Math 516
Professor Lieberman
February 14, 2005
HOMEWORK #4 SOLUTIONS
Chapter 10
38. (a) If xn → x weakly, then f (xn ) → f (x) for any f ∈ X ∗ . It follows that {ϕ[xn ]} is a
family of bounded linear operators from the Banach space X ∗ to the normed linear space
R such that, for each f ∈ X ∗ , there is a number Mf with |ϕ[xn ](f )| ≤ Mf for all xn . (In
other words, a convergent sequence is bounded.) The uniform boundedness principle then
implies that there is a number M such that kϕ[xn ]k ≤ M for all n. But ϕ is an isometry,
so kxn k ≤ M for all n.
(b) =⇒: From part (a) hkxn ki is bounded. In addition, for each integer n, the function fn ,
defined by fn (hξi i) = ξn is clearly linear and
p 1/p
|ξn | = (|ξn | )
≤
∞
X
!1/p
p
|ξi |
= khξi ik.
i=1
It follows that fn is a bounded linear function on lp and hence ξn,m = fn (xm ) → fn (x) = ξm .
⇐=: From Fatou’s Lemma, we know that kxk ≤ M , where M is any number such that
hkxn ki ≤ M . Next, we fix a bounded linear functional f on lp . By the Riesz Representation
P
Theorem, there is a sequence η ∈ lq (q = p/(p − 1)) such that f (y) =
yn ηn for all
y = hyn i ∈ lp . So, for any positive integer N , we have
∞
X
|f (xn ) − f (x)| = (ξm,n − ξm )ηm m=1
!1/q
!1/p N
N
X
X
|ηm |q
|ξm,n − ξm |p
≤
m=1
m=1
∞
X
+
!1/p
|ξm,n − ξm |p
m=N +1
≤
N
X
!1/q
∞
X
|ηm |q
m=N +1
!1/p
|ξm,n − ξn |p
kηkq + 2M
m=1
∞
X
!1/q
|ηm |q
.
m=N +1
P∞
q 1/q < ε/(4M + 1), and then J so
Given ε > 0, we can choose N so that
m=N +1 |ηm |
that |ξm,n − ξm | ≤ ε/(2N 1/p kηkq ) for n ≥ J and m ≤ N . Hence, for n ≥ J, we have
|f (xn ) − f (x)| < ε. Since f and ε are arbitrary, it follows that hxn i converges to x.
(c) Let f be a bounded linear functional on Lp . Then the Riesz Representation Theorem
gives a function g ∈ Lq (with q = p/(p − 1)) such that
Z
f (x) = xg
1
2
for all x ∈ Lp . The solution of Problem 6.17 given on the web last semester shows that
Z
Z
xg = lim xn g
for any g ∈ Lq (because it only needed that xn → x in measure) and hence xn → x weakly.
(d) As in Problem 6.17, let
Z
f (x) = x
R
(so f (x) = xg for g ≡ 1) and set xn = nχ(0,1/n) . Then kxn k1 = 1 and xn to0 in measure,
but f (xn ) = 1 for all n while f (x) = 0.
(e) If n 6= m, then kxn − xm k = 21/p , so hxn i is not a Cauchy sequence and hence it doesn’t
converge in the strong topology. On the other hand, kxn kp = 1 for all n and ξm,n → 0 as
n → ∞ for each fixed m, so x → 0 weakly by part (b).
(f) To show that the distance between any two points in F is at least 1, we consider two
cases. Let the two points be yn,m and yν,µ . First, if n = ν, then
yn,m − yν,µ = n(xm − xµ ),
so the distance between these two points is (2n)1/p ≥ 1. Second, if n 6= ν, then without
loss of generality, we may assume that n < ν. Then the n-th term of yn,m − yν,µ must be
1, so the distance between the points is at least 1. Hence any nonconstant sequence in F
isn’t Cauchy and therefore can’t converge. It follows that F is strongly closed.
(g) To show that 0 is a weak closure point of F , we need to show that, for any ε > 0 and
any finite number of functions f1 , . . . , fN in (lp )∗ , there is a choice of n and m such that
|fi (yn,m )| < ε for all i. To proceed, we write zi = hζm,i i for the sequence in lq associated
with fi . The n-th term test for convergence of sequences shows that there is an integer n
such that |ζn,i | < ε/2 for all i and (from this n) a integer m such that |ζm,i | < ε/(2n) for
all i. It follows that |fi (yn,m )| < ε for this choice of n and m.
To show that there is no sequence hzj i from F that converges weakly to zero, we write
zj = hζj,k i with ζj,k = yn(j,k),m(j,k) . If {n(j, k)} is bounded, then there is a an increasing
sequence of positive integers hj(i)i such that n(j(i), k) is constant, say equal to ν. It
follows that ζj(i),ν = 1 for all i and hence ζj,ν can’t converge to zero, so the sequence
doesn’t converge weakly to zero by part (b).
(h) No one actually gave a good solution, so I didn’t grade this one. The answer is rather
tricky. If hxn i is a sequence that converges weakly in l1 to zero, then it converges strongly!
To prove this, suppose to the contrary that hkxn k1 i does not converge to zero. Then there
is a subsequence hxnj i and a positive number ε so that kxnj k1 > ε for all j. Let us set
yj = xnj and write yj = hηm,j i. For any sequence z = hζm i in l∞ , we have
∞
∞
X
1 X
ηm,nj ζm ≤ ξm,nj ζm ,
ε
m=1
m=1
and hence hyj i converges to
with kyj k1 = 1 for all j. Since ky1 k = 1, there is a positive
Pzero
m1
integer m1 > 1 such that m=1 |ηm,1 | > 3/4. From the proof of part (b), for each m, the
sequence
P 1of numbers hηm,j i converges to zero. Hence there is a positive integer n1 > 1 such
that m
m=1 |ηm,n1 | < 1/4. Then, there is a positive integer m2 > m1 such that
m2
X
m=m1 +1
3
|ηm,n1 | > ,
4
3
and a positive integer n2 > n1 such that
m2
X
m=1
1
|ηm,n2 | < .
4
We now proceed by induction to obtain two increasing sequences of positive integers hmj i
and hnj i so that
mk+1
X
3
|ηm,nk | > ,
4
m=mk +1
and
mk+1
X
m=1
1
|ηm,nk+1 | < .
4
Now we consider the sequence z = hζm i ∈ l∞ with ζm = sgn xm,nk−1 for m in the interval
(mk−1 , mk . Then
∞
X
mk+1
X
ηm,nk ζm =
m=1
|ηm,nk | +
mk
X
m=mk +1
m=1
mk+1
mk
X
X
≥
|ηm,nk | −
m=mk +1
X
m=mk +1
∞
X
ηm,nk ζm
m=mk+1 +1
|ηm,nk | −
m=1
mk+1
=2
ηm,nk ζm +
∞
X
|ηm,nk |
m=mk+1 +1
3
1
|ηm,nk | − kynk k1 > 2
−1= .
4
2
Hence
lim
k→∞
and therefore
lim
n→∞
∞
X
ηm,nk ζm 6= 0
m=1
∞
X
ηm,n ζm 6= 0,
m=1
which contradicts the weak convergence of hxn i to zero.
Now, we turn to weak∗ convergence. Since the functionals fn from part (b) come from
sequences in c0 , it follows that, if a sequence hxn i in l1 converges weak∗ to 0, then hkxn k1 i
is bounded and ξm,n → 0 for each m. Conversely, suppose hxn i is a sequence in l1 such
that hkxn k1 i is bounded and ξm,n → 0 for each m. We want to show that
∞
X
ξm,n ηm → 0
m=1
as n → ∞ for any y = hηm i in c0 , so let y ∈ c0 and ε > 0 be given. Also, suppose that M is
a positive number such that kxn k1 ≤ M for all n. It follows that |ξn,m | ≤ M for all n and
m. Then there is a positive integer N such that |ηm | ≤ ε/(2M ) for all m ≥ N . It follows
that
∞
N
X
X
ε
|
ξm,n ηm | ≤
|ξm,n ηm | +
2
m=1
m=1
4
P
for all n. Now set M1 = N
m=1 |ξm | and choose N1 so large that |ξm,n | < ε/(2M1 + 1) for
n ≤ N and m ≥ N1 . It follows that
∞
X
|
ξm,n ηm | ≤ ε
m=1
for m ≥ N1 , which implies that hxn i converges weak∗ to zero.
43. Let x and y be points in K and let N be an open neighborhood of 0. By Proposition 16,
there is an open neighborhood U of 0 such that U + U ⊂ N and αU ⊂ U if 0 < α ≤ 1.
Then there are points x0 and y 0 in K such that x − x0 and y − y 0 are elements of U . It
follows that
(λx + (1 − λ)y) − (λx0 + (1 − λ)y 0 )
= λ(x − x0 ) + (1 − λ)(y − y 0 ) ∈ λU + (1 − λ)U ⊂ U + U ⊂ N
for any λ ∈ (0, 1). It follows that λx + (1 − λ)y ∈ K and hence K is convex.