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f (0) 2
t + (t)|t|n
t + ··· +
f (t) = f (0) + f (0)t +
2! n!
where (t) ’ 0 as t ’ 0.
This exercise introduces several ideas which we use repeatedly in this
chapter so the reader should do it carefully.
Exercise 7.1.1. In this exercise we consider functions f, g : (’a, a) ’ R
where a > 0.
(i) If f and g are di¬erentiable with f (t) ¤ g (t) for all 0 ¤ t < a and
f (0) = g(0), explain why f (t) ¤ g(t) for all 0 ¤ t < a.
(ii) If |f (t)| ¤ |t|r for all t ∈ (’a, a) and f (0) = 0, show that |f (t)| ¤
|t|r+1 /(r + 1) for all |t| < a.

141
142 A COMPANION TO ANALYSIS

(iii) If g is n times di¬erentiable with |g (n) (t)| ¤ M for all t ∈ (’a, a)
and g(0) = g (0) = · · · = g (n’1) (0) = 0, show that
M |t|n
|g(t)| ¤
n!
for all |t| < a.
(iv) If g is n times di¬erentiable in (’a, a) and g(0) = g (0) = · · · =
g (n) (0) = 0, show, using (iii), that, if g (n) is continuous at 0, then
·(t)|t|n
|g(t)| ¤
n!
where ·(t) ’ 0 as t ’ 0.
(v) If f is n times di¬erentiable with |f (n) (t)| ¤ M for all t ∈ (’a, a),
show that
n’1
f (j) (0) j M |t|n
f (t) ’ t¤
j! n!
j=0

for all |t| < a.
(vi) If f is n times di¬erentiable in (’a, a), show that, if f (n) is contin-
uous at 0, then
n
f (j) (0) j ·(t)|t|n
f (t) ’ t¤
j! n!
j=0

where ·(t) ’ 0 as t ’ 0.
Restating parts (v) and (vi) of Exercise 7.1.1 we get two similar looking
but distinct theorems.
Theorem 7.1.2. (A global Taylor™s theorem.) If f : (’a, a) ’ R is n
times di¬erentiable with |f (n) (t)| ¤ M for all t ∈ (’a, a), then
n’1
f (j) (0) j M |t|n
f (t) ’ t¤ .
j! n!
j=0

Theorem 7.1.3. (The local Taylor™s theorem). If f : (’a, a) ’ R is n
times di¬erentiable and f (n) is continuous at 0, then
n
f (j) (0) j
t + (t)|t|n
f (t) =
j!
j=0

where (t) ’ 0 as t ’ 0.
143
Please send corrections however trivial to twk@dpmms.cam.ac.uk

We shall obtain other and more precise global Taylor theorems in the
course of the book (see Exercise K.49 and Theorem 8.3.20) but Theorem 7.1.2
is strong enough for the following typical applications.

Exercise 7.1.4. (i) Consider a di¬erentiable function e : R ’ R which
obeys the di¬erential equation e (t) = e(t) with the initial condition e(0) = 1.
Quote a general theorem which tells you that, if a > 0, there exists an M
with |e(t)| ¤ M for |t| ¤ a. Show that
n’1
tj M |t|n
e(t) ’ ¤
j! n!
j=0


for all |t| < a. Deduce that
n’1
tj
’ e(t)
j!
j=0


as n ’ ∞, and so

tj
e(t) =
j!
j=0


for all t.
(ii) Consider di¬erentiable functions s, c : R ’ R which obey the di¬er-
ential equations s (t) = c(t), c (t) = ’s(t) with the initial conditions s(0) = 0,
c(0) = 1. Show that

(’1)j t2j+1
s(t) =
(2j + 1)!
j=0


for all t and obtain a similar result for c.

However, in this chapter we are interested in the local behaviour of func-
tions and therefoe in the local Taylor theorem. The distinction between local
and global Taylor expansion is made in the following very important example
of Cauchy.

Example 7.1.5. Consider the function F : R ’ R de¬ned by

F (0) = 0
F (x) = exp(’1/x2 ) otherwise.
144 A COMPANION TO ANALYSIS

(i) Prove by induction, using the standard rules of di¬erentiation, that F
is in¬nitely di¬erentiable at all points x = 0 and that, at these points,

F (n) (x) = Pn (1/x) exp(’1/x2 )

where Pn is a polynomial which need not be found explicitly.
(ii) Explain why x’1 Pn (1/x) exp(’1/x2 ) ’ 0 as x ’ 0.
(iii) Show by induction, using the de¬nition of di¬erentiation, that F is
in¬nitely di¬erentiable at 0 with F (n) (0) = 0 for all n. [Be careful to get this
part of the argument right.]
(iv) Show that

F (j) (0) j
F (x) = x
j!
j=0


if and only if x = 0. (The reader may prefer to say that ˜The Taylor expansion
of F is only valid at 0™.)
(v) Why does part (iv) not contradict the local Taylor theorem (Theo-
rem 7.1.3)?
[We give a di¬erent counterexample making use of uniform convergence in
Exercise K.226.]

Example 7.1.6. Show that, if we de¬ne E : R ’ R by

if x ¤ 0
E(x) = 0
E(x) = exp(’1/x2 ) otherwise,

then E is an in¬nitely di¬erentiable function with E(x) = 0 for x ¤ 0 and
E(x) > 0 for x > 0

Cauchy gave his example to show that we cannot develop the calculus
algebraically but must use , δ techniques. In later courses the reader will
see that his example encapsulates a key di¬erence between real and complex
analysis. If the reader perseveres further with mathematics she will also ¬nd
the function E playing a useful rˆle in distribution theory and di¬erential
o
geometry.
A simple example of the use of the local Taylor theorem is given by the
proof of (a version of) L™Hˆpital™s rule in the next exercise.
o

Exercise 7.1.7. If f, g : (’a, a) ’ R are n times di¬erentiable and

f (0) = f (0) = · · · = f n’1 (0) = g(0) = g (0) = · · · = g (n’1) (0) = 0
145
Please send corrections however trivial to twk@dpmms.cam.ac.uk

but g (n) (0) = 0 then, if f (n) and g (n) are continuous at 0, it follows that

f (n) (0)
f (t)
’ (n)
g(t) g (0)
as t ’ 0.

It should be pointed out that the local Taylor theorems of this chapter
(and the global ones proved elsewhere) are deep results which depend on
the fundamental axiom. The fact that we use mean value theorems to prove
them is thus not surprising ” we must use the fundamental axiom or results
derived from it in the proof.
(Most of my readers will be prepared to accept my word for the statements
made in the previous paragraph. Those who are not will need to work through
the next exercise. The others may skip it.)

Exercise 7.1.8. Explain why we can ¬nd a sequence of irrational numbers
an such that 4’n’1 < an < 4’n . We write I0 = {x ∈ Q : x > a0 } and

In = {x ∈ Q : an < x < an’1 }

[n = 1, 2, 3, . . . ]. Check that, if x ∈ In , then 4’n’1 < x < 4’n+1 [n ≥ 1].
We de¬ne f : Q ’ Q by f (0) = 0 and f (x) = 8’n if |x| ∈ In [n ≥ 0]. In
what follows we work in Q.
(i) Show that

f (h) ’ f (0)
’0
h
as h ’ 0. Conclude that f is di¬erentiable at 0 with f (0) = 0.
(ii) Explain why f is everywhere di¬erentiable with f (x) = 0 for all x.
Conclude that f is in¬nitely di¬erentiable with f (r) = 0 for all r ≥ 0.
(iii) Show that

f (h) ’ f (0)
’∞
h2
as h ’ 0. Conclude that, if we write

f (0) 2
h + (h)h2 ,
f (h) = f (0) + f (0)h +
2!
0 as h ’ 0. Thus the local Taylor theorem (Theorem 7.1.3) is
then (h)
false for Q.
146 A COMPANION TO ANALYSIS

7.2 Some many dimensional local Taylor the-
orems
In the previous section we used mean value inequalities to investigate the
local behaviour of well behaved functions f : R ’ R. We now use the same
ideas to investigate the local behaviour of well behaved functions f : Rn ’ R.
It turns out that, once we understand what happens when n = 2, it is easy
to extend the results to general n and this will be left to the reader.
Here is our ¬rst example.
Lemma 7.2.1. We work in R2 and write 0 = (0, 0).
(i) Suppose δ > 0, and that f : B(0, δ) ’ R has partial derivatives f ,1
and f,2 with |f,1 (x, y)|, |f,2 (x, y)| ¤ M for all (x, y) ∈ B(0, δ). If f (0, 0) = 0,
then
|f (x, y)| ¤ 2M (x2 + y 2 )1/2
for all (x, y) ∈ B(0, δ).
(ii) Suppose δ > 0, and that g : B(0, δ) ’ R has partial derivatives
g,1 and g,2 in B(0, δ). Suppose that g,1 and g,2 are continuous at (0, 0) and
g(0, 0) = g,1 (0, 0) = g,2 (0, 0) = 0. Then writing
g((h, k)) = (h, k)(h2 + k 2 )1/2
we have (h, k) ’ 0 as (h2 + k 2 )1/2 ’ 0.
Proof. (i) Observe that the one dimensional mean value inequality applied
to the function t ’ f (x, t) gives
|f (x, y) ’ f (x, 0)| ¤ M |y|
whenever (x, y) ∈ B(0, δ) and the same inequality applied to the function
s ’ f (s, 0) gives
|f (x, 0) ’ f (0, 0)| ¤ M |x|
whenever (x, 0) ∈ B(0, δ). We now apply a taxicab argument (the idea
behind the name is that a New York taxicab which wishes to get from (0, 0)
to (x, y) will be forced by the grid pattern of streets to go from (0, 0) to (x, 0)
and thence to (x, y)) to obtain
|f (x, y)| = |f (x, y) ’ f (0, 0)| = |(f (x, y) ’ f (x, 0)) + (f (x, 0) ’ f (0, 0))|
¤ |f (x, y) ’ f (x, 0)| + |f (x, 0) ’ f (0, 0)| ¤ M |y| + M |x|
¤ 2M (x2 + y 2 )1/2
147
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for all (x, y) ∈ B(0, δ).
(ii) Let > 0 be given. By the de¬nition of continuity, we can ¬nd a δ1 ( )
such that δ > δ1 ( ) > 0 and

|g,1 (x, y)|, |g,2 (x, y)| ¤ /2

for all (x, y) ∈ B(0, δ1 ( )). By part (i), this means that

|g(x, y)| ¤ (x2 + y 2 )1/2

for all (x, y) ∈ B(0, δ1 ( )) and this gives the desired result.

Theorem 7.2.2. (Continuity of partial derivatives implies di¬eren-
tiability.) Suppose δ > 0, x = (x, y) ∈ R2 , B(x, δ) ⊆ E ⊆ R2 and that
f : E ’ R. If the partial derivatives f,1 and f,2 exist in B(x, δ) and are
continuous at x, then, writing

f (x + h, y + k) = f (x, y) + f,1 (x, y)h + f,2 (x, y)k + (h, k)(h2 + k 2 )1/2 ,

we have (h, k) ’ 0 as (h2 + k 2 )1/2 ’ 0. (In other words, f is di¬erentiable
at x.)

Proof. By translation, we may suppose that x = 0. Now set

g(x, y) = f (x, y) ’ f (0, 0) ’ f,1 (0, 0)x ’ f,2 (0, 0)y.

We see that g satis¬es the hypotheses of part (ii) of Lemma 7.2.1. Thus g
satis¬es the conclusions of part (ii) of Lemma 7.2.1 and our theorem follows.


Although this is not one of the great theorems of all time, it occasionally
provides a useful short cut for proving functions di¬erentiable1 . The following
easy extensions are left to the reader.

Theorem 7.2.3. (i) Suppose δ > 0, x ∈ Rm , B(x, δ) ⊆ E ⊆ Rm and that
f : E ’ R. If the partial derivatives f,1 , f,2 , . . . f,m exist in B(x, δ) and are
continuous at x, then f is di¬erentiable at x.
(ii) Suppose δ > 0, x ∈ Rm , B(x, δ) ⊆ E ⊆ Rm and that f : E ’ Rp . If
the partial derivatives fi,j exist in B(x, δ) and are continuous at x [1 ¤ i ¤
p, 1 ¤ j ¤ m], then f is di¬erentiable at x.
1
I emphasise the word occasionally. Usually, results like the fact that the di¬erentiable
function of a di¬erentiable function is di¬erentiable give a faster and more satisfactory
proof.
148 A COMPANION TO ANALYSIS

Similar ideas to those used in the proof of Theorem 7.2.2 give our next
result which we shall therefore prove more expeditiously. We write

f,ij (x) = (f,j ),i (x),

or, in more familiar notation,

‚2f
f,ij = .
‚xi ‚xj

Theorem 7.2.4. (Second order Taylor series.) Suppose δ > 0, x =
(x, y) ∈ R2 , B(x, δ) ⊆ E ⊆ R2 and that f : E ’ R. If the partial derivatives
f,1 , f,2 , f,11 , f,12 , f,22 exist in B(x, δ) and f,11 , f,12 , f,22 are continuous at x,
then writing

f ((x + h, y + k)) =f (x, y) + f,1 (x, y)h + f,2 (x, y)k
+ (f,11 (x, y)h2 + 2f,12 (x, y)hk + f,22 (x, y)k 2 )/2 + (h, k)(h2 + k 2 ),

we have (h, k) ’ 0 as (h2 + k 2 )1/2 ’ 0.

Proof. By translation, we may suppose that x = 0. By considering

f (h, k) ’ f (0, 0) ’ f,1 (0, 0)h ’ f,2 (0, 0)k ’ (f,11 (0, 0)h2 + 2f,12 (0, 0)hk + f,22 (0, 0)k 2 )/2,

we may suppose that

f (0, 0) = f,1 (0, 0) = f,2 (0, 0) = f,11 (0, 0) = f,12 (0, 0) = f,22 (0, 0).

If we do this, our task reduces to showing that

f (h, k)
’0
h2 + k 2

as (h2 + k 2 )1/2 ’ 0.
To this end, observe that, if > 0, the continuity of the given partial
derivatives at (0, 0) tells us that we can ¬nd a δ1 ( ) such that δ > δ1 ( ) > 0
and

|f,11 (h, k)|, |f,12 (h, k)|, |f,22 (h, k)| ¤

for all (h, k) ∈ B(0, δ1 ( )). Using the mean value inequality in the manner
of Lemma 7.2.1, we have

|f,1 (h, k) ’ f,1 (h, 0)| ¤ |k|
149
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and
|f,1 (h, 0) ’ f,1 (0, 0)| ¤ |h|
and a taxicab argument gives
|f,1 (h, k)| = |f,1 (h, k) ’ f,1 (0, 0)| = |(f,1 (h, k) ’ f,1 (h, 0)) + (f,1 (h, 0) ’ f,1 (0, 0))|
¤ |f,1 (h, k) ’ f,1 (h, 0)| + |f,1 (h, 0) ’ f,1 (0, 0)| ¤ (|k| + |h|)
for all (h, k) ∈ B(0, δ1 ( )). (Or we could have just applied Lemma 7.2.1 with
f replaced by f,1 .) The mean value inequality also gives
|f,2 (0, k)| = |f,2 (0, k) ’ f,2 (0, 0)| ¤ |k|.
Now, applying the taxicab argument again, using the mean value inequal-
ity and the estimates of the ¬rst paragraph, we get
|f (h, k)| = |f (h, k) ’ f (0, 0)| = |(f (h, k) ’ f (0, k)) + (f (0, k) ’ f (0, 0))|
¤ |f (h, k) ’ f (0, k)| + |f (0, k) ’ f (0, 0)|
¤ sup |f,1 (sh, k)||h| + sup |f,2 (0, tk)||k|
0¤s¤1 0¤t¤1

¤ (|k| + |h|)|h| + |k|2
¤ 3 (h2 + k 2 ).
Since was arbitrary, the result follows.
Exercise 7.2.5. Set out the proof of Theorem 7.2.4 in the style of the proof
of Theorem 7.2.2.
We have the following important corollary.
Theorem 7.2.6. (Symmetry of the second partial derivatives.) Sup-
pose δ > 0, x = (x, y) ∈ R2 , B(x, δ) ⊆ E ⊆ R2 and that f : E ’ R.
If the partial derivatives f,1 , f,2 , f,11 , f,12 , f,21 f,22 exist in B(x, δ) and are
continuous at x, then f,12 (x) = f,21 (x).
Proof. By Theorem 7.2.4, we have
f (x + h, y + k) =f (x, y) + f,1 (x, y)h + f,2 (x, y)k
+ (f,11 (x, y)h2 + 2f,12 (x, y)hk + f,22 (x, y)k 2 )/2 + 2
+ k2)
1 (h, k)(h

with 1 (h, k) ’ 0 as (h2 + k 2 )1/2 ’ 0. But, interchanging the rˆle of ¬rst
o
and second variable, Theorem 7.2.4 also tells us that
f (x + h, y + k) =f (x, y) + f,1 (x, y)h + f,2 (x, y)k
+ (f,11 (x, y)h2 + 2f,21 (x, y)hk + f,22 (x, y)k 2 )/2 + 2
+ k2)
2 (h, k)(h
150 A COMPANION TO ANALYSIS

with 2 (h, k) ’ 0 as (h2 + k 2 )1/2 ’ 0.
Comparing the two Taylor expansions for f (x + h, y + k), we see that
2
+ k2) = 2
+ k2)
f,12 (x, y)hk ’ f,21 (x, y)hk = ( 1 (h, k) ’ 2 (h, k))(h 3 (h, k)(h

’ 0 as (h2 + k 2 )1/2 ’ 0. Taking h = k and dividing by h2 we
with 3 (h, k)
have

f,12 (x, y) ’ f,21 (x, y) = 2 3 (h, h) ’ 0

as h ’ 0, so f,12 (x, y) ’ f,21 (x, y) = 0 as required.

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