Solutions to Exercise 2 on Sheet 4 in
Functional Analysis,
LMU Munich, Summer Semester 2017
Peter Philip, Sabine Bögli
May 18, 2017
2. Let (X, T ) and P := (pi )i∈I be as in Theorem 1.40(a). Prove that if (X, T ) is T1 and
I = N, then
ci pi (x − y)
+
d : X × X −→ R0 , d(x, y) := max
: i∈N ,
(1)
1 + pi (x − y)
with (ci )i∈N an arbitrary sequence in R+ converging to 0 (e.g., ci := 1i ), is a metric on
X that induces T . Furthermore, show that d is translation-invariant and the set of open
metric balls D := {Bd,r (0) : r ∈ R+ } forms a convex balanced local base for T at 0.
Solution:
+
Since each pi is R+
0 -valued, d is well-defined and R0 -valued as well. Since pi (0) = 0 for each
i ∈ I, d(x, x) = 0 for each x ∈ X. If x 6= y, then x − y 6= 0, i.e. there exists i ∈ N such that
pi (x − y) > 0 (as P is separating due to (X, T ) being T1 ). Thus, d(x, y) > 0, showing d to
be positive definite. Since pi (x−y) = pi (y −x) for each x, y ∈ X, we have d(x, y) = d(y, x)
and d is symmetric. For the triangle inequality, one first notices that f : ] − 1, ∞[−→ R,
f (x) := x/(1 + x), is strictly increasing due to f 0 (x) = (1)/(1 + x)2 > 0. Thus, for each
x, y, z ∈ X and each i ∈ N, pi (x, z) ≤ pi (x, y) + pi (y, z) implies
pi (x, y) + pi (y, z)
pi (x, y)
pi (y, z)
pi (x, z)
≤
=
+
1 + pi (x, z)
1 + pi (x, y) + pi (y, z)
1 + pi (x, y) + pi (y, z) 1 + pi (x, y) + pi (y, z)
pi (x, y)
pi (y, z)
≤
+
.
1 + pi (x, y) 1 + pi (y, z)
In consequence,
ci pi (x − z)
d(x, z) = max
: i∈N
1 + pi (x − z)
ci pi (x − y)
ci pi (y − z)
≤ max
: i ∈ N + max
: i∈N
1 + pi (x − y)
1 + pi (y − z)
= d(x, y) + d(y, z),
completing the proof that d is a metric. That d is translation-invariant is immediate from
(1). Let S be the topology on X induced by d. Next, we fix r ∈ R+ . Since limi→∞ ci = 0,
∃
K∈N
∀
i>K
ci ≤ r, i.e.
ci p i
< r.
1 + pi
Let Ir := {i ∈ N : ci > r}. Then Ir is finite and, for each x ∈ X,
ci pi (x)
ci pi (x)
d(0, x) < r ⇔
∀
<r
⇔
∀
<r
i∈I 1 + pi (x)
i∈Ir 1 + pi (x)
r
⇔
∀ pi (x) < ri :=
.
i∈Ir
ci − r
Thus,
(
X
Bd,r (0) = T
for Ir = ∅,
6 ∅.
i∈Ir Bpi ,ri (0) for Ir =
Since each Bpi ,ri (0) is convex and balanced, so is Bd,r (0). Since each Bpi ,ri (0) is T -open
(as pi is T -continuous) and Ir is finite, Bd,r (0) ∈ T , showing S ⊆ T . Finally, let U ∈ U(0)
and B ⊆ U as in the proof of Th. 1.40(a)(i), where we may assume n > 1. If r :=
1
min{cik n−1 : k ∈ {1, . . . , N }} and x ∈ Bd,r (0), then
2
cik n>1
1
cik pik (x)
<r≤
∀
⇒ pik (x) <
,
k∈{1,...,N }
1 + pik (x)
2n
n
showing Bd,r (0) ⊆ B ⊆ U . In consequence, the Bd,r (0) form a local base for T at 0,
T ⊆ S, and the proof is complete.