, y ∈ (c, ∞), 1 1 y − x |x − y|
1 =
|f (x) − f (y)| = − = ≤ 2 |x − y| . x y
xy
xy
c
Thus, f is Lipschitz continuous with Lipschitz constant c−2 . Exercise 3.4.7
Let f : (0, 1) → R be a bounded continuous function. Show that the function
g(x) := x(1 − x)f (x)
is uniformly continuous.
Proof. By Proposition 3.4.6, g is uniformly continuous if and only if limx→0 g(x) and limx→1 g(x)
exist . Thus, we only need to show the limits exist .
We begin with limx→0 g(x). Since f is bounded, we have that |f (x)| < B for all x ∈ [0, 1].
Hence,
−Bx ≤ x(1 − x)f (x) ≤ Bx.
By the Squeeze theorem (Lemma 2.2.1), limx→0 g(x) = 0 .
A similar argument shows that limx→1 g(x) = 0 . In this case,
−B(1 − x) ≤ x(1 − x)f (x) ≤ B(1 − x)
so we can again use the Squeeze theorem to conclude. 1 Exercise 3.4.10
a. Find a continuous f : (0, 1) → R and a sequence {xn } in (0, 1) that is Cauchy, but such
that {f (xn )} is not Cauchy.
b. Prove that if f : R → R is continuous, and {xn } is Cauchy, then {f (xn )} is Cauchy.
a. Let f (x) = x1 , xn =
Cauchy, but 1
n for each n. Then, f is a continuous function on (0, 1) and {xn } is
f (xn ) = n so {f (xn )} is not Cauchy.
b. Here are two proofs:
Proof. By Theorem 2.4.5, it is enough to show {f (xn )} is convergent . Since {xn } is
Cauchy, it converges to some x. By the Bridge Theorem (Prop 3.2.2), f (xn ) → f (x) , so
{f (xn )} is convergent and hence Cauchy.
Proof. Suppose {xn } is Cauchy. Then, it is bounded, so {xn : n ∈ N} ⊂ [−R, R] for
some positive R. The restriction of f to [−R, R], f |[−R,R] , is a continuous function on a
bounded set, so by Theorem 3.4.4 it is uniformly continuous . Lemma 3.4.5 then assures
us that {f (xn )} = {f |[−R,R] (xn )} is Cauchy. Exercise 3.4.16
Suppose f : S → R and g : [0, ∞) → [0, ∞) are functions, g is continuous at 0, g(0) = 0, and
whenever x and y are in S we have |f (x) − f (y)| ≤ g(|x − y|). Prove that f is uniformly
continuous.
We remark that the function g is called the modulus of continuity of f . The case g(x) = Lx
corresponds the f being Lipschitz with Lipschitz constant L.
Proof. Since S is an arbitrary set, we cannot use the theorems about uniform continuity of
functions defined on intervals. Thus, we argue by the definition (3.4.1).
Consider > 0, arbitrary. We must show that there exists a δ such that for any x, y ∈ S
with |x − y| < δ, |f (x) − f (y)| < . Since g is continuous at 0 and g(0) = 0, there exists a
δ > 0 such that for every z with 0 ≤ z < δ ,
|g(z)| <
Then, for any x, y ∈ S with |x − y| < δ, we have that
|f (x) − f (y)| ≤ g(|x − y|) <
as was to be shown. 2 Exercise 3.5.1
Show that the limit at ∞ or −∞ is unique if it exists.
The argument follows that same line of reasoning as Proposition 2.1.6 and Proposition 3.1.4.
Proof. Suppose L1 and L2 are both limits of f (x) as x → ∞. Then, for any > 0 there exists
an N1 such that for x ≥ N1 , |f (x) − L1 | < 2 , and similarly there exists an N2 such that for
x ≥ N2 , |f (x) − L2 | < 2 . Choose x ≥ max{N1 , N2 }. Then, by the Triangle inequality,
|L1 − L2 | ≤ |f (x) − L1 | + |f (x) − L2 | <
Since this holds for all > 0, it follows that L1 = L2 . This show the limit as x → ∞ is unique.
If we define g(x) = f (−x), then
lim f (x) = lim g(x), x→−∞ x→∞ so we have also shown that the limit as x → −∞ is unique. Exercise 4.1.5
Define f : R → R by (
x2
f (x) =
0 if x ∈ Q
else Prove that f is differentiable at x = 0, but discontinuous at all points except 0.
Proof. By Definition 4.1.1, the derivative at 0 is
f (x) − f (0)
x→0
x−0
f (x)
= lim
x→0 x f 0 (0) = lim Observe that
−|x| ≤ f (x)
≤ |x|.
x So by the squeeze theorem, f 0 (0) = 0 .
To see that f is not continuous away from 0, let x 6= 0 be arbitrary. Either x is rational, or
it is irrational.
If x is rational, let {xn } be a sequence of irrational numbers converging to x . Then,
lim f (xn ) = 0 6= f (x) = x2 n→∞ so, by the (contrapositive of) the bridge theorem, f is discontinuous at x.
Similarly, if x is irrational, we let {xn } be a sequence of rational numbers converging to x .
Then,
lim f (xn ) = lim x2n = n→∞ n→∞ lim xn n→∞ so f is discontinuous at x by the bridge theorem.
3 2 = x2 6= f (x) = 0 Exercise 4.1.8
Let f : I → R be differentiable. Given n ∈ Z, define f n by f n (x) = (f (x))n . (If n < 0, assume
f (x) 6= 0). Prove that (f n )0 (x) = n[f (x)]n−1 f 0 (x).
Proof. This is a result of the chain (Prop. 4.1.10) and product (Prop. 4.1.8) rules. We recall
and prove the following result from Exercise 4.1.3:
Proposition . For n ∈ Z, (xn )0 = nxn−1 (if n < 0, we require that x 6= 0).
Proof. For n ≥ 0, we proceed by induction. If n = 0, then the proposition reads 10 = 0, which
is certainly true. Now, assume that for some n, (xn )0 = nxn−1 . Then, by the product rule,
0
xn+1 =x0 (xn ) + x (xn )0
=1 (xn ) + x nxn−1
=(n + 1)xn
which proves the result for n ≥ 0. For n < 0, the quotient rule states that
0
1
−(−n)x−n−1
n 0
(x ) =
=
= nxn−1
−n
−2n
x
x
which is the desired result.
We now return to the problem. Let g(x) = xn . Then, f n (x) = (g ◦ f )(x), so
(f n )0 (x) = (g ◦ f )0 (x) = g 0 (f (x))f 0 (x) = n[f (x)]n−1 f 0 (x)
as desired. Exercise 4.1.10
Let I1 , I2 be intervals, f : I1 → I2 be a bijective function, and g : I2 → I1 be its inverse.
Suppose both that f is differentiable at c and g is differentiable at f (c). Use the chain rule to
find a formula for g 0 (f (c)).
On one hand, by the chain rule,
(g ◦ f )0 (c) = g 0 (f (c))f 0 (c)
On the other hand, g ◦ f is the identity function g ◦ f (x) = x for all x ∈ I1 , so
(g ◦ f )0 (c) = 1.
Equating these two expressions and simplifying, we find
g 0 (f (c)) = 4 1
.
f 0 (c) Exercise 4.1.11
Suppose f : I → R is bounded and g : I → R is differentiable at c ∈ I with g(c) = g 0 (c) = 0.
Show that h(x) = f (x)g(x) is differentiable at c.
Proof. As stated in the hint, we cannot apply the product rule since f is not known to be
differentiable. However, we can attempt the same manipulations from the proof of the product
rule.
Consider the difference quotient
h(x) − h(c)
x−c
f (x)g(x) − f (0)g(0)
= lim
x→c
x−c
f (x)g(x) − f (x)g(c) + f (x)g(c) − f (c)g(c)
= lim
x→c
x−c h0 (c) = lim x→c Now, we observe that g(c) = 0, so we can drop the last two terms in the numerator. This leaves
us with
h0 (c) = lim f (x)
x→c g(x) − g(c)
x−c Now, since f is bounded, |f (x)| ≤ M for all x ∈ I. Thus, g(x) − g(c) g(x)
−
g(c) ≤M 0 ≤ f (x) x−c x−c But, since g 0(c) = 0, the limit on the right goes to 0 as x → c. By the squeeze theorem, it follows that f (x) g(x)−g(c) → 0 as x → c, and hence that
x−c
h0 (c) = lim f (x)
x→c g(x) − g(c)
= 0.
x−c Since the limit of the difference quotient exists, h is differentiable at 0. Exercise 4.1.15
Prove the following simple version of L’Hôpital’s rule. Suppose f : (a, b) → R and g : (a, b) → R
are differentiable functions whose derivatives f 0 and g 0 are continuous functions. Suppose that
at c ∈ (a, b), f (c) = 0, g(c) = 0, and g 0 (x) 6= 0 for all x ∈ (a, b), and suppose the limit as x → c
0 (x)
of fg0 (x)
exists. Prove that
f (x)
f 0 (x)
lim
= lim 0
.
x→c g(x)
x→c g (x) 5 Proof. Notice that since f 0 and g 0 are continuous with g 0 (c) 6= 0, the limit on the right is just
f 0 (c)
. Thus, it suffices to prove that
g 0 (c)
f (x)
f 0 (c)
= 0 .
x→c g(x)
g (c)
lim We proceed as follows:
f (x)
f (x) − f (c)
= lim
x→c g(x)
x→c g(x) − g(c)
lim = lim
x→c
=
= f (x)−f (c)
x−c
f (x)−f (c)
x−c (c)
limx→c f (x)−f
x−c limx→c
f (x)−f (c)
x−c f 0 (c)
g 0 (c) where we used the arithmetic rules for limits to change the limit of the quotient to a quotient
of limits in the third line. 6 ...

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