chapter one
3. Prove that if f is a real function on a measurable space X such that
{x : f (x) ≥ r} is measurable for every rational r, then f is measurable.
[proof]:For each real number α, there exists an descending sequence {rn }
of rational numbers such that lim rn = α. Moreover,we have
n→∞
(α, +∞) =
∞
[
[rn , +∞).
∞
[
f −1 ([rn , +∞)).
n=1
Hence,
f
−1
((α, +∞)) =
n=1
Since sets f
−1
([rn , +∞)) are measurable for each n, the set f −1 ((α, +∞)) is
also measurable. Then f is measurable.
4. Let {an } and {bn } be sequences in [−∞, +∞], and prove the following
assertions:
(a)
lim sup(−an ) = − lim inf an .
n→∞
n→∞
(b)
lim sup(an + bn ) ≤ lim sup an + lim sup bn .
n→∞
n→∞
n→∞
provided none of the sums is of the form ∞ − ∞.
(c) If an ≤ bn for all n, then
lim inf an ≤ lim inf bn .
n→∞
n→∞
Show by an example that strict inequality can hold in (b).
[proof]: (a) Since
sup(−ak ) = − inf ak , n = 1, 2, · · · .
k≥n
k≥n
Therefore, let n → ∞, it obtains
lim sup(−ak ) = − lim inf ak .
n→∞ k≥n
n→∞ k≥n
1
By the definations of the upper and the lower limits, that is
lim sup(−an ) = − lim inf an .
n→∞
n→∞
(b) Since
sup(ak + bk ) ≤ sup ak + sup bk , n = 1, 2, · · · .
k≥n
k≥n
k≥n
Hence
lim sup(ak + bk ) ≤ lim [sup ak + sup bk ] = lim sup ak + lim sup bk .
n→∞ k≥n
n→∞ k≥n
n→∞ k≥n
k≥n
n→∞ k≥n
By the definations of the upper and the lower limits, that is
lim sup(an + bn ) ≤ lim sup an + lim sup bn .
n→∞
n→∞
n→∞
example: we define
an = (−1)n , bn = (−1)n+1 ,
n = 1, 2, · · · .
Then we have
an + bn = 0, n = 1, 2, · · · .
But
lim sup an = lim sup bn = 1.
n→∞
n→∞
(c)Because an ≤ bn for all n, then we have
inf (ak ) ≤ inf bk , n = 1, 2, · · · .
k≥n
k≥n
By the definations of the lower limits, it follows
lim inf an ≤ lim inf bn .
n→∞
n→∞
5. (a)Supose f : X → [−∞, +∞] and g : X → [−∞, +∞] are measurable.
Prove that the sets
{x : f (x) < g(x)}, {f (x) = g(x)}
are measurable.
(b)Prove that the set of points at which a sequence of measurable realvalued functions converges (to a finite limit) is measurable.
2
[proof]: (a)Since for each rational r, the set
{x : f (x) < r < g(x)} = {x : f (x) < r} ∩ {x : r < g(x)}
is measurable. And
{x : f (x) < g(x)} =
[
{x : f (x) < r < g(x)}
r∈Q
Therefore the set {x : f (x) < g(x)} is measurable. Also beacause |f − g| is
measurable function and the set
{x : f (x) = g(x)} =
∞
\
{x : |f (x) − g(x)| <
n=1
1
}
n
is also measurable.
(b) Let {fn (x)} be the sequence of measurable real-valued functions, A
be the set of points at which the sequence of measurable real-valued functions
{fn (x)} converges (to a finite limit). Hence
A=
∞ [
∞ \
∞
\
1
{x : |fn (x) − fm (x)| < }.
r
r=1 k=1 n,m≥k
Therefore, A is measurable.
7. Suppose that fn : X → [0, +∞] is measurable for n = 1, 2, 3, · · · , f1 ≥
f2 ≥ f3 ≥ · · · ≥ 0, fn (x) → f (x) as n → ∞,for every x ∈ Xand f1 ∈ L1 (µ).
Prove that then
lim
n→∞
Z
fn dµ =
X
Z
f dµ.
X
and show that this conclusion does not follow if the condition ”f1 ∈ L1 (µ)”is
omitted.
[proof]: Define gn = f1 −fn , n = 1, 2, · · ·, then gn is increasing measurable
on X for n = 1, 2, 3, · · ·. Moreover, gn (x) → f1 (x) − f (x) as n → ∞, for every
x ∈ X. By the Lebesgue’s monotone convergence theorem, we have
Z
Z
lim
gn dµ =
(f1 − f )dµ.
n→∞
X
Since f1 ∈ L1 (µ), it shows
lim
n→∞
X
Z
fn dµ =
X
3
Z
f dµ.
X
[counterexample]: Let X = (−∞, +∞), and we define
fn = χEn , En = [n, +∞), n = 1, 2, · · · .
8. Suppose that E is measurable subset of measure space (X, µ) with
µ(E) > 0; µ(X − E) > 0, then we can prove that the strict inequality in the
Fatou’s lemma can hold.
10. Suppose that µ(X) < ∞, {fn } is a sequence of bounded complex
measurable functions on X ,and fn → f uniformly on X. Proved that
Z
Z
lim
fn dµ =
f dµ,
n→∞
X
X
and show that the hypothesis ”µ(X) < ∞”cannot be omitted.
[proof]: There exist Mn > 0, n = 1, 2, · · · such that |fn (x)| ≤ Mn for
every x ∈ X, n = 1, 2, · · ·. Since fn → f uniformly on X, there exists M > 0
such that
|fn (x)| ≤ M, |f (x)| ≤ M on X, n = 1, 2, · · · .
By the Lebesgue’s dominated convergence theorem, we have
Z
Z
lim
fn dµ =
f dµ.
n→∞
X
X
[counterexample]: Let X = (−∞, +∞), and we define
1
fn = on X, n = 1, 2, · · · .
n
12. Suppose f ∈ L1 (µ). Proved that to each ε > 0 there exists a δ > 0
R
such that E |f |dµ < ε whenever µ(E) < δ.
[proof]: Since
Z
|f |dµ = sup
X
Z
sdµ,
X
the supremum being taken over all simple measurable functions s such that
0 ≤ s ≤ |f |. Hence, for each ε > 0, there exists a simple measurable function
s with 0 ≤ s ≤ |f | such that
Z
Z
ε
sdµ + .
2
X
X
Suppose that M > 0, satisfying 0 ≤ s(x) ≤ M on X, and let δ =
Z
Z
ε
ε
|f |dµ ≤
sdµ + ≤ M m(E) + < ε
2
2
E
E
when m(E) < δ.
|f |dµ ≤
4
ε
,
2M
we have
chapter two
1. Let {fn } be a sequence of real nonnegative functions on R1 , and consider the following four statements:
(a)If f1 and f2 are upper semicontinuous, then f1 + f2 is upper semicontinuous.
(b)If f1 and f2 are lower semicontinuous, then f1 + f2 is lower semicontinuous.
(c)If each {fn } is upper semicontinuous, then
∞
P
fn is upper semicontin-
1
uous.
(d)If each {fn } is lower semicontinuous, then
∞
P
fn is lower semicontinu-
1
ous.
Show that tree of these are true and that one is false. What happens if
the word ”nonnegative” is omitted? Is the truth of the statements affected if
R1 is replaced by a general topological space?
[proof]: (a) Because for every real number α,
[
{x : f1 (x) + f2 (x) < α} =
[{x : f1 (x) < r} ∩ {x : f2 (x) < α − r}]
r∈Q
and f1 and f2 are upper semicontinuous, therefore f1 + f2 is upper semicontinuous.
(b) Because for every real number α,
[
{x : f1 (x) + f2 (x) > α} =
[{x : f1 (x) > r} ∩ {x : f2 (x) > α − r}]
r∈Q
and f1 and f2 are upper semicontinuous, therefore f1 + f2 is upper semicontinuous.
(c) This conclusion is false. counterexample: for each n ∈ N, define
(
1; x = rn
fn =
0; others
1
which {rn } are the all rationals on R1 . Then {fn } is a sequence of real nonnegative upper semicontinuous functions on R1 , moreover we have
f (x) =
∞
X
fn (x) = χQ .
1
But f (x) is not upper semicontinuous.
(d) According to the conclusion of (b), it is easy to obtain.
If the word ”nonnegative” is omitted, (a) and (b) are still true but (c)
and (d) is not. The truth of the statements is not affected if R1 is replaced by
a general topological space.
2. Let f be an arbitrary complex function on R1 , and define
φ(x, δ) = sup{|f (s) − f (t)| : s, t ∈ (x − δ, x + δ)}
φ(x) = inf{φ(x, δ) : δ > 0}
Prove that φ is upper semicontinuous, that f is continuous at a point x if and
only if φ(x) = 0, and hence that the set of points of continuity of an arbitrary
complex function is a Gδ .
[proof]: It is obvious that
1
φ(x) = inf{φ(x, ) : k ∈ N} ≥ 0
k
and
1
1
), k ∈ N
φ(x, ) ≥ φ(x,
k
k+1
For any α > 0, we will prove that the set {x : φ(x) < α} is open. Suppose
that φ(x) < α, then there exists a N > 0 such that
1
φ(x, ) < α, k > N.
k
1
1
Take y ∈ U (x, 2k
), since U (y, 2k
) ⊂ U (x, k1 ) and then
φ(y) ≤ φ(y,
1
1
) ≤ φ(x, ) < α
2k
k
So we have y ∈ {x : φ(x) < α}. Therefore,
U (x,
1
) ⊂ {x : φ(x) < α}
2k
2
Thus, it show that {x : φ(x) < α} is open. Then it follows that φ is upper
semicontinuous. In the following, we first suppose that φ(x) = 0. for any
> 0, it exists δ > 0 such that
φ(x, δ) < And so for any y ∈ (x − δ, x + δ) we have
|f (x) − f (y)| ≤ φ(x, δ) < That is f is continuous at x. Next, provided that f is continuous at x. Then
for any > 0 it exists k ∈ N satifying
1
1
|f (x) − f (y)| ≤ , ∀y ∈ (x − , x + )
2
k
k
and so φ(x, k1 ) ≤ . It shows
1
φ(x) = inf{φ(x, ) : k ∈ N} = 0
k
Finally, since
{x : φ(x) = 0} =
∞
\
{x : φ(x) <
n=1
1
}
n
then it shows that the set of points of continuity of an arbitrary complex
function is a Gδ .
3. Let X be a metric space, with metric ρ. For any nonempty E ⊂ X,
define
ρE (x) = inf{ρ(x, y) : y ∈ E}
Show that ρE is a uniformly continuous function on X. If A and B are disjoint
nonempty closed subsets of X, examine the relevance of the function
f (x) =
ρA (x)
ρA (x) + ρB (x)
to Urysohn’s lemma.
[Proof]: Since for any x1 , x2 ∈ X and y ∈ E, we have
ρ(x1 , y) ≤ ρ(x1 , x2 ) + ρ(x2 , y)
Hence
ρE (x1 ) ≤ ρ(x1 , x2 ) + ρE (x2 )
3
and so
|ρE (x1 ) − ρE (x2 )| ≤ ρ(x1 , x2 )
Therefore, it shows that ρE is a uniformly continuous function on X. In the
following, suppose that K ⊂ V ⊂ X, K is compact subset, and V is open
subset. Define
ρV c (x)
ρV c (x) + ρK (x)
then 0 ≤ f ≤ 1 and f is continuous on X. Moreover, K ≺ f ≺ V .
f (x) =
4. Examine the proof of the Riesz theorem and prove the following two
statements:
(a) If E1 ⊂ V1 and E2 ⊂ V2 , where V1 and V2 are disjoint open sets, then
µ(E1 ∪ E2 ) = µ(E1 ) + µ(E2 ),
even if E1 and E2 are not in M.
(b) If E ∈ MF , then E = N ∪ K1 ∪ K2 ∪ · · ·, where {Ki } is a disjoint
countable collection of compact sets and µ(N ) = 0.
[proof]: (a) By the proof of the Riesz theorem, there exists a Gδ set G
such that
E1 ∪ E2 ⊂ G and µ(E1 ∪ E2 ) = µ(G).
Set G1 = V1 ∩ G, G2 = V2 ∩ G, then E1 ⊂ G1 , E2 ⊂ G2 and G1 ∩ G2 = ∅.
Therefore
µ(E1 )+µ(E2 ) ≤ µ(G1 )+µ(G2 ) = µ(G1 ∪G2 ) = µ(G∩(V1 ∪V2 )) ≤ µ(G) = µ(E1 ∪E2 ).
And since
µ(E1 ∪ E2 ) ≤ µ(E1 ) + µ(E2 ),
It shows
µ(E1 ∪ E2 ) = µ(E1 ) + µ(E2 ).
(b) Since E ∈ MF , there exist a sequence of disjoint compact subsets
{Kn } such that
Kn ⊂ E −
n−1
[
Ki , m(E −
i=0
n−1
[
Ki ) < m(Kn ) +
i=0
which we order K0 = ∅.
4
1
, n = 1, 2, · · · ,
n
Set K =
∞
S
Kn . Since m(E) < +∞, we obtain that
n=1
m(E −
n
[
Ki ) <
i=1
1
, n = 1, 2, · · · .
n
Whence we have
m(E − K) ≤ m(E −
n
[
Ki ) <
i=1
1
, n = 1, 2, · · · .
n
Set N = E − K, then it shows that m(N ) = 0 and E = K ∪ N .
5. Let E be Cantor’s familiar ”middle thirds” set. Show that m(E) = 0.
even though E and R1 have the same cardinality.
[proof]: According to the construction of Cantor’s set, it shows that the
set E is closed set and
m(E) = m([0, 1]) −
∞
X
2n−1
n=1
3n
= 0.
11. Let µ be a regular Borel measure on a compact Hausdorff space
X; assume µ(X) = 1. prove that there is a compact set K ⊂ X such that
µ(K) = 1 but µ(H) < 1 for every proper compact subset H of K.
[Proof]: Define the set
Ω = {Kα : Kα ⊂ X,
Kα is compact set with µ(kα ) = 1}
then X ∈ Ω. Order
K=
\
Kα
Kα ∈Ω
Thus K is compact subset. Assume that V ⊂ X is an open set with K ⊂ V .
Since X is compact, V c is also compact. Again since
[
Vc ⊂
Kαc
Kα ∈Ω
then there exist finite sets Kα1 , · · · , Kαn such that
c
V ⊂
n
[
k=1
5
Kαc k
Because µ(X) = 1 = µ(Kα ),
µ(Kαc k ) = 0,
k = 1, 2, · · · , n
and so µ(V c ) = 0. We therefore have µ(V ) = 1. Since µ is regular, hence it
shows that µ(K) = 1. By the construction of K, it follows that µ(H) < 1
for any proper compact subset of K. If U ⊂ X is open set with µ(U ) = 0, it
follows µ(U c ) = 1. U c is compact subset since X is compact, and so K ⊂ U c
by the construction of K. Hence, U ⊂ K c . It thus shows that K c is the largest
open set in X whose measure is 0.
12. Show that every compact subset of R1 is the support of a Borel
measure.
[Proof]: Assume that µ is the Lebesgue measure on R1 and K is a compact subset of R1 . Define
φ(f ) =
Z
f dµ;
f ∈ Cc (R1 ).
K
Then φ is a positive linear functional on Cc (R1 ). By Riesz representation
theorem, there exists a σ-algebra M in R1 which contains all Borel sets in R1 ,
and there exists a unique positive measure σ on M such that
Z
φ(f ) =
f dσ, f ∈ Cc (R1 ).
R1
By theorem2.14, it shows
σ(K) = inf{φ(f ) : K ≺ f } = µ(K)
σ(X) = sup{φ(f ) : f ≺ X} ≤ µ(K)
and so σ(X) = σ(K) = µ(K) < +∞. Therefore, it follows for any measurable
set A ⊂ X,
σ(A) = σ(A ∩ K).
That is, K is the support of σ.
14. Let f be a real-valued Lebesgue measurable function on Rk . Prove
that there exist Borel functions g and h such that g(x) = h(x) a.e.[m] and
g(x) ≤ f (x) ≤ h(x) for every x ∈ Rk .
6
[Proof]: Firstly, assume that 1 > f ≥ 0. By the proof of Theorem 2.24
in text, we have
f (x) =
∞
X
tn (x), ∀x ∈ Rk
n=1
n
and 2 tn is the characteristic function of a Lebesgue measurable set Tn ⊂
Rk (n ∈ N). According to the construction of Lebesgue measurable set, there
exist Borel measurable sets Fn , En (n ∈ N) satisfying
Fn ⊂ T n ⊂ E n ;
Define
g=
∞
X
m(En − Fn ) = 0;
2−n χFn ;
h=
n=1
∞
X
n ∈ N.
2−n χEn
n=1
then g, h are Borel measurable functions on Rk and g(x) ≤ f (x) ≤ h(x).
Moreover, g(x) = h(x) a.e.[m].
Secondly, it is easy that the conclusion is correct for 0 ≤ f < M , and
hence it is also for bounded real-valued Lebesgue measurable function on Rk .
Finally, if f be a real-valued Lebesgue measurable function on Rk and if
Bn = {x : |f (x)| ≤ n},
n = 1, 2, · · ·
then we have
χBn (x)f (x) → f (x), as n → ∞.
According to the conclusion above, there exist Borel functions gn and hn such
that gn (x) = hn (x) a.e.[m] and gn (x) ≤ χBn (x)f (x) ≤ hn (x) for every x ∈ Rk .
Take
g = lim sup gn ;
h = lim inf hn
then g(x) = h(x) a.e.[m] and g(x) ≤ f (x) ≤ h(x) for every x ∈ Rk .
15. It is easy to guess the limits of
Z n
Z n
x n x
x
2
(1 − ) e dx and
(1 + )n e−2x dx
n
n
0
0
as n → ∞. Prove that your guesses are correct.
[proof]: Define
fn (x) = χ[0,n] (1 −
x n x
) e2;
n
gn (x) = χ[0,n] (1 +
7
x n −2x
) e ; n∈N
n
Then it is easy that
x
x
|fn (x)| ≤ e− 2 , and fn (x) → e− 2 ;
|gn (x)| ≤ e−x , and fn (x) → e−x .
By the Lebesgue’s Dominated Convergence Theorem, it shows
Z n
Z ∞
Z ∞
x
x n x
lim
(1 − ) e 2 dx = lim
fn (x)dx =
e− 2 dx = 2
n→∞ 0
n→∞ 0
n
0
Z ∞
Z ∞
Z n
x
gn (x)dx =
e−x dx = 1
lim
(1 + )n e−2x dx = lim
n→∞
n→∞ 0
n
0
0
17. Define the distance between points (x1 , y1 ) and (x2 , y2 ) in the plane
to be
|y1 − y2 | if x1 = x2 ;
1 + |y1 − y2 | if x1 6= x2 .
show that this is indeed a metric, and that the resulting metric space X is
locally compact.
If f ∈ Cc (X), let x1 , x2 , · · · , xn be those values of x for which f (x, y) 6= 0
for at least one y (there are only finitely many such x!), and define
n Z +∞
X
Λf =
f (xj , y)dy
j=1
−∞
Let µ be the measure associated with this Λ by Theorem2.14. If E is the
x-axis, show that µ(E) = ∞ although µ(K) = 0 for every compact K ⊂ E.
[proof]:Order
Pi = (xi , yi ), i = 0, 1, 2, 3, · · · .
(
|y1 − y2 |
if x1 = x2 ;
d(P1 , P2 ) =
1 + |y1 − y2 | if x1 6= x2
Since
d(P1 , P2 ) ≤
(
|y1 − y3 | + |y2 − y3 |
if x1 = x2 ;
1 + |y1 − y3 | + |y2 − y3 | if x1 6= x2
Therefore,
d(P1 , P2 ) ≤ d(P1 , P3 ) + d(P2 , P3 ).
That is d is metric. Take X = (R2 , d). Then
Pn → P ⇔ ∃ N > 0, such that if n > N, xn = x; and|yn − y| → 0.
8
and for any 0 < δ < 1,
U (P0 , δ) = {P ∈ X : d(P0 , P ) < δ} = {P ∈ X : x = x0 , and |y − y0 | < δ}
U (P0 , δ) = {P ∈ X : d(P0 , P ) ≤ δ} = {P ∈ X : x = x0 , and |y − y0 | ≤ δ}
Moreover, U (P0 , δ) is compact subset of X. Thus X is locally compact Hausdorff space.
In the following, we order the set for f ∈ Cc (X)
Ωf = {x ∈ R : ∃ y ∈ R such that f (x, y) 6= 0}
Let K be the support of f , K is compact. Then there exists finite P1 , P2 , · · · , Pn ∈
K for any 0 < δ < 1 such that
K⊂
n
[
U (Pj , δ)
j=1
and so Ωf = {x1 , x2 , · · · , xn }. Since
Λf =
n Z
X
j=1
+∞
f (xj , y)dy
−∞
it is easy that Λ is a positive linear functional on Cc (X). There thus exist a
measure µ associated with this Λ by Theorem2.14. Let E be the x-axis and
0 < δ < 1. Define
V =
[
U (Pi , δ); where Pi = (xi , 0);
Pi ∈E
it shows V is open and E ⊂ V . For any finite elements P1 , P2 , · · · , Pn ∈ E, the
set
F =
n
[
δ
U (Pj , )
2
j=1
is compact with F ⊂ V . By the Urysohn’s lemma, there exists f ∈ Cc (X)
satisfying F ≺ f ≺ V . So we have
µ(V ) ≥ Λf ≥ nδ.
Hence, it shows µ(V ) = ∞ as n → ∞. Therefore, µ(E) = ∞. Assume that
K ⊂ E is compact, then it is obvious that K is finite set. Take
K = {P1 , P2 , · · · , Pm }
9
For any 0 < < 1, define
G=
m
[
U (Pj , )
j=1
then g is open and K ⊂ G. By the Urysohn’s lemma, there exists g ∈ Cc (X)
satisfying K ≺ g ≺ G. Since
µ(K) ≤ Λg ≤ 2
whence µ(K) = 0.
21. If X is compact and f : X → (−∞, +∞) is upper semicontinuous,
prove that f attains its maximum at some point of X.
[proof]: for every t ∈ f (X), define
Et = {x ∈ X : f (x) < t}
Assume that f does not attain its maximum at some point of X. Then it
S
shows X =
Et . Since f is continuous, Et is open sets. Again since X is
t∈f (X)
compact, there exists finite ti (i = 1, 2, · · · , n) ∈ f (X) such that X =
n
S
E ti .
i=1
Take
t0 = max{t1 , t2 , · · · , tn }
then we have
f (x) < t0 , ∀x ∈ X.
But it is in contradiction with t0 ∈ f (X). It therefore is proved that f attains
its maximum at some point of X.
10
chapter three
1. Prove that the supremum of any collection of convex functions on (a, b)
is convex on (a, b) and that pointwise limits of sequences of convex functions
are convex. What can you say about upper and lower limits of sequences of
convex functions?
[Proof]: Suppose that α ≥ 0, β ≥ 0 and α + β = 1. Let {fi }i∈I be any
collection of convex functions on (a, b). for any x, y ∈ (a, b), we have
fi (αx + βy) ≤ αfi (x) + βfi (y), ∀i ∈ I.
It prove that
sup fi (αx + βy) ≤ α sup fi (x) + β sup fi (y).
i∈I
i∈I
i∈I
So sup fi is convex.
i∈I
Assume I = N and f is pointwise limits of sequences of convex functions {fi }.
according to the properties of limit, we have
lim fi (αx + βy) ≤ α lim fi (x) + β lim fi (y)
i→∞
i→∞
i→∞
That is f (αx + βy) ≤ αf (x) + βf (y) and so f is convex.
By the definition of upper limits and the conclusion above, it is easy that the
upper limits of sequences of convex functions is convex. But the lower limits
of sequences of convex functions is false.
2. If φ is convex on (a, b) and if ψ is convex and nondecreasing on the
range of φ, prove that ψ ◦ φ is convex on (a, b). For φ > 0, show that the
convexity of ln φ implies the convexity of φ , but not vice versa.
[Proof]: Assume that
x, y ∈ (a, b); α, β ≥ 0and α + β = 1.
Then
ψ ◦ φ(αx + βy) ≤ ψ(αφ(x) + βφ(y)) ≤ αψ ◦ φ(x) + βψ ◦ φ(y).
1
and so ψ ◦ φ is convex on (a, b). Since et is nondecreasing convex on R and
φ = eln φ , it shows from the conclusion above that the convexity of ln φ implies
the convexity of φ for φ > 0.
[counterexample]: φ(x) = x2 is convex on (0, ∞) but ln φ(x) = 2 ln x is
not convex. Moreover, for φ > 0, it can show that the convexity of log c φ(c > 1)
implies the convexity of φ.
3. Assume that φ is a continuous real function on (a, b) such that
φ(
1
1
x+y
) ≤ φ(x) + φ(y)
2
2
2
for all x and y ∈ (a, b). Prove that φ is convex.
[Proof]: According to the definition of convex function,we only need to
prove the case 0 < λ < 12 . Take
E={
k
| k = 1, 2, · · · , 2n − 1; n ∈ N}.
2n
For n = 1, it is obvious that
φ(
x+y
1
1
) ≤ φ(x) + φ(y)
2
2
2
for all x and y ∈ (a, b).
For n = 2, suppose that λk = k4 , k = 1.
φ(λ1 x + (1 − λ1 )y) = φ( 12 ( x+y
+ y))
2
≤ 12 φ( x+y
) + 12 φ(y)
2
≤ 21 [ 12 φ(x) + 21 φ(y)] + 21 φ(y)
= 41 φ(x) + 43 φ(y)
= λ1 φ(x) + (1 − λ1 )φ(y)
Assume that the case
λ=
k
2n−1
, k = 1, 2, · · · , 2n−1 − 1
is correct. For any λ ∈ E with λ =
k
;1
2n
≤ k ≤ 2n−1 , n ≥ 2, and for all x and
y ∈ (a, b), it shows that
(i)
φ(λx + (1 − λ)y) = φ(
x+y
1
1
) ≤ φ(x) + φ(y) = λφ(x) + (1 − λ)φ(y)
2
2
2
2
when k = 2n−1 .
(ii)for 1 ≤ k < 2n−1 ,
λx + (1 − λy) =
Since
k
x
2n−1
k
k
1 k
k
x + (1 − n )y = ( n−1 x + (1 − n−1 )y + y)
n
2
2
2 2
2
k
)y
2n−1
+ (1 −
∈ (a, b) and so
k
x + (1 −
φ(λx + (1 − λ)y) ≤ 21 φ( 2n−1
≤
k
φ(x)
2n
+ 21 (1 −
k
)y)
2n−1
k
2n−1
+ 21 φ(y)
)φ(y) + 21 φ(y)
= λφ(x) + (1 − λ)φ(y)
For any 0 < λ < 1, there exists a sequence {λn } ⊂ E with lim λn = λ.
n→∞
Thus
φ(λx + (1 − λ)y) = lim φ(λn x + (1 − λn )y)
n→∞
≤ lim (λn φ(x) + (1 − λn )φ(y))
n→∞
= λφ(x) + (1 − λ)φ(y)
Therefore, it shows that φ is convex on (a, b).
4. Suppose f is a complex measurable function on X, µ is a positive
measure on X, and
φ(p) =
Z
|f |p dµ = kf kpp (0 < p < ∞).
X
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 ln φ is convex in the interior of E and that φ is continuous
on E.
(c) By (a), E is connected. Is E necessarily open? losed? Can E consist
of a single point? Can E be any connected subset of (0, ∞)?
(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).
(e) Assume that kf kr < ∞ for some r < ∞ and prove that
kf kp → kf k∞ as p → ∞.
[Proof]: (a) If r < p < s, there exist 0 < α < 1 and 0 < β < 1 such that
α + β = 1 and p = αr + βs. By Hölder’s inequality, it follows that
Z
Z
Z
Z
p
p
αr+βs
r
α
βs
kf kp = φ(p) =
|f | dµ =
|f |
dµ ≤ { |f | dµ} ·{ |f |s dµ}β = kf kαr
r ·kf ks .
X
X
X
3
X
As r ∈ E and s ∈ E, it prove that p ∈ E.
(b) By (a), it reduces that E is convex. Moreover, E is connected. Since
E ⊂ (0, ∞), the interior of E is an interval and denoted by (a, b). For any s, t
in the interior of E and 0 < α < 1 and 0 < β < 1, it follows from (a) that
φ(αs + βt) ≤ φ(s)α · φ(t)β .
So it proves
ln φ(αs + βt) ≤ ln[φ(s)α · φ(t)β ] = α ln φ(s) + β ln φ(t).
That is, ln φ is convex in the interior of E. Moreover, ln φ is continuous on
(a, b). Thus, φ is also continuous on (a, b). Provided that a ∈ E. For any
decreasing sequence {sn } of (a, b) with lim sn = a, it is easy to show that
n→∞
lim |f (x)|
sn
n→∞
= |f (x)|a , ∀x ∈ X.
Since |f (x)|sn ≤ max |f (x)|a , |f (x)|s1 , ∀x ∈ X and |f (x)|a , |f (x)|s1 ∈ L1 (µ),
by the Lebesgue control convergence theorem, it follows that
lim φ(sn ) = φ(a).
n→∞
Similarly, the other cases can be proved. Hence φ is continuous on E.
(c) By (a), it shows that E is connected and convex. E can be any interval
on (0, ∞). For examples: take X = (0, ∞) and µ is the Lebesgue measure on
X.
(i) For E = (a, b), (0 < a < b), take
(
1
x− b , x ∈ (0, 1);
f (x) =
1
x− a , x ∈ [1, ∞)
(ii)For E = [a, b], (0 < a < b), take
(
x,
f (x) =
1
x− a −1 ,
x ∈ (0, 1);
(iii)For E = [a, b), (0 < a < b), take
(
1
x− b ,
f (x) =
1
x− a −1 ,
x ∈ (0, 1);
4
x ∈ [1, ∞)
x ∈ [1, ∞)
(iv)For E = (a, b], (0 < a < b), take
(
1
x− b+1 ,
f (x) =
1
x− a ,
x ∈ (0, 1);
x ∈ [1, ∞)
(v)For E = ∅, take f ≡ 1.
(d) If r < p < s,by (a), there exist 0 < α < 1 and 0 < β < 1 such that
α + β = 1 and p = αr + βs. And it is easy to know
φ(p) = φ(αr + βs) ≤ φ(r)α · φ(s)β ≤ max φ(r), φ(s).
Thus, kf kp ≤ max kf kr , kf ks. Moreover, it shows that Lr (µ) ∩ Ls (µ) ⊂ Lp (µ).
(e)Firstly since kf k∞ > 0, it shows that kf kp > 0 for any 0 < p < ∞.
secondly, assume that kf k∞ = +∞. Hence,µ({x ∈ x : |f (x)| > 2}) > 0. For
any p > r,
p
0 < 2 µ({x ∈ x : |f (x)| > 2}) ≤
Z
|f |p dµ = kf kpp .
X
Therefore, lim kf kp = +∞.
n→∞
Thirdly, assume that kf k∞ = 1 and kf kr > 0. Without generality, assume
that |f (x)| ≤ 1, ∀x ∈ X. Thus For any p > r, it shows that
kf kpp ≤ kf krr .
r
That is kf kp ≤ kf krp . And so
lim kf kp = 1 = kf k∞ .
p→∞
The general case can reduce to the case kf k∞ = 1 and kf kr > 0. This is
completed the proof.
5. Assume, in addition to the hypotheses of Exercise 4, that
µ(X) = 1.
(a) Prove that kf kr ≤ kf ks , if 0 < r < s ≤ ∞.
(b) Under what conditions does it happen that 0 < r < s ≤ ∞ and kf kr =
kf ks < ∞?
(c) Prove that Lr (µ) ⊃ Ls (µ) if 0 < r < s. Under what conditions do these
5
two spaces contain the same functions?
(d) Assume that kf kr < ∞ for some r > 0, and prove that
Z
lim kf kp = exp{ log |f |dµ}
p→0
X
if exp{−∞} is defined to be 0.
[Proof]: (a) When s = ∞, it is easy that
Z
|f |r dµ ≤ kf kr∞ µ(X)
X
s
and so kf kr ≤ kf k∞ . If s < ∞, φ(x) = x r is convex on (0, ∞) and by the
Jensen’s Inequality, it shows
Z
Z
Z
s
s
r rs
|f | dµ =
(|f | ) dµ ≥ ( (|f |r )dµ) r
X
X
X
and whence kf ks ≥ kf kr .
(b) Assume that s = ∞, and kf kr = kf k∞ < ∞(0 < r < s). Then for
any 1 < p < ∞, it shows by (a) that
kf k1 = kf kp = kf kq < ∞
and so f is constant almost everywhere by the Holder’s inequality. Secondly,
assume 0 < r < s < ∞. Take α = rs > 1, then by the Holder’s inequality
Z
Z
r
r
|f | · 1dµ ≤ ( |f |s dµ) s = kf krs = kf krr
X
X
Therefore, |f |r = constant a.e. on X. So it happen that 0 < r < s ≤ ∞ and
kf kr = kf ks < ∞ if and only if f = constant a.e. on X.
(c)If 0 < r < s, then by (a) it is easy to prove Ls (µ) ⊂ Lr (µ). Moreover,
we will prove the following:
Lr (µ) = Ls (µ) ⇔ there exist finite pairwise disjoint measurable sets at most in
Ω = {E ⊂ X : µ(E) > 0}.
Firstly, assume that there are infinte pairwise disjoint members in Ω. Then it
exists a sequence {En } ⊂ X of pairwise disjoint measurable sets such that
0 < µ(En ) < 1, n = 1, 2, · · · .
6
Take
1
s
< t < 1r , and define
f=
∞
X
n−t χEn
n=1
Then
R
X
|f |sdµ =
=
≤
P∞
R P∞
−ts
χEn dµ
n=1 n
P∞ −ts
n µEn
Pn=1
∞
−ts
n=1 n
X
n−t < +∞ and so f ∈ Ls (µ). But
R
R P∞ −tr
r
|f
|
dµ
=
χEn dµ
n=1 n
X
X
P
∞
−tr
= n=1 n µEn
P
−tr
≥ ∞
· n−α
n=1 n
P∞
= n=1 n−(tr+α)
P
−(tr+α)
. Since 0 < α + tr < 1, it shows that ∞
= +∞.
which α = 1−tr
n=1 n
2
Since st > 1, it has
n=1
Hence, f ∈
/ Lr (µ).
Next, we assume that there exist finite pairwise disjoint measurable sets at
most in Ω. Provided that f ∈ Lr (µ), then there exists an increasing seqnence
{hn } of simple measurable functions such that
0 ≤ hn (x) ≤ hn+1 (x) ≤ · · · ≤ |f (x)|, f orallx ∈ X
and
hn (x) → |f (x)|, n → ∞, ∀x ∈ X.
Put
t1 (x) = h1 (x); tn (x) = hn (x) − hn−1 (x), n = 2, 3, · · ·
then 2n tn is the characteristic function of a measurable set Tn ⊂ X, and {Tn }
are pairwise disjoint. Moreover,
f (x) =
∞
X
tn (x), ∀x ∈ X.
n=1
According to the assumption, there are finite members of {Tn } with nonzero
measure at most. Thus f equal to a simple function almost everywhere and
so f ∈ Ls (µ). Therefore, Lr (µ) = Ls (µ).
7
(d) Assume that kf kr < ∞ for some r > 0, then
kf kp ≤ kf kr < +∞; kf kp ≥ e
R
X
ln |f |dµ
; 0 < p < r.
and {kf kp } is increasing relative to p. Therefore, the limit lim kf kp = c exists.
p→0
If c=0, it is easy that e
R
X
ln |f |dµ
= 0. In the following, we assume c > 0. Since
kf kr < +∞, we may suppose that f is finite everywhere. Let
E = [|f | > 1]; F = [|f | < 1]; G = [|f | = 1]
So we have
kf kp = (
R
R
R
1
(|f |p − 1 + 1)dµ + F (|f |p − 1 + 1)dµ + G (|f |p − 1 + 1)dµ) p
R
R
1
= (1 + E (|f |p − 1)dµ + F (|f |p − 1)dµ) p
=(
Take α > 1;
R
1
|f |pdµ) p
X
E
0 < β < 1, because
x−1
= 1;
x→1 ln x
lim
|f (x)|p → 1, p → 0
there exists δ > 0 such that
|f (x)|p − 1 < α ln |f (x)|p , ∀x ∈ E; |f (x)|p − 1 < β ln |f (x)|p , ∀x ∈ F
when p < δ. Thus
R
1
p
p
p
α
ln
|f
|
dµ
+
β
ln
|f
|
dµ)
E
F
R
R
1
= (1 + p(α E ln |f |dµ + β F ln |f |dµ) p
kf kp ≤ (1 +
Then
R
0 < c = lim kf kp ≤ eα
p→0
R
E
ln |f |dµ+β
R
F
ln |f |dµ
Let
α → 1+ ; β → 1−
then
c = lim kf kp ≤ e
p→0
R
E
ln |f |dµ+
R
F
ln |f |dµ
Hence
lim kf kp = e
p→0
8
R
X
ln |f |dµ
.
=e
R
X
ln |f |dµ
.
11. Suppose µ(Ω) = 1, and suppose f and g are positive measurable
functions on Ω such that f g ≥ 1. Prove that
Z
Z
f dµ · gdµ ≥ 1.
Ω
Ω
[Proof]: Since f and g are positive measurable functions on Ω such that
1
1
1
f g ≥ 1, f 2 and g 2 are positive measurable functions on Ω such that (f g) 2 ≥ 1.
By Hölder’s inequality, it follows that
Z
Z
Z
1
2
2
f dµ · gdµ.
1 ≤ ( (f g) dµ) ≤
ω
Ω
Ω
12. Suppose µ(Ω) = 1 and h : Ω → [0, ∞] is measurable. If
Z
A=
hdµ,
Ω
prove that
√
1 + A2 ≤
Z √
1 + h2 dµ ≤ 1 + A.
Ω
[Proof]: It is obvious correct when A = ∞ or 0 and so we assume 0 <
√
A < +∞ and h : Ω → (0, ∞). Since φ(x) = 1 + x2 is convex on (0, ∞), it
shows by the Jensen’s Inequality that
Z √
√
1 + A2 ≤
1 + h2 dµ.
Ω
Because
√
1 + h2 ≤ 1 + h,
Z √
1 + h2 dµ ≤ 1 + A.
Ω
This complete the proof.
14. Suppose 1 < p < ∞, f ∈ Lp = Lp ((0, ∞)), relative to Lebesgue
measure and
1
F (x) =
x
Z
x
f (t)dt (0 < x < ∞).
0
(a) Prove Hardy’s inequality
kF kp ≤
p
kf kp
p−1
which shows that the mapping f → F carries Lp into Lp .
9
(b) Prove that equality holds only if f = 0a.e.
(c) Prove that the constant p/(p − 1) cannot be replaced by a smaller one
.
(d) If f > 0 and f ∈ L1 , prove that F ∈
/ L1 .
[Proof]: (a) Firstly, assume that f ≥ 0 and f ∈ Cc ((0, ∞)). Then
Z
1 x
F (x) =
f (t)dt (0 < x < ∞)
x 0
is differential on (0, ∞) and so xF 0 (x) = f (x) − F (x), ∀x ∈ (0, ∞). Therefore
R∞ p
R ∞ p−1
F (x)dx = F p (x)x |∞
(x)F 0 (x)dx
0 −p 0 F
0
R ∞ p−1
= −p 0 F (x)xF 0 (x)dx
R∞
= −p 0 F p−1 (x)(f (x) − F (x))dx
R∞
R∞
= −p 0 F p−1 (x)f (x)dx + p 0 F p (x)dx
Thus
(p−1)
Z
∞
p
F (x)dx = p
0
Z
∞
F
p−1
(x)f (x)dx ≤ p(
0
and so
kF kp ≤
Z
∞
F
(p−1)q
1
q
(x)dx) (
0
Z
∞
0
p
kf kp .
p−1
Moreover, it is easy that
kF kp ≤
p
kf kp , ∀f ∈ Cc ((0, ∞)).
p−1
Given f ∈ Lp ((0, ∞)), since Cc ((0, ∞)) is dense in Lp ((0, ∞)), there exist a
sequence {fn } ⊂ Cc ((0, ∞)) such that
kfn − f kp → 0, as n → ∞.
That is, for any > 0, there exists N > 0 such that kfn − f kp < if n > N .
It shows by the Jensen’s Inequality that
Rx
|Fn (x) − F (x)| ≤ x1 0 |fn (t) − f (t)|dt)
Rx
1
1
≤ x1 ( 0 |fn (t) − f (t)|p dt) p · x q
1
≤ x− p kfn − f kp
and thus
Fn (x) → F (x), n → ∞, ∀x ∈ (0, ∞).
10
1
f p (x)dx) p
Since
p
kfn kp , n = 1, 2, · · ·
p−1
it shows that Fn ∈ Lp and {Fn } ⊂ Lp is Cauchy sequence. By the completion
kFn kp ≤
of Lp , it exists G(x) ∈ Lp satisfying
kFn − Gkp → 0, n → ∞
and so kFn kp → kGkp < +∞. By the Fatou’s Lemma,
kF kp ≤ lim kFn kp = kGkp < +∞
n→∞
whence F ∈ Lp . Moreover,
p
p
kfn kp =
kf kp .
n→∞ p − 1
p−1
kF kp ≤ kGkp = lim kFn kp ≤ lim
n→∞
Therefore, the map f → F is continuous from Lp (0, ∞) to Lp (0, ∞).
(b) Firstly, assume that f ≥ 0 and f ∈ Cc (0, ∞). Then, by (a), we have
R ∞ p−1
R∞ p
p
F (x)dx = p−1
F (x)f (x)dx
0
R0 ∞ (p−1)q
1 R∞
1
p
≤ p−1 ( 0 F
(x)dx) q ( 0 f p (x)dx) p
R∞
1
1 R∞
p
≤ p−1
( 0 F p (x)dx) q ( 0 f p (x)dx) p
and so
kF kp =
p
kf kp
p−1
⇔ ∃ α, β ∈ R and αβ 6= 0 satisfying αF p = βf p a.e.
⇔ αF = βf a.e.
.
⇔ f = 0 a.e.
Therefore, it shows the conclusion f = 0 a.e. if f ∈ Cc (0, ∞) and kF kp =
p
kf kp .
p−1
(c)
R∞
(d)Suppose that f > 0 and f ∈ L1 , then G =
Z
exists N > 0 such that
when x > N . Thus
Z ∞
Z
F (x)dx ≥
0
∞
N
x
f (t)dt >
0
1
x
Z
x
0
f dµ < +∞ and so it
G
2
G
f (t)dt ≥
2
11
0
Z
∞
N
1
dx = +∞
x
and therefore F ∈
/ L1 .
18. Let µ be a positive measure on X. A sequence {fn } of complex
measurable functions on X is said to converge in measure to the measurable
function f if to every ε > 0 there corresponds an N such that
µ({x : |fn (x) − f (x)| > ε}) < ε
for all n > N . (This notion is of importance in probability theory.) Assume
µ(X) < ∞ and prove the following statements:
(a) If fn (x) → f (x)a.e., then fn → f in measure.
(b) If fn ∈ Lp (µ) and kfn − f kp → 0, then fn → f in measure; here
1 ≤ p ≤ ∞.
(c) If fn → f in measure, then {fn } has a subsequence which converges
to f a.e.
Investigate the converses of (a) and (b). What happen to (a), (b),and (c)
if µ(X) = ∞, for instance, if µ is Lebesgue measure on R1 ?
[Proof]: (a) Set E = {x ∈ X : lim fn (x) 6= f (x)}, then µ(E) = 0. For
n→∞
every ε > 0, we define
Ek = {x ∈ X : |fn (x) − f (x)| > ε}, k = 1, 2, · · · ;
Fn =
∞
[
Ek , n = 1, 2, · · · ; F =
∞
\
Fn .
n=1
k≥n
It is easy that F ⊂ E and so µ(F ) = 0. Since {Fn } is a sequence of decreasing
measurable sets and µ(X) < +∞, it shows that
lim µ(Fn ) = µ(F ) = 0.
n→∞
As En ⊂ Fn , n = 1, 2, · · ·, it is trival that
lim µ(En ) = 0.
n→∞
It is then proved that fn → f in measure.
(b) For every ε > 0, suppose that
Ek = {x ∈ X : |fn (x) − f (x)| > ε}, k = 1, 2, · · · .
Then it follows that for 1 ≤ p < ∞,
Z
p
ε µ(En ) ≤
|fn − f |p dµ = kfn − f kpp .
X
12
Since kfn − f kp → 0, it shows that
lim µ(En ) = 0.
n→∞
It is then proved that fn → f in measure. If p = +∞ and kfn − f kp → 0,
there exists a national N > 0 satisfying
kfn − f kp < ε, whenever n > N.
Hence it shows
µ(En ) = 0, whenever n > N.
It is then proved that fn → f in measure.
(c)Suppose that fn → f in measure. For every national k,there exists
national nk such that
µ(Ek ) <
1
, k = 1, 2, · · · .
2k
where Ek = {x ∈ X : |fnk (x) − f (x)| >
Define
F =
1
}.
2k
∞ [
∞
\
Ek
n=1 k≥n
then we have
µ(F ) ≤ µ(
∞
[
Ek ) ≤
k≥n
∞
X
µ(Ek ) =
k=n
1
2n−1
, n = 1, 2, · · · .
It is obvious that µ(F ) = 0. For any x ∈ X − F , it follows that there exists
national N such that
|fnk (x) − f (x)| ≤
1
, for each national k > N.
2k
Therefore, we obtain
lim fnk (x) = f (x).
k→∞
It then shows that {fn } has a subsequence which converges to f a.e.
20. Suppose φ is a real function on R such that
Z 1
Z 1
φ(
f (x)dx) ≤
φ(f )dx
0
0
for every real bounded measurable f . Prove that φ is then convex.
13
[Proof]: Assume that
x, y ∈ R;
Then
λx + (1 − λ)y =
Thus
Z
0 < λ < 1.
1
(xχ[0,λ] (t) + yχ[λ,1] )(t)dt.
0
R1
φ(λx + (1 − λ)y) = φ( 0 (xχ[0,λ] (t) + yχ[λ,1] )(t)dt)
R1
≤ 0 φ ◦ (xχ[0,λ] (t) + yχ[λ,1] )(t)dt
= λφ(x) + (1 − λ)φ(y).
Therefore, φ is convex.
14
chapter four
In this set of exercises, H always denotes a Hilbert space.
1. If M is a closed subspace of H, prove that M = (M ⊥ )⊥ . Is there a
similar true statement for subspaces M which are not necessarily closed?
[Proof]: It is easy M ⊂ (M ⊥ )⊥ . on the other hand, for any z ∈ (M ⊥ )⊥ ,
it exist x ∈ M and y ∈ M ⊥ with z = x + y when M is a closed subspace of
H. Thus,
0 = (z, y) = (x, y) + (y, y) = (y, y)
and so y = 0. Therefore, z = x ∈ M . It thus follows that M ⊃ (M ⊥ )⊥ and
whence M = (M ⊥ )⊥ .
Next, assume that M is not closed subspace. Since M ⊂ (M ⊥ )⊥ and
(M ⊥ )⊥ is closed, it is obvious that M ⊂ (M ⊥ )⊥ . since M ⊂ M , it shows
M
⊥
⊥
⊥
⊂ M ⊥ . Moreover, (M ⊥ )⊥ ⊂ (M )⊥ . Again since M = (M )⊥ , it is
proved that M = (M ⊥ )⊥ .
2. Let {xn : n = 1, 2, 3, · · ·} be a linearly independent set of vectors in
H. Show that the following construction yields an orthonormal set {un } such
that {x1 , x2 , · · · , xN } and {u1 , u2 , · · · , uN } have the same span for all N .
[Proof]: Let {xn : n = 1, 2, 3, · · ·} be a linearly independent set of vectors
in H. Put
X
vn
x1
; vn = xn −
(xn , uk )uk ; un =
; n = 2, 3, · · ·
u1 =
kx1 k
kvn k
k=1
n−1
Then it is easy that
(v2 , u1 ) = 0; (u2 , u1 ) = 0
and so (vn , uk ) = 0, k = 1, 2, · · · , n − 1. Therefore, {un } is an orthonormal set.
Moreover,
span{xn } = span{un }.
3. Show that Lp (T ) is separable if 1 ≤ p < ∞, but that L∞ (T ) is not
separable.
1
[Proof]: Firstly, assume that 1 ≤ p < ∞, then C(T ) is dense in Lp (T ).
Let P be the set of all trigonometric polynomials and P̃ be the set of all
trigonometric polynomials with rational coefficients, it is obvious that P is
dense in C(T )and P̃ is also. Since
kf − gkp ≤ kf − gk∞ , ∀f, g ∈ C(T )
and so it shows that P̃ is countable dense subset in Lp (T ). This proves that
Lp (T ) is separable.
Next, we prove that L∞ (T ) is not separable. Assume that L∞ (T ) is
separable and let {un } is the countable dense subset of L∞ (T ). Put
ft (s) = χ[0,t] (s); 0 < t < 2π
and extend it to the real axis R with period 2π. Thus,
ft ∈ L∞ (T ) and kft k∞ = 1.
Take 0 < < 21 , according to the assumption, it exists some uk satisfying
U (uk , ) ∩ {un } is infinite set.
Put ft , ft0 ∈ U (uk , ); 0 < t < t0 ≤ 2π, then
1 = kft − ft0 k∞ ≤ kft − uk k∞ + kft0 − uk k∞ < 2 < 1
It is contradiction. Therefore, L∞ (T ) is not separable.
4. Show that H is separable if and only if H contains a maximal orthonormal system which is at most countable.
[Proof]: First, there exists a countable dense subset M of H if H is
separable. Let N is the maximal linearly independent subset of M . By the
conclusion of Exercise 2, it shows the existence of a maximal orthonormal
system of H which is at most separable.
Second, suppose that {un } is a countabe maximal orthonormal system of
H. Put M = span{un }. If M 6= H, ther exists an element v ∈ H and v ∈
/ M.
Furthermore,
v=
∞
X
(v, un )un ∈ M ;
n=1
2
v − v 6= 0.
Let u0 =
v−v
,
kv−vk
then
ku0 k = 1; (u0 , un ) = 0, n = 1, 2, · · · .
Therefore, {u0 , u1 , u2 , · · ·} ⊂ H is a orthonormal set but this contradict the
maximality of {u1 , u2 , · · ·} ⊂ H. It thus follows that M = H. So H is
separable.
5. If M = {x : Lx = 0}, where L is a continuous linear functional on H,
prove that M ⊥ is a vector space of dimension 1(unless M = H).
[Proof]: If L = 0, then M = H and M ⊥ = {0}. in the following, assume
that L 6= 0. Then it exists unique element y 6= 0 of H such that
Lx = (x, y); ∀x ∈ H.
Thus, y ⊥ M and so M = span{y}⊥ . That is, M ⊥ = span{y} since M is a
closed subspace by the continuity of L. This complete the proof.
P
7. Suppose that {an } is a sequence of positive numbers such that an bn <
P 2
P 2
∞ whenever bn ≥ 0 and
bn < ∞. Prove that
an < ∞.
∞
P
[Proof]: Assume that
a2n = +∞. For any k ∈ N, it then exists nk ∈ N
such that
n=1
nk
X
a2n > k; 1 < n1 < n2 < n3 < · · ·
n=1
Hence
X
nk+1
c2k
=
a2n > 1; k = 1, 2, · · · .
nk +1
Put
bn =
Then
(
∞
X
n=1
but
∞
X
b2n
an
;
kck
nk ≤ n < nk+1 , k = 1, 2, · · ·
0;
1 ≤ n < n1
=
X 1
a2n
=
< +∞
2
2c
2
k
k
k
+1
k=1
nk+1
∞
X
X
k=1 n=nk
∞
nk+1
∞
X
X
∞
X
a2n
ck
a n bn =
=
= +∞
kck
k
n=1
k=1 n=nk +1
k=1
P 2
Therefore, the assumption is error and
an < ∞.
3
9. If A ⊂ [0, 2π] and A is measurable, prove that
Z
Z
lim
cos nxdx = lim
sin nxdx = 0.
n→∞
A
n→∞
A
[Proof]: If A ⊂ [0, 2π] and A is measurable, then χA ∈ L2 ([0, 2π]). Since
L2 ([0, 2π]) is Hilbert space and {eint : n ∈ Z} is the maximal orthonormal set
of L2 ([0, 2π]), and so we have
X
|(χA , eint )|2 = kχA k2 < +∞
n∈Z
Therefore, it shows that
|(χA , eint )| → 0, |n| → ∞
Furthermore, it follows that
Z
Z
sin nxdx = 0.
cos nxdx = lim
lim
n→∞
A
n→∞
A
16. If x0 ∈ H and M is a closed linear subspace of H, prove that
min{kx − x0 k : x ∈ M } = max{|(x0 , y)| : y ∈ M ⊥ , kyk = 1}.
[Proof]: Suppose that P and Q are the projections on M , M ⊥ respectively. For x0 ∈ H, it shows that x0 = P x0 + Qx0 and
kQx0 k = min{kx − x0 k : x ∈ M }.
On the other hand, for any y ∈ M ⊥ with kyk = 1, then
(x0 , y) = (P x0 + Qx0 , y) = (Qx0 , y)
and whence
|(x0 , y)| = |(Qx0 , y)| ≤ kQx0 k · kyk = kQx0 k.
Since Qx0 ∈ M ⊥ , we may put y0 =
Qx0
kQx0 k
when Qx0 6= 0. Then (x0 , y0 ) =
kQx0 k. Thus
max{|(x0 , y)| : y ∈ M ⊥ , kyk = 1} = kQx0 k.
This complet the proof.
4
chapter five
2. Prove that the unit ball(open or closed) is convex in every normed
linear space.
[Proof]: Let X be a normed linear space and S be the open nuit ball of
X. Assume that 0 ≤ α, β ≤ 1 and α + β = 1. For any x, y ∈ S,
kαx + βyk ≤ kαxk + kβyk = αkxk + βkyk < α + β = 1
and so αx + βy ∈ S. It shows that S is convex.
4. Let C be the space of all continuous functions on [0, 1], with the
supremum norm. Let M consist of all f ∈ C for which
Z 1
Z 1
2
f (t)dt −
f (t)dt = 1.
1
2
0
Prove that M is closed convex subset of C which contains no element of minimal norm.
[Proof]: By use of the Lebesgue dominated convergence theorem, it is
easy that M is closed convex set of C. Moreover, for any f ∈ M ,
Z 1
Z 1
Z 1
2
f (t)dt −
1=
f (t)dt ≤
|f (t)|dt = kf k1 ≤ kf k∞ .
1
2
0
0
If f ∈ C and kf k∞ = 1, it holds that
Z 1
Z 1
|f (t)|dt = 1;
(1 − |f (t)|)dt = 0; 1 − |f (t)| ≥ 0, ∀t ∈ [0, 1]
0
0
and so |f (t)| = 1, ∀t ∈ [0, 1]. But in this case, f ∈
/ M . Therefore, M is closed
convex subset of C which contains no element of minimal norm.
5. Let M be the set of all f ∈ L1 ([0, 1]), relative to Lebesgue measure,
such that
Z
1
f (t)dt = 1.
0
Show that M is a closed convex subset of L1 ([0, 1]) which contains infinitely
many elements of minimal norm.
1
[Proof]: Assume that 0 ≤ α, β ≤ 1 and α + β = 1. For any f, g ∈ M ⊂
1
L ([0, 1]),
Z
1
[αf + βg](t)dt = α
0
Z
1
f (t)dt + β
0
Z
1
g(t)dt = α + β = 1
0
whence αf + βg ∈ M . So M is convex. Next, given
{fn } ⊂ M ; and kfn − f k1 → 0, n → ∞.
Then
Z
1
fn (t)dt −
0
Z
1
f (t)dt =
0
and so
lim
n→∞
Z
Z
1
≤
0
Z
1
|fn (t) − f (t)|dt = kfn − f k1 → 0
0
1
fn (t)dt =
0
Z
1
f (t).
0
It follows that f ∈ M since {fn } ⊂ M and so M is closed convex subset.
Moreover,
kf k1 ≥ 1, ∀f ∈ M.
Since
ntn−1 ∈ M, ∀n ∈ N; and kntn−1 k1 = 1
it shows that M contains infinitely many elements of minimal norm.
6. Let f be a bounded linear functional on a subspace M of a Hilbert
space H. Prove that f has a unique norm-preserving extension to a bounded
linear functional on H, and that this extension vanishes on M ⊥ .
[Proof]: If f = 0, we only need to take the bounded linear functional
F = 0 on H. In the following, we assume f 6= 0. Since f is a bounded
linear functional on a subspace M , it is easy to extend f to a bounded linear
functional f˜ on M and kf k = kf˜k. In fact, for any x ∈ M , there exists a
sequence {xn } ⊂ M such that kxn −xk → 0 and then define f˜(x) = lim f (xn ).
n→∞
So we can define a linear functional of H such that
(
f˜(x); x ∈ M
F (x) =
0;
x ∈ M⊥
and then it is proved that F is the norm-preserving linear extension on H of f
and this extension vanishes on M ⊥ . Moreover, for any norm-preserving linear
2
⊥
extension g on H of f , g |M = f˜. Since M ⊥ = M and H = M ⊕ M ⊥ , it shows
that it is unique that the norm-preserving extension of f to a bounded linear
functional on H which vanishes on M ⊥ .
8. Let X be a normed linear space, and let X ∗ be its dual space, as defined
in Sec. 5.21, with the norm
kf k = sup{|f (x)| : kxk ≤ 1}.
(a)Prove that X ∗ is a Banach space.
(b) Prove that the mapping f → f (x) is, for each x ∈ X, a bounded linear
functional on X ∗ , of norm kxk.
(c) Prove that {kxn k} is bounded if {xn } is a sequence in X such that {f (xn )}
is bounded for every f ∈ X ∗ .
[Proof]: (a) It is obvious that X ∗ is a normed linear space. Given {fn } ⊂
X ∗ is a cauchy sequence, then for any > 0, there is a N > 0 such that
kfn − fm k < ; ∀n, m > N
and so
|fn (x) − fm (x)| ≤ kfn − fm k · kxk < kxk · ; ∀n, m > N, ∀x ∈ X.
Thus {fn (x)} ⊂ C is also a cauchy sequence and so we can define
f (x) = lim fn (x); ∀x ∈ X.
n→∞
It is easy that f is a linear functional on X. Since {kfn k} is a cauchy sequence,
ther is M > 0 satisfying
kfn k ≤ M, n = 1, 2, · · · .
Hence, for , N as above, it shows for n > N the following
|f (x)| ≤ |fn (x)| + kxk ≤ kxk(M + ); ∀x ∈ X.
It forces
|f (x)| ≤ M kxk; ∀x ∈ X
and so kf k ≤ M . That is, f ∈ X ∗ . Therefore, X ∗ is a Banach space.
3
(b) For each x ∈ X, define
λ(f ) = f (x); ∀f ∈ X ∗
Then for any f, g ∈ X ∗ and α, β ∈ C,
Λ(αf + βg) = (αf + βg)(x) = αf (x) + βg(x) = αΛ(f ) + βΛ(g)
and we see that Λ is a linear functional on X ∗ . Since
|Λ(f )| = |f (x)| ≤ kf k · kxk; ∀f ∈ X ∗
whence kΛk ≤ kxk. From the Hahn-Banach Theorem, there is g ∈ X ∗ such
that g(x) = kxk and kgk = 1. It implies that kΛk = kxk.
(c) Suppose that {xn } ⊂ X such that {f (xn )} is bounded for every f ∈
∗
X . By (a),(b) and the Banach-Steinhaus Theorem, it is easy to prove that
{kxn k} is bounded.
9. Let c0 , l1 , l∞ be the Banach spaces consisting of all complex sequences
x = {ξi }, i = 1, 2, · · ·, define as follows:
x ∈ l1 if and only ifkxk1 =
X
|ξi | < ∞.
x ∈ L∞ if and only ifkxk∞ = sup |ξi | < ∞.
c0 is the subspace of l∞ consisting of all x ∈ l∞ for which ξi → 0 as i → ∞.
Prove the following four statements.
P
(a) If y = {ηi } ∈ l1 and Λx =
ξi ηi for every x ∈ c0 , then Λ is a bounded
linear functional on c0 , and kΛk = kyk1 . Moreover, every Λ ∈ (c0 )∗ is obtained
in this way. In brief, (c0 )∗ = l1 .
(b) In the same sense, (l1 )∗ = l∞ .
(c) Every y ∈ l1 induces a bounded linear functional on l ∞ , as in (a). However,
this does not give all of (l∞ )∗ , since (l∞ )∗ contains nontrivial functionals that
vanish on all of c0 .
(d) c0 and l1 are separable but l∞ is not.
[Proof]: (a) For each y = {ηi } ∈ l1 , we define
Λx =
∞
X
ξi ηi ; ∀x = {ξi } ∈ c0 .
i=1
4
Then the above series converges absolutely and so Λ is linear mapping. Since
|Λx| ≤ kxk∞ · kyk1
it follows that kΛk ≤ kyk1 . Thus, Λ ∈ (c0 )∗ . On the other hand, put
i
z }| {
ei = {0, 0, · · · , 1, 0, 0, · · · , }; i = 1, 2, · · · .
It is obvious that ei ∈ c0 and kei k∞ = 1, i = 1, 2, · · ·. For any Λ ∈ (c0 )∗ ,take
ηi = Λei , i = 1, 2, · · · .
Then
|ηi | = |Λei | ≤ kΛk · kei k∞ = kΛk.
Let y = {ηi } and then y ∈ l∞ with kyk∞ ≤ kΛk. Assume xn = {ξin }, where
(
ηi
; ηi 6= 0, i = 1, 2, · · · , n;
n
|ηi |
ξi =
0;
ηi = 0, or i > n.
then xn ∈ c0 with kxn k∞ = 1, n = 1, 2, · · ·. Moreover,
Λxn =
n
X
|ηi | ≤ kΛk · kxn k∞ = kΛk
i=1
whence
∞
X
|ηi | ≤ kΛk
i=1
Thus y ∈ l1 and kyk1 ≤ kΛk. Since for any x = {ξi } ∈ c0 , x =
so Λx =
∞
P
∞
P
ξi ei and
i=1
ξi ηi . From the proof above, it implies that kΛk ≤ kyk1 and so
i=1
kΛk = kyk1 . Therefore, (c0 )∗ = l1 .
(b) For each y = {ηi } ∈ l∞ , define
Λx =
∞
X
ξi ηi ; ∀x = {ξi } ∈ l1 .
i=1
Then the above series converges absolutely and so Λ is linear mapping. Since
|Λx| ≤ kyk∞ · kxk1
5
it shows that kΛk ≤ kyk∞ . Thus, Λ ∈ (l1 )∗ . On the other hand, suppose that
Λ ∈ (l1 )∗ . Put
ηi = Λei , i = 1, 2, · · · .
Then
|ηi | = |Λei | ≤ kΛk · kei k1 = kΛk.
Let y = {ηi } and then y ∈ l∞ with kyk∞ ≤ kΛk. Since for any x = {ξi } ∈ l1 ,
∞
∞
P
P
x=
ξi ei and so Λx =
ξi ηi . It implies that kΛk ≤ kyk∞ and so kΛk =
i=1
i=1
kyk∞. Therefore, (l1 )∗ = l∞ .
(c) For each y = {ηi } ∈ l∞ , define
Λx =
∞
X
ξi ηi ; ∀x = {ξi } ∈ l∞ .
i=1
Then the above series converges absolutely and so Λ is linear mapping. Since
|Λx| ≤ kxk∞ · kyk1
it shows that kΛk ≤ kyk1 . Thus, Λ ∈ (l∞ )∗ . Since c0 ⊂ l∞ is closed subspace,
by the Hahn-Banach theorem, there is f ∈ (l ∞ )∗ such that f |c0 = 0. By (a),
(c0 )∗ = l1 and so f ∈
/ l1 . Therefore, l1 6= (l∞ )∗ .
(d) Since
l1
c0
span{ei } = l1
span{ei } = c0 ;
it is easy to prove that c0 , l1 are separable.
P
10. If αi ξi converges for every sequence (ξi ) such that ξi → 0 as i → ∞,
P
prove that
|αi | < ∞.
[Proof]: Put
x = {αi }; xn = {α1 , α2 , · · · , αn , 0, 0, · · ·}; ξ = {ξn } ∈ c0 .
and define
ρ(ξ) =
∞
X
αi ξi ; ρn (ξ) =
n
X
αi ξ i .
i=1
i=1
Then ρ is linear and ρn (n ∈ N) is bounded linear functionals on c0 . Moreover,
kρn k =
n
X
|αi |, n ∈ N.
i=1
6
Since
lim ρn (ξ) = ρ(ξ); ∀ξ ∈ c0
n→∞
exists, by the Banach-Steinhaus Theorem, there is M > 0 satisfying
kρn k ≤ M ; n ∈ N
That is,
It thus forces that
n
X
|αi | ≤ M, n ∈ N.
i=1
P∞
i=1 |αi | < +∞.
11. For 0 < α ≤ 1, let Lipα denote the space of all complex functions f
on [a, b] for which
Mf = sup
s6=t
|f (s) − f (t)|
< ∞.
|s − t|α
Prove that Lipα is a Banach space, if kf k = |f (a)| + Mf ; also, if
kf k = Mf + sup |f (x)|.
x∈[a,b]
[Proof]: It is easy to prove that Lipα is a normed linear space with the
corresponding norm. Let {fn } be a cauchy sequence of Lipα.
(a) Assume
kf k = |f (a)| + Mf .
There thus exists G > 0 such that Mfn ≤ G. For any > 0, there is N > 0
such that
|fn (a) − fm (a)| < ; Mfn −fm < when n, m > N . For every s ∈ (a, b],
|fn (s)−fm (s)|
|s−a|α
≤
≤
|fn (a)−fm (a)|
n (a)+fm (a)|
+ |fn (s)−fm (s)−f
|s−a|α
|s−a|α
m (a)|
Mfn −fm + |fn (a)−f
|s−a|α
It implies that {fn (s)} is cauchy sequence for each s ∈ [a, b]. Therefore, we
can define
lim fn (s) = f (s); ∀s ∈ [a, b].
n→∞
Since for any s, t ∈ [a, b] and s 6= t, for each n ∈ N,
|f (s)−f (t)|
|s−t|α
≤
≤
|fn (s)−fn (t)|
(s)−f (s)|
(t)−f (t)|
+ |fn |s−t|
+ |fn|s−t|
α
α
|s−t|α
|fn (s)−f (s)|
|fn (t)−f (t)|
+ |s−t|α + G
|s−t|α
7
Let n → ∞, then
sup
s6=t
|f (s) − f (t)|
≤ G < +∞.
|s − t|α
That is, f ∈ Lipα and so Lipα is a Banach space.
(b) Assume
kf k = Mf + sup |f (x)|.
x∈[a,b]
There thus exists G > 0 such that Mfn ≤ G. Put kf k∞ = supx∈[a,b] |f (x)| and
so {kfn k∞ } is cauchy sequence. Then it exists a complex function f on [a, b]
such that
sup |fn (x) − f (x)| → 0; n → 0.
x∈[a,b]
Moreover, for any s, t ∈ [a, b] and s 6= t, for each n ∈ N,
|f (s)−f (t)|
|s−t|α
≤
≤
|fn (s)−fn (t)|
(s)−f (s)|
(t)−f (t)|
+ |fn |s−t|
+ |fn|s−t|
α
α
|s−t|α
|fn (s)−f (s)|
|fn (t)−f (t)|
+ |s−t|α + G
|s−t|α
Let n → ∞, then
sup
s6=t
|f (s) − f (t)|
≤ G < +∞.
|s − t|α
That is, f ∈ Lipα and so Lipα is a Banach space.
16. Suppose X and Y are Banach spaces, and suppose Λ is a linear
mapping of X into Y , with the following property: For every sequence {xn }
in X for which x = lim xn and y = lim Λxn exist, it is true that y = Λx. Prove
that Λ is continuous.
[Proof]: Provided that X, Y are Banach spaces. Define
X ⊕ Y = {(x, y) | x ∈ X, y ∈ Y }
The addition and scalar multiplication on X ⊕ Y are difined in the obvious
manner as follows:
(x1 , y1 )+(x2 , y2 ) = (x1 +x2 , y1 +y2 ); α(x, y) = (αx, αy); ∀x, xi ∈ X, y, yi ∈ Y, i = 1, 2.
If put
k(x, y)k = kxk + kyk; ∀x ∈ X, y ∈ Y
8
it then implies that X ⊕ Y is normed linear space. Given {(xn , yn )} ⊂ X ⊕ Y
is cauchy sequence, and so {xn } ⊂ X and {yn } ⊂ Y are also cauchy sequences.
Therefore, it exist x ∈ X and y ∈ Y such that
kxn − xk → 0; kyn − yk → 0; if n → ∞.
Hence,
k(xn , yn ) − (x, y)k = k(xn − x, yn − y)k = kxn − xk + kyn − yk → 0; n → ∞.
This implies that X ⊕ Y is Banach space. Put
G = {(x, Λx) | x ∈ X}
it is easy that G is a subspace of X⊕Y . Assume that {(xn , Λxn )} ⊂ G is cauchy
sequence, and then {xn } ⊂ x and {Λxn } ⊂ Y are aslo cauchy sequences. So
there exist x ∈ X and y ∈ Y such that
kxn − xk → 0; kΛxn − yk → 0; n → ∞.
According to the property of Λ, it holds that Λx = y and (x, Λx) ∈ G. It thus
shows that G is closed and so G is a Banach space. Take T (x, Λx) = x for
each x ∈ x, then T is linear mapping and
kT (x, Λx)k = kxk ≤ kxk + kΛxk = k(x, Λx)k; ∀x ∈ X
whence kT k ≤ 1. Moreover, T is injective and surjective bounded linear
mapping. By the open mapping theorem, it show that T −1 is continuous and
so there is M > 0 such that kT −1 k ≤ M . Thus
kxk + kΛxk = k(x, Λx)k = kT −1 xk ≤ M kxk; ∀x ∈ X
and so
kΛxk ≤ M kxk; ∀x ∈ X.
It follows that Λ is continuous.
17. If µ is a positive measure, each f ∈ L∞ (µ) defines a pultiplication
operator Mf on L2 (µ) into L2 (µ), such that Mf g = f g. Prove that kMf k ≤
kf k∞ . For which measure µ is it true that kMf k = kf k∞ for all f ∈ L∞ (µ)?
For which f ∈ L∞ (µ) does Mf map L2 (µ) onto L2 (µ)?
9
[Proof]: For each f ∈ L∞ (µ), define
Mf g = f g; ∀g ∈ L2 (µ)
Then it is easy that Mf is a linear operator from L2 (µ) into L2 (µ). Furthermore,
kMf gk2 = kf gk2 ≤ kf k∞ · kgk2 ; ∀g ∈ L2 (µ)
Thus it shows that kMf k ≤ kf k∞ .
If f ∈ L∞ (µ) with
1
f
∈ L∞ (µ), then Mf maps L2 (µ) onto L2 (µ).
18. Suppose {Λn } is a sequence of bounded linear transformations from
a normed linear space X to a Banach space Y , suppose kΛn k ≤ M < ∞ for
all n, and suppose there is a dense set E ⊂ X such that {Λn x} converges for
each x ∈ E. Prove that {Λn x} converges for each x ∈ X.
[Proof]: For every x ∈ X, there is a sequence {xk } ⊂ E such that
kxk − xk → 0(k → ∞) since E ⊂ X is a dense subset. Moreover, {Λn xk } ⊂ Y
converges for each k ∈ N. It thus exists a sequence {yk } ⊂ Y such that
Λn xk → yk , n → ∞; k = 1, 2, · · · .
For any > 0, there exists N > 0 such that
kxk − xk0 k <
; ∀k, k 0 > N.
M
and so
kyk − yk0 k = lim kΛn (xk − xk0 )k ≤ lim kΛn k · kxk − xk0 k ≤ M ·
n→∞
n→∞
=
M
It hence implies that {yk } ⊂ Y is a cauchy sequence. Since Y is a Banach
space, there is y ∈ Y with kyk − yk → 0. Therefore,
kΛn x − yk = kΛn x − Λn xk + Λn xk − yk + yk − yk
≤ kΛn k · kxk − xk + kΛn xk − yk k + kyk − yk
≤ M · kxk − xk + kΛn xk − yk k + kyk − yk
This implies that lim Λn x = y.
n→∞
10
chapter six
2. Prove that the example given at the end of Sec. 6.10 has stated
properties.
[Proof]:
3. Prove that the vector space M (X) of all complex regular Borel measures on a locally compact hausdorff space X is a Banach space if kµk = |µ|(X).
[Proof]: Firstly, assume µ, λ ∈ M (X) and α ∈ C. Then αµ ∈ M (X) and
|µ|, |λ| are regular Borel measures. For any measurable set E in X and > 0,
it exist compact sets K1 , K2 ⊂ E and open sets G1 , G2 ⊃ E satisfying:
|µ|(E) < |µ|(K1 ) + ; |λ(E)| < |λ|(K2 ) +
2
2
|µ|(G1 ) < |µ|(E) + ; |λ|(G2 ) < |λ|(E) +
2
2
Take
K = K 1 ∪ K2 ; g = G 1 ∩ G 2
it implies that Kis compact and G is open. Moreover,
(|µ| + |λ|)(E) < (|µ| + |λ|)(K) + ; (|µ| + |λ|)(G) < (|µ| + |λ|)(E) + and so |µ|+|λ| is regular Norel measure. Since |µ+λ| and |µ|+|λ| are bounded
positive measures and
|µ + λ| ≤ |µ| + |λ|
it shows that |µ + λ| is regular Borel measure. In fact, we have the following
conclusion:
Suppose that µ is bounded positive measure, then
µ is regular ⇔ for any measurable set E and > 0, there are compact subset
K ⊂ E and open set G ⊃ E such that
µ(E − K) < ; µ(G − E) < .
Therefore, it shows that M (X) is a linear space. Define
kµk = |µ|(X), ∀µ ∈ M (X)
1
it is easy to show that kµk is norm and M (X) is normed linear space. By
Theorem 6.19, it is obtained that M (X) ∼
= (C0 (X))∗ . Since (C0 (X))∗ is a
Banach space and so M (X) is also.
4. Suppose 1 ≤ p ≤ ∞, and q is the exponent conjugate to p. Suppose µ is
a positive σ-finite measure and g is a measurable function such that f g ∈ L1 (µ)
for every f ∈ Lp (µ). Prove that then g ∈ Lq (µ).
[Proof]: Since µ is a positive σ-finite measure, there is a sequence of
disjoint measurable sets {Xn } of X such that
X=
∞
[
Xn ;
µ(Xn ) < +∞, n = 1, 2, · · · .
n=1
It thus shows that
χXn ∈ Lp (µ), n = 1, 2, · · · .
Therefore,
g ∈ L1 (Xn , µ), n = 1, 2, · · · .
It implies that g is finite almost everywhere on X and so we may assume that
g is finite on X. Let
En = {x ∈ X | |g(x)| ≤ n}, n = 1, 2, · · · .
Then {En } is a sequence of increasing measurable sets and X =
gn = χEn · g and then {gn } are measurable on X. Moreover,
S∞
n=1
En . Put
(a) gn (x) → g(x); ∀x ∈ X.
(b) |gn (x)| ≤ n; ∀x ∈ X, n = 1, 2, · · ·.
(c) gn ∈ Lq (µ), n = 1, 2, · · ·.
Define
Tn (f ) =
Z
f gn dµ; ∀f ∈ Lp (µ)
X
it is easy to prove that {Tn } is bounded linear functionals on Lp (µ) with
kTn k = kgn kq ; n = 1, 2, · · · .
Since
|Tn (f )| ≤
Z
|f gn |dµ =
X
Z
|f g|dµ ≤
En
2
Z
|f g|dµ ≤ kf gk1 ; n = 1, 2, · · ·
X
By use of the Banach-Steinhaus Theorem, there is M > 0 such that
kTn k ≤ M ; n = 1, 2, · · · .
It hence implies that
kgn kq ≤ M ; n = 1, 2, · · · .
S
If q = ∞, it shows from X = ∞
n=1 En that there is n > 0 for any x ∈ X with
x ∈ En and so
|g(x)| = |gn (x)| ≤ kgn k∞ ≤ M.
it shows that kgk∞ ≤ M and g ∈ L∞ (µ).
In the case 1 ≤ q < ∞, because
|gn (x)| ≤ |gn+1 (x)|, n = 1, 2, · · · ; |gn (x)| → |g(x)|, ∀x ∈ X
By the Lebesgue’s monotone convergence Theorem, it shows
Z
Z
q
|gn | dµ →
|g|q dµ, n → ∞
x
X
and so
(
Z
1
|g|q dµ) q ≤ M.
X
q
Therefore, kgkq ≤ M and g ∈ L (µ).
6. Sppose 1 < p < ∞ and prove that Lq (µ) is the dual space of Lp (µ)
even if µ is not σ-finite.
[Proof]: Sppose 1 < p < ∞ and µ is not σ-finite.
(a) Since for any f ∈ Lp (µ) and g ∈ Lq (µ), it is easy that f · g ∈ L1 (µ). Then
we can define
Tg (f ) =
p
Z
f · gdµ, ∀f ∈ Lq (µ)
x
∗
and Tg ∈ (L (µ)) with kTg k ≤ kgkq . Assume
En = {x ∈ X : |g| ≥
1
}, n = 1, 2, · · · .
n
It is obvious that {En } is an increasing sequence of measurable sets and
∞
[
En = E = {x ∈ X : |g| > 0}.
n=1
3
Put
fn = χEn |g|q−1 α
where αg = |g|; |α| = 1. Then fn ∈ Lp (µ) and
Z
1
kfn kp = (
|g|q dµ) p ; n = 1, 2, · · · .
En
Moreover, {|fn |} converges increasingly to |g|q−1 . Therefore,
Z
Z
Tg (fn ) =
fn · gdµ =
|g|q dµ; n = 1, 2, · · · .
X
Z
and so
En
q
|g| dµ ≤ kTg k · kfn kp = kTg k · (
En
Z
It shows that
Z
1
|g|q dµ) p
En
|g|q dµ ≤ kTg kq ; n = 1, 2, · · · .
En
By the Lebesgue’s monotone convergence Theorem, it shows that
Z
Z
Z
q
|g| dµ =
|fn | · |g|dµ →
|g|q dµ; n → ∞.
Hence,
R
En
X
X
X
|g|q dµ ≤ kTg kq and so kgkq = kTg k.
(b) Assume Φ ∈ (Lp (µ))∗ and kΦk 6= 0. At first, for each f ∈ Lp (µ),
define
1
Df = {x ∈ X : f (x) 6= 0}; En = {x ∈ X : |f (x)| ≥ }.
n
S∞
Since Df = n=1 En , and
Z
Z
1 p
p
µ(En )( ) ≤
|f | dµ ≤
|f |pdµ < +∞
n
En
X
it thus follows that µ(En ) < +∞, n = 1, 2, · · ·; and so Df is σ-finite measurable set.
Secondly, given E ⊂ X is σ-finite measurable set, and put
M = {f ∈ Lp (µ) : Df ⊂ E}
then M ⊂ Lp (µ) is a Banach subspace. In fact, provided that {fn } ⊂ M is a
cauchy sequence, there is f ∈ Lp (µ) such that
kfn − f kp → 0, n → ∞.
4
So it exists a subsequence of {fn } ⊂ M which converges to f almost everywhere. It implies that f ∈ M . Therefore, Φ |M is a bounded linear functional
on M . Since it also that M ⊂ Lp (E, µ), by the Hahn-Banach Theorem, it can
extend Φ |M to a bounded linear functional Φ |E of Lp (E, µ) and so there is
unique hE ∈ Lq (E, µ) such that
Z
Φ(f ) =
f · hE dµ, ∀f ∈ M ; khE kq = kΦ |E k = kΦ |M k ≤ kΦk.
E
Take gE = hE on E and gE = 0 when x ∈ X − E. It is easy that
gE ∈ Lq (µ); DgE ⊂ E; kgE kq = khE kq ≤ kΦk
and
Φ(f ) =
Z
f · hE dµ =
E
Z
f · gE dµ; ∀f ∈ M.
X
Order
α = sup{kgE kq : E ⊂ X is σ − finite measurable subset}
it is obvious that 0 ≤ α ≤ kΦk. Because there is a sequence {fn } ⊂ Lp (µ)
satisfying:
kfn kp = 1, n = 1, 2, · · · ; |Φ(fn )| → kΦk, n → ∞.
Take F =
S∞
n=1
Dfn , then F is a σ-finite measurable set. Moreover,
|Φ(fn )| ≤ kΦ |F k = kgF kq ≤ α ≤ kΦk, n = 1, 2, · · · .
Therefore, it holds that kΦk = kgF kq = α. Write g = gF , then g ∈ Lq (µ) and
kΦk = kgkq .
Thirdly, assume that A ⊂ X is σ-finite measurable set and write B = A ∪ F ,
B is also σ-finite measurable set. Then
kΦk = α = kgF kq ≤ kgB kq ≤ α ≤ kΦk.
That is,
kΦk = kgF kq = kgB kq .
Since Lp (F, µ) ⊂ Lp (B, µ), it show sthat hB = hF on F almost everywhere
5
and so gB = gF a.e. on F . Again since
R
kΦkq = X |gB |q dµ
R
= B |gB |q dµ
R
R
= A−F |gB |q dµ + F |gB |q dµ
R
R
= A−F |gB |q dµ + F |gF |q dµ
R
≥ F |gF |q dµ
= kΦkq
Hence, it holds that
R
A−F
|gB |q dµ = 0 and so gB = 0 a.e. on A − F . It implies
that g = gF = gB a.e. on X.
finally, from the conclusion above, it shows that
Z
Z
Φ(f ) =
gDf ∪F f dµ =
gf dµ.
X
p
∗
X
q
Therefore, (L (µ)) = L (µ).
6