MATH 104: INTRODUCTORY ANALYSIS
SPRING 2009/10
PROBLEM SET 7 SOLUTIONS
5.31: Prove that if f 0 is continuous in a neighborhood of x = a then
f (a + h/2) − f (a − h/2)
= f 0 (a).
h→0
h
lim
Solution.
Let δ > 0 be such that f is differentiable and f 0 is continuous in Nδ (a). Define
F (h) = f (a + h/2) − f (a − h/2)
and G(h) = h.
Then
(i) F 0 (h) = 12 [f 0 (a + h/2) + f 0 (a − h/2)] and G0 (h) = 1 exist in Nδ∗ (a);
(ii) G0 (h) = 1 6= 0 in Nδ∗ (a);
(iii) limh→a F (h) = limh→a G(h) = 0;
(iv) limh→a [F 0 (h)/G0 (h)] exists since
f 0 (a + h/2) + f 0 (a − h/2)
f 0 (a) + f 0 (a)
F 0 (h)
=
lim
=
= f 0 (a)
h→a
h→a G0 (h)
2
2
lim
by the continuity of f 0 at a.
By l’hôpital rule, limh→a [F (h)/G(h)] exists and
F 0 (h)
F (h)
= lim 0
.
h→a G (h)
h→a G(h)
lim
Hence
f (a + h/2) − f (a − h/2)
= f 0 (a).
h→0
h
lim
5.32: Prove that if f 00 exists and is continuous in a neighborhood of x = a then
f (a + h) − 2f (a) + f (a − h)
= f 00 (a).
h→0
h2
lim
Solution. Let δ1 > 0 be such that f is twice differentiable and f 00 is continuous in Nδ1 (a).
Let δ = min{δ1 , |a|/2} if a 6= 0 and δ = δ1 if a = 0 (so that condition (ii) below is satisfied).
Define
F (h) = f (a + h) − 2f (a) + f (a − h) and G(h) = h2 .
Then
(i) F 0 (h) = f 0 (a + h) − f 0 (a − h) and G0 (h) = 2h exist in Nδ∗ (a);
(ii) G0 (h) = 2h 6= 0 in Nδ∗ (a);
(iii) limh→a F (h) = limh→a G(h) = 0;
(iv) limh→a [F 0 (h)/G0 (h)] exists since
F 0 (h)
f 0 (a + h) − f 0 (a − h) t=2h
f 0 (a + t/2) − f 0 (a − t/2)
=
lim
=
lim
= f 00 (a)
t→0
h→a G0 (h)
h→a
2h
t
lim
where the last equality follows from part (a).
Date: May 4, 2010 (Version 1.0).
1
By l’hôpital rule, limh→a [F (h)/G(h)] exists and
F 0 (h)
F (h)
= lim 0
.
h→a G (h)
h→a G(h)
lim
Hence
f (a + h) − 2f (a) + f (a − h)
= f 00 (a).
h→0
h2
lim
5.33: Prove Theorem 5.12 using the mean-value theorem. If f is continuous at x = a and if
f is differentiable in Nδ∗ (a) for some δ > 0 and if limx→a f 0 (x) exists then f 0 (a) exists and
f 0 (a) = limx→a f 0 (x).
Solution. Note that if 0 < h < δ, then f is continuous on [a, a + h] and differentiable on
(a, a + h) so mean value theorem implies that there exists ch ∈ (a, a + h) such that
f (a + h) − f (a)
.
h
Now note that a < ch < a + h implies that limh→0+ ch = 0. Hence
f 0 (ch ) =
lim f 0 (ch ) = lim
h→0+
h→0+
f (a + h) − f (a)
.
h
Applying the same argument to f on the interval [a − h, a], we get
lim f 0 (ch ) = lim
h→0−
h→0−
f (a + h) − f (a)
.
h
But since limx→a f 0 (x) exists, we have
lim f 0 (ch ) = lim f 0 (x) = lim f 0 (ch ).
x→a
h→0−
h→0+
Hence
lim
h→0−
f (a + h) − f (a)
f (a + h) − f (a)
= lim f 0 (x) = lim
+
x→a
h
h
h→0
and so
f (a + h) − f (a)
h→0
h
0
exists and equals limx→a f (x). In other words, f 0 (a) exists and f 0 (a) = limx→a f 0 (x).
lim
4.46: Prove that if f and g are each uniformly continuous on I then the sum f +g is uniformly
continuous on I.
Solution. Let ε > 0. There exists δ1 > 0 such that |f (x) − f (y)| < ε/2 whenever
|x − y| < δ1 . There exists δ2 > 0 such that |g(x) − g(y)| < ε/2 whenever |x − y| < δ2 . Let
δ = min(δ1 , δ2 ). Then whenever |x − y| < δ, we have
ε ε
|[f (x) + g(x)] − [f (y) + g(y)]| ≤ |f (x) − f (y)| + |g(x) − g(y)| < + = ε.
2 2
4.48: Prove that if f is uniformly continuous on I and k ∈ R then k ·f is uniformly continuous
on I.
Solution. Let ε > 0. If k = 0, any δ > 0 would work. So assume k 6= 0. There exists
δ > 0 such that |f (x) − f (y)| < ε/|k| whenever |x − y| < δ. Then whenever |x − y| < δ, we
have
ε
|k · f (x) − k · f (y)| = |k||f (x) − f (y)| < |k| ×
= ε.
|k|
4.50: Prove that if f is uniformly continuous on a bounded interval I then f is bounded on
I.
2
Solution. Pick ε = 1. Then there exists δ > 0 such that |f (x) − f (y)| < 1 whenever
|x − y| < δ. Now since I is bounded, we may pick points y1 , . . . , yn ∈ I so that
I⊆
n
[
Nδ (yk ).
k=1
Since |f (x) − f (yk )| < 1 whenever x ∈ Nδ (yk ) for k = 1, . . . , n, using triangle inequality, we
get |f (x)| < |f (yk )| + 1 whenever x ∈ Nδ (yk ) for k = 1, . . . , n. Hence
|f (x)| < max{|f (yk )| + 1 | k = 1, . . . , n}
for all x ∈ I. Therefore f is bounded on I.
4.54: Show by example that a continuous, bounded function on the bounded, open interval
(a, b) need not be uniformly continuous on (a, b).
Solution. Consider the function f (x) = sin(1/x) on the open interval (0, 1). Note that
|f (x)| ≤ 1 for all x ∈ (0, 1) and so is bounded. Also since 1/x is continuous on (0, 1) and
sin is continuous on (1, ∞), f is continuous on (0, 1) by Theorem 4.3. However f is not
uniformly continuous on (0, 1) since
1
1
=1
f
−f
2nπ
2nπ + π/2 for all n ∈ N (pick ε < 1 and note that |1/2nπ − 1/(2nπ + π/2)| < δ for any δ > 0 when n
is large enough).
6.5: Prove that if f is bounded on [a, b] and has exactly one discontinuity in [a, b] then f is
Riemann-integrable on [a, b].
Solution. Suppose c ∈ (a, b). Let ε > 0. Since f is bounded, there exists M > 0 such that
|f (x)| ≤ M for all x ∈ [a, b]. Let δ = ε/12M . Since f is continuous on [a, c − δ] and [c + δ, b],
f is Riemann-integrable on [a, c − δ] and [c + δ, b] by Theorem 6.2, so there exists partitions
P1 = {a = x0 < · · · < xn = c − δ} of [a, c − δ] and P2 = {c + δ = y0 < · · · < ym = c − δ} of
[c + δ, b] such that
U (P1 , f ) − L(P1 , f ) <
ε
3
ε
and U (P2 , f ) − L(P2 , f ) < .
3
Consider the partition of [a, b] given by P = P1 ∪ P2 . Then
U (P, f ) =
L(P, f ) =
n
X
i=1
n
X
i=1
Mi ∆xi + 2δ
sup
f (x) +
x∈[c−δ,c+δ]
mi ∆xi + 2δ
inf
f (x) +
x∈[c−δ,c+δ]
m
X
j=1
m
X
Mj ∆yj ≤ U (P1 , f ) + 2M δ + U (P2 , f ),
mj ∆yj ≥ L(P1 , f ) − 2M δ + L(P2 , f ),
j=1
and so
U (P, f ) − L(P, f ) ≤ [U (P1 , f ) − L(P1 , f )] + 4M δ + [U (P2 , f ) − L(P2 , f )]
ε
ε
ε
< + 4M ×
+ = ε.
3
12M
3
Hence f is Riemann integrable by Theorem 6.1.
If c = a, a similar argument applied with a choice of δ = ε/4M and a partition P2 of the
interval [a + δ, b] shows that
U (P2 , f ) − L(P2 , f ) <
3
ε
2
and that the partition of [a, b] given by P = {a} ∪ P2 would have
U (P, f ) − L(P, f ) ≤ 2M δ + [U (P2 , f ) − L(P2 , f )]
ε
ε
+ = ε.
< 2M ×
4M
2
If c = b, a similar argument applied with a choice of δ = ε/4M and a partition P1 of the
interval [a, b − δ] shows that
ε
U (P1 , f ) − L(P1 , f ) <
2
and that the partition of [a, b] given by P = P1 ∪ {b} would have
U (P, f ) − L(P, f ) ≤ [U (P1 , f ) − L(P1 , f )] + 2M δ
ε
ε
+ = ε.
< 2M ×
4M
2
6.6: Prove that if f is bounded on [a, b] and f has only finitely many discontinuities in [a, b]
then f is Riemann-integrable on [a, b].
Solution. Since we have not proved Theorem 6.8 in this section, we will prove it without
assuming Theorem 6.8. Let c1 < · · · < ck be all the discontinuities of f in [a, b]. We will
assume that a < c1 and ck < b. The cases a = c1 ro ck = b can be dealt with as in
Problem 6.5. The proof here is also similar to Problem 6.5. Let ε > 0. Since f is bounded,
there exists M > 0 such that |f (x)| ≤ M for all x ∈ [a, b]. Let δ = ε/8M k. Since f is
continuous on [a, c1 − δ], [c1 + δ, c2 − δ], . . . , [ck + δ, b], f is Riemann-integrable on these
intervals by Theorem 6.2, so there exists partitions P0 = {a = x0 < · · · < xn = c1 − δ} of
[a, c1 − δ], P1 = {c1 + δ = y0 < · · · < ym = c2 − δ} of [c1 + δ, c2 − δ], . . . , Pk = {ck + δ =
z0 < · · · < zl = b} of [ck + δ, b] such that
ε
for i = 0, . . . , k.
U (Pi , f ) − L(Pi , f ) <
2(k + 1)
Consider the partition of [a, b] given by P = P0 ∪ P1 ∪ · · · ∪ Pk . Then
U (P, f ) ≤ U (P0 , f ) + 2M δ + U (P1 , f ) + 2M δ + · · · + 2M δ + U (Pk , f ) = 2M kδ +
k
X
U (Pi , f ),
i=0
k
X
L(P, f ) ≥ L(P0 , f ) − 2M δ + L(P1 , f ) − 2M δ + · · · − 2M δ + L(Pk , f ) = −2M kδ +
i=0
and so
U (P, f ) − L(P, f ) ≤ 4M kδ +
k
X
[U (Pi , f ) − L(Pi , f )
i=0
ε
ε
+ (k + 1) ×
= ε.
8M k
2(k + 1)
Hence f is Riemann integrable by Theorem 6.1.
< 4M k ×
4
L(Pi , f ),