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Arguing informally, this suggests that, to ¬rst order,
d(z + δz, z) = g(z)δs = g(z)|(z + δz) ’ z|.
If T is one of the transformations we are considering, then we must have, to
¬rst order,
g(z)|(z + δz) ’ z| = d(z + δz, z) = d(T (z + δz), T z)
= g(T z)|T (z + δz) ’ T (z)| = g(T z)|T (z)||δz|,
and so
g(z) = g(T z)|T (z)|.
Taking T = Ta , we obtain g(z) = g(z + a) for all real a so g(x + iy) = h(y)
for some well behaved function h : (0, ∞) ’ (0, ∞).
Taking T = D» (z), we obtain g(z) = »g(»z) for all real strictly positive ».
Thus h(y) = »h(»y) for all », y > 0. Taking » = 1/y, we obtain h(y) = Ay
for some A > 0. The factor A merely scales everything, so we take A = 1
and decide to experiment with g(x + iy) = 1/y.
Exercise 10.5.5. Verify that, if g(x + iy) = 1/y and T = J, then equation
holds.
Exercise 10.5.6. Suppose that γ : [0, 1] ’ H is well behaved and T ∈ H.
˜
Let γ = T —¦ γ (that is, γ (z) = T (γ(z))). If “ is the path described by γ and “
˜ ˜
is the path described by γ show, using whatever de¬nition of the line integral
˜
that you wish, that

inf g(z) ds = inf g(z) ds
˜
“ “

for T = Ta , T = D» and T = J. Conclude that the equality holds for all
T ∈ H.
258 A COMPANION TO ANALYSIS

Exercise 10.5.6 shows that, if we set
1
d(z1 , z2 ) = inf ds : “ joining z1 and z2 ,
y


then we do, indeed, get an invariant metric (that is, a metric with d(T z 1 , T z2 ) =
d(z1 , z2 ) for all z1 , z2 ∈ H whenever T ∈ H).
To ¬nd out more about this metric we need to ¬nd the geodesics and to do
this we use the methods of the calculus of variation described in Section 8.4.
We shall use the methods in a purely exploratory manner with no attempt
at rigour. (Notice that, even if we did things rigorously, we would only
get necessary conditions.) Suppose that we wish to ¬nd the path “0 which
1
minimises “ y ds among all paths “ from z1 = x1 + iy1 to z2 = x2 + iy2 with
x1 < x2 and y1 , y2 > 0. It seems reasonable to look for a path given by
z = x + iy(x) where y : [x1 , x2 ] ’ (0, ∞) is well behaved. We thus wish to
minimise
x2 x2
1
(1 + y (x)2 )1/2 dx = G(y(x), y (x)) dx
y(x)
x1 x1

where G(v, w) = v ’1 (1 + w2 )1/2 . The Euler-Lagrange equation gives, via
Exercise 8.4.9 (i),

G(y(x), y (x)) ’ y (x)G,2 (y(x), y (x)) = c,

where c is a constant. Writing the equation more explicitly, we get

(1 + y 2 )1/2 y2
’ = c,
y(1 + y 2 )1/2
y
so that
2
1 = cy(1 + y )1/2 .

Setting a = c’1 , we obtain
2
a2 = y 2 (1 + y )

so that, solving formally,
y dy
= dx
(a2 ’ y 2 )1/2
and

’(a2 ’ y 2 )1/2 = x ’ b,
259
Please send corrections however trivial to twk@dpmms.cam.ac.uk

for some constant b ∈ R. Thus

(x ’ b)2 + y 2 = a2 .

We expect the geodesic to be the arc of a circle with its centre on the real
axis passing through the two points z1 and z2 .
Exercise 10.5.7. Suppose (x1 , y1 ), (x2 , y2 ) ∈ R2 and x1 = x2 . Show that
there is one and only one circle with its centre on the real axis passing through
the two points (x1 , y1 ) and (x2 , y2 ).
What happens if z1 = z2 ? One way of guessing is to consider the
geodesic path between z1 and z2 + δ where δ is real and non-zero. If we let
δ ’ 0, the appropriate circle arcs approach a straight line joining z1 and z2
so we would expect this to be the solution.
Exercise 10.5.8. Attack the geodesic problem by considering paths given by
z = x(y) + iy where y : [y1 , y2 ] ’ R is well behaved. (The di¬culties are
mainly notational, the formulae of the variational calculus assume that y is
a function of x and it requires a certain amount of clear headedness to deal
with the case when x is a function of y. You should be able to make use
of Exercise 8.4.9 (ii).) Check, by choosing appropriate constants, that your
solutions include straight lines perpendicular to the x-axis.
Now that we know (or at least guess) what the geodesics are we can see
(at least if we know a little about M¨bius maps) a di¬erent way of showing
o
this.
Exercise 10.5.9. We work in C— = C ∪ {∞}.
(i) Show that, if “ is a circle with centre on the real axis, there is a
T ∈ H such that T (“) is a circle with centre on the real axis passing through
the origin.
(ii) Show that if “ is a circle with centre on the real axis passing through
the origin, there is a T ∈ H such that T (“) is a line perpendicular to the real
axis.
(iii) Show that if “ is a circle with centre on the real axis or a line per-
pendicular to the real axis, there is a T ∈ H such that T (“) is the imaginary
axis.
Exercise 10.5.6 and Exercise 10.5.9 show that the following two theorems
are equivalent.
Theorem 10.5.10. The geodesic path between iy1 and iy2 [y1 , y2 real, un-
equal and strictly positive] is a straight line.
260 A COMPANION TO ANALYSIS

Theorem 10.5.11. If z1 , z2 ∈ H with z1 = z2 , the geodesic path between
them is an arc of the circle through z1 and z2 , unless z1 = z2 , in which
case, it is the straight line between the two points.
Exercise 10.5.12. Explain why Theorem 10.5.10 implies Theorem 10.5.11.
We now turn to the proof of Theorem 10.5.10.
Sketch proof of Theorem 10.5.10. Let Z : [0, 1] ’ H be a well behaved
function such that Z(0) = y1 and Z(1) = y2 with y2 > y1 . We write
Z(t) = X(t) + iY (t) with X(t) and Y (t) real and take “ to be the path
described by Z. We observe that
1 1
(Y (t)2 )1/2
1 1 2 2 1/2
(X (t) + Y (t) ) dt ≥
ds = dt
y 0 Y (t) Y (t)
“ 0
1 1
|Y (t)| Y (t)
dt = [log Y (t)]1
dt ≥
= 0
Y (t) 0 Y (t)
0
y2
1 1
= log y2 ’ log y1 = dt = ds,
t y
y1 “0

where “0 is the straight line path from y1 to y2 . We observe that the argument
above shows that
1 1
ds = ds
y y
“ “0

only if X (t)2 = 0 and Y (t) ≥ 0 for all t ∈ [0, 1] and so, by simple arguments,
X(t) = 0 and Y (t) ≥ 0 for all t ∈ [0, 1]. Thus “0 is the unique geodesic.
Remark: This is only a sketch of a proof because we have not really decided
which curves will be eligible for paths. The proof strategy of ˜project the path
onto the y-axis and compare™ ought to work for any reasonable de¬nition of
line integral and any eligible path.
Exercise 10.5.13. We work in R2 . Suppose that we want a metric de¬ned
by

d(A, B) = inf g(x, y) ds : “ joining A and B ,


which is invariant under translation and rotation. Copy the investigation
above going through the following steps.
(i) By considering points which are close and working to ¬rst order, show
that we should try g constant. Without loss of generality take g = 1.
(ii) Use the calculus of variations to suggest the form of the geodesics.
(iii) Prove that your guess is correct.
(iv) Show that our metric is the usual Euclidean metric.
261
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Exercise 10.5.14. In this section we de¬ned a metric d on H invariant
under H but did not calculate it. We could ¬nd d by using the strategy of
Exercise 10.5.9, but, in this exercise, we follow the easier path of allowing
someone else to ¬nd the answer and then verifying it.
If z, w ∈ H, we set

|z ’ w— | + |z ’ w|
ρ(z, w) = log .
|z ’ w— | ’ |z ’ w|

(i) Show that |z ’ w — | ’ |z ’ w| > 0 for all z, w ∈ H, so ρ is well de¬ned.
(ii) Show that ρ(T z, T w) = ρ(z, w) for all T ∈ H.
(iii) Show that d(iy1 , iy2 ) = ρ(iy1 , iy2 ) for all y1 , y2 > 0.
(iv) Deduce that d = ρ.
Chapter 11

Complete metric spaces

11.1 Completeness
If we examine the arguments of Section 10.3 we see that they are all mere
algebra. What must we introduce to do genuine analysis on metric spaces?
We cannot use a variant of the fundamental axiom because there is no order
on our spaces1 . Instead, we use a generalisation of the general principle of
convergence.

De¬nition 11.1.1. If (X, d) is a metric space, we say that a sequence of
points xn ∈ X is Cauchy if, given any > 0, we can ¬nd n0 ( ) such that
d(xp , xq ) < for all p, q ≥ n0 ( ).

De¬nition 11.1.2. A metric space (X, d) is complete if every Cauchy se-
quence converges.

In this context, Theorem 4.6.3 states that Rn with the Euclidean metric
is complete.
The contraction mapping theorem (Theorem 12.1.3) and its applications
will provide a striking example of the utility of this concept. However, this
section is devoted to providing background examples of spaces which are and
are not complete.
If you want to see completeness in action immediately you should do the
next example.

Exercise 11.1.3. We say that a metric space (X, d) has no isolated points
if, given y ∈ X and > 0, we can ¬nd an x ∈ X such that 0 < d(x, y) < .
1
There is an appropriate theory for objects with order (lattices) hinted at in Ap-
pendix D, but we shall not pursue this idea further.


263
264 A COMPANION TO ANALYSIS

Show by the methods of Exercise 1.6.7 that a complete non-empty metric
space with no isolated points is uncountable.
Give an example of a countable metric space. Give an example of an
uncountable metric space all of whose points are isolated.

The next lemma gives a good supply of metric spaces which are complete
and of metric spaces which are not complete.

Lemma 11.1.4. Let (X, d) be a complete metric space. If E is a subset of
X and we de¬ne dE : E 2 ’ R by dE (u, v) = d(u, v) whenever u, v ∈ E, then
(E, dE ) is complete if and only if E is closed in (X, d).

Proof. This is just a matter of de¬nition chasing.
Observe that any Cauchy sequence xn in E is a Cauchy sequence in
X and so converges to a point x in X. If E is closed, then x ∈ E and
dE (xn , x) = d(xn , x) ’ 0 as n ’ ∞. Thus (E, dE ) is complete whenever E
is closed.
Suppose now that (E, dE ) is complete. If xn ∈ E and d(xn , x) ’ 0 for
some x ∈ X, we know (by the argument of Lemma 4.6.2 if you need it)
that xn is a Cauchy sequence in X and so a Cauchy sequence in E. Since
(E, dE ) is complete, we can ¬nd a y ∈ E such that dE (xn , y) ’ 0. Now
d(xn , y) = dE (xn , y) ’ 0 so, by the uniqueness of limits, y = x and x ∈ E.
Thus E is closed in (X, d).
Thus, for example, the closed interval [a, b] is complete for the usual metric
but the open interval (a, b) is not.

Exercise 11.1.5. Let (X, d) be a metric space and E a subset of X. De¬ne
dE : E 2 ’ R as in Lemma 11.1.4.
(i) Show that, if E is not closed in (X, d), then (E, dE ) is not complete.
(ii) Give an example where E is closed in (X, d) but (E, dE ) is not com-
plete.
(iii) Give an example where (X, d) is not complete but (E, dE ) is.

The reader is warned that, at least in my opinion, it is harder than it
looks to prove that a metric space is or is not complete and it is easy to
produce plausible but unsatisfactory arguments in this context.
If we wish to show that a metric space (X, d) is incomplete, the natural
way to proceed is to ¬nd a Cauchy sequence xn and show that it does not
converge. However, we must show that xn does not converge to any point in
X and not that ˜xn does not converge to the point that it looks as though it
ought to converge to™. Among the methods available for a correct proof are
the following.
265
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˜˜ ˜˜
(1) Embed (X, d) in a larger metric space (X, d) (that is, ¬nd (X, d) such
˜
˜
that X ⊇ X and d(x, y) = d(x, y) for x, y ∈ X) and show that there is an
˜
˜
x ∈ X \ X such that d(xn , x) ’ 0. (See Exercise 11.1.5.)
(2) For each ¬xed x ∈ X show that there is a δ(x) > 0 and an N (x) (both
depending on x) such that d(xn , x) > δ(x) for n > N (x).
(3) Show that the assumption d(xn , x) ’ 0 for some x ∈ X leads to a
contradiction.
Of course, no list of this sort can be exhaustive. None the less, it is good
practice to ask yourself not simply whether your proof is correct but also
what strategies it employs.
Here are a couple of examples.

Example 11.1.6. Consider s00 , the space of real sequences a = (an )∞ such
n=1
that all but ¬nitely many of the an are zero, introduced in Exercise 10.4.8.
The norm de¬ned by

|an |
a =
1
n=1

is not complete.

Proof. Set

a(n) = (1, 2’1 , 2’2 , . . . , 2’n , 0, 0, . . . ).

We observe that, if m ≥ n,
m
2’j ¤ 2’n ’ 0
d(a(n), a(m)) =
j=n+1


as n ’ ∞ and so the sequence a(n) is Cauchy.
However, if a ∈ s00 , we know that there is an N such that aj = 0 for all
j ≥ N . It follows

d(a(n), a) ≥ 2’N

whenever n ≥ N and so the sequence a(n) has no limit.

[The proof above used method 2. For an alternative proof using method 1,
see page 269. For a generalisation of the result see Exercise K.187.]
The next example needs a result which is left to the reader to prove. (You
will need Lemma 8.3.2.)
266 A COMPANION TO ANALYSIS

Exercise 11.1.7. Let b > a. Consider C([a, b]) the set of continuous func-
tions f : [a, b] ’ R. If we set
b
f ’g |f (x) ’ g(x)| dx,
=
1
a

show that is a norm.
1

Lemma 11.1.8. The normed space of Exercise 11.1.7 is not complete.

Proof. This proof uses method 3. With no real loss of generality, we take
[a, b] = [’1, 1]. Let

fn (x) = ’1 for ’1 ¤ x ¤ ’1/n,
for ’1/n ¤ x ¤ 1/n,
fn (x) = nx
for 1/n ¤ x ¤ 1.
fn (x) = 1

If m ≥ n,
1 1/n
2
fn ’ fm |fn (x) ’ fm (x)| dx ¤ ’0
= 1 dx =
1
n
’1 ’1/n

as n ’ ∞ and so the sequence fn is Cauchy.
Suppose, if possible, that there exists an f ∈ C([’1, 1]) such that f ’ fn ’0
1
as n ’ ∞. Observe that, if n ≥ N , we have
1 1 1
|f (x) ’ 1| dx = |f (x) ’ fn (x)| dx ¤ |f (x) ’ fn (x)| dx = fn ’ f ’0
1
’1
1/N 1/N

as n ’ ∞. Thus
1
|f (x) ’ 1| dx = 0
1/N


and, by Lemma 8.3.2, f (x) = 1 for x ∈ [1/N, 1]. Since N is arbitrary,
f (x) = 1 for all 0 < x ¤ 1. A similar argument shows that f (x) = ’1 for all
’1 ¤ x < 0. Thus f fails to be continuous at 0 and we have a contradiction.
By reductio ad absurdum, the Cauchy sequence fn has no limit.

Lemma 11.1.8 is important as an indication of the unsatisfactory results
of using too narrow a class of integrable functions.
The next exercise goes over very similar ground but introduces an inter-
esting set of ideas.
267
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Exercise 11.1.9. Let b > a. Consider C([a, b]) the set of continuous func-
tions f : [a, b] ’ R. If f , g ∈ C([a, b]) and we de¬ne
b
f, g = f (t)g(t) dt
a

show that (C([a, b]), , ) is an inner product space. More formally, show
that
(i) f, g ≥ 0 with equality if and only if f = 0,
(ii) f, g = g, f ,
(iii) »f, g = » f, g ,
(iv) f, g + h = f, g + f, h
for all f , g, h ∈ C([a, b]) and all » ∈ R.
Use the arguments of Lemmas 4.1.2 and 4.1.4 to show that setting
1/2
b
1/2 2
f = f, f = f (t) dt

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