Math 515
Professor Lieberman
October 18, 2004
HOMEWORK #8 SOLUTIONS
Chapter 5
16. (a) Define g by
x
Z
g(x) =
f 0 (t) dt.
a
Then g is absolutely continuous and g 0 = f 0 a.e. by Theorem 63. It follows that h = f − g
has zero derivative a.e., and, for any x > y, we have
Z x
f 0 (t) dt − f (x) + f (y) ≥ 0
h(x) − h(y) =
y
by Proposition 51, so h is increasing.
(b) Let ε > 0 be given and set η = ε/(b − a). Then set
I = {[x, x + h] : |f (x) − f (x + h)| < ηh}.
Then I is a Vitali covering of the set on which f 0 = 0, so, for a given δ > 0, there is a finite
collection {[xn , xn + hn ]} of disjoint intervals from I such that the sums of their lengths is
at least [a, b] − δ. We write them in increasing order (as in the proof of Lemma 62). It
follows that
X
X
(f (xn + hn ) − f (xn )) < η
hn ≤ η(b − a) = ε.
Setting y0 = a and yk = xk−1 + hk−1 , we infer that
X
X
(f (xk ) − f (yk )) = f (b) − f (a) −
(f (xn + hn ) − f (xn )) > f (b) − f (a) − ε.
(c) For each n, set En = {x ∈ (a, b) : f 0 (x) > 1/n}. We shall show that mEn = 0. Since
{x ∈ (a, b) : f 0 (x) 6= 0} = ∪En , it will follow that f 0 = 0 a.e. as desired.
Let η > 0 be given and let n be a given positive integer. Then Property (S) gives a finite
collection of disjoint intervals {Ik } with Ik = [xk , yk ] for xk and yk satisfying
a < x1 < y1 < x2 < · · · < b,
and
X
η
η
`(Ik ) > (b − a) − ,
f (yk ) − f (xk ) <
.
2
2n
Proposition 51 then gives
Z
Z
XZ
X
η
0
0
f ≤
f =
f0 ≤
f (yk ) − f (xk ) <
.
2n
En ∩∪Ik
∪Ik
Ik
X
But
Z
En ∩∪Ik
f0 ≥
1
m(En ∩ ∪Ik ),
n
and
η
m(En ∩ ∪Ik ) = m(En ) − m(En ∼ ∪Ik ) ≥ m(En ) − m([a, b] ∼ ∪Ik ) ≥ m(En ) − .
2
1
2
It follows that
Z
f0 ≥
En ∩∪Ik
1
η
m(En ) −
.
n
2n
Hence
1
η
η
m(En ) −
<
,
n
2n
2n
so mEn < η. Since η is arbitrary, it follows that mEn = 0.
(d) Let ε and δ be given. For each integer j, set
Fj =
j
X
fn ,
G j = f − Fj .
n=1
P
Now choose j so that Gj (b) < ε/2. Since Fj0 = jn=1 fn0 , it follows that Fj is also singular.
P
Hence, from part (b), there is a finite collection of disjoint intervals [yk , xk ] with yk −xk <
δ and
X
ε
Fj (xk ) − Fj (yk ) > Fj (b) − Fj (a) − ≥ f (b) − f (a) − ε.
2
(Note that f (a) ≥ Fj (a).) Also,
f (xk ) − f (yk ) = Fj (xk ) − Fj (yk ) + Gj (xk ) − Gj (yk ) ≥ Fj (xk ) − Fj (yk )
because Gj is a sum of increasing functions, so it’s also increasing. It follows that
X
f (xk ) − f (yk ) > f (b) − f (a) − ε,
so f satisfies Property (S) and hence is singular.
(e) Let hri i be an enumeration of the rational numbers, define
(
0
if x < ri ,
fi (x) =
−i
2
if x ≥ ri .
then fi isPincreasing and fi0 = 0 everywhere except at ri , so each fi is singular. We then
set f =
fi . By part (d), f is singular. To see that f is strictly increasing, let x < y.
Then there is a rational number rj in the interval (x, y), so f (y) − f (x) ≥ 2−j > 0, so f is
strictly increasing.
20. (a) Let ε > 0 be given and set δ = ε/M (assuming without loss of generality that M > 0).
If {(xi , x0i )} is a collection of nonoverlapping intervals with
X
|x0i − xi | < δ,
then
X
|f (x0i ) − f (xi )| ≤
X
M |x0i − xi | = M
X
|x0i − xi | < M δ = ε,
so f is absolutely continuous.
(b) =⇒: Since f satisfies a Lipschitz condition, we have
f (x + h) − f (x) 0
|f (x)| = lim
≤M
h→0
h
because (f (x + h) − f (x))/h ≤ M .
⇐=: Suppose |f 0 | ≤ M . Then Corollary 64 implies that
Z x
f (x) − f (y) =
f 0 (t) dt
y
3
for any x > y. Hence
Z
x
|f (x) − f (y)| ≤
Z
0
x
|f | ≤
M = M |x − y|.
y
y
(c) Suppose |D+ f | ≤ M . Define g and h by g(x) = f (x) + M x and h(x) = f (x) − M x.
Then D+ g ≥ 0 so g is increasing and D+ h ≤ 0 so h is decreasing. If x > y, it follows that
f (x) − f (y) = g(x) − g(y) − M (x − y) ≥ −M (x − y)
and
f (x) − f (y) = h(x) − h(y) + M (x − y) ≤ M (x − y).
Combining these two inequalities gives |f (x) − f (y)| ≤ M |x − y|.
21. (a) Since O is open, we can write it as the union of disjoint open intervals In . Since g is increasing, g −1 [In ] is an open interval, which we denote by (an , bn ). Then In = (g(an ), g(bn )),
so
Z
Z
bn
m(In ) = g(bn ) − g(an ) =
g 0 (t) dt =
g 0 (x) dx.
g −1 [In ]
an
It follows that
m(O) =
X
Z
m(In ) =
Z
0
g 0 (x) dx.
g (x) dx =
∪g −1 [In ]
g −1 [O]
(b) For each positive integer n, let Hn = {x : g 0 (x) > 1/n}. We shall show that En =
g −1 (E) ∩ Hn has measure zero. Since H = ∪Hn , it follows that g −1 (E) ∩ H also has
measure zero.
Hence, we fix n and let ε > 0 be given. Then there is an open set O such that E ⊂ O
and mO < ε/n. Setting On = g −1 [O] ∩ Hn , we have
Z
Z
ε
1
> mO =
g0 ≥
g 0 ≥ m(On ).
n
n
g −1 [O]
On
It follows that m(On ) < ε and hence, because En ⊂ On , mEn < ε. Since ε > 0 is arbitrary,
it follows that mEn = 0.
(b) From Proposition 28, there is an Fσ set, E1 , and a set E0 with measure zero such that
E = E0 ∪ E1 . Then
[
F = g −1 [E0 ] ∩ H
g −1 [E1 ] ∩ H.
From part (a), g −1 [E0 ] ∩ H is measurable. In addition g −1 [E1 ] is a Borel set (by problem
3.26) so it’s measurable, and H is measurable because g 0 is measurable by virtue of Theorem
52, Theorem 54 and Lemma 60. It follows that g −1 [E] ∩ H is also measurable.
Part (a) also implies that, for any closed set C, we have
Z b
Z
Z
0
0
mC = (d − c) − mO =
g −
g =
g0
g −1 [O]
a
g −1 [C]
for O = [c, d] ∼ C. (This set O may be open or it may be the union of an open set
with a one- or two-point set. This second possibility isn’t a problem because points have
measure zero.) If S is an arbitrary Fσ set, then we have S = ∪Cn , with each Cn closed and
Cn ⊂ Cn+1 , so
Z
mCn =
g0.
g −1 [Cn ]
4
Applying Proposition 22 (with En = [c, d] ∼ Cn ) shows that mS = lim mCn and the
monotone convergence theorem (with fn = χg−1 [Cn ] ) implies that
Z
Z
lim
g0 =
g0,
g −1 [Cn ]
and hence
g −1 [S]
Z
g0
mS =
g −1 [S]
for any Fσ set S. Hence
Z
mE = mE1 =
g0.
g −1 [E1 ]
But g 0 = 0 on g −1 [E1 ] ∼ H and m(g −1 [E0 ] ∩ H) = 0, so
Z
Z
0
g0.
g =
mE =
g −1 [E1 ]∩H
F
(d) First, f ◦ g is the composition of a measurable function with a continuous function, so
it’s measurable, and we showed in part (b) that g 0 is measurable.
P To prove the integral equality, we first suppose that f is a simple function. Then f =
ai χEi for some finite number of (nonzero) constants ai and disjoint measurable sets Ei .
It follows in this case that
Z d
Z b
X
X Z b
0
f (y) dy =
ai mEi =
ai
χEi (g(x))g (x) dx =
f (g(x))g 0 (x) dx.
c
a
a
In general f is the limit of an increasing limit of simple functions fn . It follows that
h(fn ◦ g)g 0 i is an increasing sequence of functions, so the Monotone Converge Theorem
shows that
Z
Z
d
b
f (y) dy =
c
f (g(x))g 0 (x) dx
a
in this case as well.
25. (a) ϕ00 (t) = p(p − 1)(a + tb)p−2 b2 , which is nonnegative if 1 ≤ p < ∞ and nonpositive if
0 < p ≤ 1. Corollary 68 completes the proof.
(b) If p > 1, then ϕ00 > 0. It follows that ϕ0 is strictly increasing. Now let x < y be given,
and set
ψ(t) = ϕ[ty + (1 − t)x] − tϕ(y) − (1 − t)ϕ(x)
as in the proof of Proposition 67. We see that ψ 0 is strictly increasing and (as in that
proof), we can’t have ψ 0 > 0 on the whole interval (0, 1). It follows that ψ is initially
strictly decreasing, then is zero at one point (the differentiability of ψ 0 implies that ψ 0 is
continuous), and is positive after that, so it can’t be zero anywhere in the interval (0, 1).
Therefore ϕ is strictly convex. A similar argument applies if 0 < p < 1.
Chapter 6
2. Set M0 = kf k∞ . then
Z
kf kp =
p
|f |
1/p
Z
≤
M0p
It follows that
limkf kp ≤ M0 .
1/p
= M0 .
5
On the other hand, if M < M2 , then m{x : |f (x)| > M } = ε is positive, so
Z
|f |p ≥ M p ε,
so
kf kp ≥ M ε1/p .
Since ε1/p → 1 as p → ∞, it follows that
limkf kp ≥ M.
But M < M0 is arbitrary, so
limkf kp ≥ M0 .
It follows that lim kf kp = M0 .
8. (a) Set t = p ln a and s = q ln b, so a = et/p and b = es/q . Since ex is a convex function of
x, it follows that
et es
q
b
ab = e(t/p)+(s/q) ≤
+
= + .
p
q
p q
Since ex is strictly convex, it follows that equality holds if and only if t = s, which means
ap = bq .
(b) Without loss of generality, kf kp and kgkq are positive. Set F = f /kf kp and G = g/kgkq .
Then Young’s inequality implies that
Z
Z
Z
1
1 1
1
p
|F | +
|G|q = + = 1.
|F ||G| ≤
p
q
p q
It follows that
Z
Z
|f g| =
|F ||G| kf kp kgkq ≤ kf kp kgkq .
If equality holds, then we must have |F |p = |G|q a.e. and hence α|f |p = β|g|q a.e.
(c) If 0 < p < 1, then q < 0. Let us set P = 1/p and Q = 1/(1 − p) so that P1 + Q1 = 1.
Then we take α = (ab)p and β = b−p . Young’s inequality gives
αβ ≤
αP
βQ
+
.
P
Q
Since q = −pQ, it follows that
p
ap ≤ pab − bq ,
q
so (after dividing both sides by p and then adding bq /q to both sides)
ap bq
+ .
ab ≥
p
q
(d) (We follows the proof of part (b) with the change from Young’s inequality) Without
loss of generality, kf kp and kgkq are positive. Set F = f /kf kp and G = g/kgkq . Then
Young’s inequality implies that
Z
Z
Z
1
1
1 1
p
|F ||G| ≥
|F | +
|G|q = + = 1.
p
q
p q
It follows that
Z
Z
|f g| =
|F ||G| kf kp kgkq ≥ kf kp kgkq .