Homework 4 Solutions, Real Analysis I, Fall, 2010.
(14) Let X be a topological space and M be a σ-algebra on X which
contains all Borel sets. Let m, µ be two positive measures on M.
Assume there is a constant c ∈ [0, ∞) so that µ(E) = c m(E)
for each Borel set E. Assume that m is regular. Let A ∈ M
and m(A) = 0. Prove µ(A) = 0.
Solution: Assume A ∈ M and m(A) = 0. Let > 0. Then
since m is (outer) regular, there is an open set V ⊃ A so that
m(V ) < /c. By assumption, then µ(V ) = c m(V ) < . Then
since A ⊂ V , µ(A) ≤ µ(V ) < . Since is arbitrary, we must
have µ(A) = 0.
(16) The Cantor set.
(a) Define subsets of the unit interval On , Xn inductively by
O1 = ( 13 , 23 ),
On+1 = 13 On ∪ ( 13 On + 23 ),
X0 = [0, 1],
Xn+1 = Xn − On .
Show that Xn is a disjoint union of closed intervals of each
of length 3−n .
Solution: We first prove the following inductive definition
for Xn alone:
(1)
X0 = [0, 1],
Xn+1 = 31 Xn ∪ ( 13 Xn + 23 ).
To show this, compute
Xn+1 =
=
=
=
=
=
Xn − On
[0, 1] − (O1 ∪ O2 ∪ · · · ∪ On )
([0, 1] − O1 ) − (O2 ∪ · · · ∪ On )
([0, 13 ] ∪ [ 23 , 1]) − { 13 O1 ∪ ( 13 O1 + 23 ) ∪ · · · ∪ 13 On−1 ∪ 13 (On−1 + 23 )}
1
([0, 1] − O1 ∪ · · ·
3
1
X ∪ ( 13 Xn + 23 ).
3 n
∪ On−1 ) ∪ { 13 ([0, 1] − O1 ∪ · · · ∪ On−1 ) + 23 }
Now we prove by induction that Xn is a disjoint union of
2n closed subintervals of [0, 1], each of length 3−n . This is
clearly true for n = 0.
Now we assume the statement is true for Xn . To prove it for
Xn+1 , note that since Xn ⊂ [0, 1], Xn+1 = 13 Xn ∪ ( 13 Xn + 23 )
consists of two disjoint copies of Xn , each scaled down by 13 ,
contained in the intervals [0, 13 ] and [ 23 , 1]. Thus the number
of intervals in Xn+1 is double that of Xn : 2 · 2n = 2n+1 ,
1
2
and each of these intervals has length 13 of the length of
those in Xn : 13 · 3−n = 3−(n+1) . Thus the inductive case is
checked and (1) is proved.
(b) Define the Cantor set
!
∞
∞
\
[
K=
Xn = X0 −
On .
n=1
n=1
Show K is a compact subset of R with Lebesgue measure
0.
Solution: By (1), m(Xn ) = 2n · 3−n = ( 32 )n . Therefore,
since Xn are nested measurable sets of finite measure,
m(K) = lim m(Xn ) = lim ( 23 )n = 0.
n→∞
n→∞
As an intersection of closed sets, K is closed. It is also
bounded in R. Thus it is compact.
(c) Show that K consists of all numbers in the unit interval
[0, 1] which have a base-3 expansion which only uses 0’s
and 2’s. (One special point to recall is that (0.2̄)3 = (1.0̄)3 ,
where the bar signifies infinite repetition of the ternary
digits. Therefore, the endpoint 1 is legitimately in K.)
Solution: We prove by induction that Xn consists of ternary
numbers of the form (0.d1 d2 · · · dn dn+1 · · · )3 so that di ∈
{0, 2} for i ≤ n. For the base case, we consider X1 =
[0, 31 ] ∪ [ 23 , 1]. The first interval consists of all ternary numbers between 0 = (0.00̄)3 to 13 = (0.02̄)3 , while the second
interval consists all ternary numbers between 23 = (0.20̄)3
and 1 = (0.22̄)3 . This proves the base case.
For the inductive case, assume the statement is true for Xn .
Then if x = (0.d1 d2 · · · dn dn+1 · · · )3 ∈ Xn , then di ∈ {0, 2}
for i ≤ n. So the elements of Xn+1 are all of the form
1
3
· x = (0.0d1 d2 · · · dn dn+1 · · · )3 and 13 · x + 23 = (0.2d1 d2 · · · dn dn+1 · · · )3 .
This clearly shows that criterion for Xn+1 is true. Thus
the inductive proof is complete.
Now since K is the intersection of all the Xn , we have
that the elements of K can all be represented by ternary
expansions all of whose digits are 0 or 2.
(d) Construct a surjective map from K onto [0, 1]. (Hint: Relate the base-3 expansions of elements of K to base-2 expansions of elements of [0, 1].) Conclude that K is uncountable.
3
Solution: First of all we show that every element x of K
can be uniquely determined by x = (0.d1 d2 · · · )3 , where
each di ∈ {0, 2}. The only case of non-uniqueness in a
ternary expansion is that
(0.d1 d2 · · · dn−1 dn 2̄)3 = (0.d1 d2 · · · dn−1 en 0̄)3 ,
where dn 6= 2 and en = dn + 1. But when all the di ’s are
0 or 2, this case can only happen if dn = 0 and so en = 1,
which is not allowed in our ternary expansions of elements
of K. Thus the expansions involving only 0’s and 2’s are
unique for elements of K.
Now we define our surjective map from φ : K → [0, 1] by
φ : (0.d1 d2 · · · )3 7→ (0.f1 f2 · · · )2 ,
where fi = 0 if di = 0 and fi = 1 if di = 2. Since every element of [0, 1] can be represented by such a base-2
representation, φ is surjective. (φ is not injective, since binary representations are not unique, for the same reasons
as above.)
Using the fact that [0, 1] is uncountable, since φ : K → [0, 1]
is surjective, then K must be uncountable as well. (Since if
K were countable, there would be a one-to-one correspondence with [0, 1] with a subset of K, which contradicts the
fact that [0, 1] is uncountable.)
(17) Let f be an L1 function with respect to Lebesgue measure on
R. Use the result of Chapter 1, Problem 12, to show that
Z x
f (t) dt
g(x) =
0
is continuous.
Solution: Let > 0. Then by Chapter 1, Problem 12, in
Rudin,
R there is a δ > 0 so that if E is measurable and µ(E) < δ,
then E |f | dµ < .
For |x − y| < δ, let I be the interval from x to y. Then
Z
Z
|g(x) − g(y)| = f (t) dt ≤ |f (t)| dt < .
I
I
This is exactly the condition that g is (uniformly) continuous.
(19) Rudin, chapter 3, problem 4.
Suppose f is a complex measurable function on X, µ is a
positive measure on X, and
Z
ϕ(p) =
|f |p dµ = kf kpp (0 < p < ∞).
X
4
Let E = {p : ϕ(p) < ∞}. Assume kf k∞ > 0.
(a) If r < p < s, r ∈ E, and s ∈ E, prove that p ∈ E.
(b) Prove that if log ϕ is convex in the interior of E and that
ϕ is continuous on E.
Solution: The main idea is to use Hölder’s inequality on
integrals of the form
Z
Z
p
|f | dµ =
|f |γ · |f |p−γ dµ.
X
X
So for 1 < α < ∞ and α1 + β1 = 1, we have
Z
ϕ(p) =
|f |p dµ
ZX
=
|f |γ · |f |p−γ dµ
X
≤ k|f |γ kα · k|f |p−γ kβ
β1
Z
α1 Z
(p−γ)β
γα
=
|f | dµ
·
|f |
dµ
X
X
1
1
α
= ϕ(γα) · ϕ((p − γ)β) β ,
log ϕ(p) ≤ α1 log ϕ(γα) + β1 log ϕ((p − γ)β).
Now put r = γα and s = (p−γ)β. (Notice given r < p < s,
we can solve for α ∈ (1, ∞) and γ ∈ (0, p).) In this case,
1
α
·r+
1
β
· s = γ + (p − γ) = p.
Thus we have
log ϕ( α1 · r +
1
β
· s) ≤
1
α
log ϕ(r) +
1
β
log ϕ(s)
for α1 ∈ (0, 1) and α1 + β1 = 1. This is the definition of the
convexity of ϕ. ϕ is continuous on the interior of E by
Theorem 3.2
Note that we have proved (a) along the way, since if φ(r)
and φ(s) are finite, then the inequality implies
1
1
φ(p) ≤ φ(r) α φ(s) β < ∞.
(c) By (a), E is connected. Is E necessarily open? Closed?
Can E consist of a single point? Can E be any connected
subset of (0, ∞)?
Solution: For Lebesgue measure on R, E can be any connected subset of (0, ∞). The basic functions to consider
are of the form x` and x` | log x|m near x = 0 and x = ∞.
5
Thus we will multiply them by characteristic functions of
the form χ(0, 1 ] and χ[2,∞) . So define
2
g` (x) = x` χ(0, 1 ] (x),
2
h` (x) = x` χ[2,∞) (x),
g`,m (x) = x` | log x|m χ(0, 1 ] (x),
2
h`,m (x) = x` | log x|m χ[2,∞) (x).
R
It is easy to
check
that
g < ∞ if and only if ` > −1.
R `
R
Similarly, R h` < ∞ if and only if ` < −1. By comparing
+
−
Ras x → 0 1 ≤ | log x| ≤ x for all > 0, we find that
g
is finite for ` > −1 and infinite for ` < −1. For
R `,m
` = −1, compute by substituting u = log x
Z
Z 1
Z − ln 2
2
−1
m
g−1,m =
x | log x| dx =
|u|m du,
R
0
−∞
whichRis finite if and only if m < −1. Similarly, we can
show R h`,m is finite if ` < −1 and infinite for ` > −1. If
` = −1, this integral is finite if and only if m < −1.
p
Note that g`p = g`p , hp` = h`p , g`,m
= g`p,mp , hp`,m = h`p,mp .
For f = g−1,−2 + h−1,−2 , then E = {1}. Replacing this by
1
f p gives E = {p}.
To achieve E = ∅, one may take f = g−1 + h−1 .
To achieve E = [1, p), one may take f = g− 1 + h−1,−2 . To
p
1
get an interval of the form E = [a, b) for b = pa, take f a
instead.
1
To get E = (1, p), take f = g− 1 + h−1 , and f a gives
p
E = (a, pa).
1
To get E = [1, p], take f = g− 1 ,− 2 + h−1,−2 . Then f a gives
p
p
E = [a, pa].
E = (1, ∞) comes from f = h−1 . E = [1, ∞) comes from
f = h−1,−2 . E = (0, 1) comes from f = g−1 . E = (0, 1]
comes from f = g−1,−2 . All other intervals with 0 or ∞ as
1
one endpoint can be achieved by taking f a as above.
To get E = (0, ∞) for kf k∞ > 0, we may take f = e−|x| .
This exhausts all the connected subsets of (0, ∞).
(For different measure spaces, the same is not true. For
example, if µ(X) < ∞, then Hölder’s inequality applied to
|f |r = |f |r · 1 shows that s ∈ E =⇒ r ∈ E for all r < s.)
6
(d) If r < p < s, prove that kf kp ≤ max(kf kr , kf ks ). Show
that this implies the inclusion Lr (µ) ∩ Ls (µ) ⊂ Lp (µ).
Solution: By the solution to (b) above, we have
r
1
s
kf kpp = ϕ(p) ≤ ϕ(r) α · ϕ(s) = kf krα · kf ksβ
and so
r
s
kf kp ≤ kf krpα · kf kspβ .
Compute by our definition of r, s that
r
s
γ p−γ
+
= +
= 1.
pα pβ
p
p
So if M = max(kf kr , kf ks ), we have
s
r
kf kp ≤ M pα · M pβ = M.
So if f ∈ Lr (µ) ∩ Ls (µ), then kf kr and kf ks are finite. So
for r < p < s, we have
kf kp ≤ max(kf kr , kf ks ) < ∞,
and f ∈ Lp (µ). Therefore, Lr (µ) ∩ Ls (µ) ⊂ Lp (µ).
(e) Assume that kf kr < ∞ for some r < ∞ and prove that
kf kp → kf k∞
as p → ∞.
Solution: First of all, let α < kf k∞ . We’ll show that
there is P so that if p ≥ P , then kf kp > α. First of all,
choose β ∈ (α, kf k∞ ).
By the definition
of kf k∞ as the essential supremum, if
E = {x ∈ X |f (x)| > β}, we have µ(E) > 0. Then
kf kp =
p1 Z
p1 Z
p1
1
p
p
|f | dµ
≥
|f | dµ
≥
β dµ
= β(µ(E)) p .
Z
p
X
E
E
1
But as p → ∞, β(µ(E)) p → β > α, and so there is a P so
that if p > P , we have kf kp > α.
This allows us to settle the case of kf k∞ = ∞ immediately,
as the definition of limp→∞ kf kp = ∞ = kf k∞ is satisfied.
7
In the case kf k∞ < ∞, we need to use Hölder’s inequality
as above. Compute
kf kp =
Z
=
Z
p1
|f | dµ
p
X
r
p−r
|f | · |f |
p1
dµ
X
≤
p1
|f |r dµ · ess sup(|f |p−r )
Z
X
p−r
r
= kf krp · kf k∞p
p−r
r
Now as p → ∞, kf krp · kf k∞p → kf k∞ . So for any > 0,
there is a P 0 so that if p > P 0 , kf kp < kf k∞ +. The opposite inequality kf kp > kf k∞ − for p > P is proved above.
Therefore, limp→∞ kf kp = kf k∞ in the case kf k∞ < ∞ as
well.
(20) Rudin, chapter 3, problem 10.
Suppose fn ∈ Lp (µ) for n = 1, 2, 3, . . . , and kfn − f kp → 0
and fn → g a.e., as n → ∞. What relation exists between f
and g?
Solution: f = g a.e. To prove this, note that Theorem 3.12,
together with the assumption that kfn − f kp → 0, shows that
there is a subsequence fni so that fni → f a.e. as i → ∞.
But we also assume that fn → g a.e., and so the subsequence
fni → g a.e. also. Therefore, f = g a.e.
(22) Rudin, chapter 3, problem 12.
Suppose µ(Ω) = 1 and h : Ω → [0, ∞] is measurable. If
Z
A=
h dµ,
Ω
prove that
√
1+
A2
≤
Z √
1 + h2 dµ ≤ 1 + A.
Ω
If µ is Lebesgue measure on [0, 1] and if h is continuous, h = f 0 ,
the above inequalities have a simple geometric interpretation.
From this, conjecture (for general Ω) under what conditions on
h equality can hold in either of the above inequalities, and prove
your conjecture.
8
Solution: To prove the right-hand inequality, note since µ(Ω) =
1 that
Z
1+A=
(1 + h) dµ.
Ω
√
√
Since h ≥ 0, we have 1 + h2 ≤ 1 + 2h + h2 = 1 + h. Thus
integrating this inequality, we have
Z √
Z
2
1 + h dµ ≤ (1 + h) dµ = 1 + A.
Ω
Ω
Thus, as long as the integrals are finite, they are equal if and
only if
Z
√
[(1 + h) − 1 + h2 ] dµ = 0.
Ω
√
Since (1 + h)√− 1 + h2 ≥ 0, the inequality is true if and only
if (1 + h) − 1 + h2 = 0 a.e., which is equivalent by the computation above to h = 0 a.e. If h is assumed to be continuous,
equality is satisfied if and only if h = 0 everywhere. So if h = f 0 ,
this equality is satisfied if and only if f is constant.
For the
√ other inequality, we will use Jensen’s inequality. If
φ(t) = 1 + t2 , compute
φ0 (t) =
1
(1
2
1
1
+ t2 )− 2 (2t) = t(1 + t2 )− 2 ,
1
3
3
φ00 (t) = (1 + t2 )− 2 + t(− 12 )(1 + t2 )− 2 (2t) = (1 + t2 )− 2 .
Thus φ00 (t) > 0 for all real t and so φ is strictly convex on
(−∞, ∞). Thus, in the case h(Ω) ⊂ [0, ∞), Jensen’s inequality
implies
Z
Z
Z √
√
2
1+A =φ
h dµ ≤
φ(h) dµ =
1 + h2 dµ.
Ω
Ω
Ω
(If h has infinite values, then there are two cases: if h−1 (∞)
has positive measure, then all the relevant integrals are infinite
and the inequalities hold. On the other hand, if h−1 (∞) has
measure zero, then we may redefine h to be 0 on this set without affecting the integrals. Then Jensen’s inequality applies as
above.)
The proof of Jensen’s inequality relies simply on integrating
the pointwise inequality
Z
φ(t) − φ(s)
φ(h(x))−φ(t)−β(h(x)−t) ≥ 0, where t =
h dµ and β = sup
.
t−s
s<t
Ω
9
Thus, when the integrals in Jensen’s inequality are both finite,
they are equal if and only if the pointwise statement
φ(h(x)) − φ(t) − β(h(x) − t) = 0 a.e.
In our case φ is differentiable and β = φ0 (t). The condition is
then that
φ(h(x)) = φ(t) + φ0 (t)(h(x) − t) a.e.
This is the condition that φ(h(x)) is equal to the linear approximation of φ at t, evaluated at h(x). Since φ is strictly convex,
this is satisfied only when
Z
h(x) = t =
h dµ a.e.
Ω
Thus h is constant almost everywhere.
For h continuous, the condition on equality is just that h is
constant, and if h = f 0 , this is equivalent to f being a linear
function.
For the geometric interpretation
of the inequality if Ω = [0, 1]
R √
0
and h = f is continuous, then Ω 1 + h2 dµ is the arc-length
of the graph of f , A = f (1) − f (0) is the height of a right
triangle with hypotenuse from (0, f (0)) to (1, f (1)). 1 + A is
the
√ sum of the lengths of the two sides of this right triangle.
1 + A2 is the length of the hypotenuse.
Thus the left inequality is that the arc-length of the graph is
≥ the length of the secant line segment. We have proved above
that this inequality is an equality if and only if f is a linear
function.
The right inequality is that the sum of the lengths of the two
sides of the triangle from (0, f (0)) to (1, f (0)) to (1, f (1)) is ≥
the arc-length of the graph of f . This is an equality if and only
if f is constant.