Solutions to Exercises 2, 4 on Sheet 3 in
Functional Analysis,
LMU Munich, Summer Semester 2017
Peter Philip, Sabine Bögli
May 11, 2017
2. Prove Theorem 1.29.
Solution:
(a): It suffices to show the equivalence between (i) and (iii). Since d1 is translationinvariant, we have
∀
d1 (xn , xm ) = d1 (xn − xm , 0).
m,n∈N
If (xn )n∈N is d1 -Cauchy and U ∈ U(0), then there exists r ∈ R+ such that Br (0) ⊆ U and
∃
N ∈N
∀
m,n>N
d1 (xn , xm ) < r.
Thus, for each m, n > N , xn − xm ∈ Br (0) ⊆ U , showing (xn )n∈N to be Cauchy in the
sense of Def. 1.27. Conversely, if (xn )n∈N is Cauchy in the sense of Def. 1.27 and r ∈ R+ ,
then Br (0) ∈ U(0) implies
∃
N ∈N
∀
m,n>N
xn − xm ∈ Br (0).
Thus, for each m, n > N , d1 (xn , xm ) = d1 (xn − xm , 0) < r, showing (xn )n∈N to be d1 Cauchy.
(b) is an immediate consequence of (a).
(c): By definition of an F -space, TY is induced by a complete translation-invariant metric
d. We consider the corresponding balls B 1 (0) ⊆ Y , which are open in Y , i.e. there exist
n
Un ∈ T ∩ U(0) such that B 1 (0) = Y ∩ Un . Using Prop. 1.5(a), we may choose a sequence
n
(Vn )n∈N in T ∩ U(0) such that
∀
Vn = −Vn ∧ Vn + Vn ⊆ Un ∧ Vn+1 ⊆ Vn .
(1)
n∈N
We let x ∈ Y and have to show x ∈ Y . Consider an arbitrary W ∈ T ∩ U(0). Letting clY
denote the closure in Y , we show
\
∃
{yW } =
clY Y ∩ x + (W ∩ Vn ) :
(2)
yW ∈Y
n∈N
Since x ∈ Y and x + (W ∩ Vn ) ∈ U(x), there exists yn ∈ Fn := Y ∩ x + (W ∩ Vn ) . Then
(yn )n∈N is a sequence in Y and we verify the sequence to be Cauchy: Given > 0, choose
N ∈ N such that N1 < . Then
(1)
∀
m,n>N
(1)
yn − ym ∈ Y ∩ (Vn − Vm ) ⊆ Y ∩ (VN − VN ) ⊆ Y ∩ UN = B 1 (0) ⊆ B (0),
N
proving (yn )n∈N to be Cauchy. As Y is assumed to be complete,
there exists yW ∈ W such
T
that limn→∞ yn = yW . Since Fn+1 ⊆ Fn , we know yW ∈ n∈N clY (Fn ), proving ⊆ in (2).
Since, for each n ∈ N, Y ∩ (Vn − Vn ) ⊆ Y ∩ Un = B 1 (0), we have limn→∞ diam(Fn ) = 0. In
n
T
consequence, there can not be y0 6= yW such that y0 ∈ n∈N clY (Fn ) (d(y0 , yW ) > 0 would
be in contradiction to limn→∞ diam(Fn ) = 0). This proves ⊇ in (2) and, thus, equality.
Since
∀
Y ∩ x + (W ∩ Vn ) ⊆ Y ∩ x + (X ∩ Vn ) = Y ∩ (x + Vn ),
W ∈T ∩U(0)
we have yW = yX for each W ∈ T ∩ U(0). Moreover, for each such W , we have Y ∩ x +
(W ∩ Vn ) ⊆ x + W , showing yW ∈ x + W . Since (X, T ) is T1 , this shows yW = x, i.e.
x ∈ Y and Y is closed in X.
4. Consider the following two metrics on C[0, 1] :
Z 1
d1 (f, g) :=
|f (t) − g(t)| dt, d∞ (f, g) := max |f (t) − g(t)|.
0
t∈[0,1]
(a) Prove that (C[0, 1], d1 ) is not complete (in contrast to (C[0, 1], d∞ )).
(b) Prove that d1 and d∞ are not equivalent. Is it true that one metric generates a finer
topology than the other?
Solution:
(a): We construct a Cauchy sequence in (C[0, 1], d1 ) that is not convergent. For n > 2
define


t ∈ 0, 21 − n1 ,
−1,
fn (t) := n t − 12 , t ∈ 12 − n1 , 12 + n1 ,


1,
t ∈ 12 + n1 , 1 .
Note that fn (t) ∈ [−1, 0] for t ∈ 0, 12 , and
fn (t) ∈ [0, 1] for t ∈ 21 , 1 . Let
n ≥ N .
1 m,
1
1
1
Then
fm (t) = fn (t)
= −1 for t ∈ 0, 2 − N and fm (t) = fn (t) = 1 for t ∈ 2 + N , 1 . For
t ∈ 21 − N1 , 12 + N1 we estimate |fn (t) − fm (t)| ≤ 1. Thus
Z 1+ 1
2
N
2
d1 (fn , fm ) =
|fn (t) − fm (t)| dt ≤ .
1
1
N
−N
2
In the limit N → ∞ the latter converges to 0, which proves that (fn )n is a Cauchy sequence
with respect to d1 . Assume that there exists f ∈ C[0, 1] such that limn→∞ d(fn , f ) = 0.
Since f is continuous, there exists δ > 0 such that |f (t) − f ( 21 )| < 12 for t ∈ 12 − δ, 12 + δ .
Then, for n > 1δ ,
1
+δ
2
1
+δ
2
Z 1 +δ 2
1 1 d1 (fn , f ) ≥
|fn (t) − f (t)| dt ≥
fn (t) − f
f (t) − f
dt −
dt
1
1
1
2
2
−δ
−δ
−δ
2
2
2
Z 1 +δ Z 1 +δ Z 1− 1 2
n 2
2
1
1
dt +
dt −
−1 − f
1 − f
f (t) − f 1 dt
≥
1
1
1
2 2 2 −δ
+1
−δ
2
2 n 2
1
−1 − f 1 + 1 − f 1 − δ ≥ δ − 1 2 − δ = δ − 2 .
≥ δ−
n
2
2 n
n
Z
Z
Since in the limit n → ∞ the right hand side converges to δ > 0, we obtain a contradiction.
So no such f can exist.
(b): One may check that both metrics are translation-invariant. If they were equivalent,
then Th. 1.29(b) would imply that (C[0, 1], d1 ) is complete since (C[0, 1], d∞ ) is complete.
The obtained contradiction shows that the metrics cannot be equivalent.
However, since d1 (f, g) ≤ d∞ (f, g), we have Bd1 ,r (f ) ⊆ Bd∞ ,r (f ) and hence d1 generates
a finer topology than d∞ .