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0


g,2 (x, 0) dx = 0.
0

We can certainly choose v with v (0) = 0 (for example v(y) = y). Why does
this not contradict Theorem 11.4.21?

Exercise K.217. [11.4, H] In this exercise we modify the ideas we used in
Exercise K.216 to obtain an example relevant to Theorem 8.4.3.
Suppose that u, v : R ’ R have continuous derivatives, that v(t)t1/2 ’ 0
as t ’ 0, that u(t) ≥ 0 for all t ∈ R, that u(t) = 0 if t ¤ 1 or t ≥ 2 and

that 0 u(x) dx = 1. In this question we shall be interested in the function
g : R2 ’ R de¬ned by

g(x, y) = |y|’1/2 v(y)u(x|y|’1/2 ) for y = 0, g(x, 0) = 0.

(i) Just to get an idea of what is going on, sketch g in the particular case
when v(y) = y
(ii) Show that g is everywhere continuous.
(iii) Show that g has partial derivatives g,1 and g,2 everywhere and that
these derivatives are continuous except, possibly, at (0, 0).
547
Please send corrections however trivial to twk@dpmms.cam.ac.uk

1
(iv) Show that G(y) = g(x, y) dx then G (0) = v (0), but
0


g,2 (x, 0) dx = 0.
0

We can certainly choose v with v (0) = 0 (for example v(y) = y). Why does
this not contradict Theorem 8.4.3?

Exercise K.218. (A dominated convergence theorem for integrals.)
[11.4, P] Suppose that g : [0, ∞) ’ R is a continuous function such that

g(t) dt converges. If fn : [0, ∞) ’ R is a continuous function with
0
|fn (t)| ¤ g(t) for all t ∈ [0, ∞) and fn ’ f uniformly on [0, ∞) as n ’ ∞,
∞ ∞
show that all the integrals 0 fn (t) dt [n ≥ 1] and 0 f (t) dt converge and
∞ ∞
fn (t) dt ’ f (t) dt.
0 0

[Hints. If you cannot do this at once, you can adopt one or more of the
following strategies. (1) Consider the special case h(t) = (1 + t2 )’1 . (2) Ask
how the hypotheses prevent us using Example 11.4.12 as a counterexample.
(3) Reread the proof of Lemma 5.3.3 where a similar result is obtained.]

Exercise K.219. [11.4, P] (i) Suppose that d1 , d2 , . . . are metrics on a
space X. Show that

2’j dj (x, y)
d(x, y) =
1 + dj (x, y)
j=1

is a metric on X. Show that if each dj is complete, then so is d.
(ii) Consider C ∞ , the space of in¬nitely di¬erentiable functions on [0, 1].
Show that

2’j supt∈[0,1] |f (j) (t) ’ g (j) (t)|
d(f, g) =
1 + supt∈[0,1] |f (j) (t) ’ g (j) (t)|
j=0


is a complete metric on C ∞ . (Be careful, supt∈[0,1] |f (j) (t) ’ g (j) (t)| is not a
metric on C ∞ if j = 0.)
(j)
Show that d(fn , f ) ’ 0 as n ’ ∞ if and only if fn (x) ’ f (x) uniformly
on [0, 1] for each j.
Let gn (x) = n’n sin(n + 1)2 πx. Show that d(gn , 0) ’ 0 but
(j)
sup sup |gn (x)| = ∞
j≥0 x∈[0,1]
548 A COMPANION TO ANALYSIS

(j)
(more formally, the set {|gn (x)| : x ∈ [0, 1], j ≥ 0} is unbounded) for all n.
(iii) Consider the space C(D) of continuous (but not necessarily bounded)
functions f : D ’ C, where D is the open unit disc given by

D = {z ∈ C : |z| < 1}.

Find a complete metric d such that d(fn , f ) ’ 0 as n ’ ∞ if and only if

|f (z) ’ g(z)| ’ 0
sup
|z|¤1’n’1


for all n ≥ 1. Prove that you have, indeed, found such a metric.

Exercise K.220. (Pointwise convergence cannot be given by a met-
ric.) [11.4, H] The object of this exercise is to show that the notion of a
metric is not su¬cient to cover all kinds of convergence. More speci¬cally,
we shall show that there does not exist a metric d on the space C([0, 1])
of continuous functions such that d(fn , f ) ’ 0 as n ’ ∞ if and only if
fn (x) ’ f (x) for all x ∈ [0, 1]. The proof is quite complicated and simpler
proofs can be obtained once we have studied topology, so the reader may
prefer to omit it.
The proof is by contradiction, so we assume that d is a metric such that
d(fn , f ) ’ 0 as n ’ ∞ if and only if fn (x) ’ f (x) for all x ∈ [0, 1].
(i) Let gn : [0, 1] ’ R be a ˜witch™s hat™ given by

gn (x) =1 ’ n(x ’ n’1 ’ 2’1 ) for |x ’ n’1 ’ 2’1 | ¤ n’1 ,
gn (x) =0 otherwise.

Show that d(gn , 0) ’ 0 as n ’ ∞.
(ii) By using the ideas of (i), or otherwise, show that, given any closed
interval I ⊆ [0, 1] and any > 0, we can ¬nd a g ∈ C([0, 1]) and a closed
interval J ⊆ I such that d(g, 0) ¤ but g(x) ≥ 1/2 on I.
(iii) Show that we can ¬nd a sequence of closed intervals I1 ⊇ I2 ⊇ I3 ⊇
. . . and continuous functions fn such that d(fn , 0) ¤ 2’n but fn (x) ≥ 1/2 on
In .
(iv) Use Exercise 4.3.8 to derive a contradiction.

Exercise K.221. [11.4, T, S] (This sketches some background for the next
exercise.) Let V be a vector space. Suppose that is a norm on V and d a
metric on V such that d(xn , x) ’ 0 as n ’ ∞ if and only if xn ’ x ’ 0.
Show that
(a) d(a + xn , a + x) ’ 0 if and only if d(xn , x) ’ 0.
(b) d(»xn , 0) ’ 0 whenever d(xn , 0) ’ 0.
549
Please send corrections however trivial to twk@dpmms.cam.ac.uk

(c) d(»n x, 0) ’ 0 whenever »n ’ 0.
These remarks make it easy to ¬nd metrics which do not have the con-
vergence properties of norms. As an example, check which of properties (a),
(b) and (c) hold for the discrete metric and deduce that the discrete metric
does not have the convergence properties of a norm. Do the same for the
British Railway non-stop metric of Exercise 10.3.7.
Show however, that the metric of part (ii) of Exercise K.219 has all the
properties (a), (b) and (c). In the next exercise we show that, even so, it
does not have the convergence properties of a norm.
Exercise K.222. (An interesting metric not derivable from a norm.)
[11.4, H, ‘ ] In this question we shall show that there does not exist a norm

D on the space C ([0, 1]) of continuous functions such that fn D ’ 0 as
(j)
n ’ ∞ if and only if fn (x) ’ 0 uniformly on [0, 1] for each j. This shows
that the metric d de¬ned in part (ii) of Exercise K.219 cannot be replaced
by a norm.
The proof proceeds by contradiction. We assume that D has the prop-
erties given in the previous paragraph.
(i) By reductio ad absurdum, or otherwise, show that, for each integer j ≥
0, there exists an j > 0 such that f D ≥ j whenever supx∈[0,1] |f (j) (x)| ≥ 1.
(ii) Deduce that g D ≥ j supx∈[0,1] |g (j) (x)| for all g ∈ C ∞ ([0, 1]) and all
j ≥ 0.
(iii) Show that, by choosing δk and Nk appropriately, and setting fk (x) =
δk sin πNk x, or otherwise, that we can ¬nd fk ∈ C ∞ ([0, 1]) such that
(j)
sup |fk (x)| ¤ 2’k when 0 ¤ j ¤ k ’ 1,
x∈[0,1]
(k)
sup |fk (x)| ≥ 2j j .
j
x∈[0,1]

(j)
(iv) Show that fn D ’ ∞ as n ’ ∞, but fn (x) ’ 0 uniformly on
[0, 1] for each j. This contradiction gives the desired conclusion.
Exercise K.223. (A continuous nowhere di¬erentiable function.) [11.4,
T](i) Suppose that f : [0, 1] ’ R is di¬erentiable with continuous derivative.
Explain why we can ¬nd a K, depending on f , such that
|f (x) ’ f (y)| ¤ K|x ’ y|
for all x, y ∈ [0, 1].
(ii) Suppose f : [0, 1] ’ R is di¬erentiable with continuous derivative.
By considering
g(x) = f (x) + sin N x
550 A COMPANION TO ANALYSIS

with N very large, show that, given any > 0 and any · > 0, we can ¬nd a
g : [0, 1] ’ R with the following properties.
(a) g is di¬erentiable with continuous derivative.
(b) g ’ f ∞ ¤ .
(c) Given any x ∈ [0, 1], we can ¬nd a y ∈ [0, 1] such that |x ’ y| < ·
but |g(x) ’ g(y)| > /2.
(iii) Set g0 = 0. Show that we can ¬nd a sequence of functions gn :
[0, 1] ’ R such that, whenever n ≥ 1,
(a)n gn is di¬erentiable with continuous derivative.
(b)n gn ’ gn’1 ∞ ¤ 2’3n .
(c)n Given any x ∈ [0, 1], we can ¬nd a y ∈ [0, 1] such that |x’y| < 2’4n
but |gn (x) ’ gn (y)| > 2’3n’1 .
(iv) Show that gn converges uniformly to a continuous function g. Show
further that

2’3j = 2’3n /7.
g ’ gn ¤

j=n+1


Use this fact, together with (c)n , to show that, given any x ∈ [0, 1], we can
¬nd a y ∈ [0, 1] such that |x ’ y| < 2’4n but

|g(x) ’ g(y)| > 2’3n’1 ’ 2’3n+1 /7 = 3 — 2’3n’1 /7.

Deduce that
3 — 2n’1
g(x) ’ g(y)

sup
x’y 7
y∈[0,1], y=x


for all n ≥ 1 and all x. Conclude that g is nowhere di¬erentiable.
[At the beginning of the 20th century continuous, nowhere di¬erentiable func-
tions were considered to be playthings of pure mathematicians of uncertain
taste. Classes of continuous nowhere di¬erentiable functions now form the
main subject of study in ¬nancial mathematics and parts of engineering.]

Exercise K.224. (A continuous space ¬lling curve.) [11.4, T] This
exercise uses Exercise 11.3.9. We work in R2 and R with the usual Euclidean
norms.
(i) Let (x0 , y0 ), (x1 , y1 ) ∈ [0, 1]2 . Show, by specifying f piece by piece,
or otherwise, that we can ¬nd a continuous function f : [0, 1] ’ [0, 1]2 such
that f (0) = (x0 , y0 ), f (1) = (x1 , y1 ) and (rN ’1 , sN ’1 ) ∈ f ([0, 1]) for each
0 ¤ r, s ¤ N . (In other words the curve described by f starts at (x0 , y0 ),
ends at (x1 , y1 ) and passes through all points of the form (rN ’1 , sN ’1 ).)
551
Please send corrections however trivial to twk@dpmms.cam.ac.uk

(ii) Suppose that h : [0, 1] ’ [0, 1]2 is continuous and 0 = t0 < t1 < t2 <
· · · < tM = 1. Show that given any > 0 we can ¬nd an · > 0 such that

t0 + · < t 1 ’ · < t 1 + · < t 2 ’ · < t 2 + · < t 3 ’ · < t 3 + · < · · · < t M ’ ·

and h(t) ’ h(s) < whenever t, s ∈ [0, 1] and |t ’ s| ¤ 2·.
(iii) Let g0 : [0, 1] ’ [0, 1]2 be a continuous function such that g0 (0) =
(0, 0), g0 (1/3) = (0, 1), g0 (2/3) = (1, 0) and g0 (1) = (1, 1) and let E0 =
{0, 1/3, 2/3, 1}. Show that we can ¬nd a sequence of functions gn : [0, 1] ’
[0, 1]2 and ¬nite sets En ‚ [0, 1] such that, if n ≥ 1,
(a)n gn ’ gn’1 ∞ ¤ 2’n+4 .
(b)n If a ∈ En then gn (a) = (r2’n , s2’n ) for some integers r and s with
0 ¤ r, s ¤ 2n .
(c)n If r and s are integers with 0 ¤ r, s ¤ 2n then there exists an
a ∈ En with gn (a) = (r2’n , s2’n ).
(d)n If r and s are integers with 0 ¤ r, s ¤ 2n’1 and a ∈ En’1 satis¬es
gn’1 (a) = (r2’n+1 , s2’n+1 ) then, if u and v are integers with 0 ¤ u, v ¤ 2n
and

|u2’n ’ r2n’1 |, |v2’n ’ s2n’1 | ¤ 2’n ,

we can ¬nd b ∈ En such that |a ’ b| ¤ 2’n and gn (b) = (r2’n , s2’n ).
[Conditions (a)n and (d)n are the important ones. The other two are included
to help keep track of what is going on. Note that gn will not be injective for
n ≥ 1 and that En will contain several points x with gn (x) = (r2’n , s2’n ).]
(iv) Show that gn converges uniformly to a continuous function g :
[0, 1] ’ [0, 1]2 .
(v) Using the fact that every x = [0, 1] can be written x = ∞ κj 2’j
j=1
with κj ∈ {0, 1}, or otherwise, use condition (d)n to show that if (x, y) ∈ [0, 1]
we can ¬nd tn ∈ En [n ≥ 0] such that |tn ’ tn’1 | ¤ 2’n for n ≥ 1 and
gn (tn ) ’ (x, y) ’ 0 as n ’ 0. Explain why we can ¬nd a t ∈ [0, 1] with
tn ’ t as n ’ ∞ and use the inequality

g(t) ’ (x, y) ¤ g(t) ’ g(tn ) + g(tn ) ’ gn (tn ) + gn (tn ) ’ (x, y)
¤ g(t) ’ g(tn ) + g ’ gn ∞ + gn (tn ) ’ (x, y)

to show that g(t) = (x, y). Conclude that g : [0, 1] ’ [0, 1]2 is a continuous
surjective map.
[I cannot claim any practical use for space ¬lling curves, but the popularity
of fractals means that mathematicians do come across continuous maps h :
[0, 1] ’ [0, 1]2 where the image h([0, 1]) is ˜unexpectedly large™.]
552 A COMPANION TO ANALYSIS

Exercise K.225. (The devil™s staircase.) [11.4, T] This result is less
spectacular than the previous two and probably harder to explain. If you do
not understand my presentation, the example can be found in most books
on measure theory. (The set E de¬ned in (iii) is called the Cantor set or the
˜Cantor middle third set™.)
(i) We work on [0, 1]. Write E(0) = (3’1 , 2.3’1 ) for the ˜middle third™
open interval of [0, 1]. We have [0, 1] \ E(0) = [0, 3’1 ] ∪ [2.3’1 , 1] and we
write E0 = (3’2 , 2.3’2 ) for the middle third open interval of [0, 3’1 ] and
E1 = (2.3’1 + 3’2 , 2.3’1 + 2.3’2 ) for the middle third interval of [2.3’1 , 1].
More generally, we set
n n
2 (j)3’j + 3’n’1 , 2 (j)3’j + 2.3’n’1 ),
E (1), (2),..., (n) = (
j=1 j=1

where (j) = 0 or (j) = 1. Sketch

E(n) = E (1), (2),..., (n)
(j)∈{0,1}

for n = 1, 2, 3 and verify that E(n) is the union of 2n intervals each of
length 3’n’1 and such that the intervals are the middle third open intervals
of the closed intervals which make up [0, 1] \ n’1 Ek . (You should worry
k=0
more about understanding the pattern than about rigour.)
(ii) We take fn : [0, 1] ’ R to be the simplest piecewise linear function
which satis¬es fn (0) = 0, fn (1) = 1 and
fn (x) = 2’1 for x ∈ E(0),
m
fn (x) = 2’(m+1) + 2’j (j)
for x ∈ E (1), (2),..., (m) and 1 ¤ m ¤ n.
j=1

I sketch f2 in Figure K.4. Sketch f3 for yourself. Explain why fn is an
increasing function. (iii) Show that fn ’ fm ∞ ¤ 2’n , and deduce that fn
converges uniformly to a continuous function f . Show that f is increasing
and that f (0) = 0, f (1) = 1 and
f (x) = 2’1 for x ∈ E(0),
m
f (x) = 2’(m+1) + 2’j (j)
x ∈ E (1), (2),..., (m) and all m ≥ 1.
j=1

Explain why f is di¬erentiable on E(m) with f (x) = 0 for each x ∈ E(m).
Conclude that f is di¬erentiable on E = ∞ E(m) with f (x) = 0 for each
m=0
x ∈ E.
553
Please send corrections however trivial to twk@dpmms.cam.ac.uk




Figure K.4: The second approximation to the devil™s staircase

(iv) Show that the total length of the intervals making up n E(m)
m=0
is 1 ’ (2/3) . If IE : [0, 1] ’ R is given by IE (x) = 1 if x ∈ E and
n+1

IE (x) = 0 otherwise, show, by ¬nding appropriate dissections, that IE is
Riemann integrable and
1
IE (x) dx = 1.
0

Informally, we have shown that ˜f has zero derivative on practically the whole
of [0, 1] but still manages to increase from 0 to 1™ !
(v) This part takes us back to Section 9.4, and, in particular, to the
promise made on page 223 to exhibit a real-valued random variable which
is not a simple mix of discrete and continuous. Our argument is informal
(partly because we have not de¬ned what we mean by a ˜mix of discrete and
continuous™). De¬ne F : R ’ R by F (x) = 0 if x < 0, F (x) = f (x) if
0 ¤ x ¤ 1 and F (x) = 1 otherwise. Explain why F is continuous. Let X be
the real-valued random variable with Pr{X ¤ x} = F (x).
Suppose, if possible, that X is a ˜mix of discrete and continuous™. Since
F is continuous, X must, in fact, be a continuous random variable and so
x
F (x) = Pr{X ¤ x} = g(t) dt
’∞

for some well behaved g. If g is Riemann integrable, then it is bounded on
554 A COMPANION TO ANALYSIS

[0, 1] with |g(x)| ¤ K for all x ∈ [0, 1]. Deduce that, on this assumption,

|F (x) ’ F (y)| ¤ K(y ’ x)

for 0 ¤ x ¤ y ¤ 1. By showing that this last inequality is false for some x
and y, obtain a contradiction.
Thus X must be a novel type of random variable.
[Graphs very similar to that of F turn up in the theory of di¬erential equa-
tions, in particular for processes intended to mimic transition to turbulence
in ¬‚uid ¬‚ows.]
Exercise K.226. [11.4, H] (i) By considering fn (x) = An sin Bn (x + Cn )
where An , Bn and Cn are to be stated explicitly, show that we can ¬nd an
in¬nitely di¬erentiable function fn : R ’ R such that (writing g (0) (x) =
g(x))

|fn (x)| ¤ 2’n’1 for all n > r ≥ 0 and all x ∈ R
(r)


yet
n’1
(n) 2 (n)
≥ (n!) + 1 + |fr (0)|.
fn (0)
r=0

Explain why fn tends uniformly to an in¬nitely di¬erentiable function
f : R ’ R with f (n) (0) ≥ (n!)2 . Show that the Taylor series

f (n) (0) n
x
n!
n=0

diverges for all x = 0.
(ii) The result of (i) can be extended as follows. Show that we can ¬nd a
sequence y1 , y2 . . . of rational numbers such that each rational number occurs
in¬nitely often in the sequence. (In other words, given a rational number q
and an integer N , we can ¬nd an n ≥ N with yn = q.)
By considering fn (x) = An sin Bn (x + Cn ) where An , Bn and Cn are to
be stated explicitly, show that we can ¬nd an in¬nitely di¬erentiable function
fn : R ’ R such that

|fn (x)| ¤ 2’n’1 for all n > r ≥ 0 and all x ∈ R
(r)


yet
n’1
(n) 2 (n)
≥ (n!) + 1 + |fr (yn )|.

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