Math 516
Professor Lieberman
March 6, 2009
HOMEWORK #4 SOLUTIONS
Chapter 10
20. Define p on l∞ by p(hξn i) = lim ξn . Problem 16 of Chapter 2 tells us that p(x + y) ≤
p(x) + p(y) for all x and y in l∞ and problem 17 of Chapter 2 (with xn replaced by
a constant sequence) shows that p(αx) = αp(x) for all α ≥ 0 and all x ∈ l∞ . We
then define the operator A on l∞ by A(hξn i) = hηn i with ηn = ξn+1 for all n, and we
take G to be the set of all operators of the form Ak with k a nonnegative integer. (In
this notation, A0 is the identity, that is, A0 x = x for all x ∈ l∞ .) We then write S
for the collection of all sequences that have a limit (This set is denoted c in Problem
15 of Chapter 6.), and we define f on S by f (hξn i) = lim ξn . By the properties of
limits, we know that f (Ak x) = f (x) for all nonnegative integers k and all x ∈ S,
so Proposition 10.5 gives a linear functional F on l∞ such that F (x) ≤ p(x) for all
x ∈ l∞ and F (Bx) = F (x) for all x ∈ l∞ and all B ∈ G. Hence F satisfies properties
(ii), (iii), and (iv). In addition, we have
F (hξn i) ≤ lim ξn
because F (x) ≤ p(x) and
F (hξn i) = −F (h−ξn i) ≥ − lim(−ξn ) = lim ξn ,
and hence property (i) is also true.
25. For x ∈ X, we define F (x) = [kxk + 1]f (x/[kxk + 1]). Our first step is to show that
f = F|S . First, we observe that θ = 0θ + 0θ, so
f (θ) = 0f (θ) + 0f (θ) = 0.
Hence f (θ) = F (θ). In addition, if α ∈ [−1, 1] and x ∈ S, then αx ∈ S and
f (αx) = f (αx + 1θ) = αf (x) + 1f (θ) = αx.
If kxk ≤ 1 and x 6= 0, then x = [kxk + 1](x/[kxk + 1]), so f (x) = F (x). In addition,
for any x ∈ X and any nonzero B ≥ kxk, we have
F (x) = Bf (x/B).
Now we show that F is linear. Let x and y be vectors, and let α and β be real
numbers, and set B = |α|kxk + |β|kyk + 1. Then
µ ¶¸
¶
· ³ ´
µ
βy
αx
αx + βy
+f
=B f
.
F (αx + βy) = Bf
B
B
B
Now, because B > |α|kxk, there is a positive number C ≥ kxk such that |αC| < B,
so
³ αx ´ αC ³ x ´ α
f
=
f
= F (x),
B
B
C
B
1
2
and, similarly,
µ
f
βy
B
¶
=
β
F (y),
B
so
F (αx + βy) = αF (x) + βF (y),
and therefore F is linear.
30. (a) First, suppose that B is a base at θ for a translation invariant topology. If U
and V are elements of B, then U ∩ V is an open set containing θ, so there is
W ∈ B such that W ⊂ U ∩ V . Moreover, if U ∈ B and x ∈ U , then translation
invariance says that −x + U is an open set. Since θ ∈ −x + U , there is a V ∈ B
such that V ⊂ −x + U and hence x + V ⊂ U .
Conversely, if B satisfies (i) and (ii), we take B 0 to consist of all translates of
sets of B, that is, B 0 is the collection of all sets of the form x + U with x ∈ X
and U ∈ B. Property (i) implies that B 0 is the base for some topology on X by
virtue of Proposition 8.5. To show that this topology is translation invariant,
let O be any open set in this topology and let x ∈ X. We want to show that
x + O is also open. So let y ∈ x + O and set z = y − x. Then z ∈ O, so there
is U ∈ B 0 such that z ∈ U ⊂ O. Hence y ∈ x + U ⊂ x + O. But U = w + V
for some V ∈ B, so x + U = x + w + V ∈ B. Hence the topology is translation
invariant.
(b) Suppose first that addition is continuous and let U ∈ B. Then addition is
continuous at (θ, θ), so there is an open subset W of X × X such that (θ, θ) ∈ W
and (x, y) ∈ W implies that x + y ∈ U . Since W is open in X × X, there are
open subsets V1 and V2 of X such that θ ∈ V1 , θ ∈ V2 and V1 × V2 ⊂ W . For
V = V1 ∩ V2 , it follows that x ∈ V and y ∈ V implies that (x, y) ∈ W , so
x + y ∈ U . In other words V + V ⊂ U .
Conversely, if property (iii) is true, then we only have to show that addition is
continuous at (θ, θ), so let O be an open set containing θ. Then there is U ∈ B
such that U ⊂ O. If V ∈ B satisfies V + V ⊂ U , then we set W = V × V . This
is an open subset of X × X such that (x, y) ∈ W implies that x + y ∈ U ⊂ O.
Hence addition is continuous at (θ, θ).
(c) Let x ∈ X and U ∈ B. Because multiplication by scalars is continuous at (0, θ)
and because the topology is translation invariant, it follows that multiplication
by scalars is also continuous at (0, x). Hence, for any U ∈ B, there are an
open subset V containing θ and a positive number ε such that λy ∈ U for all
λ ∈ (−ε, ε) and all y ∈ V . We now choose λ = ε/2 (in fact, any nonzero number
in the interval (−ε, ε) would work) and set α = 1/λ. then λx ∈ U because
x ∈ V and hence x ∈ αU .
(d) The phrase “local base” really means “base at θ”. Hence, we just have to show
that, if O is any open set containing θ, then there is U ∈ B with V ⊂ O. Because
scalar multiplication is continuous at (0, θ), there are an open set V containing
θ and a positive number ε such that λy ∈ U for all λ ∈ (−ε, ε) and all y ∈ V .
3
(As the book says, λV ⊂ O for |λ| < ε.) We now set U =
[
λV . Then U is
|λ|<ε
open and θ ∈ U ⊂ O. If |α| ≤ 1 and y ∈ U , then y = λx for some x ∈ V and
some λ with |λ| < ε. Hence αy = αλx and |αλ| < ε, so αy ∈ αλV ⊂ U . It
follows that U ∈ B and V ⊂ O. Hence this B is a base at θ for the topology.
In addition, if U ∈ B and |α| ≤ 1, then we have shown that αU ⊂ U . Furthermore, if β ∈ (−1, 1) and x ∈ αU , then β(αx) = αβx. Since βU ⊂ U , it follows
that βx ∈ U , so βαx ∈ αU , so βαU ⊂ αU . Therefore αU ∈ B.
(e) First, we show that scalar multiplication is continuous at h0, xi for any x 6= θ.
Let O be an open subset containing θ. Then there is U ∈ B such that U ⊂ O.
From (iv), there is α ∈ R so that x ∈ αU . Let ε be a positive number such that
ε|α| ∈ (0, 1) and set V = αU . Then V is an open set containing x and, for all
λ ∈ (−ε, ε), we have λV = αλU ⊂ U . Hence scalar multiplication is continuous
at h0, xi for any x 6= θ.
Next, we show that scalar multiplication is continuous at hα, θi for any α ∈ R.
Again, we let O be an open subset containing θ, and then there is U ∈ B
such that U ⊂ O. Now, we choose ε > 0 so that ε|α| < 1 and then δ > 0 so
that ε|α − δ|) < 1 and ε|α + δ| < 1. Finally, we set V = εU . Then, for any
β ∈ (α − δ, α + δ), we have that βV = βεU ⊂ U because |β|ε < 1. Hence scalar
multiplication is continuous at hα, θi for any α ∈ R.
To show that scalar multiplication is continuous at hα, xi with α 6= 0 and x 6= θ,
let O be an open set containing αx, so there is a U ∈ B such that αx + U ⊂ O.
From (iii), there is V ∈ B such that V + V ⊂ U . From the continuity of scalar
multiplication at h0, xi, we conclude that there is ε1 > 0 such that λx ∈ V for
all λ with |λ| < ε1 . Also the continuity of scalar multiplication at hα, θi implies
that there are ε2 > 0 and W ∈ B so that βz ∈ V for |β − α| < ε2 and z ∈ W .
We now set ε = min{ε1 , ε2 } If β ∈ (α − ε, α + ε) and y ∈ x + W , it follows that
(α − β)x ∈ V and β(x − y) ∈ V . Hence αx − βy ∈ V + V ⊂ U because
αx − βy = (α − β)x + β(x − y).
(It was also acceptable to prove this last part by using the phrase “translation
invariance”.
(f) If X is T1 , then for any x 6= θ, there is an open set Ox which contains θ but not
x. Hence, there is Ux ∈ B with x ∈
/ Ux . It follows that
\
\
{U ∈ B} ⊂
Ux = {θ}.
x∈X\{θ}
Now suppose conditions (iii) and (vi) hold. Then, from (vi), there is U ∈ B
such that x ∈
/ U , and (iii) gives a V ∈ B such that V + V ⊂ U . In addition, (v)
implies that −V ∈ B. If there were a point y ∈ V = ∩(x + V ), then we would
have a z ∈ V such that y = x + z. Hence −z ∈ V , so x = y + (−z) ∈ V + V ⊂ U ,
which contradicts our assumption that x ∈
/ U . It follows that V and x + V are
disjoint. But V is an open set containing θ and x + V is an open set containing
x, so X is Hausdorff.
4
41. (a) Recall that the sequence (fn ) must be in S. If fn → f weakly in L2 , then
T (fn ) → T (f ) for all bounded linear functionals T on L2 . In particular, if we
fix y ∈ [0, 1], then Problem 28 gives a function ky ∈ L2 such that
Z
g(y) = ky g
for all g ∈ S, and we know from Chapter 6 that Ty , defined by
Z
Ty g = k y g
is a bounded linear functional on L2 . It follows that Ty fn → Ty f . Hence, at each
y ∈ [0, 1], fn (y) converges to some limiting value g(y). To see that g = y, we use
Problem 38c, which says that fn converges to f in measure. Then Corollary 4.19
implies that there is a subsequence (fnk ) which converges almost everywhere to
f . But we have just shown that the whole sequence (fn ) converges everywhere to
g and therefore f = g. (Alternatively, you can use Proposition 10.16 to conclude
that S is weakly closed, so f ∈ S.)
(b) From Problem 38a, kfn k2 is bounded, so Problem 28a implies that kfn k∞ is
bounded. Therefore |fn − f |2 is bounded and |fn − f |2 → 0 almostR everywhere.
Hence, the Lebesgue dominated convergence theorem implies that |fn − f |2 →
0. In other words fn → f strongly in L2 .
(c) To show that S is locally compact, we note first that L2 is reflexive, so the
weak topology and the weak∗ topology are the same on L2 . Then Alaoglu’s
Theorem says that the closed unit ball in S is weakly compact because S is a
closed subset of L2 , so S is a Banach space with the L2 norm. Now let O be
the open unit ball in S and write O for its closure in the strong topology. Then
O is weakly compact, so if (fn ) is any sequence in O, it follows that there is a
weakly convergent subsequence (fnk ) with limit f ∈ O. Part (b) tells us that fnk
converges strongly to f , so O is strongly compact. Since S is a Banach space, it
follows that the closure of any open ball is compact, and therefore S is locally
compact. Problem 37 now implies that S is finite-dimensional.