Math 515
Professor Lieberman
October 11, 2004
HOMEWORK #7 SOLUTIONS
Chapter 5
6. As a preliminary, we’ll show that
lim f (x) = A,
lim g(x) = B
imply that
lim f (x)g(x) = AB
provided B > 0 using Problem 2.49. By 2.49a and the definition of limit, for any ε > 0,
there is a number δ such that, for all x with 0 < |x − y| < δ, we have f (x) ≤ A + ε/(2B)
and
ε
|g(x) − B| < B min 1,
.
(A + 1)B
It follows that
ε
ε
)B 1 + min 1,
≤ AB + ε,
f (x)g(x) ≤ (A +
2B
(A + 1)B
so lim f (x)g(x) ≤ AB. A similar argument using 2.49b gives lim f (x)g(x) ≥ AB and hence
lim f (x)g(x) = AB.
(a) Note first that there is a positive constant h0 such that g(γ + h) > g(γ) if 0 < h < h0 .
Hence
f ◦ g(γ + h) − f ◦ g(γ)
g(γ + h) − g(γ)
D+ f ◦ g(γ) = lim
lim
= D+ f (c)g 0 (γ).
+
+
g(γ + h) − g(γ)
h
h→0
h→0
(b) Now we have g(γ + h) < g(γ), so
lim
h→0+
f ◦ g(γ + h) − f ◦ g(γ)
f (c − η) − f (c)
f (c) − f (c − η)
= lim
= lim
= D− f (c).
+
g(γ + h) − g(γ)
−η
η
η→0
η→0+
Now we use Problem 2.49c and d as well to finish.
(c) Note that the finiteness of the derivates implies that |f (c + h) − f (c)|/h is bounded, say
by M , in some neighborhood of c. Then
|f ◦ g(γ + h) − f ◦ g(γ)| ≤ M |g(γ + h) − g(γ)|
and, for any ε > 0, there is a δ > 0 such that |g(γ + h) − g(γ)| ≤
that
f ◦ g(γ + h) − f ◦ g(γ) <ε
h
for 0 < |h| < δ and hence (f ◦ g)0 (γ) = 0.
1
ε
Mh
for |h| ≤ δ. It follows
2
7. (a) Let f = h − g with h and g both increasing. Then
lim h(x) = inf h(x),
x→c+
x>c
lim h(x) = inf h(x)
x→c−
x<c
and similarly for g, so the one-sided limits for f also exist.
If f is increasing, set M = f (b) − f (a). Let n be an integer and suppose we have k points
c1 , . . . ck such that c1 < · · · < ck with |f (ci +) − f (ci −)| > 1/n. Then, by monotonicity,
f (c1 −) ≥ f (a), f (ci −) ≥ f (ci−1 ), f (ci +) ≥ f (ci ).
Hence
k X
<
|f (ci +) − f (ci )| ≤ f (b) − f (a) = M.
n
It follows that k < M n, so there are only finitely many points at which the inequality
occurs. Since every discontinuity must have this form, the number of discontinuities is
countable.
(b) Write the rationals as hri i and define fi by
(
0 if x < ri ,
fi (x) =
1 if x ≥ ri .
P −i
Now set f (x) = 2 fi (x). Since f is a sum of increasing functions, it’s increasing as well.
To see where f is discontinuous, we examine two cases. At a rational number r, we note
that r = rj for some j. If x < r, then f (r) − f (x) ≥ 2−j , so f is discontinuous at r. If x is
irrational, then, given ε > 0, there is a number δ such that the interval [x−δ, x+δ] contains
none of the rational numbers r1 , . . . , rj with 2−j < ε. It follows that, on this interval, f
can increase by no more than
∞
X
2−i = 2−j < ε,
i=j+1
so f is continuous at x.
10. (a) NO. Set yj = ((j + 12 )π)−1/2 and take the xi ’s in the subdivision to be yj for all j’s less
than or equal to some integer J (so n = J + 1 and xi = yJ−i+1 for 1 ≤ i < J, x0 = −1, and
xJ = 1). Then
J+1
J X
X
1
1
+
|f (xi ) − f (xi−1 | ≥
.
(j + 12 )π (j + 23 )π i=1
j=1
This is the partial sum of a divergent series (essentially the harmonic series), so we can
make it arbitrarily large by choosing J large enough.
(b) YES. Now take yj = 1/((j + 12 )π). Notice that
(
2x sin x1 − cos x if x 6= 0
0
g (x) =
0
ifx = 0
so g 0 (x) ≥ −2. Hence h(x) = g(x) + 2x has a nonnegative derivative, so it is increasing. It
follows that g is the difference of two increasing functions: h and k(x) = 2x.
3
12. The continuity of f does not imply that f is absolutely continuous on [0, 1]. Take f to be
the function from problem 10(a). Since this function is not of bounded variation on [0, 1],
it isn’t absolutely continuous there. On the other hand, it’s continuously differentiable on
[ε, 1] for any ε ∈ (0, 1) and hence it’s the indefinite integral of its derivative there by the
Fundamental Theorem of Calculus.
R1
If f is continuous and of bounded variation, then f 0 is integrable, so 0 f 0 exists. The
Lebesgue Convergence Theorem (applied to fn = f χ[1/n,1] implies that
Z 1
Z 1
0
lim
f =
f.
1/n
0
Since f absolutely continuous on [1/n, 1], it follows that
continuity of f then implies that
Z 1
lim
f 0 = f (1) − f (0).
R1
1/n f
0
= f (1) − f (1/n). The
1/n
Hence, for any x ∈ [0, 1], we have
Z
f (x) = f (1) −
1
0
Z
f = f (0) +
x
x
f 0,
0
so f is an indefinite integral of its derivative and hence f is absolutely continuous.