20122 Week 2: Solutions
Problem 9.
Explain why neither of the following functions d defines a metric on the set
of continuous functions [0, 1] → R:
1. d(f, g) = |f (1/2) − g(1/2)|
2. d(f, g) = supx∈[0,1] |f 0 (x) − g 0 (x)|, where f 0 (x) := df (x)/dx.
Solution 9.
1. Let f (x) = x and g(x) = 1 − x. Then d(f, g) = |1/2 − (1 − (1/2)| = 0; so
the first axiom of Definition 1.1 fails.
2. If one of f or g fails to be differentiable for some x ∈ [0, 1], then d(f, g) is
not defined! And even if we restrict attention to the subset of differentiable
functions, the examples f (x) = 1 and g(x) = 0 show that d(f, g) = |0−0| =
0; so the first axiom fails again.
Problem 10.
Assume that 1 < a < b in R.
In the space of polynomials P[a, b], compute dsup (x3 , 2 − x2 ); what do your
calculations imply about dsup (x3 , 2 − x2 ) in the space of continuous functions
C[a, b]?
Compute d1 (x3 , 2 − x2 ) in the space L1 [a, b].
Solution 10.
In P[a, b],
dsup (x3 , 2 − x2 ) = sup |x3 + x2 − 2| .
x∈[a,b]
Since 1 < a < b, it follows that x3 + x2 > 2 on [a, b], and attains its maximum
value at x = b. So dsup (x3 , 2 − x2 ) = b3 + b2 − 2. The same holds in C[a, b], via
Definition 1.2.
In L1 [a, b],
Z b
b
3
2
d1 (x , 2 − x ) =
|t3 + t2 − 2|dt = t4 /4 + t3 /3 − 2t a .
a
1
This reduces to (b4 − a4 )/4 + (b3 − a3 )/3 − 2(b − a).
Problem 11.
Consider the set X of closed intervals in the Euclidean line, with the interval
metric dH . Describe the elements of the closed ball B 1 ([0, 1]), and explain its
∞
relationship with the closed ball B 1 ((0, 1)) ⊂ R2 under the metric d∞ .
Solution 11.
The interval [0, 1] is a point of (X, dH ). For any other point [r, s],
dH ([0, 1], [r, s]) = max{|r|, |s − 1|}
so points in the ball B 1 ([0, 1]) satisfy max{|r|, |s − 1|} ≤ 1. These are precisely
the intervals [r, s] for which −1 ≤ r ≤ 1 and 0 ≤ s ≤ 2. Of course, r ≤ s is also
required. Examples include the intervals
√
[−1, 0], [−1, 2], {0}, [0, 2], [1/2, 1], {1}, [1, π/2], [1, 2] .
. . . and infinitely many others!
The function (X, dH ) → (R2 , d∞ ) defined by [x1 , x2 ] 7→ (x1 , x2 ) is an isometry
onto the half-space H := {(x1 , x2 ) : x1 ≤ x2 }. In particular, it identifies B 1 ([0, 1])
∞
with B 1 ((0, 1)) ∩ H. The latter is given by
{(r, s) : r ≤ s, −1 ≤ r ≤ 1, 0 ≤ s ≤ 2} .
This is obtained from a closed square in R2 with vertices (1, 2), (−1, 2), and
(−1, 0); its fourth vertex (1, 0) is deleted, along with the entire lower right corner
r > s. The result is a pentagon with two pairs of parallel edges.
Problem 12.
Let (X, d1 ) be the space of sequences a = (ai : i ≥ 0) of Example 1.30,
equipped with the l1 metric. Find a distance preserving injection of the space
({0, 1}∞ , d∗ ) of Problem 8 into (X, d1 ).
/ B1/10 (0).
Find an example of a point a in X such that a ∈ B 1/10 (0), but a ∈
Solution 12.
Consider the function i : {0, 1}∞ → X defined by
i(x0 x1 . . . xj . . . ) = (x0 , x1 /2, . . . , xj /2j , . . . ) ;
the right hand side is well defined, because
∞
X
j=0
|xj /2j | ≤
∞
X
1/2j = 2 ,
j=0
as required. Obviously i is injective; it also preserves distance, because
∞
∞
X
X
|xj − yj |
=
|xj /2j − yj /2j | = d1 (i(x), i(y)) ,
d (x, y) =
j
2
j=0
j=0
∗
2
as sought.
P
A point a satisfies a ∈ B 1/10 (0) and a ∈
/ B1/10 (0) iff ∞
j=0 |aj | = 1/10. Any
example such as
(1/10, 0, 0, . . . ) or (1/20, 1/40, . . . , 1/2j .10, . . . )
will do!
Problem 13.
Maurice Fréchet was the mathematician who invented the concept of a metric
space; in which year did he first publish the idea? He did not, however, use the
term metric space; which mathematician was responsible for the name?
Solution 13.
The concept appeared in Fréchet’s PhD thesis in 1906. The name Metric
Space is due to the German mathematician Felix Hausdorff. For more, see
http://www-history.mcs.st-and.ac.uk/Biographies/Frechet.html
for example.
Problem 14.
Prove that the function d(m, n) = |m−1 − n−1 | for any m, n ∈ N defines a
metric on the set of natural numbers. Does this metric extend to R+ ?
Solution 14.
Note that d is well defined because 0 ∈
/ N. The first axiom of Definition 1.1
holds because |m−1 − n−1 | = 0 iff 1/m = 1/n, and therefore m = n; the second
is automatic because |m−1 − n−1 | = |n−1 − m−1 |. For the third axiom,
|m−1 − n−1 | = |(m−1 − p−1 ) − (n−1 − p−1 )| ≤ |m−1 − p−1 | + |n−1 − p−1 |
implies that d(m, n) ≤ d(m, p) + d(p, n) for all m, n, p ∈ N.
Every step in this analysis applies just as well to positive reals as positive
integers, so the metric does extend to R+ .
Problem 15.
Show that
1. the Euclidean and taxicab metrics d2 and d1 on Rn
2. the metrics dmin and d∗ on {0, 1}∞
are Lipschitz equivalent. Deduce that, in either case, if a set is open with one
metric it is open with the other.
Solution 15.
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1. For any x, y ∈ Rn , note that
0 ≤ d2 (x, y)2 =
n
X
(xi − yi )2 ≤
n
X
i=1
2
|xi − yi | = d1 (x, y)2 ,
i=1
so d2 (x, y) ≤ d1 (x, y). Moreover,
2 1/2
|xi − yi | = ((xi − yi ) )
≤
n
X
(xi − yi )2
1/2
i=1
for every 1 ≤ i ≤ n, so d1 (x, y) ≤ nd2 (x, y), as required.
P
j
2. For any x, y ∈ {0, 1}∞ , note that d∗ (x, y) = 1/2n + ∞
j=n+1 |xj − yj |/2 by
n
∗
definition, where dmin (x, y) = 1/2 . Thus dmin (x, y) ≤ d (x, y). Moreover
∞
X
|xj − yj |/2j ≤
j=n+1
∞
X
1/2j = 1/2n ,
j=n+1
so d∗ (x, y) ≤ 1/2n + 1/2n = 2dmin (x, y), as required.
If two metrics d and d0 on X satisfy
hd0 (x, y) ≤ d(x, y) ≤ kd0 (x, y)
for any x, y ∈ X, then their respective open balls satisfy Bh (x) ⊆ B0 (x) and
0
B/k
(x) ⊆ B (x).
2
(x) ⊆ B1 (x), and
In the cases given, we deduce that B1 (x) ⊆ B2 (x) and B/2
min
(x) ⊆ B∗ (x) respectively. In either case, the
that B∗ (x) ⊆ Bmin (x) and B/2
existence of an open ball B (x) ⊆ U with respect to one of the metrics implies
the existence of an open ball (perhaps smaller) with respect to the other.
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