Real Analysis
Chapter 5 Solutions
Jonathan Conder
4. Note that kT x − Tn xn k ≤ kT x − T xn k + kT xn − Tn xn k ≤ kT kkx − xn k + kT − Tn kkxn k, and the limit as n → ∞ of
the right hand side is 0, so limn→∞ Tn xn = T x.
P
6. (a) Clearly kxk1 ≥ 0 for all x ∈ X. If nk=1 ak ek ∈ X is non-zero then am 6= 0 for some m ∈ {1, 2, . . . , n}. This
P
P
implies that k nk=1 ak ek k1 ≥ |am | > 0. Moreover, if nk=1 ak ek ∈ X and b ∈ K then
n
n
n
X
X
X
ak ek =
|bak | = |b| ak ek .
b
k=1
If
Pn
1
k=1
k=1
1
Pn
∈ X then
n
n
n
n
n
n
n
X
X
X
X
X
X
X
ak ek +
bk ek =
|ak + bk | ≤
|ak | +
|bk | = ak ek + bk ek .
k=1 ak ek ,
k=1 bk ek
k=1
k=1
1
k=1
k=1
k=1
k=1
k=1
1
1
This shows that k · k1 is a norm on X.
P
(b) Let (a1 , a2 , . . . , an ) ∈ K n and suppose that k(a1 , a2 , . . . , an )k2 = 1. Then nk=1 |ak |2 = 1, and in particular
|ak |2 ≤ 1 for all k ∈ {1, 2, . . . , n}. It follows that |ak | ≤ 1 for all k ∈ {1, 2, . . . , n}, in which case
n
n
X
X
|ak | ≤ n,
ak e k =
k=1
1
k=1
P
and hence the linear map (a1 , a2 , . . . , an ) 7→ nk=1 ak ek is bounded, thus continuous.
P
(c) Define C := {(a1 , a2 , . . . , an ) ∈ K n | k nk=1 ak ek k1 = 1}, so that the image of C under the map from (b) is the
given set. Hence, it suffices to show that C is compact. Note that k · k1 : X → R is (Lipschitz) continuous, so
P
the map (a1 , a2 , . . . , an ) 7→ k nk=1 ak ek k1 from K n to R is continuous, implying that C (the preimage of {1}) is
closed. Moreover C is bounded, thus compact, because for each (a1 , a2 , . . . , an ) ∈ K n
k(a1 , a2 , . . . , an )k2 ≤
n
X
k(0, . . . , 0, ak , 0, . . . , 0)k2 =
k=1
n
X
|ak | = 1.
k=1
Pn
(d) Let k · k : X → R be a norm on X. If k=1 ak ek ∈ X then
n
n
n
n
X
X
X
X
|ak |kek k ≤ max{kek k}nk=1
|ak | = max{kek k}nk=1 ak e k .
ak ek ≤
k=1
k=1
k=1
k=1
1
This proves one half of the equivalence and shows that k · k is (Lipschitz) continuous on (X, k · k1 ). It follows that
k · k has a minimum at some u ∈ B, where B is the compact set {x ∈ X | kxk1 = 1}. Since 0 ∈
/ B, u 6= 0 and
hence kuk > 0. If x ∈ X is non-zero then
x ≥ kxk1 kuk,
kxk = kxk1 kxk1 which also holds if x = 0. This proves the other half of the equivalence, so every norm on X is equivalent to k · k1 .
P
k
7. (a) Since X is a Banach space, L(X, X) is also a Banach space. The series ∞
k=0 (I − T ) is absolutely convergent
P∞
P
∞
k
k
because
k=0 k(I − T ) k ≤
k=0 kI − T k , which is a geometric series with ratio kI − T k < 1. Therefore
1
Real Analysis
P∞
Chapter 5 Solutions
Jonathan Conder
some S ∈ L(X, X). Given ε ∈ (0, ∞), there exists N ∈ N such that kS −
n ≥ N, in which case
N
+1
N
+1
X
X
k
k
kS − I − S(I − T )k = S −
(I − T ) +
(I − T ) − S(I − T )
k=0
k=1
N +1
N
+1
X
X
k
k−1
≤ S −
(I − T ) + (I − T ) (I − T ) − S(I − T )
k=0
k=1
!
N
X
<ε+
(I − T )k − S (I − T )
k=0
N
X
k
≤ ε + (I − T ) − S kI − T k
k
k=0 (I − T ) converges to
T )k k < ε for all n ∈ N with
Pn
k=0 (I
−
k=0
< ε + εkI − T k
< 2ε.
This implies that kS − I − S(I − T )k = 0, so S − I = S(I − T ) = S − ST and hence ST = I. A very similar
calculation shows that kS − I − (I − T )Sk = 0 and hence T S = I.
(b) Since kI − T −1 Sk = kT −1 S − Ik = kT −1 S − T −1 T k ≤ kT −1 kkS − T k < kT −1 kkT −1 k−1 = 1, exercise (a)
implies that T −1 S has an inverse R ∈ L(X, X). Then RT −1 is an inverse for S, as (RT −1 )S = R(T −1 S) = I
and S(RT −1 ) = T (T −1 S)RT −1 = T T −1 = I. Thus every element of BkT −1 k−1 (T ) is invertible for all invertible
T ∈ L(X, X), so the set of invertible elements of L(X, X) is open.
8. Clearly M (X) is a C-space. For each ν ∈ M (X) there is a finite measure µ on X and a function f ∈ L1 (µ) such that
R
dν = f dµ. Note that kνk = |ν|(X) = |f | dν is zero iff f vanishes almost everywhere, or equivalently ν = 0. If a ∈ C
R
R
then kaνk = |aν|(X) = |af | dµ = |a| |f | dµ = |a||ν|(X) = |a|kνk because d(aν) = (af ) dµ. The triangle inequality
follows from Proposition 3.14. Therefore k · k is a norm on M (X).
P∞
P∞
If
n=1 νn (A) also converges absolutely
n=1 νn is an absolutely convergent series in M (X), and A ∈ M, then
P
because |νn (A)| ≤ |νn |(A) ≤ |νn |(X) for all n ∈ N. Hence, we may define ν : M → C by ν(A) := ∞
n=1 νn (A). Clearly
∞
ν(∅) = 0; if (Ak )k=1 is a sequence of disjoint elements of M whose union is A ∈ M then
∞
X
|ν(Ak )| ≤
∞ X
∞
X
|νn (Ak )| =
k=1 n=1
k=1
∞ X
∞
X
|νn (Ak )| ≤
n=1 k=1
∞ X
∞
X
|νn |(Ak ) =
n=1 k=1
∞
X
|νn |(A) ≤
n=1
∞
X
kνn k < ∞
n=1
by Tonelli’s theorem, and Fubini’s theorem implies that
∞
X
ν(Ak ) =
∞ X
∞
X
νn (Ak ) =
k=1 n=1
k=1
∞ X
∞
X
νn (Ak ) =
n=1 k=1
∞
X
νn (A) = ν(A).
n=1
P∞
In other words k=1 ν(Ak ) converges absolutely to ν(A), which shows that ν ∈ M (X). It remains to show that
P∞
n=1 νn converges to ν in M (X). If N, M ∈ N and E1 , . . . , EM ∈ M are disjoint sets covering X, then
∞
!
M N
−1
M X
∞
∞ X
M
∞
M X
X
X
X
X
X
X
νn (Em ) =
νn (Em ) ≤
|νn |(Em ) =
|νn |(Em ) =
kνn k.
ν−
m=1
Therefore kν −
n=1
PN −1
n=1
νn k ≤
m=1 n=N
P∞
n=N
m=1 n=N
n=N m=1
n=N
kνn k by Exercise 3.21 (Homework 6). Now take the limit N → ∞.
2
Real Analysis
Chapter 5 Solutions
Jonathan Conder
9. (a) Let f ∈ C([0, 1]) and suppose that f is k times continuously differentiable on (0, 1) with limx&0 f (j) (x) and
limx%1 f (j) (x) existing for all j ∈ {0, 1, . . . , k}. Fix j ∈ {1, 2, . . . , k} and assume (using induction) that f (j−1)
is continuous on [0, 1]. Let ε ∈ (0, ∞) and define m := limx&0 f (j) (x). Then there exists δ ∈ (0, ∞) such that
| Re(f (j) (x))−Re(m)| < ε for all x ∈ (0, δ). Now let x ∈ (0, δ). By the mean value theorem there exists x0 ∈ (0, x)
such that Re(f (j) (x0 ))x = Re(f (j) (x0 ))(x − 0) = Re(f (j−1) (x)) − Re(f (j−1) (0)). Thus
Re(f (j−1) (x)) − Re(f (j−1) (0))
Re(f (j) (x ))x
0
−
Re(m)
=
−
Re(m)
= | Re(f (j) (x0 )) − Re(m)| < ε
x
x
because x0 ∈ (0, δ). This shows that Re(f (j−1) ) has a one-sided derivative at 0, since
Re(f (j−1) (x)) − Re(f (j−1) (0))
= Re(m).
x&0
x
lim
Similarly Im(f (j−1) ) has one-sided derivative Im(m) at 0, so f (j) (0) = m. Therefore f (j) is defined and continuous
on [0, 1). A similar argument shows that f (j) (1) = limx%1 f (j) (x), so by induction f ∈ C k ([0, 1]).
The converse is clear because f (j) is continuous for all f ∈ C k ([0, 1]) and j ∈ {0, 1, . . . , k}.
(b) Clearly kf k ≥ 0 for all f ∈ C k ([0, 1]). If f ∈ C k ([0, 1]) is non-zero then f (0) u > 0 and hence kf k > 0. Also, if
f ∈ C k ([0, 1]) and a ∈ C then
k k k
X
X
X
(j) (j) kaf k =
|a| f (j) = |a|kf k.
(af ) =
af =
u
j=0
u
j=0
u
j=0
Moreover, if f, g ∈ C k ([0, 1]) then
k k k k X
X
X
X
(j)
(j) (j) (j) (j) kf + gk =
(f + g) =
f + g ≤
f +
g = kf k + kgk.
u
j=0
u
j=0
u
j=0
u
j=0
This shows that k · k is a norm on C k ([0, 1]). We assume (using induction) that C k−1 ([0, 1]) is complete under the
corresponding norm (this is known for k = 1 because C 0 ([0, 1]) = C([0, 1])). Let hfn i∞
n=1 be a Cauchy sequence
k
in C ([0, 1]). For each ε ∈ (0, ∞) there exists N ∈ N such that
k k−1 X
X
(j)
(j)
(j) (j) fn − fm
fn − fm
= kfn − fm k < ε
≤
u
j=0
u
j=0
∞
k−1 ([0, 1]), so it converges to some
for all m, n ∈ N with m ≥ n
D ≥ N.
E∞This shows that hfn in=1 is Cauchy in C
(k)
f ∈ C k−1 ([0, 1]). Similarly fn
is Cauchy in C([0, 1]), so it converges to some g ∈ C([0, 1]). Note that
n=1
D
E∞
P
(k−1)
(j)
(k−1)
(j) for all n ∈ N.
fn
converges in C([0, 1]) to f (k−1) , because fn
− f (k−1) ≤ k−1
f
−
f
n
j=0
n=1
u
u
E∞
D
(k) Since |g| is bounded above by a constant function, fn is eventually dominated by a slightly larger
n=1
constant function (in L1 ([0, 1])). By the dominated convergence theorem, it follows that
Z x
Z x
Z
(k−1)
(k)
(k)
(k−1)
(k−1)
(k−1)
f
(x) − f
(0) = lim fn
(x) − fn
(0) = lim
fn =
lim fn =
n→∞
n→∞ 0
0 n→∞
x
g
0
for all x ∈ [0, 1], and hence f (k) = g by exercise
(a). This implies
that f ∈ C k ([0, 1]). Moreover hfn i∞
n=1 converges
P
(j)
(k)
k−1
to f in C k ([0, 1]), because kfn − f k = j=0 fn − f (j) + fn − g for all n ∈ N. This shows (by induction)
u
that C k ([0, 1]) is a Banach space for all k ∈ N.
3
u
Real Analysis
10.
Chapter 5 Solutions
Jonathan Conder
R 1 (j)
(j) 1
Let hfn i∞
n=1 be a Cauchy sequence in Lk ([0, 1]). If j ∈ {0, 1, . . . , k} then 0 fn − fm ≤ kfn − fm k for all m, n ∈ N,
D
E
(j) ∞
is (modulo equality almost everywhere) a Cauchy sequence in the Banach space L1 ([0, 1]), so it has
so fn
n=1
R x (j+1)
(j)
(j)
a limit gj ∈ L1 ([0, 1]). Fix j ∈ {0, 1, . . . , k − 1}, n ∈ N, and note that fn (x) − fn (0) = 0 fn
for all x ∈
D
E
(j)
(j+1)
(j) ∞
which
[0, 1], either because fn is absolutely continuous or fn
is continuous. Choose a subsequence fni
i=1
D
E∞
(j+1)
converges in L1 ([0, 1]) to gj+1 . Taking a
converges to gj pointwise almost everywhere, and note that fni
i=1
D
E∞
(j+1)
further subsequence, we may assume that fni
converges to gj+1 pointwise almost everywhere. This implies
i=1
E∞
D
(j+1)
converges to |gj+1 | pointwise almost everywhere, and clearly
− gj+1 + |gj+1 |
that fni
i=1
Z
lim
i→∞ 0
1 Z
(j+1)
+
|g
|
=
0
+
lim
f
−
g
ni
j+1
j+1 1
i→∞ 0
Z
|gj+1 | =
1
|gj+1 |,
0
(j+1) (j+1)
while fni ≤ fni
− gj+1 + |gj+1 | for all i ∈ N. Hence, by the generalised dominated convergence theorem
Z x
Z x
(j)
(j+1)
gj (x) − gj (0) = lim fn(j)
(x)
−
f
(0)
=
lim
f
=
gj+1
ni
ni
i
i→∞
i→∞ 0
0
for almost all x ∈ [0, 1]. This shows that gj is almost everywhere equal to the absolutely continuous function hj :
Rx
Rx
[0, 1] → C defined by hj (x) := gj (0) + 0 gj+1 . If j < k − 1, it follows that hj (x) − hj (0) = 0 hj+1 for all x ∈ [0, 1],
(j)
so that h0j = hj+1 , by the fundamental theorem of calculus and exercise 9(a). By induction h0 = hj for all
R x (k)
(k−1)
j ∈ {0, 1, . . . , k − 1}. Since h0
= hk−1 is absolutely continuous, it follows that h0 ∈ L1k ([0, 1]) and 0 h0 =
R (k)
R
Rx
(k−1)
(k−1)
h0
(x) − h0
(0) = 0 gk for all x ∈ [0, 1]. The Borel measures E 7→ E h0 and E 7→ E gk agree on intervals, so
(k)
they are equal
and hence
h0R = gk almost
everywhere. Given ε ∈ (0, ∞) and j ∈ {0, 1, . . . , k}, there exists Nj ∈ N
R 1 (j)
(j) 1 (j)
ε
such that 0 fn − h0 = 0 fn − gj < k+1
for all n ∈ N with n ≥ Nj . It follows that kfn − h0 k < ε for all n ∈ N
1
1
with n ≥ max{Nj }kj=0 . so hfn i∞
n=1 converges to h0 in Lk ([0, 1]). Therefore Lk ([0, 1]) is a Banach space.
11. (a) Clearly kf kΛα ≥ 0 for all f ∈ Λα ([0, 1]). If f ∈ Λα ([0, 1]) is non-zero, then there exists x ∈ [0, 1] such that
(0)|
f (x) 6= 0, and either f (x) = f (0) or x 6= 0 and |f (x)−f
> 0, in which case kf kΛα > 0. Moreover,
|x−0|α
kaf kΛα = |af (0)| +
sup
x,y∈[0,1], x6=y
= |a||f (0)| +
|af (x) − af (y)|
|x − y|α
sup
x,y∈[0,1], x6=y
= |a| |f (0)| +
|f (x) − f (y)|
|x − y|α
!
|f (x) − f (y)|
|x − y|α
|a|
sup
x,y∈[0,1], x6=y
= |a|kf kΛα
for all a ∈ C and f ∈ Λα ([0, 1]). Finally, if f, g ∈ Λα ([0, 1]) then
|f (x0 ) + g(x0 ) − f (y 0 ) − g(y 0 )|
|f (x0 ) − f (y 0 )| |g(x0 ) − g(y 0 )|
≤
+
|x0 − y 0 |α
|x0 − y 0 |α
|x0 − y 0 |α
|f (x) − f (y)|
|g(x) − g(y)|
≤
sup
+
sup
α
|x
−
y|
|x − y|α
x,y∈[0,1], x6=y
x,y∈[0,1], x6=y
for all x0 , y 0 ∈ [0, 1] with x0 6= y 0 , and hence
kf + gkΛα = |f (0) + g(0)| +
sup
x,y∈[0,1], x6=y
4
|f (x) + g(x) − f (y) − g(y)|
|x − y|α
Real Analysis
Chapter 5 Solutions
≤ |f (0)| + |g(0)| +
sup
x,y∈[0,1], x6=y
Jonathan Conder
|g(x) − g(y)|
|f (x) − f (y)|
+
sup
α
|x − y|
|x − y|α
x,y∈[0,1], x6=y
= kf kΛα + kgkΛα .
This shows that k · kΛα is a norm on Λα ([0, 1]). Given f ∈ Λα ([0, 1]) and x ∈ (0, 1], it is clear that
|f (x)| ≤ |f (x) − f (0)| + |f (0)| = |f (0)| +
|f (x) − f (0)| α
|f (x) − f (0)|
|x| ≤ |f (0)| +
≤ kf kΛα ,
α
|x − 0|
|x − 0|α
which implies that kf ku ≤ kf kΛα because |f (0)| ≤ kf kΛα by definition. It follows that every sequence hfn i∞
n=1
which is Cauchy in Λα ([0, 1]) is also Cauchy in B([0, 1]), and therefore has a limit f ∈ B([0, 1]). Given x, y ∈ [0, 1]
with x 6= y, there exists n ∈ N such that kf − fn ku < |x − y|α , and hence
|f (x) − f (y)|
|f (x) − fn (x)| + |fn (x) − fn (y)| + |fn (y) − f (y)|
|fn (x) − fn (y)|
≤
<1+
+ 1 ≤ kfn kΛα + 2.
α
α
|x − y|
|x − y|
|x − y|α
This implies that f ∈ Λα ([0, 1]) because every Cauchy sequence is bounded (in Λα ([0, 1])). Given ε ∈ (0, ∞)
there exists N ∈ N such that kfm − fn kΛα < 4ε for all m, n ∈ N with m ≥ n ≥ N. If n ∈ N and x, y ∈ [0, 1] with
n ≥ N and x 6= y, there exists m ∈ N such that m ≥ n and kf − fm ku < 6ε |x − y|α ≤ 6ε , so
|f (x) − fn (x) − f (y) + fn (y)|
|f (x) − fm (x) + fm (x) − fn (x) + fn (y) − fm (y) + fm (y) − f (y)|
=
|x − y|α
|x − y|α
|f (x) − fm (x)| + |fm (x) − fn (x) + fn (y) − fm (y)| + |fm (y) − f (y)|
≤
|x − y|α
ε
ε
< + kfm − fn kΛα +
6
6
ε ε
< +
4 3
and hence
|f (x) − fn (x) − f (y) + fn (y)|
|x − y|α
x,y∈[0,1], x6=y
ε ε
≤ |f (0) − fm (0)| + |fm (0) − fn (0)| + +
4 3
ε ε ε ε
< + + +
6 4 4 3
= ε.
kf − fn kΛα = |f (0) − fn (0)| +
sup
Therefore hfn i∞
n=1 converges to f in Λα ([0, 1]), which shows that Λα ([0, 1]) is a Banach space.
(b) Clearly λ1 ([0, 1]) consists of all differentiable functions f : [0, 1] → C such that f 0 = 0. By the mean value
theorem these functions have constant real and imaginary parts, so λ1 ([0, 1]) consists only of constant functions.
Now suppose that α ∈ (0, 1). Clearly 0 ∈ λα ([0, 1]). If f ∈ λα ([0, 1]) and a ∈ C then
lim
x→y
|f (x) − f (y)|
|af (x) − af (y)|
= lim |a|
= |a| · 0 = 0
x→y
|x − y|
|x − y|
for all y ∈ [0, 1], so af ∈ λα ([0, 1]). Given f, g ∈ λα ([0, 1]) and x, y ∈ [0, 1] with x 6= y,
0≤
|f (x) + g(x) − f (y) − g(y)|
|f (x) − f (y)| |g(x) − g(y)|
≤
+
|x − y|
|x − y|
|x − y|
5
Real Analysis
Chapter 5 Solutions
Jonathan Conder
so f + g ∈ λα ([0, 1]) by the squeeze theorem. This shows that λα ([0, 1]) is a subspace of Λα ([0, 1]). Let hfn i∞
n=1
be a sequence in λα ([0, 1]) which converges to f ∈ Λα ([0, 1]). Given ε ∈ (0, ∞) and y ∈ [0, 1], there exists n ∈ N
such that kf − fn kΛα < 2ε and δ ∈ (0, ∞) such that
ε
|fn (x) − fn (y)|
<
α
|x − y|
2
for all x ∈ [0, 1] with 0 < |x − y| < δ. It follows that
|f (x) − f (y)|
|f (x) − fn (x) − f (y) + fn (y)| |fn (x) − fn (y)|
ε
≤
+
< kf − fn kΛα + < ε,
|x − y|α
|x − y|α
|x − y|α
2
which shows that f ∈ λα ([0, 1]). Therefore λα ([0, 1]) is a closed subspace of Λα ([0, 1]). For each β ∈ (α, 1), define
fβ : [0, 1] → R by fβ (x) := xβ . Then fβ0 (x) = βxβ−1 and hence fβ00 (x) = β(β − 1)xβ−2 for all x ∈ (0, 1]. This
implies that fβ00 ≤ 0 on (0, 1], so fβ0 is decreasing on (0, 1] by the mean value theorem. Given x, y ∈ (0, 1] with
x < y, there exist a ∈ (0, x), b ∈ (x, y), c ∈ (0, y − x) and d ∈ (y − x, y) such that
fβ (y) − fβ (x)
fβ (x)
fβ (x) − fβ (0)
=
= fβ0 (a) ≥ fβ0 (b) =
x
x−0
y−x
and
fβ (y) − fβ (y − x)
fβ (y − x)
fβ (y − x) − fβ (0)
fβ (y) − fβ (y − x)
=
= fβ0 (c) ≥ fβ0 (d) =
=
,
y−x
y−x−0
y − (y − x)
x
again by the mean value theorem. Therefore fβ (x)(y − x) ≥ (fβ (y) − fβ (x))x, so that fβ (x)y ≥ fβ (y)x, and
fβ (y − x)x ≥ fβ (y)(y − x) − fβ (y − x)(y − x), whence fβ (y − x)y ≥ fβ (y)(y − x). These inequalities combine to
give fβ (y − x)y + fβ (x)y ≥ fβ (y)y, and hence fβ (y − x) ≥ fβ (y) − fβ (x). It follows that
0≤
fβ (y) − fβ (x)
|xβ − y β |
|xβ − y β |
=
|x − y|β−α ≤ |x − y|β−α ,
|x − y|β−α =
α
β
|x − y|
fβ (y − x)
|x − y|
which also holds if x = 0 and y ∈ (0, 1]. This shows that f ∈ Λα ([0, 1]) since |x − y|β−α ≤ 2β−α , and the
squeeze theorem implies that f ∈ λα ([0, 1]). It remains to show that {fβ }β∈(α,1) is linearly independent. To this
end, let n ∈ N, a1 , a2 , . . . , an ∈ C and β1 , β2 , . . . , βn ∈ (α, 1). Suppose a1 fβ1 + a2 fβ2 + · · · + an fβn = 0. Then
Pn
Qk−1
(k)
(k)
(k)
i=1 ai
j=0 (βi − j) = a1 fβ1 (1) + a2 fβ2 (1) + · · · + an fβn (1) = 0 for all k ∈ N ∪ {0}. Therefore






1
β1
..
.
Qn−2
j=0 (β1
···
···
..
.
1
β2
..
.
− j)
Qn−2
j=0 (β1
− j) · · ·

  
a1
0
   
  a2  0
  .  = . ,
  .  .
  .  .
1
βn
..
.
Qn−2
j=0 (β1
− j)
an
0
which has non-trivial solutions iff the rows of the above n × n matrix are linearly dependent. If this were the
Q
case, some linear combination of the polynomials k−1
j=0 (x − j) would be a non-zero polynomial of degree n − 1
with n distinct roots β1 , β2 , . . . , βn . This is impossible, so it must be the case that a1 = a2 = · · · = an = 0.
12. (a) If X ∈ X/M then kXk = inf y∈X kyk is independent of the representation of X. It is clear that kMk = 0.
Conversely, if x ∈ X and kx + Mk = 0 then there is a sequence (yn )∞
n=1 in M such that limn→∞ kx + yn k = 0. It
follows that x = limn→∞ (−yn ) ∈ M, i.e. x + M = M. If x ∈ X and a ∈ K × then
ka(x + M)k = kax + Mk = inf kax + yk = |a| inf kx + a−1 yk = |a|kx + Mk.
y∈M
y∈M
6
Real Analysis
Chapter 5 Solutions
Jonathan Conder
The case a = 0 follows from above. Finally, if x, y ∈ X then
k(x + y) + Mk = inf kx + y + zk
z∈M
= inf kx + y + w + zk
w,z∈M
≤ inf (kx + yk + kw + zk)
w,z∈M
= inf kx + wk + inf ky + zk
w∈M
z∈M
= kx + Mk + ky + Mk.
1
, so kx + Mk < kx+Mk
(b) Since M 6= X, there exists x ∈ X \ M. If ε ≥ 1 we are done, otherwise 1 < 1−ε
1−ε and there
kx+Mk
exists y ∈ M such that kx − yk < 1−ε . The idea is that y is close to x, so x − y is close to being “normal” to
M. Thus we define z := kx − yk−1 (x − y), and note that kzk = 1 while
kz + Mk =
k(x − y) + Mk
kx + Mk
=
> 1 − ε.
kx − yk
kx − yk
(c) Clearly kπ(x)k = inf y∈M kx + yk ≤ kxk for all x ∈ X, which means (π is bounded and) kπk ≤ 1. Conversely, if
ε ∈ (0, 1) then 1 − ε < kπ(x)k ≤ kπkkxk for some x ∈ X with kxk = 1. This implies that 1 ≤ kπk.
P
(d) Let ∞
n=1 Xn be an absolutely convergent series in X/M. For each n ∈ N, note that kXn k = inf x∈Xn kxk, and
P
hence there exists xn ∈ Xn such that kxn k < kXn k + 2−n . Clearly ∞
n=1 xn is an absolutely convergent series in
P
P∞
P∞
X, so it has a limit x ∈ X. Since π is bounded, n=1 Xn = n=1 π(xn ) = π( ∞
n=1 xn ) = π(x).
(e) If U ⊆ X/M is norm-open, then π −1 (U ) is open and hence U is open in the quotient topology. Conversely, if
U ⊆ X/M is open in the quotient topology, then π −1 (U ) is open. Given x + M ∈ U, there exists r ∈ (0, ∞) such
that Br (x) ⊆ π −1 (U ). If y +M ∈ Br (x+M) then kx−y +Mk < r, so there exists z ∈ M such that kx−y +zk < r.
In particular y − z ∈ π −1 (U ), which implies that y + M = π(y − z) ∈ U. Therefore Br (x + M) ⊆ U, which shows
that U is norm-open.
13. Clearly 0 ∈ M. Moreover, if a ∈ K and x, y ∈ M then kax + yk ≤ |a|kxk + ky| = 0, so ax + y ∈ M. Therefore
M is a subspace of X. If x, y ∈ X and x + M = y + M then kx + Mk = kxk ≤ kx − yk + kyk = 0 + ky + Mk, so
kx + Mk = ky + Mk by symmetry. Clearly kMk = 0. Conversely, if x ∈ X and kx + Mk = 0 then kxk = 0, which
means x ∈ M and hence x + M = M. If x ∈ X and a ∈ K then kax + Mk = kaxk = |a|kxk = |a|kx + Mk. Finally, if
x, y ∈ X then k(x + y) + Mk = kx + yk ≤ kxk + kyk = kx + Mk + ky + Mk.
18. (a) Let M be a proper closed subspace of X and let x ∈ X \ M. There exists f ∈ X∗ such that f (x) 6= 0 and
f (M ) = {0}. Let hun + an xi∞
n=1 be a sequence in M + Kx that converges to y ∈ X. Then
f (y) = lim f (un + an x) = lim an f (x),
n→∞
n→∞
han i∞
n=1
since f is continuous, so
converges to a := f (y)/f (x), which implies that han xi∞
n=1 converges to ax.
∞
∞
Therefore hun in=1 = h(un + an x) − an xin=1 converges to y − ax, which lies in M because M is closed. It follows
that y ∈ M + Kx, which shows that M + Kx is closed.
P
(b) Every finite-dimensional subspace of X is of the form nk=1 Kxk for some n ∈ N ∪ {0} and x1 , x2 , . . . , xn ∈ X,
so the result follows from exercise (a) by induction, starting from M = {0}. Alternatively, observe that every
n-dimensional subspace M of X has an induced norm equivalent to the usual norm on K n , so M is a Banach
space and therefore closed in X.
7
Real Analysis
Chapter 5 Solutions
Jonathan Conder
19. (a) Choose x1 ∈ X with kx1 k = 1. Given n ∈ N we construct xn+1 inductively, assuming we already have
Pn
x1 , x2 , . . . , xn . Since M :=
k=1 Kxk is finite-dimensional, it is a proper closed subspace of X. There exists
y ∈ X \ M, and r := inf u∈M ky − uk > 0 because otherwise we could find a sequence in M converging to y. Choose
y−u
. Then kxn+1 k = 1, and for all v ∈ M
u ∈ M with ky − uk < 2r, and define xn+1 := ky−uk
y−u
1
1
1
=
kxn+1 − vk = −
v
y
−
u
−
ky
−
ukv
> r= .
ky − uk
ky − uk
2r
2
In particular kxn+1 − xk k ≥
1
2
for all k ∈ {1, 2, . . . , n}.
(b) Suppose that 0 ∈ X had a compact neighbourhood C, and choose r ∈ (0, ∞) such that B2r (0) ⊆ C. Since scalar
multiplication is continuous, 1r C is compact, and contains all x ∈ X with kxk = 1. By exercise (a), we can
construct a sequence in 1r C with no convergent subsequence (every ball of radius 14 contains at most one point
from the sequence). This is a contradiction because 1r is sequentially compact. Therefore 0 does not have a
compact neighbourhood, so X is not locally compact.
20. Let x1 , . . . , xm ∈ M form a basis, and T : K m → M the corresponding isomorphism (of K-spaces). Clearly kT (·)k is
a norm on K m , and it is equivalent to k · k1 by Exercise 6. In other words T and T −1 are bounded. The projections
π1 , . . . , πm : K m → K induce linear functionals f1 , . . . , fm : M → K such that
|fi (x)| = |πi (T −1 (x))| ≤ kT −1 (x)k1 ≤ kT −1 kkxk
for all i ∈ {1, . . . , m} and x ∈ M. By Hahn-Banach, there exist linear functionals F1 , . . . , Fm : X → K extending
f1 , . . . , fm such that each is bounded with norm at most kT −1 k. Let N be the closed subspace ∩m
ker(Fi ). Note
Pm i=1
m
m
that M ∩ N = ∩i=1 ker(fi ) = T (∩i=1 ker(πi )) = T ({0}) = {0}. Moreover, if x ∈ X then i=1 Fi (x)xi ∈ M and
Pm
P
x− m
i=1 Fi (x)xi ) = 0 for all j ∈ {1, . . . , m}. This shows that X = M + N.
i=1 Fi (x)xi ∈ N, because Fj (x −
22. (a) By definition T † is linear. If f ∈ Y∗ and kf k = 1 then
kT † f k = kf ◦ T k ≤ kf kkT k = kT k,
so that T † ∈ L(Y∗ , X∗ ) with
kT † k = sup{kT † f k | f ∈ Y∗ , kf k = 1} ≤ kT k.
If kT k = 0 it follows that kT † k = kT k. Otherwise, given ε ∈ (0, kT k) there exists x ∈ X such that kxk = 1 and
kT xk > kT k − ε. By a corollary of Hahn-Banach there exists f ∈ X∗ such that kf k = 1 and f (T x) = kT xk.
Therefore |(T † f )(x)| = |f (T x)| = |kT xk| > kT k − ε, in which case kT † f k > kT k − ε and hence kT † k > kT k − ε.
This implies that kT † k ≥ kT k, so kT † k = kT k.
(b) The image of x ∈ X in X∗∗ is the map x
b : f 7→ f (x). Therefore T †† (b
x)(f ) = x
b(T † f ) = (T † f )(x) = f (T x) for all
f ∈ Y∗ . But Tcx : f 7→ f (T x), which implies that T †† (b
x) = Tcx. Therefore T †† |Xb = T modulo the identifications
b and Y = Y.
b
X=X
(c) Since T (X) is a subspace of Y, so is T (X). Suppose T (X) is not dense in Y. Then there exists y ∈ Y \ T (X), and
hence f ∈ Y∗ such that f (y) 6= 0 but f (T (X)) = {0}. It follows that T † f = f ◦ T = 0, so f is a non-zero element
of ker(T † ), which shows that T † is not injective.
Conversely, suppose that T † is not injective. Then there exists f ∈ ker(T † ) such that f 6= 0. Hence, there
exists y ∈ Y such that f (y) 6= 0. Since f is continuous, y has a neighbourhood U on which |f | > 0. Since
f ◦ T = T † f = 0, it follows that U is disjoint from T (X). Therefore T (X) ⊆ U c , so T (X) is not dense in Y.
8
Real Analysis
Chapter 5 Solutions
Jonathan Conder
(d) Suppose T is not injective. Then T x = 0 for some non-zero x ∈ X. Since T is linear, we may assume kxk = 1.
There exists f ∈ X∗ such that kf k = 1 and f (x) = kxk = 1. If g ∈ X∗ and kf − gk < 1 then
|1 − g(x)| = |f (x) − g(x)| = |(f − g)(x)| ≤ kf − gkkxk < 1,
which implies that g(x) 6= 0. Since g(x) = 0 for all g ∈ T † (Y∗ ) = {h ◦ T | h ∈ Y∗ }, it follows that B1 (f ) is disjoint
from T † (Y∗ ). Therefore T † (Y∗ ) ⊆ B1 (f )c , so T † (Y∗ ) is not dense in X∗ .
Now suppose that X is reflexive and that T † (Y∗ ) is not dense in X∗ . Then there exists f ∈ X∗ \ T † (Y∗ ), and hence
x
b ∈ X∗∗ such that x
b(f ) 6= 0 but x
b(T † (Y∗ )) = {0}. As the name suggests, x
b corresponds to some x ∈ X. Therefore
f (x) 6= 0 but g(x) = 0 for all g ∈ T † (Y∗ ). In particular x 6= 0. If T x 6= 0 then there exists g ∈ Y∗ such that
g(T x) 6= 0. This is a contradiction because g ◦ T = T † g ∈ T † (Y∗ ). Therefore x is a non-zero element of ker(T ),
in which case T is not injective.
23. (a) M0 = ∩x∈M ker(b
x) and N⊥ = ∩f ∈N ker(f ).
(b) If x ∈ M, then f (x) = 0 for all f ∈ M0 , so x ∈ (M0 )⊥ . Conversely, given x ∈ X\M there exists (by Hahn-Banach)
f ∈ X∗ such that f |M = 0 (i.e. f ∈ M0 ) but f (x) 6= 0, so x ∈
/ (M0 )⊥ .
If f ∈ N then f (x) = 0 for all x ∈ N⊥ , so f ∈ (N⊥ )0 . Conversely, given f ∈
/ N there exists (by Hahn-Banach,
⊥
assuming X is reflexive) x ∈ X such that x̂|N = 0 (i.e. x ∈ N ) but f (x) = x̂(f ) 6= 0, so f ∈
/ (N⊥ )0 .
(c) Note that α = π † , so α ∈ L((X/M)∗ , X∗ ) and kαk = 1 by Exercises 22(a) and 12(c). Since π|M = 0 it is clear
that Im(α) ⊆ M0 . Conversely, if f ∈ M0 then f factors through π by the universal property of quotient spaces
(alternatively, check that x + M 7→ f (x) is a well-defined element of (X/M)∗ ), and hence f ∈ Im(α). It remains
to show that kα(f )k ≥ kf k for all f ∈ (X/M)∗ . If X ∈ X/M and x ∈ X then
|f (X)| = |f (π(x))| ≤ kf ◦ πkkxk
and hence |f (X)| ≤ kf ◦ πk inf x∈X kxk = kα(f )kkXk, so we are done.
(d) Clearly β|M0 = 0, so (by the universal property of quotient spaces) β = β ◦ π for some β ∈ L(X∗ /M0 , M∗ ), where
π : X∗ → X∗ /M0 is the quotient map. If F ∈ X∗ /M0 and f ∈ F then
kβ(F )k = kβ(f )k =
|f (x)|
|f (x)|
≤ sup
= kf k
kxk
x∈M\{0}
x∈X\{0} kxk
sup
and hence kβ(F )k ≤ inf f ∈F kf k = kF k. Conversely, choose f ∈ F and note that x 7→ kβ(F )kkxk is a seminorm
on X such that
|f (x)| = |β(f )(x)| = |β(F )(x)| ≤ kβ(F )kkxk
for all x ∈ M. By Hahn-Banach there exists g ∈ X∗ such that kgk ≤ kβ(F )k and g|M = f |M . In particular g ∈ F,
so kF k ≤ kgk ≤ kβ(F )k. Since β is surjective (Hahn-Banach), so is β.
b 0 . Then F (X)
b = {0} and there exists f ∈ X∗ such that F (g) = g(f ) for all g ∈ X∗∗ . If x ∈ X
24. (a) Let F ∈ (X∗ )b ∩ X
then f (x) = x
b(f ) = F (b
x) = 0, so f = 0 and hence F (g) = g(0) = 0 for all g ∈ X∗∗ . Therefore F = 0, which
b 0 = {0}.
shows that (X∗ )b ∩ X
Now let F ∈ X∗∗∗ . Define f : X → K by f (x) := F (b
x). Note that f ∈ X∗ because it is the composition of two
b 0 and hence
bounded linear maps. Now fb ∈ (X∗ )b and fb(b
x) = x
b(f ) = f (x) = F (b
x) for all x ∈ X, so F − fb ∈ X
b 0 . This shows that (X∗ )b + X
b 0 = X∗∗∗ .
F = fb + (F − fb) ∈ (X∗ )b + X
9
Real Analysis
Chapter 5 Solutions
Jonathan Conder
b 0 = {F ∈ X∗∗∗ | F (X∗∗ ) = {0}} = {0}, so X∗∗∗ = (X∗ )b by exercise (a). This
(b) Suppose that X is reflexive. Then X
shows that X∗ is reflexive.
b 0 = X∗∗∗ ∩ X
b 0 = {0} by exercise (a). Since X is a Banach space, X
b is a closed
Conversely, if X∗ is reflexive, then X
b then there exists g ∈ X∗∗∗ such that g(f ) 6= 0 but g(X)
b = {0}. This
subspace of X∗∗ . If there exists f ∈ X∗∗ \ X
b 0 , which is a contradiction. Therefore X∗∗ \ X
b = ∅, which shows that
implies that g is a non-zero element of X
X is reflexive.
∗
25. If X = {0} then it is clearly separable. Otherwise, let {fn }∞
n=1 be a countable dense subset of X . For each n ∈ N
either kfn k = 0 or kfn k > 0. In the first case choose xn ∈ X with kxn k = 1, so that |fn (xn )| = 0 = 21 kfn k. Otherwise
1
1
2 kfn k < kfn k = sup{|fn (x)| | x ∈ X, kxk = 1}, so there exists xn ∈ X such that kxn k = 1 and |fn (xn )| ≥ 2 kfn k.
Let M be the closed subspace of X generated by {xn }∞
n=1 , and suppose that there exists x ∈ X \ M. Then there
∗
exists f ∈ X such that f (x) 6= 0, f (M) = {0} and kf k = 1. Since {fn }∞
n=1 is dense there exists n ∈ N such that
1
kf − fn k < 3 . It follows that
1
1
1
1 2
+ 2|fn (xn )| = + 2|fn (xn ) − f (xn )| ≤ + 2kfn − f kkxn k < + = 1,
3
3
3
3 3
which is a contradiction. Therefore M = X. Now let Q be a countable dense subset of K, and let x ∈ X. Given
P
ε ∈ (0, ∞), there exist n ∈ N ∪ {0} and a1 , a2 , . . . , an ∈ K such that nk=1 ak xk ∈ Bε/2 (x). For each k ∈ {1, 2, . . . , n}
ε
choose qk ∈ Q with |ak − qk | < 2n
. Then
n
n
n
n
n
X
X
X
X
X
ε
ε
= ,
ak xk −
qk xk = (ak − qk )xk ≤
|ak − qk |kxk k =
|ak − qk | ≤ n
2n
2
k=1
k=1
k=1
k=1
k=1
P
which implies that nk=1 qk xk ∈ Bε (x). Therefore the set of linear combinations of {xn }∞
n=1 with coefficients in Q is
∞
n
dense in X. This set is countable because it is an image of the countable set ∪n=0 Q . It follows that X is separable.
1 = kf k ≤ kf − fn k + kfn k <
29. (a) Clearly 0 ∈ X. If f, g ∈ X and a ∈ K then
∞
X
n|(f + ag)(n)| =
n=1
∞
X
n|f (n) + ag(n)| ≤
n=1
∞
X
n|f (n)| +
n=1
P∞
∞
X
n|ag(n)| =
n=1
P∞
∞
X
n|f (n)| + |a|
n=1
∞
X
n|g(n)| < ∞,
n=1
does not, the map n 7→ n−2 is in Y \ X. Let
converges but n=1
so X is a subspace of Y. Since n=1
P∞
f ∈ Y and ε ∈ (0, ∞). There exists N ∈ N such that n=N |f (n)| < ε. It follows that
∞
X
n−2
|f (n) − n−1 f (n)| =
∞
X
n−1
(1 − n−1 )|f (n)| ≤
n=N
n=N
which implies that kf − gk < ε, where g ∈ X is defined by

f (n)
g(n) :=
n−1 f (n)
∞
X
|f (n)| < ε,
n=N
if n < N
if n ≥ N.
This shows that X is a proper dense subset of Y, so X = Y 6= X and hence X is not complete.
∞
(b) Let hfn i∞
n=1 be a sequence in X which converges to f ∈ X such that hT fn in=1 converges to g ∈ Y. Given ε ∈ (0, ∞),
P∞
P
ε
there exists N ∈ N such that n=N n|f (n)| < 4ε and ∞
n=N |g(n)| < 4 . Moreover, there exists M ∈ N such that
kg − T fm k < 4ε and kf − fm k < 4ε N −1 for all m ∈ N with m ≥ M. It follows that
∞
X
n=1
|T f (n) − T fm (n)| =
N
−1
X
|nf (n) − nfm (n)| +
n=1
∞
X
n=N
10
|nf (n) − T fm (n)|
Real Analysis
Chapter 5 Solutions
≤
<
N
−1
X
n=1
N
−1
X
∞
X
n|f (n) − fm (n)| +
Jonathan Conder
n|f (n)| +
n=N
N |f (n) − fm (n)| +
n=1
< N kf − fm k +
ε
+
4
∞
X
|T fm (n)|
n=N
∞
X
|T fm (n) − g(n)| +
n=N
∞
X
|g(n)|
n=N
ε
ε
+ kT fm − gk +
4
4
<ε
for all m ∈ N with m ≥ M. Therefore hT fn i∞
n=1 converges to T f, so T is closed.
For each m ∈ N define fm ∈ X by

1 if m = n
fm (n) :=
0 if m =
6 n.
Then kfm k = 1 but
kT fm k =
∞
X
|T fm (n)| =
n=1
∞
X
|nfm (n)| =
n=1
∞
X
nfm (n) = m
n=1
for all m ∈ N. This shows that T is not bounded.
(c) Clearly Sf (n) = n−1 f (n) for all f ∈ Y and n ∈ N. It follows that
kSf k =
∞
X
|Sf (n)| =
n=1
∞
X
|n−1 f (n)| =
n=1
∞
X
n=1
n−1 |f (n)| ≤
∞
X
|f (n)| = kf k
n=1
for all f ∈ Y, so S is bounded. Since S = T −1 , it is obvious that S is surjective. If S were open then T would
be continuous, contradicting part (b).
32. The identity map from (X, k · k2 ) to (X, k · k1 ) is a bijective bounded linear map between Banach spaces, so it is
a homeomorphism by the open mapping theorem. It follows that the identity map from (X, k · k1 ) to (X, k · k2 ) is
bounded, so there exists M ∈ [0, ∞) such that k · k2 ≤ M k · k1 . This shows that the two norms are equivalent.
1
33. Suppose there is such a sequence (an )∞
n=1 , and let T : B(N) → L (µ) be the corresponding map. For each n ∈ N
multiplication by an is a linear map C → C, and the product of these maps gives a linear map T : CN → CN . As each
∞
1
∞
an is positive, it is clear that T is injective. If (cn )∞
n=1 is bounded then (an cn )n=1 is in L (µ). Conversely, if (cn )n=1
P
cn ∞
cn
1
is in L1 (µ), then ∞
n=1 an | an | < ∞, so ( an )n=1 is bounded. This shows that T restricts to a bijection B(N) → L (µ).
P∞
P∞
The restriction is bounded because n=1 |an cn | ≤ n=1 an kck∞ = kak1 kck∞ for every bounded sequence (cn )∞
n=1
(this is a special case of Hölder’s inequality; we know that kak1 < ∞ because (1)∞
is
bounded).
Hence,
by
the
open
n=1
mapping theorem, T −1 : L1 (µ) → B(N) is bounded. For each k ∈ N
∞
X
δkn ∞ 1
≤ kT −1 k
=
|δkn | = kT −1 k,
an
ak
n=1 u
n=1
which implies that
P∞
k=1 ak
≥
P∞
k=1 kT
−1 k
= ∞. This is impossible, so no such sequence exists.
1
1
1
34. (a) Let hfn i∞
n=1 be a sequence in Lk ([0, 1]) which converges in Lk ([0, 1]) to f ∈ Lk ([0, 1]) and also converges in
C k−1 ([0, 1]) to g ∈ C k−1 ([0, 1]). Suppose that f 6= g. There exists x ∈ [0, 1] such that f (x) 6= g(x), and hence
11
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Jonathan Conder
there exists ε ∈ (0, ∞) such that |f (x) − g(x)| > 2ε. Since f and g are continuous, there exists δ ∈ (0, ∞) such
that |f (y) − g(y)| > ε for all y ∈ [0, 1] ∩ (−δ, δ). Clearly m([0, 1] ∩ (−δ, δ)) ≥ δ, so
Z 1
|f − g| ≥ δε.
0
However, there exists N ∈ N such that kf − fN kL1 <
k
Z
1
1
Z
Z
δε
2,
and hence
δε
+ kg − fN kC k−1 < δε.
2
|fN − g| ≤ kf − fN kL1 + kg − fN ku <
k
0
0
0
and kg − fN kC k−1 <
1
|f − fN | +
|f − g| ≤
δε ≤
δε
2
This is impossible, so in fact f = g. This shows that the inclusion L1k ([0, 1]) ,→ C k−1 ([0, 1]) is closed, so it is
continuous by the closed graph theorem.
(b) Let f ∈ L1k ([0, 1]) and suppose that kf kL1 = 1. Also let j ∈ {0, . . . , k − 1}. Then f (j) is absolutely continuous,
k
so for all x ∈ [0, 1]
Z x
Z x
Z 1
(j)
(j)
(j+1)
(j+1)
≤
|f (j+1) | ≤ kf kL1 = 1.
|f (x) − f (0)| = |
≤
f
|f
k
0
0
0
Moreover,
|f
(j)
1
Z
|f
(0)| =
(j)
Z
(0)| dx ≤
0
1
|f
(j)
(0) − f
(j)
Z
|f
(x)| dx +
0
1
(j)
Z
1
(x)| dx ≤
0
0
1 dx + kf kL1 = 2
k
and hence
|f (j) (x)| ≤ |f (j) (x) − f (j) (0)| + |f (j) (0)| ≤ 1 + 2 = 3
for all x ∈ [0, 1]. This implies that kf (j) ku ≤ 3, in which case kf kC k−1 ≤ 3(k − 1). This shows that the inclusion
L1k ([0, 1]) ,→ C k−1 ([0, 1]) is bounded as a linear map, so it is continuous.
37. Define T 0 : Y∗ → X∗ by T 0 f := f ◦ T. Note that T 0 is well-defined because f ◦ T ∈ X∗ for all f ∈ Y∗ . Moreover, T 0 is
linear for the same reason that the adjoint of a bounded linear map is linear. Set A := {b
x ◦ T 0 | x ∈ X, kxk = 1}. If
x ∈ X, f ∈ Y∗ and kxk = kf k = 1 then
|b
x(T 0 f )| = |T 0 f (x)| = |f (T x)| ≤ kf kkT (x)k = kT (x)k,
which implies that x
b ◦ T 0 is bounded. Therefore A is a subset of L(Y∗ , K). Since
sup |Sf | =
|b
x(T 0 f )| =
sup
x∈X, kxk=1
S∈A
sup
|T 0 f (x)| = kT 0 f k < ∞
x∈X, kxk=1
for all f ∈ Y∗ , the uniform boundedness principle implies that M := supS∈A kSk < ∞. Let x ∈ X with kxk = 1 and
T x 6= 0. By Hahn-Banach there exists g ∈ Y∗ such that kgk = 1 and g(T x) = kT xk. It follows that
kT xk = g(T x) = |g(T x)| ≤
sup
|f (T x)| =
f ∈Y∗ , kf k=1
sup
|T 0 f (x)| =
f ∈Y∗ , kf k=1
sup
|b
x(T 0 f )| = kb
x ◦ T 0 k ≤ M.
f ∈Y∗ , kf k=1
This also holds if we allow T x = 0, which shows that T is bounded.
38. Since addition and scalar multiplication respect limits, T is linear. Every convergent sequence in a metric space is
bounded, so supn∈N kTn xk < ∞ for each x ∈ X. Therefore M := supn∈N kTn k < ∞ by the uniform boundedness
principle. It follows that T is bounded, since for each x ∈ X
kT xk = lim Tn x = lim kTn xk ≤ sup kTn xk ≤ sup kTn kkxk = M kxk.
n→∞
n→∞
n∈N
12
n∈N
Real Analysis
Chapter 5 Solutions
Jonathan Conder
40. If k ∈ N, then supj∈N kTjk k = ∞ (otherwise kTjk xk ≤ supj∈N kTjk kkxk < ∞ for all j ∈ N and x ∈ X), and hence (by
the uniform boundedness principle) the set Ek := {x ∈ X | supj∈N kTjk xk < ∞} is meager. It follows that ∪∞
k=1 Ek is
∞
meager, so ∪k=1 Ek 6= X. Hence, there exists x ∈ X such that supj∈N kTjk xk = ∞ for all k ∈ N.
∞
N
41. Fix a norm on X. Let {xn }∞
n=1 be a (Hamel) basis for X, so that X = ∪N =1 span{xn }n=1 . Given N ∈ N, the subspace
span{xn }N
n=1 is closed by exercise 18(b). If Y is a subspace of X with non-empty interior, there exists r ∈ (0, ∞) and
y ∈ Y such that Br (y) ⊆ Y. Since Y is closed under subtraction, it follows that Br (0) ⊆ Y. This implies that Y = X,
because Y is closed under scalar multiplication. Therefore, the proper closed subspace span{xn }N
n=1 of X is nowhere
dense. This shows that X is the countable union of nowhere dense sets, so X is not complete by the Baire category
theorem.
43. (a) Suppose there is a nonzero x ∈ X such that pα (x) = 0 for all α ∈ A. If α ∈ A and ε ∈ (0, ∞) then 0 ∈ Uxαε ,
so 0 lies in every finite intersection of sets of this form. By Theorem 5.14(a), it follows that 0 lies in every
neighbourhood of x, so X is not Hausdorff (or even T0 , by a similar argument).
Conversely, suppose that, for each nonzero x ∈ X, there exists α ∈ A such that pα (x) 6= 0. If x, y ∈ X are distinct,
, and note that Uxαε ∩ Uyαε = ∅; indeed
then x − y 6= 0 and hence pα (x − y) 6= 0 for some α ∈ A. Set ε := pα (x−y)
2
pα (x − y) ≤ pα (x − z) + pα (z − y) for all z ∈ X, so pα (x − z) and pα (z − y) cannot both be less than ε. This
implies that X is Hausdorff.
(b) Replace A by N (if A is finite, extend it by adding copies of the zero seminorm) and define ρ : X2 → [0, ∞)
P
−n Φ(p (x − y)), where Φ is the function from Exercise 4.56. If x, y ∈ X and ρ(x, y) = 0
by ρ(x, y) := ∞
n
n=1 2
then Φ(pn (x − y)) = 0 and hence pn (x − y) = 0 for all n ∈ N, so x − y = 0 by part (a). It is clear that ρ
has all the other properties of a translation-invariant metric. If x ∈ X, n ∈ N, ε ∈ (0, ∞) and y ∈ Uxnε , then
δ := 2−n Φ(ε − pn (y − x)) is positive and Bδ (y) ⊆ Uxnε by the triangle inequality. This shows that Uxnε is
open with respect to ρ. Conversely, let x ∈ X, ε ∈ (0, ∞) and y ∈ Bε (x). If ε ≥ 1 then Bε (x) = X, which is
P
−n < ε − ρ(x, y), set δ := Φ−1 ( ε−ρ(x,y) ) and note that
open in X. Otherwise choose N ∈ N such that ∞
n=N 2
2
∩N
n=1 Uynδ ⊆ Bε (x). This shows that Bε (x) is open in X, so X is metrisable (with respect to ρ).
44. Let hxα iα∈A be a Cauchy net in X, and {Un }∞
n=1 a nested neighbourhood base of 0 ∈ X. For each n ∈ N there exist
αn , βn ∈ A such that xα − xβ ∈ Un for all α, β ∈ A with α & αn and β & βn . Without loss of generality assume
that αn−1 . αn = βn (take an upper bound of {αn−1 , αn , βn }, and ignore the condition α0 . α1 ). If U ⊆ X is a
neighbourhood of 0, then UN ⊆ U for some N ∈ N, and xαm − xαn ∈ UN for all m, n ∈ N with m, n ≥ N. This shows
2
that (xαn )∞
n=1 is a Cauchy sequence, so it has a limit x ∈ X. If U ⊆ X is a neighbourhood of x, its preimage in X
under addition is a neighbourhood of (0, x). In particular, there exists N ∈ N and a neighbourhood V of x such that
UN × V maps into U under addition. Without loss of generality xαn ∈ V for all n ∈ N with n ≥ N. If α ∈ A and
α & αN , then xα − xαN ∈ UN and xαN ∈ V, so xα = xα − xαN + xαN ∈ U. This shows that hxα iα∈A converges to x.
45. For each n, k ∈ N ∪ {0} set Kn := [−n, n] and define a seminorm pn,k on C ∞ (R) by pn,k (f ) := kf (k) |Kn ku (this is
indeed a seminorm because differentiation and restriction are linear maps). If f ∈ C ∞ (R) is nonzero then f |Kn 6= 0
for sufficiently large n ∈ N, so pn,0 (f ) 6= 0. By Exercise 43(a) C ∞ (R), with the topology defined by these seminorms,
is Hausdorff. Every sequence in C ∞ (R) whose derivatives of every order converge uniformly on compact sets to the
derivatives of some f ∈ C ∞ (R) also converges (to f ) in this topology, by Theorem 5.14(b). The converse holds because
every compact subset of R is contained in some Kn . By Exercise 43(b) C ∞ (R) is metrisable, hence first countable.
It therefore remains to show that every Cauchy sequence in C ∞ (R) converges. If (fj )∞
j=1 is a Cauchy sequence, then
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Real Analysis
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Jonathan Conder
(k)
(fj |Kn )∞
n=1 is uniformly Cauchy by Theorem 5.14(b), and hence uniformly convergent, for each n, k ∈ N ∪ {0}.
By taking pointwise limits it is clear that the limits of these sequences agree, namely for each k ∈ N ∪ {0} there
(k)
exists gk ∈ C(R) such that (fj )∞
n=1 converges to gk uniformly on compact sets. If x ∈ R then, by the dominated
convergence theorem,
Z
Z
x
g0 (x) = lim fj (x) = lim
j→∞
j→∞ 0
x
fj0 =
g1 .
0
(k)
By the fundamental theorem of calculus, it follows that g00 = g1 ; similarly g0
converges to g0 ∈ C ∞ (R), and C ∞ (R) is complete.
= gk for all k ∈ N. Therefore (fj )∞
n=1
47. (a) If hTn i∞
n=1 converges to T strongly, then it converges to T weakly. Hence, it suffices to consider only the weak
case. Fix x ∈ X, and let f ∈ Y∗ . Then hf (Tn x)i∞
n=1 converges (to f (T x)), and hence supn∈N |f (Tn x)| < ∞. In
†
other words supn∈N |b
x(Tn f )| = supn∈N |b
x(f ◦ Tn )| < ∞. By the uniform boundedness principle, it follows that
†
M := supn∈N kb
x ◦ Tn k < ∞. Given n ∈ N, either Tn x = 0 or (by Hahn-Banach) there exists g ∈ Y∗ such that
kgk = 1 and g(Tn x) = |g(Tn x)| = kTn xk. In the first case kTn xk = 0 ≤ M, and otherwise
kTn xk = |g(Tn x)| ≤
sup
f ∈Y∗ , kf k=1
|f (Tn x)| =
sup
f ∈Y∗ , kf k=1
|Tn† f (x)| =
sup
f ∈Y∗ , kf k=1
|b
x(Tn† f )| = kb
x ◦ Tn† k ≤ M.
This shows that supn∈N kTn xk ≤ M < ∞, so supn∈N kTn k < ∞ by the uniform boundedness principle.
∗
∞
(b) Let hxn i∞
n=1 by a weakly convergent sequence in X, and let f ∈ X . Then hf (xn )in=1 converges in K, and
hence supn∈N |f (xn )| < ∞. In other words supn∈N |b
xn (f )| < ∞, in which case supn∈N kb
xn k < ∞ by the uniform
∞
boundedness principle. Since b· is an isometry, this shows that hxn in=1 is bounded.
The weak* topology on X∗ is the same as the strong operator topology, so every weak*-convergent sequence in
X∗ is bounded by part (a).
48. (a) Let hxα iα∈A be a net in B which converges weakly to x ∈ X. If f ∈ X∗ and kf k = 1 then
|b
x(f )| = |f (x)| = lim f (xα ) = lim |f (xα )| ∈ [0, 1]
α∈A
α∈A
because | · | is continuous and 0 ≤ |f (xα )| = |b
xα (f )| ≤ kb
xα kkf k = kxα k ≤ 1 for all α ∈ A. This shows that
kxk = kb
xk ≤ 1, so x ∈ B and hence B is weakly closed.
(b) Choose r ∈ (0, ∞) such that E ⊆ Br [0]. Then r−1 E ⊆ B, so r−1 E ⊆ B where r−1 E is the weak closure of
r−1 E. Since the weak topology makes X a toplogical vector space, multiplication by r−1 is weakly continuous
and hence r−1 E ⊆ r−1 E ⊆ B. This implies that E ⊆ Br [0], which shows that E is bounded.
(c) Choose r ∈ (0, ∞) such that F ⊆ Br [0]. Let f be a functional in the weak* closure of F. There exists a net
hfα iα∈A in F which weak*-converges to f. If x ∈ X and kxk = 1 then
|f (x)| = lim fα (x) = lim |fα (x)| ∈ [0, r]
α∈A
α∈A
because | · | is continuous and 0 ≤ |fα (x)| ≤ kfα kkxk = kfα k ≤ r for all α ∈ A. This implies that
kf k =
sup
x∈X, kxk=1
which shows that the weak* closure of F is bounded.
14
|f (x)| ≤ r,
Real Analysis
Chapter 5 Solutions
Jonathan Conder
∗
(d) Let hfn i∞
n=1 be a weak*-Cauchy sequence in X . Then hfm − fn i(m,n)∈N2 weak*-converges to 0. Given x ∈ X,
it follows that hfm (x) − fn (x)i(m,n)∈N2 converges to 0, which means that for every ε ∈ (0, ∞) there exists
(M, N ) ∈ N2 such that |fm (x) − fn (x)| < ε for all m, n ∈ N with m ≥ M and n ≥ N. This implies that
hfn (x)i∞
n=1 is a Cauchy sequence in the usual sense (by taking max{M, N }), so it has a limit f (x) ∈ K. By
exercise 38, it follows that f ∈ X∗ . Since hfn i∞
n=1 converges to f pointwise, it weak*-converges to f.
49. (a) Let E ⊆ X be non-empty and weakly open. Choose x ∈ E. Then there exist ε1 , . . . , εn ∈ (0, ∞) and f1 , . . . , fn ∈
X∗ such that ∩nk=1 {y ∈ X | |fk (x)−fk (y)| < εk } ⊆ E. Since X is infinite-dimensional, we may choose x1 , . . . , x2n ∈
n
X so that {xk }2k=1 is linearly independent. If k ∈ {1, . . . , 2n−1 } then
f1 (x2k )x2k−1 − f1 (x2k−1 )x2k ∈ ker(f1 )
is zero only if f1 (x2k ) = f1 (x2k−1 ) = 0. In either case there exists x0k ∈ span{x2k−1 , x2k } such that x0k 6= 0 and
f1 (x0k ) = 0. Thus {x01 , . . . , x02n−1 } is a linearly independent subset of ker(f1 ). Applying this construction again
gives non-zero vectors x001 , . . . , x002n−2 ∈ ker(f2 ) such that x00k ∈ span{x02k−1 , x02k } ⊆ ker(f1 ) for all k ∈ {1, . . . , 2n−2 }.
(n)
Continuing inductively, we eventually obtain x1
(n)
∈ ∩nk=1 ker(fk ) such that x1
(n)
6= 0. Set y :=
then fk (x − ny) = fk (x) − nfk (y) = fk (x) for all k ∈ {1, . . . , n}, which implies that
x1
(n)
kx1 k
. If n ∈ N
x − ny ∈ ∩nk=1 {y ∈ X | |fk (x) − fk (y)| < εk } ⊆ E.
Moreover n = knyk ≤ kny − xk + kxk and hence kx − nyk ≥ n − kxk, which implies that E is unbounded.
Now let F ⊆ X∗ be non-empty and weak*-open. Every sequence in F c which converges weakly to f ∈ X∗
also weak*-converges to f, so f ∈ F c and hence F c is weakly closed. Therefore F is weakly open, so it is
unbounded by the previous argument (note that X∗ is infinite-dimensional, because otherwise X∗∗ would be a
b
finite-dimensional space with an infinite-dimensional subspace X).
(b) Let E ⊆ X be bounded. The weak closure of E is bounded by exercise 48(b), so it has empty interior by part
(a). Similarly, if F ⊆ X∗ is bounded, its weak* closure is bounded by exercise 48(c), and thus has empty interior
by part (a). Therefore E and F are nowhere dense in the weak and weak* topologies.
(c) This follows immediately from part (b), because X = ∪n∈N Bn [0] and X∗ = ∪n∈N Bn [0].
(d) Suppose the weak* topology on X∗ was defined by some translation-invariant metric ρ. Let hfn i∞
n=1 be a Cauchy
∗
sequence in the metric space (X , ρ). Given a weak*-neighbourhood U of 0, there exists ε ∈ (0, ∞) such that
Bερ (0) ⊆ U. Moreover, there exists N ∈ N such that ρ(fm , fn ) < ε for all m, n ∈ N with m ≥ N and n ≥ N.
This implies that ρ(fm − fn , 0) < ε, and hence fm − fn ∈ Bερ (0) ⊆ U, for all (m, n) ∈ N2 with m ≥ N and
∗
n ≥ N. This means that hfm − fn i(m,n)∈N2 weak*-converges to 0, so hfn i∞
n=1 is a weak*-Cauchy sequence in X ,
∗
∗
which weak*-converges by exercise 48(d). Therefore hfn i∞
n=1 converges in (X , ρ), which shows that (X , ρ) is a
complete metric space. This contradicts the Baire category theorem, by part (c). Therefore, the weak* topology
on X∗ is not defined by a translation-invariant metric.
50. Let B be the closed unit ball in X∗ , and choose dense subsets D ⊆ X and E ⊆ K. The collection of finite intersections
of sets of the form x
b−1 (Bq (c)) ∩ B, where x ∈ D, q ∈ Q ∩ (0, ∞) and c ∈ E, is clearly countable. If U ⊆ B is open
and f ∈ U then f ∈ ∩nk=1 x
b−1
k (Uk ) ∩ B ⊆ U for some x1 , . . . , xn ∈ X and open sets U1 , . . . , Un ⊆ K (by definition of
the weak* topology). It therefore suffices to show the following: given x ∈ X and U ⊆ K open such that f (x) ∈ U,
there exist y ∈ D, q ∈ Q ∩ (0, ∞) and c ∈ E such that f ∈ yb−1 (Bq (c)) ∩ B ⊆ x
b−1 (U ) ∩ B.
15
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Jonathan Conder
To this end, choose r ∈ (0, ∞) such that B3r (f (x)) ⊆ U, y ∈ Br (x) ∩ D, q ∈ Q ∩ (0, r) and c ∈ Bq (f (x)) ∩ E. Clearly
f ∈ yb−1 (Bq (c)). Moreover, if g ∈ yb−1 (Bq (c)) and kgk ≤ 1 then
|g(x) − f (x)| ≤ |g(x) − g(y)| + |g(y) − c| + |c − f (x)| < kgkkx − yk + q + r < 3r
and hence g ∈ x
b−1 (U ). This shows that f ∈ yb−1 (Bq (c)) ∩ B ⊆ x
b−1 (U ) ∩ B, so B is second countable. It is also normal,
because it is compact and Hausdorff, and hence metrisable.
51. Every weakly closed subset of X is norm-closed, because the weak topology is weaker than the norm topology (every
element of X∗ is norm-continuous, by definition). Conversely, let M ⊆ X be a norm-closed subspace. If x ∈ X \ M,
there exists (by Hahn-Banach) fx ∈ X∗ such that fx |M = 0 and fx (x) 6= 0. It is clear that M = ∩x∈X\M ker(fx ), which
is weakly closed because each fx is weakly continuous.
52. (a) Obviously M ⊆ N0 . If f ∈ N0 , then the image of (T, f ) : X → Cn+1 is a (closed) subspace not containing en+1 .
By Hahn-Banach there exists λ ∈ (Cn+1 )∗ such that λ|Im(T,f ) = 0 but λ(en+1 ) 6= 0. If x ∈ X then
0 = λ(f1 (x), . . . , fn (x), f (x)) =
n
X
fi (x)λ(ei ) + f (x)λ(en+1 ),
i=1
and hence f = −
λ(ei )
i=1 λ(en+1 ) fi
Pn
∈ M. It follows from Exercise 23 that M ∼
= (X/N)∗ .
(b) Let ε ∈ (0, ∞) and π : X → X/N the quotient map. By Exercise 23 π † : (X/N)∗ → M is an isometric
b
isomorphism, so 𠆆 : M∗ → (X/N)∗∗ is also an isomorphism († is a functor). Note that 𠆆 (F |M ) = X
for some X ∈ X/N (the embedding X/N → (X/N)∗∗ is between spaces of the same dimension). For each
i ∈ {1, . . . , n} there exists f i ∈ (X/N)∗ such that f i ◦ π = fi (by the universal property of quotient spaces), so
F (fi ) = F (π † (f i )) = 𠆆 (F |M )(f i ) = f i (X) = fi (x) for all x ∈ X. It remains to show that kxk ≤ (1 + ε)kF k for
b = kF ◦ π † k ≤ kF kkπ † k = kF k.
some x ∈ X; this follows from the fact that kXk = kXk
(c) If f ∈ X∗ and x ∈ X then, by definition f (b
x) = f (x) = x
b(f ), so we identify f with fb|Xb , which is continuous in the
b because fb is continuous in the weak* topology on X∗∗ . Thus, the weak topology
relative weak* topology on X,
b is weaker than the relative weak* topology.
on X
Conversely, note that the weak* topology on X∗∗ is generated by sets of the form fb−1 (U ), where f ∈ X∗ and
b is generated by sets of the
U ⊆ K is open. It follows from Proposition 4.4 that the relative weak* topology on X
b In particular, it suffices to show that fb| b is continuous in the weak topology, for all f ∈ X∗ . If
form fb−1 (U ) ∩ X.
X
x ∈ X then, by definition fb(x) = fb(b
x) = x
b(f ) = f (x), so we identify fb| b with the continuous functional f.
X
X∗∗ ,
X∗ .
(d) Let F ∈
ε1 , . . . , εn ∈ (0, ∞) and f1 , . . . , fn ∈
By part (b), there exists x ∈ X such that F (fi ) = fi (x) for
all i ∈ {1, . . . , n} (so far we have not used the independence of {f1 , . . . , fn }). Equivalently
|fbi (b
x) − fbi (F )| = |b
x(fi ) − F (fi )| = 0
for all i ∈ {1, . . . , n}, in which case x
b ∈ ∩ni=1 UF fbi εi . Sets of this form give a neighbourhood base at F (in the
b so X
b is weak*-dense in X∗∗ .
weak* topology), by Theorem 5.14(a). It follows that F ∈ X,
If kF k < 1 we may choose x with kxk < 1, and if kF k = 1 then every weak*-neighbourhood U of F contains
G ∈ X∗∗ such that kGk < 1 (U is also a neighbourhood in the norm topology), and again we may choose x with
b is weak*-dense in the closed unit ball of X∗∗ .
kxk < 1. Thus, the closed unit ball of X
16
Real Analysis
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Jonathan Conder
(e) If X is reflexive, its closed unit ball corresponds to the closed unit ball of X∗∗ , which is weak*-compact by
Alaoglu’s theorem, and hence weakly compact (in X) by part (c). Conversely, if the closed unit ball of X is
weakly compact, its image in X∗∗ is compact in the (relative) weak* topology, which is Hausdorff by Proposition
b for some
5.16, so its image is the closed unit ball of X∗∗ by part (d). Thus, if F ∈ X∗∗ \ {0} then kF1 k F = x
\
x ∈ X, in which case F = kF
kx. This shows that X is reflexive.
∞
53. (a) Let hxn i∞
n=1 be a sequence in X which converges to x ∈ X. Then hTn xin=1 converges to T x, and by exercise 47(a)
there exists M ∈ (0, ∞) such that kTn k ≤ M for all n ∈ N. Given ε ∈ (0, ∞), there exists N ∈ N such that
ε
kxn − xk < 2M
and kTn x − T xk < 2ε for all n ∈ N with n ≥ N. It follows that
kTn xn − T xk ≤ kTn xn − Tn xk + kTn x − T xk < kTn (xn − x)k +
ε
ε
ε
ε
≤ kTn kkxn − xk + < M
+ =ε
2
2
2M
2
for all n ∈ N with n ≥ N, which implies that hTn xn i∞
n=1 converges to T x.
∞
∞
(b) Let x ∈ X. Then hSn xi∞
n=1 converges to Sx. By part (a), it follows that h(Tn Sn )xin=1 = hTn (Sn x)in=1 converges
to T Sx. This shows that hTn Sn i∞
n=1 converges to T S in the strong operator topology.
56. If x ∈ E, and y ∈ E ⊥ , then x ⊥ y and hence x ∈ (E ⊥ )⊥ . It is clear that (E ⊥ )⊥ is a closed subspace of H. If F is
another closed subspace of H containing E, then F ⊥ ⊆ E ⊥ because, for each x ∈ F ⊥ , x ⊥ y for all y ∈ F and hence
all y ∈ E. Similarly (E ⊥ )⊥ ⊆ (F ⊥ )⊥ . Given x ∈ (F ⊥ )⊥ , write x = y + z for some y ∈ F and z ∈ F ⊥ (this works
because H = F ⊕ F ⊥ ). Note that 0 = hx, zi = hy, zi + hz, zi = kzk2 , so z = 0 and hence x ∈ F. This shows that
(F ⊥ )⊥ ⊆ F, so (E ⊥ )⊥ is a subset of every closed subspace of H containing E, whence it is the smallest one.
57. (a) The definition of T ∗ is given, and T ∗ ∈ L(H, H) because conjugation is an involution. If x, y ∈ H then
hx, T ∗ yi = hx, V −1 T † V yi = (V V −1 T † V y)(x) = (V y)(T x) = hT x, yi,
and if T 0 ∈ L(H, H) is another map with this property then hx, T 0 yi = hT x, yi = hx, T ∗ yi for all x, y ∈ H, so
T 0 y = T ∗ y for all y ∈ H and hence T 0 = T ∗ .
(b) If x, y ∈ H then hT ∗ x, yi = hy, T ∗ xi = hT y, xi = hx, T yi, which implies that T = T ∗∗ by part (a). If x ∈ H
kT xk2 = hT x, T xi = hx, T ∗ T xi ≤ kxkkT ∗ T xk ≤ kxkkT ∗ kkT xk
and hence kT xk ≤ kT ∗ kkxk, which shows that kT k ≤ kT ∗ k. Thus kT ∗ k ≤ kT ∗∗ k = kT k, so kT ∗ k = kT k. Now
kT ∗ T k ≤ kT ∗ kkT k = kT k2 ; a variant of the above calculation gives kT xk2 ≤ kT ∗ T kkxk2 for all x ∈ H and hence
kT k2 ≤ kT ∗ T k. Therefore kT ∗ T k = kT k2 .
If a, b ∈ C and S, T ∈ L(H, H) then
(aS + bT )∗ = V −1 (aS + bT )† V = V −1 (aS † + bT † )V = aV −1 S † V + bV −1 T † V = aS ∗ + bT ∗
and (ST )∗ = V −1 (ST )† V = V −1 T † S † V = V −1 T † V V −1 S † V = T ∗ S ∗ .
(c) By Theorem 5.25
Im(T )⊥ = {x ∈ H | hy, xi = 0 for all y ∈ Im(T )}
= {x ∈ H | hT y, xi = 0 for all y ∈ H}
= {x ∈ H | hy, T ∗ xi = 0 for all y ∈ H}
17
Real Analysis
Chapter 5 Solutions
Jonathan Conder
= {x ∈ H | T ∗ x = 0}
= N(T ∗ )
and hence N(T )⊥ = N(T ∗∗ )⊥ = (Im(T ∗ )⊥ )⊥ is the smallest closed subspace of H containing Im(T ∗ ). In particular
Im(T ∗ ) ⊆ N(T )⊥ . But Im(T ∗ ) is a subspace; thus Im(T ∗ ) = N(T )⊥ .
(d) If T is unitary, it is invertible. Moreover hT x, yi = hT x, T T −1 yi = hx, T −1 yi for all x, y ∈ H, which shows that
T −1 = T ∗ by part (a). Conversely, if T is invertible and T −1 = T ∗ then hT x, T yi = hx, T ∗ T yi = hx, yi for all
x, y ∈ H. In other words T is unitary.
58. (a) Let x, y ∈ H and a ∈ C. Then (ax + y) − (aP x + P y) = a(x − P x) + (y − P y) ∈ M⊥ because M⊥ is a subspace
of H, and aP x + P y ∈ M because P x, P y ∈ M and M is a subspace of H. Therefore P (ax + y) = aP x + P y,
which shows that P is linear. If x ∈ H then
kP xk2 ≤ kP xk2 + kx − P xk2 = kP x + x − P xk2 = kxk2
by Pythagoras, so kP xk ≤ kxk and hence P ∈ L(H, H). Moreover, if x, y ∈ H then
hP x, yi = hP x, y − P y + P yi = hP x, y − P yi + hP x, P yi = hx − x + P x, P yi = hx, P yi − hx − P x, P yi = hx, P yi,
so P ∗ = P. If x ∈ H then P 2 x = P (P x) ∈ M and x − P 2 x = (x − P x) + (P x − P 2 x) ∈ M⊥ , which implies
that P x = P 2 x. Therefore P 2 = P. If x ∈ M then x − x = 0 ∈ M⊥ , so P x = x and hence x ∈ Im(P ). Since
Im(P ) ⊆ M by definition, it follows that Im(P ) = M. Therefore M⊥ = Im(P )⊥ = N(P ∗ ) = N(P ).
(b) Since P 2 = P it is clear that Im(P ) = N(P − I), which is closed because P − I is bounded. If x ∈ H then
x − P x ∈ N(P ) = N(P ∗ ) = Im(P )⊥ because P 2 = P. Therefore P is the orthogonal projection onto Im(P ).
P
(c) Let x ∈ H. Clearly α∈A hx, uα iuα ∈ M. If aα ∈ C for each α ∈ A then
*
+
*
+
X
X
X
X
X
aα x −
aα (hx, uα i − hx, uα i) = 0,
hx, uβ iuβ , uα =
x−
hx, uα iuα ,
aα uα =
α∈A
which implies that x −
α∈A
P
α∈A hx, uα iuα
α∈A
α∈A
β∈A
∈ M⊥ . Therefore P x =
P
α∈A hx, uα iuα .
∞
59. Choose a sequence hxn i∞
n=1 in K such that hkxn kin=1 converges to r := inf x∈K kxk. Given ε ∈ (0, ∞), there exists
N ∈ N such that kxn k < r + 2ε for all n ∈ N with n ≥ N. If m, n ∈ N and m ≥ n ≥ N then
kxm − xn k2 + kxm + xn k2 = hxm − xn , xm − xn i + hxm + xn , xm + xn i = 2kxm k2 + 2kxn k2 < 4(r + 2ε )2
2
and kxm +xn k2 = 4k 21 xm + 12 xn k2 ≥ 4r2 (because 21 xm + 12 xn ∈ K), so kxm −xn k2 < 4 ε4 = ε2 and hence kxm −xn k < ε.
This shows that hxn i∞
n=1 is a Cauchy sequence, so it has a limit x ∈ K = K. Since k · k is continuous, it is clear that
kxk = limn→∞ kxn k = r, so x has the smallest norm among the elements of K. It is clear that x is unique, because if
y ∈ K and kyk = r then
kx − yk2 = 2kxk2 + 2kyk2 − kx + yk2 = 2r2 + 2r2 − 4k 12 x + 12 yk2 ≤ 4r2 − 4r2 = 0.
∗
63. (a) Let hun i∞
n=1 be an orthonormal sequence in H. If f ∈ H , there exists x ∈ H such that f = h·, xi. Also
∞
X
|hx, un i|2 ≤ kxk2
n=1
(this is Bessel’s inequality), so limn→∞ |f (un )|2 = limn→∞ |hun , xi|2 = limn→∞ |hx, un i|2 = 0. It follows that
limn→∞ f (un ) = 0 = f (0), so hun i∞
n=1 converges weakly to 0.
18
Real Analysis
Chapter 5 Solutions
Jonathan Conder
x
(b) Let x ∈ B. If x 6= 0, construct an orthonormal sequence {un }∞
n=0 in H such that u0 = kxk , using the GramSchmidt process and the fact that H is infinite-dimensional. Otherwise, construct an orthonormal sequence
p
{un }∞
1 − kxk2 , and for each n ∈ N define xn := aun + x.
n=1 in H. In either case un ⊥ x for all n ∈ N. Set a :=
Then (xm − x) ⊥ (xn − x) for all m, n ∈ N, and
kxn k2 = haun + x, aun + xi = |a|2 kun k2 + ahun , xi + ahx, un i + kxk2 = a2 + 0 + 0 + kxk2 = 1
for all n ∈ N. This implies that hxn i∞
n=1 is a sequence in S. Moreover, by Bessel’s inequality
∞
X
2
|hy, xn − xi| ≤
n=1
∞
X
2
kxn − xk |hy,
2
xn −x
kxn −xk i|
≤
n=1
∞
X
−x
i|2 ≤ 4kyk2 ,
(kxn k + kxk)2 |hy, kxxnn −xk
n=1
and hence limn→∞ hxn − x, yi = 0, for all y ∈ H. This implies that hxn i∞
n=1 converges weakly to x, by a similar
argument to part (a). Therefore x is in the weak closure of S, so S is weakly dense in B.
66. (a) Let T : M ,→ C([0, 1]) be the inclusion. Also let hfn i∞
n=1 be a sequence in M which converges to f ∈ M such
∞
that hT fn in=1 converges to g ∈ C([0, 1]). Suppose that T f 6= g. Then |f (x) − g(x)| > 0 for some x ∈ [0, 1], so
there exist ε, δ ∈ (0, ∞) such that
|f (y) − g(y)| > ε for all
y ∈ [0, 1] such that |x − y| < δ. Moreover, there exists
√
√
ε δ
ε δ
n ∈ N such that kf − fn kL2 < 2 and kT fn − gku < 2 . It follows that
s
s
√
Z 1
Z 1
√
√
ε
δ
2
ε δ≤
+
|f − g| = kf − gkL2 ≤ kf − fn kL2 + kT fn − gkL2 <
kT fn − gk2u < ε δ,
2
0
0
which is plainly impossible. Therefore T f = g, which shows that T is a closed linear map, so by the closed graph
theorem (note that M is closed, thus complete, and C([0, 1]) is also complete) there exists C ∈ (0, ∞) such that
kf ku = kT f ku ≤ Ckf kL2 for all f ∈ M.
(b) Given x ∈ [0, 1], define x
b : M → C by x
b(f ) := f (x). Since addition and scalar multiplication in M are defined
b ∈ M∗ . It follows that
pointwise, x
b is linear. Moreover, if f ∈ M then |b
x(f )| = |f (x)| ≤ kf ku ≤ Ckf kL2 , so x
x
b = h·, gx i for some gx ∈ M. In other words, f (x) = hf, gx i for all f ∈ M. In particular, kgx k2L2 = hgx , gx i =
gx (x) ≤ kgx ku ≤ Ckgx kL2 , which implies that kgx kL2 ≤ C (even if kgx kL2 = 0).
(c) Let hfn iN
n=1 be a finite orthonormal sequence in M. If x ∈ [0, 1] there exists gx ∈ M such that f (x) = hf, gx i for
all f ∈ M. Using Bessel’s inequality, it follows that
N
X
2
|fn (x)| =
n=1
Therefore
N=
N
X
2
|hfn , gx i| =
n=1
N
X
n=1
kfn k2L2
=
N
X
|hgx , fn i|2 ≤ kgx k2L2 ≤ C 2 .
n=1
N Z
X
1
2
Z
|fn | =
n=1 0
N
1X
2
Z
|fn | ≤
0 n=1
1
C 2 = C 2,
0
so every linearly independent subset of M has cardinality at most C 2 (otherwise one could construct an orthonormal set with more than C 2 elements using the Gram-Schmidt process).
67. Note that M = N(U − I) = N(U (I − U ∗ )) = N(I − U ∗ ) = N((I − U )∗ ) = Im(I − U )⊥ is a closed subspace of H, by
Exercise 57. Therefore M⊥ = (Im(I − U )⊥ )⊥ = Im(I − U ) by Exercise 56. If x ∈ M and n ∈ N then
n−1
n−1
k=0
k=0
1X k
1X
n
Sn x =
U x=
x = x = x,
n
n
n
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Real Analysis
Chapter 5 Solutions
Jonathan Conder
and hence limn→∞ Sn x = x. On the other hand, if x ∈ M⊥ and ε ∈ (0, ∞), there exists y ∈ Im(I − U ) such that
kx − yk < 2ε . Moreover, if z ∈ H such that y = z − U z then
n−1
1 n−1
1
X
X
1
1
2
kSn yk = kSn (z − U z)k = U kz −
U k+1 z = kz − U n zk ≤ (kzk + kU n zk) = kzk
n
n
n
n
n
k=0
k=0
for all n ∈ N, so there exists N ∈ N such that kSn yk <
ε
2
for all n ∈ N with n ≥ N. It follows that
n−1
1 n−1
ε
1X k
ε
n
ε
X k
kSn xk ≤ kSn (x − y)k + kSn yk < U (x − y) + ≤
kU (x − y)k + = kx − yk + < ε
n
2
n
2
n
2
k=0
k=0
for all n ∈ N with n ≥ N. Therefore limn→∞ Sn x = 0. If x ∈ H then P x ∈ M and x − P x ∈ M⊥ , so
lim Sn x = lim Sn (P x + x − P x) = lim Sn (P x) + lim Sn (x − P x) = P x.
n→∞
n→∞
n→∞
This shows that hSn i∞
n=1 converges to P in the strong operator topology.
20
n→∞