Math 516
Professor Lieberman
January 31, 2005
HOMEWORK #2 SOLUTIONS
Chapter 10
20. First, we let f be the linear functional defined on c (the space of all sequences with limits)
by f (hxn i) = lim xn , define p on l∞ by p(hxn i) = lim sup xn , and let G be the semigroup of
linear operators of the form A such that there is a positive integer M such that Ahxn i =
hxn+M i. (Of course, the integer M will be different for different elements of G.) Note that
f (Ax) = f (x) and p(Ax) = p(x) and that f (x) = p(x) for all x ∈ c. Proposition 10.5 then
implies that there is a linear functional F on l∞ which extends f . The second inequality
in Property i then follows because F ≤ p while the first inequality follows because
F [hξn i] = −F [h−ξn i] ≥ − lim sup(−ξn ) = lim inf ξn .
Properties ii and iii are just the linearity of F and Property iv is the statement F (Ax) =
F (x) for any A ∈ G and any x ∈ l∞ .
21. Let X be the vector space of all bounded functions defined on R which are zero outside of
some finite interval, and let S be the subspace of X consisting of all measurable functions
which are elements of X. We then define the linear functional f on S by
Z
f (x) = x(t) dt,
and we define p on X by setting
Z
p(x) = inf{
y : y ∈ L∞ , x ≤ y}.
R
(For brevity, we write x for this infimum, in analogy to the upper Riemann integral.)
Finally, we write G for the semigroup of all translation operators, so A ∈ G if there is a
number s such that Ax(t) = x(t + s). It’s easy to check that the hypotheses of the HahnBanach theorem are satisfied in this case. In particular,Rthe subadditivity of p follows by
letting ε > 0 and taking y1 and y2 so that xi ≤ yi and yi ≤ p(xi ) + ε/2 for i = 1, 2. It
follows that
Z
p(x1 + x2 ) ≤ y1 + y2 ≤ p(x1 ) + p(x2 ) + ε
and hence p(x1 + x2 ) ≤ p(x1 ) + p(x2 ). Hence, there is a linear functional F defined on X
which is an extension of f . Let us also note that
Z
Z
F (x) = −F (−x) ≥ − − x(t) dt = x(t) dt.
Hence, if x is nonnegative, then so is F (x).
We now define µ by µ(A) = F (χA ). We now check the claimed properties for µ. First,
if A ∩ B = ∅, then χA∪B = χA + χB , so
µ(A ∪ B) = F (χA ) + F (χB ) = µ(A) + µ(B).
1
2
Next, property ii. is a consequence of the invariance of F under the semigroup G. Furthermore, if A ⊂ B, then we can write B = A ∪ C with C = B ∼ A, and we see that A ∩ C = ∅.
Therefore,
µB = µA + µC ≥ µA
because χC is nonnegative and we Rjust showed that µC = F (χC ) ≥ 0. Finally, if A is
Lebesgue measurable, then µ(A) = χA , which is just the Lebesgue measure of A.
22. First, write ϕ for the natural isomorphism of X into X ∗∗ and ϕ∗ for the natural isomorphism
of X ∗ into X ∗∗∗ . If X is reflexive, let y ∈ X ∗∗∗ and define ψ : X ∗∗∗ → X ∗ by (ψy)(x) =
y(ϕx). But, for any x ∈ X, we have (ϕ∗ (ψy))(ϕx) = (ϕx)(ψy) = (ψy)(x), so y = ϕ∗ (ψy).
Hence the image of ϕ∗ is all of X ∗∗∗ and therefore X ∗ is reflexive.
If X is not reflexive, then there is x∗∗ ∈ X ∗∗ ∼ ϕ(X). Since X is a Banach space,
ϕ(X) is closed, and hence there is a positive number δ such that d(x∗∗ , ϕ(X)) ≥ δ. Then
Proposition 6 shows that there is z ∈ X ∗∗ such that z(ϕx) = 0 for all x ∈ X but z(x∗∗ ) = δ.
If y ∈ X ∗ , then (ϕ∗ y)(ϕx) = (ϕx)(y) = y(x), so, if (ϕ∗ y)(ϕx) = 0 for all x ∈ X, it follows
that y = 0. Hence z cannot be ϕ∗ y for any y ∈ X ∗ , which means that X ∗ is also not
reflexive.
26. Take T to be the family of all operators Tn . For each x, hTn xi is a convergent sequence, so it’s
bounded. Proposition 11 then implies that there is a constant M such that kTn xk ≤ M kxk
for all x ∈ X and all n. Sending n → ∞, we see that kT xk ≤ M kxk and hence T is
bounded.
29. (a) (Ignore the hint). Let X be the vector space C 1 [0, 1] with norm kxk = sup x, let Y
0
be the vector space C[0, 1] with norm kyk = sup y, and
pdefine Ax = x . Then X is not a
Banach space: The sequence hxn i defined by x+n(t) = x + 1/n converges uniformly to x,
defined by x(t), but xn ∈ X and x ∈
/ X. This sequence also shows that A is unbounded. But
if hxn i is a convergent sequence in X such that hAxn i converges in Y , then hx0n i converges
uniformly to some continuous function y and hxn i converges uniformly to some continuous
function x. It follows from advanced calculus that x0 = y and hence A is closed.
(b) (From Real Variables by Alberto Torchinsky, but that isn’t the original source of this
example) Let (V, k · k) be an infinite-dimensional Banach space. Then, as shown in 515,
there is a vector space basis B for V . (It can be obtained via the Hausdorff maximal
principle as a maximal linearly independent subset of V .) Define
H = {h : h =
b
, b ∈ B},
kbk
and define a new norm k · k1 on V by
kxk1 =
n
X
|ai |,
i=1
where
x=
n
X
ai hi ,
hi ∈ H.
i=1
(In other words, the ai ’s are the coefficients for x with respect to this basis.) Then we
write X for the Banach space (V, k · k) and Y for the Banach space (V, k · k1 ), and we define
3
A : X → Y by Ax = x. To see that A is closed, let ixn h be a sequence in V and suppose
that xn → x in X and Axn → y in Y . First, we observe that
n
n
X
X
kxk = ai hi ≤
|ai |khi k = kxk1 .
i=1
i=1
Hence
kxn − yk = kAxn − yk ≤ kAxn − yk1 ,
so xn → y in X and therefore x = y. To show that A is unbounded, we only have to
show that Y is not complete by virtue of Proposition 10. Let hhi i be a sequence of distinct
vectors in H and let
n
X
xn =
2−i hi .
i=1
Then it’s easy to check that hxn i is a Cauchy sequence in Y . Suppose now that this sequence
had a limit x. Then x must be a linear combination of the hi ’s in our sequence (because
the component of each xn with respect to any other h is zero), but the component with
respect to hi must be nonzero (in fact 2−i ) for each i, so x is not a linear combination of
the hi ’s. It follows that Y is not a Banach space, so A is unbounded.