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I a,. I for n � n0 , it is easily seen that the condition a,. + 0 does not hold, and (b) follows .
Note: The knowledge that lim a,. + 1 /a,. = 1 implies nothing about the convergence of l:a,. . The series 'I,1/n and l:1/n2 demonstrate this . 3.35
Examples (a) Consider the series 1 1 1 1 1 1 1 1 34 2 + 3 + 22 + 32 + 23 + 33 + 24 + + for which lim inf 11 + 00
a,. + 1 a,.
= lim 11 + 00
(�),. 3
�3111 2if. "
. tn . f vr:: l 1m a,. = t•1m 2 ..
ft + CIO
,.jv a,.
lim sup n + CIO
lim sup ,. ... oo
a,. + 1 a,.
n + 00
= lim II+ 00
= lim ,. ... oo
=
=
;; = 2
(�)" 2
=
· · · ,
0,
1
J3
 ,
1
,
2
J
+ ao .
The root test indicates convergence ; the ratio test does not apply. (b) The same is true for the series 1 1 I 1 I 1 1 1 2 + + 8 + 4 + 32 + 16 + 1 28 + 64 + . . . '
where
. f a,. + 1 = 1 , I 1m . tn 8 a,. ,. ... ao
I.tm sup
but
,. ... ao
a,. + 1 a,.
= 2,
lim y/a,. =
f.
68
PRINCIPLES OF MATH EMATICAL ANALYSIS
3.36
Remarks The ratio test is frequently easier to apply than the root test,
since it is usual ly easier to compute ratios than nth roots. However, the root test has wider scope. M ore precisely : Whenever the ratio test shows conver gence, the root test does too ; whenever the root test is i nconclusive, the ratio test is too. This is a conseque.nce of Theorem 3.3 7 , and is illustrated by the above examples. N either of the two tests is subtle with regard to divergence. Both deduce divergence from the fact that an does not tend to zero as n + oo .
3.37 Theorem
.�.Y:?o r any sequence {en } of positive numbers, en + 1 I l. ffi In f Cn , < Jim inf �n /n+ oo Cn n + oo .
Cn + l . I 1m sup vn /en < I.tm sup en n + oo n+ oo
·
Proof We shall prove the second inequality ; the proof of the first is quite similar. Put rx =
. en + 1 I 1m sup Cn .n + oo
I f rx = + oo , there is nothing to prove. is an integer N such that
for n > N. I n particular, for any
I f rx
·
is finite, choose
p > 0, (k
=
0, 1 , . .
.
'p
M ultiplying these inequalities, we obtai n or
Hence
< CN p Cn 
 N pn •
so that ( 1 8)
lim sup n + oo
(n � N).
y/en < p ,

1).
p > rx.
There
N U MERICAL SEQUENCES AND SERIES
69
by Theorem 3 .20(b) . Since ( 1 8) is true for every p > a, we have lim sup
n+
oc
ylc;, < a.
POWER SERIES 3.38
Definition
Given a sequence
{en} of complex numbers, the series <X>
) J....J cn z" n=O i s called a power series . The numbers en are called the coefficients of the series ; ( 1 9)
z is a complex number. In general , the series will converge or diverge, depending on the choice of z. M ore specifically, with every power series there is associated a ci rcle, the circle of convergence, such tha t f l 9) converges if z is i n the interior of the circle and diverges i f z is in the extei ior (to cover all cases , we have to consider the plane as the interior of a circle of infinite radius, and a point as a circle of radius zero). The behavior on the ci rcle of convergence is much more varied and can not be described so si mply.
3.39 Theorem
Given the power series I:c,. z", put a = lim sup
V'Tc:" l ,
(lf a = O , R = + oo ; if a = + oo , R = O . ) diverges if I z I > R.
Proof Put an
=
1 R= · ct
Then :Ecn z" converges if l z l < R, and
en z", and apply the root test :
lim sup
n +
<X>
v' l an l
= l z l lim sup
n +
<X>
v'�l
lzl R
Note: R is called the radius of convergence of I:cn z". 3.40 Examples
(a) The series :En" z" has R = 0. (b)
The series
z"
L n;. has R = + oo . ( I n this case the ratio test is easier to
apply than the root test. )
70
PRINCIPLES OF MATH EMATICAL ANALYSIS
(c) The series l:zn has R = 1 . I f l z l = 1 , the series diverges , since {z"} does not tend to 0 as n + oo . zn (d) The series L  has R = 1 . I t diverges if z = I . I t converges for all n other z with I z I = 1 . (Th e last assertion will be proved i n Theorem 3.44.) zn (e) The series L 2 has R = 1 . I t converges for all z with I z I = 1 , by n the comparison test, since I z"/n 2 1 = 1fn2 •
SUMMATION BY PARTS 3.41 Theorem
Given two sequences {a,}, {b,}, put
ifn > 0 ; put A _ 1
=
0. Then, if O < p < q, we have
q 1 q an bn = L A n (b,  bn + l) + A qbq  A p  l bp . L n n
(20)
=p
=p
Proof
q q q q l L an bn = L (An  An  1 ) b, = L A n bn  L A, bn + l ' n= p  l n=p n=p n=p and the last expression on the right is clearly equal to the right side of (20). Formula (20), the socalled "partial summation formula, " is useful i n the investigation of series of the form l:an bn , particularly when { b,} is monotonic. We shall now give applications.
3.42 Theorem
Suppose
(a) the partial sums A n of l:a, form a bounded sequence; ( b) b 0 > b 1 > b 2 > · · · ; (c) lim b, = 0. " .... 00
NUMERICAL SEQUENCES AND SERIES
71
Proof Choose M such that I An I < M for all n. Given e > 0, there is a n integer N such that bN < (ef2 M). For N < p < q, we have
q q 1 an bn = L A n(bn  bn + 1 ) + A q bq  A p  t bp nL n=p =p q 1 < M L (bn  biJ + 1 ) + bq + bp n=p = 2 MbP < 2 MbN < e .
Convergence now follows from the Cauchy criterion. We note that . the first inequality in the above chain depends of course on the fact that
bn  bn + 1 > 0.
3.43· Theorem
Suppose
(a) l c 1 l > l c2 l > l c 3 1 > · · · ; (b) C 2 m  1 > 0, C2 m < 0 (m = 1 , 2, 3, . . . ) ; (c) limn+ oo en = 0 .
Then Len converges. Series for which known to Leibnitz.
(b) holds are called "alternating series" ; the theorem was
n Proof Apply Theorem 3 .42, with an = (  1 ) + 1 , bn = I en I .
n Suppose the radius of convergence of rcn z is 1 , and suppose n Co > c 1 > c2 > . . . ' limn+oo Cn = 0 . Then rcn z converges at every point on the circle I z I = 1 , except possibly at z = 1. n Proof Put an = z , bn = en . The hypotheses of Theorem 3 . 42 are then
3.44 Theorem
satisfied, since
< if 1 z1 = 1 ,
z
:�=
2 
1 1  zl '
1.
ABSOLUTE C ONVERGENCE The series
l:an is said to converge absolutely if the series l: I a,. I converges.
3.45 Theorem If ra,. converges absolutely, then
u,.
converges.
72.
PRINCIPLES OF MATHEMATICAL ANALYSIS
Proof The assertion follows from the inequality m
m
Ln ak < nL l ak l , =k k= plus the Cauchy criterion .
3.46
Remarks
For series of positive terms, absolute convergence is the same
as convergence. I f l:an converges, but l: I an I diverges, we say that absolutely. For instance, the series
( l) L n
l:an converges non
n
converges nonabsolutely (Theorem 3.43 ) . The comparison test, as well as the root and ratio tests, is really a test for absolute convergence, and therefore can not give any information about non absolutely convergen t series. Summation by parts can sometimes be used to handle the latter. I n particular, power series converge absolutely i n the interior of the circle of convergence. We shall see that we may operate with absolutely convergent series very much as with finite �urns. We may multiply them term by term and we may change the order in which the additions are carried out, without affecting the sum of the series. But for nonabsolutely convergent series this is no longer true, and more care has to be taken when dealing with them.
ADDITION AND MULTIPLICATION OF SERIES
If Lan = A, and l:bn = B, then !'.(an Lean = cA , for an y fixed c .
3.47 Theorem
+
Proof Let
Then
n An + Bn = L (ak + bk ) . k=O Since lim n + An = A and lim n + Bn = B, we see that 00
00
n + oo The proof of the second assertion is even si mpler.
bn) = A + B, and
N UMERICAL SEQUENCES AND SERIES
73
Thus two convergent series may be added term by term, and the result i ing series converges to the sum of the two series. The situation becomes more < compl icated when we consider multiplication of two series. To begin with, we I have to define the prod uct. This can be done in several ways ; we shall consider 1 the socal led "Cauchy product. " � 3.48
Definition Given l:an and l:bn , we put
1,
n
(n = 0 , 2 , . . . ) en = 2: a, bn  k k=O ' and call l: cn the product of the two given series. This definition may be motivated as follows. I f we take tw9 power n n � series Ian z and I bn z , multiply them term by term, and collect terms contain i ing the same power of z, we get
n I an zn · I bn z = (a o + al z + a2 z 2 + · · ·> n=O n=O = a0 b 0 + (a0 h1 + a1 b0)z + (a0 b 2 + a1 b1 + a2 b0)z 2 + · · · = c0 + c 1 z + c 2 z 2 + · · · . 'X)
�
: Setting : 3.49
z=
1,
we arrive at the above definition.
Example
If
and An + A , Bn + B, then it is n ot at all clear that { Cn} will converge to AB, since we do not have Cn = An Bn . The dependence of {Cn} on {An} and {Bn} is quite a complicated one (see the proof of Theorem 3. 50). We shall now show that the prod uct of two convergent series may actually diverge. The series 00
(  I )n
I n = o Jn+l
1 1 1 = 1 ++ ··· J2 J3 J4
converges (Theorem 3.43) . We form the product of this series with itself and obtain 00
I n n = OC =
1 1 ) ( 1 1 (3 J2 J2 
+
+
� +
'J

1
+
1 )
J2J2 J3
(1
1
1
+  J4 + J3 J2 J2 J3
+
1)
··· + , J4
74
PRINCIPLES OF MATHEMATICAL ANALYSIS
so that
Since
(n  k + 1 )(k + 1 ) we have
2 ( 2 ) · ) ( 2 n n ) n ( = + + 2
1
I cn I > �
2
2
k= o n + 2
so that the condition en not satisfied.
_ 
0 be given . By (c), {3,. + 0 . Hence we can choose N such that I {3,. I < e for n > N, in which case
I Y,. I < l f3o a,. + · · · + f3N an  N I + I PN + t an  N  1 + S l f3o a,. + · · · + f3N an  N I + B !X. Keeping N fixed, and letting n + oo , we get
···
+ P,. a o l
lim sup I )',. I < e !X , since
n� oo
a" + 0 as k + oo . Since e is arbitrary, (2 1 ) follows.
Another question which may be asked is whether the series l:c,. , if con vergent, must have the sum AB. Abel showed that the answer is in the affirma tive.
Theorem If the series l:a,. , "Lb,. , l:c,. converge to A , B, C, and c,. = a0 b,. + · · · + a,. b0 , then C = AB .
3.51
Here no assumption is made concerning absolute convergence. We shall give a simple proof (which depends on the continuity of power series) after Theorem 8 .2.
REARRANGEM ENTS Definition Let {k,.}, n = 1 , 2, 3, . . . , be a sequence in which every positive integer appears once and only once (that is, {k,.} is a 1  1 function from J to J, in the notation of Definition 2.2). Putting
3.52
a = a" I
we say that
n
"
(n = 1 , 2, 3, . . . ) ,
l:a� is a rearrangement of l:a,. .
76
PRINCIPLES OF MATH EMATICAL ANALYSIS
I f {sn}, {s�} are the sequences of partial sums of l:an , l:a� , it is easily seen that, i n general , these two sequences consist of entirely different numbers . We are thus led to the problem of determining u nder what conditi ons all rearrangements of a convergent series will converge and w hether the sums are necessarily the same.
3.53 Example
Consider the convergent series
1 t+! ! +t ! + ···
(22) .
a nd one of its rearrangements
1 + !  ! + t + �  ! + � + fr  A + · · ·
(23)
i n which two positive terms are always followed by one negative. I f s is the sum of (22), then
s < 1  ! + ! = i .
Since
1 1 1 > 4k  3 + 4k 1 2 k 0 
for k > 1 , we see that s) < s� < s9 < · · · where s� is nth partial sum of (23). Hence . I 1 m SUp Sn > S 3' = _S_6 ' ,
n + oo
'
so that (23) certainly does not converge to s [we leave it to the reader to verify that (23) does, however, converge]. This example illustrates the following theorem, due to Riemann.
3.54 Theorem L et 1:an be a series of real numbers which converges, but not
absolutely. Suppose
Then there exists a rearrangement l:a� with partial sums s� such that lim inf s� =
(24)
n + oo
(X,
lim sup s� = p.
n + oo
Proof Let
(n = 1 , 2, 3, . . . ) .
77
N UM ERICAL SEQU ENCES AND SERIES
Then Pn  qn = an ' Pn + qn = I an I ' Pn > 0 , qn > 0. must both diverge. For if both were convergent, then
The series "£pn ' "£qn
would converge, contrary to hypothesis. Since N
I
n= 1
an
=
N
N
N
L ( Pn  qn) = I Pn  I qn ' n= l n=1
n= 1
divergence of Ipn and convergence of Iqn (or vice versa) implies diver gence of "£an again contrary to hypothesis. Now let P1 , P2 , P 3 , denote the nonnegative terms of Ian , in the order in which they occur, and let Q 1 , Q 2 , Q 3 , be the absol ute values of the negative terms of Lan , also i n their original order. The series IPn , I Qn d i ffer from Ipn , Iqn only by zero terms, and are therefore divergent. We shall construct sequences {Inn }, {kn }, such that the series ,
•
•
•
•
(2 5 )
P1
+ · · ·
+ P, .  Q r 
· · ·
 Qk . + P, , + 1
•
•
+ · · .
+ pm 2
 Qk , + 1
• 
•
·

Qk 2
+
wh ich clearly is a rearrangement of Lan , satisfies (24 ). Choose realval ued sequences {an}, {/3n } such that an � a, Pn an < Pn , P1 > o . Let n1 1 , k 1 be the smallest i ntegers such that P1 +
· · ·
P1 +
···
+ Pm 1 >
+ pm 1  Ql 
· · ·
let m2 , k2 be the smallest i ntegers such that P1
P1
+
+
. . . + P, .  Q t  . . .  Qk , · · · + Pm ,  Q 1  · · ·  Qk 1
+
P, , + 1 +
+ pm 1
+
1
f3 t ,  Qk , · · · +
+ · · ·
·
· ·'
�
/3,
< (.Xl ; P, 2 > f1 2 ,
+ pm 2  Q k , + 1  . . .  Qk2
< (12 ;
and continue in this way. Thi s is possible since 'LP and I Qn diverge. " I f Xn ' Yn denote the partial sums of (25) whose last terms are p  Qk" , then I
mn '
Since Pn � 0 and Qn � 0 as n . oo , we see that x" . /3, Yn . a. Final ly, it i s clea r that no number less than a or greater than fJ can be a subsequential l i mit of t he pa rtial s u 1ns of (25).
78
PRINCIPLES OF MATHEMATICAL ANALYSIS
3.55 Theorem Jf�a is a series of complex numbers which converges absolutely, then every rearrangement of �an converges, and the,. all converge to the same sum. n
Proof Let �a� be a rearrangement, with partial sums s� . Given e > 0, there exists an integer N such that m > n > N implies m
Ln l a d < 8 · i=
(26)
Now choose p such that the integers 1 , 2, . . , N are all contained in the set kb k2 , , k P (we use the notation of Definition 3. 52) . Then if n > p, the nu mbers a 1 , , aN will cancel i n the difference sn·  s� , so that I sn  s� I < e, by (26). Hence {s�} converges to the same sum as { sn} . .
•
•
•
•
•
.
EXERCISES 1.
2.
Prove that convergence of {sn} implies convergence of { I sn I }. Is the converse true ?
Calculate lim (v n2 + n  n). n + oo 3. I f S t = v2, and 4.
(n = 1 , 2, 3, . . . ),
prove that {sn} converges, and that sn 2 for < n = 1 , 2, 3, . . . . Find the upper and lower limits of the sequence {sn} defined by
5.
For any two real sequences {an}, {bn}, prove that
6.
provided the sum on the right is not of t he form oo  oo . Investigate the behavior (convergence or divergence) of Lan if
7.
lint sup (an + bn) < lim sup an + lim sup b n , n + oo n + oo n + oo
(a) an = Vn + 1  Vn ; V n + 1  V� (b) On = ; n (c) On = (V" n  1 ) n ;
1 for complex values of z. 1 + z" , Prove that the convergence of Lan implies the convergence of
(d) On =
if an > 0.
L
v� n
'
NUM ERICAL SEQUENCES AND SERIES
8. If
79
l:a" converges, and if {bn} is monotonic and bounded, prove that La" bn con
verges. 9. Find the radius of convergence of each of the following power series :
2" (b) L n ! z",
Suppose that the coefficients of the power series :La" z" are i ntegers, infin i tely many of which are d istinct from zero . Prove that the radius of convergence is at most I . 1 1 . Suppose a" > 0, Sn = a 1 + · · · + a" , and l:a,. d iverges.
10.
(a) Prove that L
a . diverges. 1 + an
(b) Prove that aN + k aN + l SN • >}+ ··+ SN + k SN + 1 SN + k and deduce that L
a" d iverges. Sn
(c) Prove that
and deduce that
""' an £.... 2 converges. Sn
(d) What can be said about
12.
Suppose a" > 0 and l:a" converges. Put 00
rn = L am . m=n (a) Prove that
if m
< n,
a n d deduce that L
an
r,.
d iverges.
80
PRINCI PLES OF MATHEMATICAL ANALYSIS
(b) Prove that �
�r,.
< 2( v r  v r
n
11
·v
and deduce that � 13.
14.
a,. �
1 
v r ,.
11
+
1)
converges.
Prove that the Cau,·hy product of two absolutely convergent series converges absolutely. If {s,.} is a complex sequence, define i ts arithmetic means a,. by So + St + . . . + Sn a,. = n+ 1 (a) (b) (c) (d)
(n
=
0, 1 , 2, . . . ).
I f lim s,. = s, prove that lim a,. = s . Construct a sequence {s,.} which does not converge, although lim a,. 0. Can it happen that s,. > 0 for all n and that lim sup s, = oo , although lim a, = 0 ? Put a,. = s,.  s,.  t , for n > 1 . Show that =
1 s,.  a,. = � kak . n + 1 k=I n
Assume that lim (na,.) = 0 and that {a,.} converges. Prove that {s,.} converges. [This gives a converse of (a), but under the additional assumption that na,. + 0.] (e) Derive the last conclusion from a weaker hypothesis : Assume M < oo , I na,. I < M for all n, and J im a,. = a. Prove that lim s, = a, by completing the following outl ine : If m < n, then m+ 1 s,.  a, = ( a,. nm

1 a m) + nm
� (s,.  s,). n
t =m+ l
For these i, I s,.  s d
(n  i)M (n  m  1 )M . < m+ 2 i+ 1
0 and associate with each n the integer m that satisfies m
E,. and if +1,
lim diam E,. = 0, n + oo
ll.
23.
then n f E,. consists of exactly one point. Suppose X is a complete metric space, and {G,.} is a sequence of dense open subsets of X. Prove Baire's theorem, namely, that n i G,. is not empty. (In fact, it is dense in X.) Hint: Find a shrinking sequence of neighborhoods E,, such that E, c G,. , and apply Exercise 21 . Suppose {p,.} and {q,.} are Cauchy sequences in a metric space X. Show that the sequence {d(p,. , q,.)} converges. Hint: For any m, n, d(p,. , q,.) < d(p,. , Pm) + d(pm , qm) + d(qm , q,. ) ; it follows that I d(p ,. , q,.)  d(pm , qm) I
24.
is small if m and n are large. Let X be a metric space. (a) Call two Cau�hy sequences {p,.}, {q,.} in X equivalent if l im d(p,. , q,) = 0. Prove that this is an equivalence relation. (b) Let X* be the set of all equivalence classes so obtained . If P E X*, Q E X*, { p } E P, {q,} E Q, define ,.
�(P, Q) = lim d(p,. , q,.) ; ,. .... 00
by Exercise 23, this limit exists. Show that the number �(P, Q) is unchanged if {p,.} and {q,.} are replaced by equivalent sequences, and hence that � is a distance function in X* . (c) Prove that the resulting metric space X* is complete. (d) For each p E X, there is a Cauchy sequence all of whose terms are p ; let Pp be the element of X* which contains this sequence. Prove that �(Pp , Pq) = d(p, q)
for all p, q E X. In other words, the mapping fP defined by cp(p) = Pp is an isometry (i.e. , a distancepreserving mapping) of A into X* . ( e) Prove that cp(X) is dense in X*, and that cp(X) = X* if X is complete. By (d) , we may identify X and cp(X) and thus regard X as embedded in the complete metric space X* . We call X* the completion of X. 25. Let X be the metric space whose points are the rational numbers, with the metric d(x, y) = I x  y I · What is the completion of this space ? (Compare Exercise 24.)
4 CONTINU ITY
The function concept and some of the related terminology were introduced i n Definitions 2. 1 and 2 . 2. Although we shall (i n later chapters) be mainly interested in real and complex functions (i .e. , i n functions whose values are real or complex numbers) we shal\ also discuss vectorvalued functions (i . e . , functions with values in Rk ) and functions with values i n an arbitrary metric space . The theo rems we shall discuss in this general setting would not become any easier i f we restricted ourselves to real functions, for instance, and it actually simplifies and clarifies the picture to discard unnecessary hypotheses and to state and prove theorems in an appropriately general context. The domains of definition of our functions will also be metric spaces, suitably specialized in various instances.
LIMITS O F FUNCTIONS 4.1 Y,
(1)
Let X and Y be metric spaces ; suppose E c X, f maps E i nto is a limit point of E. We write f(x) � q as x � p, or
Definition and
p
Iim f(x) = q
x + p
84
PRINCIPLES OF MATHEMATICAL ANALYSIS
if there is a point q E Y with the following property : For every e > 0 there exists a � > 0 such that
dy(f(x), q) < e
(2) for all points
x E E for which 0
0 be given. Then there exists � > 0 such that dy(f(x), q ) < e if x E E and 0 < dx(x, p) < �. _Adso, there exists N such that n > N implies 0 < dx(Pn , p) < �. Thus, for n > N, we have dy(f(Pn), q) < e, which shows that (5) holds. Conversely, suppose (4) is fa1se . Then there exists some e > 0 such that for every {J > 0 there exists a point x E E (depending on �), for which dy(f(x), q) > e but 0 < dx(x , p) < �. Ta k ing bn = 1 /n (n = 1 , 2, 3, . . . ), we thus find a sequence in E satisfying (6) for which (5) is false.
Corollary
Iff has a limit at p, this limit is unique.
This follows from Theorems 3.2(b) and 4.2.
CONTIN UITY
85
Suppose we have two complex functions, l and g, both defi ned on E. By I + g we mean the function which assigns to each point x of E the number l(x) + g(x) . Similarly we define the difference 1  g , the product fg, and the quotient f/g of the two functions, with the understanding that the quo tient is defi ned only at those points x of E at which g(x) =F 0 . If f assigns to each point x of E the same number c, then f is said to be a constant function, or si mply a constant, and we write I = c. If f and g are real functions, and if f(x) > g(x) for every x E £, we shall sometimes write f > g, for brevity. Similarly, if f and g map E into R k , we define f + g and f · g by
4.3
Definition
f(x) + g(x) , (f · g) (x) and if A. is a real number, ()�f)( x) = )..f(x). (f + g)(x)
=
=
f(x) · g(x) ;
Suppose E c X, a metric space, p is a limit point of E, f and g are complex functions on E, and lim g(x) = B . lim f(x) A
4.4
Theorem
x + p
Then (a) lim (f + g)(x) x + p
(b) lim (fg) (x) x + p
( c) lim
x + p
(1g) (x)
=
=
A + B;
=
AB;
=
A
B
,
,
x + p
if B =I= 0.
Proof In view of Theorem 4.2, these assertions follow immediately from the analogous properties of sequences (Theorem 3 . 3) .
Remark
(b')
If f and g map E into Rk , then ( a) remains true, and lim (f · g) (x) = A · B.
(b) becomes
x + p
(Compare Theorem 3.4.)
C O NTINUO US FUNCTIONS Suppose X and Y are metric spaces, E c X, p E £, and f maps E into Y. Then f is said to be continuous at p if for every e > 0 there exists a � > 0 such that
4.5
Definition
d y(f(x) , f(p)) < e
for all points x e E for which dx (x, p) < �. IfI is continuous at every poi nt of E, then I is said to be continuous on E. It should be noted that I has to be defined at the point p i n order to be continuous at p. (Compare this with the remark following Definition 4. 1 .)
86
PRINCIPLES OF MATHEMATICAL ANALYSIS
If p is an isolated point of E, then our definition implies that every function f which has E as its domain of definition is conti nuous at p. For, no matter which e > 0 we choose, we can pick � > 0 so that the only poi nt x E E for which dx(x, p) < � is x = p ; then
dr (f(x) ,f(p)) = 0 < e. 4.6 Theorem In the situation given in Definition 4. 5, assume also that p is a limit point of E. Then f is continuous at p if and only if lim x p f(x) = f(p) . 
Proof This is clear if we compare Definitions 4. 1 and 4. 5.
We now turn to compositions of functions. A brief statement of the following theorem is that a continuous function of a conti nuous function i s continuous.
Suppose X, Y, Z are metric spaces, E c X, f maps E into Y, g maps the range off, f(E) , into Z, and h is the mapping of E into Z defined by h(x) = g(f(x)) (x e £).
4.7
Theorem
Iff is continuous at a point p E E and if g is continuous at the point f(p) , then h is continuous at p. This function h is called the composition or the composite of f and g. The notation
h = g oj
is frequently used i n this context. Proof Let e > 0 be given. Since g is continuous at f(p) , there exists '7 > 0 such that
dz( g(y), g(f(p))) < e if d y (y, f(p)) < '1 and y e f(E) . Since f is continuous at p, there exists � > 0 such that dy(f(x) , f(p)) < '1 if dx (x, p) < � and x E £.
It follows that
dz(h(x) , h(p)) = dz (g(f(x)) , g(f(p))) < e if dx (x, p) < � and x E £. Thus h is continuous at p.
A mapping f of a metric space X into a metric space Y is con tinuous on X if and only if/  1 ( V) is open in X for every open set V in Y.
4.8 Theorem
(Inverse i mages are defined in Definition 2.2 . ) This is a very useful charac terization of continuity.
CONTINUITY
87
Proof Suppose / is continuous on X and V is an open set in Y. We have to show that every point of f  t ( V) i s an i nterior point of f  1 ( V ) . So, suppose p E X and f( p) E V. Since V is open, there exists e > 0 such that y E V if dy(f(p), y) < e ; and since f is continuous at p, there exists � > 0 such that dy (f(x), f(p)) < e if dx(x, p) < �. Thus x e f  1 ( V ) as soon as dx (x, p) < �. Conversely, suppose f  1 ( V) is open in X for every open set V in Y. Fix p E X and e > 0, let V be the set of all y E Y such that dy(y, f(p)) < e . Then V is open ; hence f  1 ( V) is open ; hence there exists � > 0 such that x ef  1 ( V ) as soon as dx(P, x) < �. But if x e f  1 ( V ), then f(x) E V, so that dy (f(x), f(p)) < e. This completes the proof. A mapping f of a nletric space X into a metric space Y is continuous if and only iff  1 ( C) is closed in X for every closed set C in Y.
Corollary
This follows from the theorem, since a set is closed if and only if its com plement is open, and since f  1 ( E c) = [f  1 ( E)]c for every E c Y. We now turn to complexvalued and vectorvalued functions, and to functi ons defined on subsets of R k . 4.9 Theorem L et f and g be complex continuous functions on a metric space X.
Then f + g, fg, andffg are continuous on X.
I n the last case, we m ust of course assume that g(x) =I= 0, for all
x
e X.
Proof At isolated points of X there i s nothing to prove. At limit points, the statement follows from Theorems 4.4 and 4.6 . 4. 10
Theorem
(a) L et .ft , . . . , .h be real functions on a metric space X, and let f be the mapping of X into Rk defined by f(x)
(7)
=
(/1 (x), . . . , J,. (x))
(x e X) ;
then f is continuous if and only if each of the functions/1 , , h. is continuous. (b) If f and g are continuous mappings of X in to R k , then f + g and f · g are continuous on X. •
g
•
, h are called the components of f. Note that The functions /1 , is a m apping i nto R k , whereas f · g i s a real function on X. •
f+
•
•
•
88
OF MATHEMATICAL ANALYSIS
PRINCIPLES
Proof Part (a) fol lows from the inequalities lfi(x)  jj(y) l < l f(x)  f(y) l =
tt1 1Ji(x)  /b) l 2 r.
for j = 1 , . . . , k. Part (b) follows from (a) and Theorem 4.9. 4. 1 1 Examples If x" . . . , x are the coordinates of the point x e Rk, the k functions tPi defined by
t/J;(x) = xi
(8)
are continuous on Rk, since the i nequality shows that we m ay take � = e in Definition 4. 5. The functions tjJ i are sometimes called the coordinate functions. Repeated application of Theorem 4.9 then shows that every monomial
xp�l
(9)
..
.
�k
where n 1 , , nk are nonnegative integers, is continuous on Rk. The same is true of constant multiples of (9), since constants are evidently conti nuous. It fol lows that every polynomial P, given by •
•
•
( 1 0)
is continuous on Rk . Here the coefficients cn . · · · nk are complex numbers, n 1 , ,n k are nonnegative i ntegers, and the sum in ( 1 0) has finitely many terms. , x , that i s, every quotient Furthermore, every rational fu nction in x 1 , k of two polynomials of the form ( 1 0) , is continuous on Rk wherever the denomi n ator is different from zero . From the triangle i nequality one sees easily that •
•
•
•
•
•
l l x i  I Y I I < I x  y l (x, y e Rk) . Hence the mapping x + I x I is a continuous real function on Rk. If now f is a continuous m apping from a metric space X into Rk , and if tjJ is defi ned on X by setting t/J(p) = I f(p) I , it follows, by Theorem 4. 7, that tjJ i s a continuous real function on X.
(1 1)
Remark We defined the notion of continuity for functions defined on a subse t E of a metric space X. However, the complement of E in X plays no role whatever i n this definition (note that the situation was somewhat different for limits of functions). Accordi ngly, we lose nothing of interest by discarding the complement of the domain off This means that we may just as wel l talk only about continuous mappings of one metric space into another, rather than
4.12
CON TI N U ITY
89
of mappings of subsets. This simpli fies statements and proofs of some theorems. We have already made use of this principle in Theorems 4 . 8 to 4. 1 0, and will continue to do so in the following section on compactness .
C O NTI N UITY A N D COM PACT NESS
4. 13 Definition A mapping f of a set E in to Rk is said to be bounded if there is a real number M such that I f(x) I < M for all x E £. 4. 14 Theorem Suppose f is a continuous n1apping of a con1pact n1etric space X into a metric space Y. Then f( X) is compact. Proof Let { V2} be an open cover off( X). Si ncefis contin uous, Theorem 4. 8 shows that each of the sets f  1 ( Vt%) is open . Si nce X is com pact, there are fi n itely many indices, say ct 1 , , ctn , such that •
( 1 2)
Si nce .f(f  1 (£)) c
•
.
X c f  1 ( Va 1 ) u · · · u f  1 ( Vt%, ) .
E
Y,
for every E c
( 1 2) implies that
f( X ) c Vt% u · · · u Vt% .
( 1 3)
This com pletes the proof.
tl
I
Note : We have used the relati on f(f  1 ( £ )) c E. valid for E c
E c X, then f  1 ((( £ ) )
E ; equality need not hold in either case. We shall n ow deduce some consequences of Theorem 4. 1 4.
Y.
If
=>
4.15 Theorem If f is a continuous mapping of a con1pact n1etric space X into Rk , then f(X) is closed and bounded. Thus, f is bounded. This follows from Theorem 2 . 4 1 . when .f is real :
4.16 Theorem space X, and ( 1 4)
The result is particularly im porta nt
Suppose f is a continuous real function on a compact
M = sup f(p), peX
n1
Then there exist points p, q E X such that f(p)
=
n1etric
inf .f(p). peX
= A1
and f(q)
=
n1.
The notation in ( 1 4) means that M is the least upper bound of the set of all numbers j (p), where p ranges over X, and that m is the greatest lower bound of this set of numbers.
96
PRINCI PLES OF MATHE MATICA L A N ALYSIS
The concl usion may also be stated as follows : There exist points p and q in X suclz that f(q) < f(x) < f(p) for all x E X � that is. f a t ta i n s i ts m a x i m u n1 (at p) and i ts m ini m u m (at q). Proof By Theorem 4. 1 5, f( X ) is a cl osed a n d bounded set of bers ; hence f( X ) contai ns
M = s u p f( X )
and
rea l
n u tn
m = i nfj( X).
by Theorem 2.28. 4. 17 Theorem Suppose f is a continuous 1  1 nzapping o.f a con1pact nzetric space X onto a metric sp ace Y. Then the inverse m app ing f  1 define d on Y by is
a
continuous n1 app ing of
Y
(x E X ) onto X.
Proof A pplying Theorem 4. 8 to .f  1 i n place off, we see t h a t i t suffices to prove that f( V ) is an open set in Y for every open set V i n .X'. F i x s uch a set V. The com plement V o f V is cl osed i n X, hence com pact (The o re m 2 . 3 5) ; hence .f( V c) is a c o m pac t su bset of Y (Theorem 4 . 1 4) a n d so i s cl osed i n Y (Theorem 2. 34). Si nce .f is onet o  one a n d onto, /( V ) is the complement off( V ) H ence / ( V) is open . c
c
.
4. 18 Definition Let f be a mapping of a metric space X in t o a metric s p a c e Y. We say t h a t f is uniform�J' continuous on X i f fo r e v e ry e > 0 t h e r e e x i sts b > 0 such that
( 1 5)
dy (f(p), f(q)) < e
for all p and q in X for wh ich dx(P, q) < b. Let us consider the differences between the concepts of continuity an d of uniform conti nuity. Fi rst, uniform continuity is a property of a function on a set, whereas continuity can be defi ned at a si ngle point. To ask whether a gi v en function is uniformly contin uous at a certain point is meaningless. Second, if f is conti nuous on X, then it is possible to fi nd, for each e > 0 and for each point p of X, a number {) > 0 having the property specified in Defi niti on 4. 5 . This {J depends on e and on p. Iff is, however, uniformly continuous on X, then it is possible, for each e > 0, to find one number 1J > 0 which wi l l do for all poi nts p of X. Evidently, every uniformly continuous functi on is continuous. That the two concepts are equivalent on compact sets follows from the next theorem .
CONTI N U ITY
4.19
91
Theorem Let f be a continuous n1apping o.f a co1npact n1etric space X
into a n1etric space
Y.
Then 1· is
unijor1nly
continuous on X.
Proof Let e > 0 be given . Since f is continuous, we can associate to each point p E X a positive number ¢ (p) such that
( 1 6)
t;
q E X, dx( P , q ) < ¢(p) implies dr (f( p), f(q)) < 2
·
Let J(p) be the set of all q E X for which ( 1 7)
dx (P , q) < !¢(p) . Si nee p E J(p ), the collect ion of all sets J(p) is an ·open cover of X; and si nce X is compact, there is a finite set of points p1 , , Pn i n X, such that • • •
( 1 8)
We put ( 1 9)
tJ = ! min
[¢( p 1 ),
• • •
, ¢ (P n ) ] .
Then 1J > 0. (Th is is one point where the finiteness of the coveri ng, in herent in the defi nition of com pactness. is essential . The minimum of a fini te set of posi tive num bers is positive, whereas the i n f of an i n finite set of positive numbers may very well be 0.) Now let q and p be points of X, such that dx ( P, q ) < f> . By ( 1 8). there is an in teger n1 , I < 111 < n , such that p E J(p'") ; hence (20) and we also have
dx(q, Pm ) < dx(p , q) + dx(P , Pm) < tJ + !c/>(Pnr ) < l/J(pm). Final ly, ( 1 6 ) shows that therefore
dl'(.f(p), f(q)) < dy(f(p) , f(Pm )) + dy(f(q) ,f(p'" )) < e. This completes the proof. An alternative proof is sketched i n Exercise 1 0. We now proceed to show that com pactness is essential in the h ypotheses of Theorems 4. 1 4, 4. 1 5, 4. 1 6, and 4. 1 9. 4.20 Theorem Let E be a noncon1pact set in R 1 • Then
(a) there exists a continuous function on E which is not bounded,· (b) there exists a continuous and bounded function on E which has no maxin1un1. If, in addition, E is bounded, then
92
PRINCIPLES OF MAT H EMATICAL ANALYSIS
(c) there exists a continuous function on E which is not unijorn1ly continuous. Proof Suppose fi rst that E is bounded, so that there exists a li mit point x0 of E which is not a poi nt of E. Consider
f(x) =
(2 1 )
1
(x E E).

x  x0
This is conti n uous on E (Theorem 4.9), but evidently unbou nded . To see that (2 1 ) is not uniform l y continuous, let e > 0 and () > 0 be arbitrary, and choose a point x E E such that I x  x0 I < f>. Taking t close enough to x0 , we can then make the difference 1/( t )  f(x) I greater than e, alth.ough I t  x I < f>. Si nce this is true for every () > 0, f is not uniformly conti nu ous on E. The fu nction g given by
g (x) =
(22)
1 l + (x  X o) 2
(x E £ )
is continuous on E, and is bounded, since 0 < g(x) < 1 . It is clear that sup g(x) = 1 , xe E
whereas g(x) < 1 for all x E E. Thus g has no maximum on £. H aving proved the theorem for bounded sets E. let us now suppose that E is unbo u nded. Then f(x) = x establishes (a), whereas x2 h(x) = 1 + x2
( 23)
(x E £ )
establishes (b), since sup h(x) = 1 xeE and h (x) < 1 for all x E E. Assertion (c) would be false if boundedness were omitted from the hypotheses . For, let E be the set of all integers. Then every fu nction defined on E is uniformly continuous o n E. To see this, we need merely take {> < 1 in Defi nition 4. 1 8 . We conclude this section by showing that compactness is also essential i n Theorem 4. 1 7 .
CONTINUITY
93
4.2 1 Example Let X be the halfopen interval [0, 2rr) on the real line, and let f be the mapping of X onto the circle Y consisting of all points whose distance from the origin is 1 , given by
(24)
f(t )
=
(cos t, sin t)
(0 < t < 2n).
The continuity of the trigonometric functions cosi ne and sine, as wel l as their periodicity properties, wi1 l be established in Chap. 8 . These results show that f is a continuous 1  1 mapping of X onto Y. However, the inverse mapping (which exists, si nce f is onetoone and o nto ) fails to be continuous at the point ( I , 0) = f(O). Of cou rse X i s no t com pact in this examp le. (It may be of interest to obser v e that f  t fails to be continuous in spite of the fact that Y is compact !) ,
CO NTI NUITY AND C O N NEC TE D N ESS Theorem Iff is a continuous mapping of a n1etric space X into a space Y, and if E is a connected subset of X, then f(E ) is connected. 4.22
n1
etric
Proof A ssume , on the contrary, that j'(E) = A u B, wh e re A and B are nonempty separated su bsets of Y. P ut G = E n /  1 (A ), H = t� n /  1 (B) . Then E = G u H, and neither G nor H is empty. Si nce A c A (the closure of A ), we have G c f  1 (A) ; the l a t ter set is closed, since f is continuous � hence G c f  1 (A). I t fol lows that f{G) c A . Since .f(H) = B and A n B is empty, we conclude that G n H is empty. The same argument shows that G n H is empty. Thus G and H are separated . This is impossi ble if E is connected. 4.23 Theorem Let f be a continuous real function on t he interval [a, b]. If f(a) < f(b) and if c is a number such that f(a) < c < f(b), then there exists a point x e (a, b) such tha t f(x) = c.
A sim ilar result holds, of course, if f(a ) > f(b). Roughly speaking, the theorem says that a continuous real function assumes all intermediate v alues on an interval.
Proof By Theorem 2.47, [a, b] is connected � hence Theorem 4. 22 shows that /([a , b ]) is a connected subset of R 1 , and the asserti on fol l ows if we appeal once more to Theorem 2.47.
4.24 Remark At first glance, it might seem that Theorem 4.23 has a converse. That is, one mi ght think that if for any two points x1 < x2 and for any number c between f(x1 ) and f(x ) there is a point x in (x 1 , x 2 ) su c h that f(x) = c, then f 2 must be continuou s. That this i s not so may be concl uded from Example 4.27 (d) .
94
PRINCIPLES OF MATHEMATICAL ANALYSIS
DISCO NTI N UITI ES
If x is a poi nt in the d omai n of defi nition of the fu nction f at which f is not continuous, we say that / is discontinuous at x, or tha t / has a discontinuity at x. If f is defi ned on an i nterval or on a segment, it is customary to divide d i scon tinuities into two types. Before gi ving this classification, we have to defi ne the righthand and the left h and lin1its of.lat x, which we denote by f(x + ) andf(x  ), respectively. 4.25 Definition Let ;· be defi ned on ( a, b) . Consider any point x such that a < x < b . We write
f(x + ) = q if f(t") +> q as n +> oo , for all sequences {t,} i n (x, b) such that t" +> x. To obtai n the defi nition o f f(x  ) , for a < x < b, we restrict ourselves to sequences { t11} i n (a, x). It is clear that any poi nt x of ( a, b), lim f(t ) exi sts if and only i f tx
f(x + ) =f(x  ) = Iim f(t ) . 4.26 Definition Let f be defined o n (a, b). I ff i s disconti nuous at a poi nt x , and if f(x + ) and j'(x  ) exist. then .l is said to have a di scontin uity of the first kind, or a simple discontinuity, at x. Otherwise the discontinuity is sai d to be of the second kind. There are two ways in which a functi on can have a si tn ple di sconti n uity : either f(x + ) =F f(x  ) [i n which case the val ue f(x) is i m materi al], or f(x + ) = f(x  ) =F .f(x). 4.27
Examples (a) Define
f(x)
=
{�
(x rational), (x i rrational).
Then f has a discontinuity of the second kind at every point x. si nce neither f(x + ) nor f(x  ) exists . (b) Define f(x)
=
{�
(x rational), (x i rrational).
CONTIN UITY
Then f is continuous at x kind at every other point. (c) Defi ne f(x)
=
=
0 and has a discontinuity of the second
x+2
(  3 < X <  2),
x  2
(  2 < x < 0)�
x+2
(0 < X < 1 ) . =
Then j' has a sin1 ple disc ontinuity at x every other p oint of (  3 , I ). (d) Define .f(x)
=
95
0 and is continuous at
. 1 stn 
(x i= 0),
0
(x
x
=
0).
Si nce nei ther f(O + ) nor f(O  ) exists, .f has a disconti nuity of the second kind at x = 0. We have not yet shown that s i n x is a continuous function. If '"'·e assume this result for the moment, �fheorem 4. 7 implies that f is contin uous at every point x =F 0.
M O NOTONIC F U N CTI O N S
W e shal l now study those functions which never decrease (or never inc r ease) on a given segment. 4.28 Definition Let f be real on (a. b). The n f is s aid to be 1nonotonical/y increasing on (a, b) if a < x < y < b i m plies .f(x) < f(y). If the last inequali ty is reversed , we obtai n the defi nition of a 1nonotonical/y de creasing function. The class of m onotonic functions consists of both the increasi ng and the decreasi ng functions. 4.29 Theorem Let f be monotonically increasing o n (a, b). Then f(x + ) and f(x  ) ex ist at every point of x of (a, b) . .\lore precisely,
(25)
sup f( t ) a 0 such that a < x  {> < x and
A  e < f(x  f>) < A .
(27)
Since f i s monotonic, we have f(x  f>) < .f(t ) < A
(28)
(x 
[,
< t < x).
Combi ni ng (27) and (28), we see that
lf(t )  A I < e
(x  {> < t < x).
Hence f(x  ) = A . The second half of (25) is proved i n precisely the same way. Next, if a < x < y < b, we see from (25) that f(x + ) = inf f(t ) = i nf f( t )
(29)
x ( x) to exist at a point x J < n 1 > (t ) must exist in a neighbor hood of x (or in a onesided neighborhood, if x is an endpoint of the interva l on which / is defined), and f< n  t ) must be differentiable at x. Si nce f < n  t ) must exist i n a neighborhood of x , j· must be differentiable i n that neighborhood . .
.

,
TAYLOR 'S THE O REM 5.15 Theorem Suppose f is a real function on [a, b ], n is a positive integer, < n  1 > is continuous on [a, b ] , J (t ) exists for every t E (a, b). Let et , f3 be distinct J points of [a, b ], and define
(23)
P( t )
=
n  l j( k){et)
L k= O
k!
(t

rx
)k .
DIFFERENTIATION
Then there exists a point x between
r:t.
111
and p such that
(2 4) For n = I , this is j ust the mean val ue theore m . In general , the theorem shows that .f ca n be a pproxi mated by a polynomial of degree n  I , and that ( 24) al lows us t o est i mate the error, if \Ve k now bou nds on I J. 2 x    I >   1 .
(38) Hence
.f '(x) g' (x)
9 (3 ) and so
I
x) f ( ' lim x  o g '(x)
( 40 )
X
 < l g'(x) l  2  x ==
0
.
By ( 36) a nd ( 40 ) , L , Hospit a r s rule fa i l s i n t h is case. Note a l so that g' (x) i= 0 on ( 0 � I ), by ( 3R ) . H oweve r. there is a conseq uence of the mean va l ue theorem wh ich , for pu rposes of a ppl ications, is a l most a s usefu l as Theorem 5. 1 0. and wh ich re m a i n s true fo r vectorva l ued fu nct ion s : From Theorem 5. I 0 i t fol l ows that
lf"'( x) i .
If( b )  j"( a ) I < ( h  a) s u p
(4 I )
a<x 0 i n (a, b). Prove that / is strictly in creasing in (a, h ) , a n d let ,q be its inverse funct ion . Prove that g is differentiable, and that g ' ( f(x) )
=
(a < x < h).
�
f' x )
3. Suppose g is a real fu nction on R 1 , with bounded derivative (say
l _q' I
0, and define f(x) = x + eg (x ) . Prove that f is onetoone if E is small en ough . (A set of admissi ble val ues of E can be determined which depends only on M. ) 4. If c1
Co +  + · · · + where Co , .
..,
2
c·n n
I
+
cn + 1
11
=
0'
Cn are real constants, prove that the eq uation C + C 1 X + . + Cn  1 X n  1 + Cn X n 0 0
.
=
•
has at least one real root between 0 and 1 . 5. Suppose / is defined and d ifferentiable for every x > 0, and ['(x) Put g(x) = f(x + 1 )  /(x). Prove that g(x) � 0 as x � + oo .
6. Suppose
� 0 as x �
+
oo .
(a) f is continuous for x > 0, (b) /'(x) exists for x > 0, ( c) f(O) 0, =
(d ) [' is monotonically increasing. Put g (x)
= f(x)
(x > 0)
X
and prove that g is monoton ically i ncreasing. 7. Suppose /'(x ), g '(x ) exist, g '( x) "* 0 , and /(x) lim r + x
= g (x)
=
0. Prove that
f(t ) = f '( x) . g( t )
g' (x)
(This holds also for con1 plex functions.) 8. Suppose /' is contin uous on [ a, b] and e > 0. Prove that there ex ists S > 0 such that /( t )  f(x) f' (x) tx
<e
DIFFERENTIATION
115
whenever 0 < I t  x I < S , a < x < b, a < t � b . (Th is could be expressed by saying that f is uniformly differentiable on [a, b] i f f is contin uous on [a, b ] . ) Does this hold for vect orval ued funct ions too ? 9. Let f be a continuous real function on R 1 , of wh ich it is kn own that /'(x) exists for all x # 0 and that /'(x) + 3 as x + 0. Does it follow that /'(0} exists ? 10. Suppose f and g are complex differen tiable funct ions on (0, I ), /( x ) � 0, _q(x) + 0, /'(x) + A , g'(x )  + B as x + 0, where A and B arc complex numbers, B # 0. Prove that '
A . im /(x) = I x + o g(x ) B
Com pare with Example 5 . 1 8 . Hint :
/(x) g(x)
={
x /( )  A \ . � + A . � . x } g(x) g(x)
A pply Theorem 5 . 1 3 to the real and imaginary parts of f(x ) /x and g(x)/x. 1 1 . Suppose f is defined in a neighborh ood of x, and suppose /"(x ) exists. Show that
. f(x + h ) + f(x  h )  2/( x ) I1m h2
h 0
= X
" / ( )
•
Show by an example that the J imit n1ay exist even if f''(x) d oes not . Hint: U se Theorem 5 . 1 3 . 12. I f /(x ) I x 1 3 , compute f'(x), f"( x ) for a l l real x, a n d show that j< 3 >(0) does not ex ist. 13. Su ppose a and c are real numbers, c > 0, and f is defined on [  1 , 1 ] by
=
f(x)
{Xa = 0
sin (x  c )
(if (if
X X=
=f 0), 0).
Prove the following statements : (a) f is continuous if and only if a > 0. (b) /'( 0 ) exists if and only if a > 1 . (c) f' is bounded if and only if a > 1 + c. (d) f' is continuous if and only if a > I + c. (e) / (0 ) exists if and only if a > 2 + c. (/) /" is bounded if and only if a > 2 + 2c. (g) f" is continuous if and only if a > 2 + 2c. 14. Let f be a differentiable real function defined in (a, b). Prove that f is convex i f and only i f f ' i s monoton ically increasing. Assume next that /"(x) ex ists for every x E (a, b), and prove that / is convex if and only if f''(x) > 0 for all x E (a, b). 15. Suppose a E R 1 , /is a twicedifferentiable real function on (a, oo ) , and Mo , M. , M2 are the least upper bou nds of 1 /(x ) I , lf'(x) I , 1/"(x) I , respectively, on (a, oo ) . Prove that "
1 16
PRINCIPLES OF MATHEMATICAL ANALYSIS
Hint: If h > 0, Taylor's theorem shows that f' (x) = for some g
E
� [f(x + 2h)  /(x) ]  hf"(g )
2
(x, x + 2h ) . Hence
(x) > 3 for some x E (  1 , 1 ) Note that equali ty holds for �( x 3 + x 2 ) . Hint: Use Theorem 5 . 1 5, with ex = 0 and f3 = ± I , to show that there exist s E (0, 1 ) and t E (  I , 0) such that .
J< 3>(s)
18. Suppose f is a real furtctio n on
+ j< 3 >(t )
= 6.
[a, b ], n is a positive in teger, and j . , . . . , 4> " , and c is the vector (c � , . . . , c�c). U se Exercise 26, for vectorvalued functions.
2 29. Special ize Exercise 28 by considering t he system
(j = 1 , . . . , k  1 ),
'
Y1 = YJ + t y � = J<x>
"
 JL ui x)yj , = l
where /, g . , . . . , g" are contin uous real functions on [a , b], and derive a uniqueness theorem for solut ions of the equation
y< " > + U�c(x)y0'  1 > + · · · + g 2 (x)y' + U• (x)y
subject to initial cond itions
y '(a) = c 2 ,
...
'
=
/(x),
6 THE RIEMANNSTIELTJES INTEGRAL
The present chapter is based on a definition of the Riemann integral which depends very explicitly on the order structure of the real line. Accordingly, we begin by discussing integration of realvalued functions on intervals. Ex tensions to complex and vectorvalued functions on intervals follow in later sections. I ntegrati on over sets other than intervals is discussed in Chaps. 1 0 and 1 1 .
DEFINITIO N AND EXISTENCE O F THE INTEGRAL Let [a, b] be a given interval . By a mean a finite set of points x0 , x 1 , x" , where
6.1
Definition
•
•
•
partition P of [a, b] we
,
a  x < x1 < ··· < xn  1 < xn  b• 
0 
We write (i = 1 ,
. . , n) . .
THE RIEMANNSTIELTJES INTEG RAL
121
Now suppose f i s a bounded real function defined on [a, b]. Corresponding to each partition P of [a, b] we put M; = sup f(x) n1 ;
= inf/(x)
(X; _ 1 < X < X ;), (X;  1 < X < X;),
n
U(P, f) = L M ; Ax; , i= 1
L(P, f) = L m ; Ax ; , i= 1 n
and finally (1)
Ia f dx = inf U(P, f),
(2)
r f dx = sup L( P, f),
b
_a
where the inf and the sup are taken over all partitions P of [a, b] . The left members of (1 ) and (2) are called the upper and lower Riemann integrals of f over [a, b ], respectively. I f the upper and lower integrals are equal, we say that f is Riemann integrable on [a, b ], we write f e §I (that is, � denotes the set of Riemann integrable functions), and we denote the common value of ( 1 ) and (2) by
J: t dx,
(3) or by
J:f(x) dx.
(4)
This is the Riemann integral of f over [a, b ]. Since f is bounded , there exist two numbers, m and M, such that
m < f(x) < M
(a < x < b).
Hence, for every P,
m(b  a) < L(P, f) < U(P,f) < M(b  a), so that the numbers L(P,f ) and U(P,f) form a bounded set. This shows that the upper and lower integrals are defined for every bounded function f The question of their equality, and hence the question of the integrability off, is a more delicate one. Instead of investigating it separately for the Riemann integral, we shall immediately consider a more general situation.
122
PRINCIPLES OF MATH EMATICAL ANALYSIS
Definition Let e< be a monotonically i ncreasing function on [a, b] (si nce a(a) and a(b ) are finite, it follows that a is bounded on [a , b]). Correspondi ng to each partition P of [a, b ], we wri te 6.2
�ai =
a(x ; )  a(x ; _ 1 ) .
I t is clea r that � a i > 0. For any real function f which i s bounded on [a, b] we put n
U ( P, J, a) = I M ; �a i , i= 1 n
L( P, J, a) = L m ; �a i , where M i ,
n1 i
i= 1
have the same meaning as in Definition 6. 1 , and we define
(5) (6)
fu J f drx. = sup L( P, J, b
f drx. = i nf U(P, J, a.),
b
_a
)
a. ,
the inf and sup again being taken over all partitions. I f the left members of (5) and (6) are equal, we denote their com mon value by (7)
or sometimes by (8)
{ f(x) drx.(x) . b
This i s the RiemannStie/tjes integral (or simply the Stieltjes integral ) of f with respect to a, over [a, b ] . I f (7) exists, i .e., if (5) and (6) are equal, we say that f is integrable with respect to a, in the R iemann sense, and write f e Bl(a) . By taking a( x) = x, the Riemann integral is seen to be a special case of the RiemannStieltjes integral . Let us mention explicitly, however, that in the general case a need not even be continuous. A few words should be said about the notation. We prefer (7) to (8), since the letter x which appears in (8) adds nothing to the content of (7) . I t i s i m material which letter we use to represent the socalled "variable of i ntegrati on." For instance, (8) is the same as
L f(y) drx.(y). b
T H E RIEMANNSTIELTJES I NTEG RAL
123
The i ntegra l depends on f, a , a and b, but not on the variable of i ntegrati on, which may as well be omitted . The role played by the vari able of i ntegration i s quite analogous to that of the index of summation : The two symbols
n C iL =l ;,
n LC k= l k
are the same, since each means c 1 + c 2 + · · · + en . Of cou rse, no hann is done by i nserting the variable of integration, and i n many cases it i s actual!y convenient to do so. We shall now investigate the existence of the i ntegral (7) . Without saying so every time, f will be assumed real and bounded , and a. monotonically increasing on [a, b] ; and , when there can be no misunderstanding, we shall write i n place of
l
b
J
·
6.3 Definition We say that the partition P* is a refinenzent of P if P * => P (that is, i f every poi nt of P is a poi nt of P *). Given two partitions, P1 and P2 , we say that P * is their common refinement if P * = P1 u P2 • 6.4
Theorem If P* is a refinement of P, then L(P, f, a) < L(P* , J, a)
(9) and
U( P * , J, a.) < U(P, f, a).
( 1 0)
Proof To prove (9), suppose fi rst that P* contains just one point more than P. Let th is ext ra point be x *, and suppose x i  t < x * < x b where X; _ 1 and x i are two consecutive points of P. Put
W1 w2 Clearly w 1 > m; and w 2 >
mi Hence
=
=
=
1n ; ,
i nff(x) inff(x)·
(x i  t
0 there ex ists a
partition P such that
( 1 3)
U(P , f, a)  L(P, f, a) < e. Proof For every P we have
I
I
L(P, J, oc) < f da. < f da. < U(P, J, a.). Thus ( 1 3) implies 0
0, we have
I1 da. = I1 da., that is, f e 91(a). Conversely, suppose f e fJt(a) , and let e > 0 be given . Then there exist partitions P 1 and P2 such that
I
;
( 1 4)
U(P2 , f, a)  f da.
f. a) + e < L(P, f, a) + e ,
so that ( 1 3) holds for this partition P .
Theorem 6.6 furnishes a convenient criterion for integrabi lity. Before we apply it, we state some closely related facts. 6.7 Theorem (a) If ( 1 3) holds for some P and some e , then ( 1 3) holds ( with the same e)
for every refinement of P. (b) If ( 1 3) holds for P {x0 , [x i _ 1 , X;], then =
•
•
•
,
xn} and if S; , I ; are arbitrary points in
L l f(s i)  f( t ;) I n
Aa.; < e .
i= 1 (c) Iff e 9l(!X) and the hypotheses of (b) hold, then
t
i l
f(t ;) Aa ; 
{ f da
< e.
Proof Theorem 6.4 implies (a). Under the assumptions made in (b), bothf(s;) andf(ti) lie i n [m ; , M;], so that If(�·;)  f(t;) I < M i  m i . Thus
L lf(s;)  f( t ;) I n
Aa.; < U(P,f, a.)  L(P, f, a.), i= 1 which proves (b). The obvious i nequalities
and pr.ove (c). 6.8 Theorem
L(P, f, a.) < 'l_ f( t ;) Aa.; < U(P, f, a.) L(P, f, a.) < J fda. < U(P, f, a.) Iff is continuous on [a, b] then f e al (a.) on [a, b].
Proof Let e > 0 be given. Choose
11
>
0 so that
[a.(b)  a.(a) ]17 < e .
Sinee f is uniformly continuous on [a , b ] (Theorem 4. 1 9), there exists a � > 0 such that ( 1 6)
lf(x)  f( t ) I < '1
126
PRINCIPLES OF MATHEMATICA L ANA LYSIS
if x e [a, b ] , t e [a, b ] and l x  t l < b. I f P is any partiti o n of [a, b] such tha t �x; < () for implies that M '·  nl l· < .n, ( 1 7) (i  1, . . . , n) ,
al l i,
then ( 1 6)

and the refore
U( P, f, a)  L( P, J: a.) < '1 L Aa. ; n
By Theorem 6.6, f E f!A(r:�.).
i= 1
=
L ( A1 ;  nzJ � 'Y. i n
i=
1
= '1 [ a( b)
 a.( a)] < e.
6.9 Theorem Iff is 1nonotonic on [a, b ], and if � is continuous on [a, b ], then f E !!A(a.). ( We still assume, of course, that a. is 1nonotonic.) Proof Let such that
e
> 0 be given . For any po s itive intege r n, choose a pa rt iti o n
�et . '
=
a( b)  a.(a) n
(i = 1 , . . , n). .
This is possi ble since a. is cont inuous (Theorem 4.23). We suppose that/is monotonically increasing (the proof is a n alogou s i n the other case) . Then Mi
so that
= f (x; ) ,
U( P, f, et )

L ( P, f, a)
(i
a(a) n
=
a.(b)
=
a( b)  �(b) n
11
I
i= 1 ·
= 1,
.
. .
[f(x ;)
, n ),
'
 ftxi  t )J
[f(b)  f(a)]
0 be given. Since uniformly continuous on [a, b ], there exists � > 0 such that
P = {x0 , X ;  1 < t < . x ; , i t fol lows that •
•
•
,
I y'( t ) l < I y' (x ; ) l + e .
Hence
f
x; 
1
I i ( t ) I dt < I y '(x i) I Ax 1 + s Ax,
 r· < r·
[y'(t ) + y'(xi)  y ' ( t )] dt + e Ax;
X;  1
y'( t ) dt +
r
Xi  I
X;  t
[y ' (x i)  y '(t )] dt + e �xi
< I Y(x i)  Y(x;  1 ) I + 2e L\x i . I f we add these inequalities, we obtain
J: I y '( t) I dt < A(P, y) + 2e(b  a) Since
e was
y' is
if l s t l < � . I y'(s)  y'(t) I < e xn } be a partition of [a, b ], with Ax; < � for all i. If 
Let
137
arbitrary,
< A(y) + 2e(b  a) .
f.. I y'(t) l dt < A(y). b
This completes the proof.
138
PRINCIPLES OF MATH EMATICAL ANALYSIS
EXERCISES 1 . Suppose
f(x)
=
increases on [a, b ], a < x o < b, ex is continuous at x o , f(x o ) 0 if x # Xo. Prove that f E Bl(ex) and t hat f f dex 0. ex
=
1 , and
=
2. Suppose f > 0, f is continuous o n [a, b], and
I: /(x) dx
=
0. Prove that /(x)
=
0
for all x E [a, b ] . (Compare this with Exercise 1 .) 3. Define three functions f3h {32 , {33 as follows : f3J {x) 0 i f x < 0, f3ix) 1 if x > 0 for j 1 , 2, 3 ; and {3 1 (0) = 0, {32(0) 1 , {33(0) !. Let f be a bounded function on [  1 ' 1 ]. (a) Prove that f E f!A(/3 . ) i f and only i f ((0 + ) = /(0) and that then =
=
=
I
1
=
=
df3,
=
/(0).
(b) State and prove a similar result for !32 . (c) Prove that f E flt(/33) i f and only i f f is continuous at 0. (d) I f f is continuous at 0 prove that
4. If f(x ) 0 for all irrational x , /(x) = 1 for all rational x, prove that f ¢ f!A o n [a , b] for any a < b. 2 5. Suppose f is a bou�ded real function on [a, b], and 1 E 8l on [a, b] . Does it 3 follow that I E fit ? Does the answer change if we assume that / E fit ? 6. Let P be the Cantor set constructed i n Sec. 2.44 . Let f be a bounded real function on [0, 1 ] which is continuous at every point outside P. Prove that f E fit on [0, 1 ] . Hint: P can be covered by finitely many segments whose total length can be made as small as desired . Proceed as in Theorem 6. 1 0. 7. Suppose I is a real function on (0, 1 ] and f E fit on [c, 1 ] for every c > 0. Define =
f l(x) dx = I 1
0
lim
C + 0
1
f(x) dx
C
if this limit exists (and is finite). (a) I f I E fit on [0, 1 ], show that this definition of the integral agrees with the old one. (b) Construct a function f such that the above limit exists, although it fails to exist with I I I in place of f. 8. Suppose f E fit on [a, b] for every b > a where a is fixed . Define
I«>f(x) dx = Ibf(x) dx lim
G
b + CX>
G
if this limit exists (and is finite). In that case, we say that the integral on the left
converges. I f it also converges after f has been replaced by I l l , it is said to con verge absolutely.
TH E RIEMANNSTIELTJES INTEGRAL
Assume that f(x ) > 0 and that that
converges i f and only
f decreases monotonically on [1 ,
oo
139
) . Prove
J.""/(x) dx
if
00
L f(n)
n= l
converges. (This is the socalled "integral test " for convergence of series.) 9. Show that integration by parts can sometimes be applied to the " improper " integral s defined in Exercises 7 and 8 . (S tate appropriate hypotheses, formulate a
oo f COS X
(£ f
theorem, and prove i t . ) For instance show that
o
1 +
x
dx =
o
sin
X dx.
+ x) 2
(1
Show that one o f these integrals converges absolutely , but that t he other does not. 1 0. Let p and q be positive real numbers such t hat 1 1 + = 1.
p
q
Prove the following statements.
(a ) If u > 0 and v > 0, then
u" vq uv <  +  .
Equality holds i f and only if u" = v4•
p
q
(b) If f E �(ex), g E � (ex), / > 0, g > 0, and
f. p da. = f. g• da., 1 =
b
then
b
r fg da. < 1 . ..
(c)
I ff and g are complex functions i n �(ex) , then
r fg da. < { f l f i • da.}" { f l u i • da.r . '
·
This is Holder' s inequality . When p = q = 2 it is usually called the Schwarz inequality. (Note that Theorem 1 .35 is a very special case o f this.)
(d) Show that Holder's inequality is also true for the " improper " i ntegrals de scribed in Exercises 7 and 8.
140
PRINCIPLES
1 1 . Let
<X
OF
MATHEMATICAL ANALYSIS
be a fixed increasing function on [a, b) . For u E !Jt(<X), define
ll u l l 2
=
{J. I u l 2 drx} 1 b
/
2
•
Suppose f, g, h E !Jt(<X), and prove the tria n gl e inequality
I I/  h l l 2 < 11 /  u lb + l lu  h l l 2
as a consequence of the Schwarz inequality, as in the proof of Theorem 1 . 37. 12. With the notations of Exercise 1 1 , suppose f E 9t{a) and e > 0. Prove that there exists a contin u o us function g on [a, b] such that I I/  g l 1 2 < e . Hint: Let P = {xo , . . . , x,.} be a suitable partition of [a, b), define g(t) =
1
if X 1 � t < Xt . 13. Define
t  x,  1 x,  t (x, ) /(x , _ t ) + � � x, x, f
f(x)
=
fx+l x
1 /{x) I < 1 /x if x > 0. Hint: Put t 2 = u a nd integrate by
(a) Prove that
sin ( t 2 )
dt.
J<x+ 1>2
parts, to show that /(x) is equal t o
(x 2) cos [(x + 1 ) 2 ] 2x 2 x + 1) .x 2
(
cos

Replace cos u by
 1.
cos u
4u 3 1 2 du
(b) Prove that
2xf(x) = cos (x 2)  cos [(x + 1 ) 2 ] + r (x) where I r(x) I < c/x
and c is a constant .
(c) Find the upper and lower limits of xf(x), as
(d)
Does
Jo"'
sin (1 2 )
14. Deal similarly with
dt converge ?
x � oo .
fx+l f(x) = x (e') dt. sin
Show that
ex ! /(x) l < 2
and that exf(x) = cos
where I r(x) I
0 there is an integer N such that n > N implies lfn (x)  f(x) I < B ( 1 2) for all x E £. It is clear that every uniformly convergent seq uence is pointwise con vergent. Quite expli c itly , the difference between the two c oncepts is this : If {/n} converges pointwise on E, then there exists a fun c tion f such that , for every B > 0, a nd for every x E £, there is an integer N, dependi ng on B and on x, such that ( 1 2) holds if n > N ; if{/,} co nverges uniformly on £, it is possible, for each B > 0, to find one integer N which will do for all x E £. We say that the series L.fn(x) converges uniformly on E if the sequence { sn} of partia l sums defi ned by
7.7
n
i= 1
L /; (x)
=
sn(x)
converges uniformly on E. The Cauchy criteri on for u niform convergence is as follows. 7.8 Theorem The sequence offunct ions {.(,.}, defined on E, converges uniformly on E if and only ij'for every B > 0 there exists an integer N such that m > N, n > N, x E E implies l /,,(x) j� (x) I < B. ( 1 3) 
Proof Suppose {/,} converges u niformly on E, and let f be the limit function. Then there is an integer N such that n > N, x E E implies l f,(x)
B , <  f(x) I 2
so that if n
>
Jf,.(x)  fm (x) I < Jfn (x)  f(x) I
N,
m
> N, x E
£.
+ lf(x)  fm(x) I < B
PRINCIPLES OF MATH EMATICAL ANALYSIS
148
Co nverse l y , suppose the Cauchy condition holds. By Theorem 3 . 1 1 , the sequence {/,.(x) } converges, for every x, to a limit which we may call f(x) . Thus the sequence {/,.} converges on E, to f. We have to prove that the convergence is uniform. Let e > 0 be given, and choose N such that ( 1 3) holds. Fix n, and let m + oo in ( 1 3) . Since fm(x) + f(x) as m + oo , this gives
l .fn(x)  f(x) I � e
( 1 4) for every n >
N and every x e £, which completes the proof.
The following criterion is sometimes useful. 7.9
Theorem
Suppose lim f,(x)
n + oo
Put
M,. =
=
f(x)
sup
xeE
(x e E).
l.fn (x)  f(x) 1 .
Then /,. + f uniformly on E if and only if M,. + 0 as r. + oo . Since this is an immediate consequence of Definition 7. 7, we omit the details of the proof. For series, there is a very convenient test for uniform convergence, due to Weierstrass . 7. 10
Theorem
Suppose {/,.} is a sequence offunctions defined on E, and suppose l.fn (x) I < M,.
(x e E, n
=
I , 2, 3, . . . ) .
Then "Lf,. converges uniformly on E if "LM,. converges. Note that the converse is not asserted (and is, in fact, not true) . Proof
If "LM,. converges, then, for arbitrary e m
m
i = lf
i=n
L f;(x) < L M; < e
> 0,
(x e E) ,
provided m and n are large enough. Uniform convergence now follows from Theorem 7.8.
SEQUENCES AND SERIES OF FU NCTIONS
149
UNIFORM C O NVERGENCE AND CONTINUITY
Suppose /, + f uniformly on a set E in a metric space. Let x be a limit point of E, and suppose that lim f,(t) = A ;. ( 1 5) (n = I , 2, 3, . . ).
7. 11
Theorem
.
t + x
Then { A,. } converges, and Iim f(t ) = lim
( 1 6)
n + oo
t + x
A ,. .
In other words, the conclusion is that ( 1 7)
lim lim f,(t) = lim lim f,(t). t + x
n + oo
n + oo
t + x
Proof Let e > 0 be given. By the uniform convergence exists N such that n > N, m > N, t e E imply
of {/,},
there
Jf,(t)  fm (t ) I < e.
( 1 8)
Letting t + x in
( 1 8), we obtain
for n > N, m > N, so converges, say to A . Next,
l A,.  Am I < t that { A ,.} is a Cauchy
sequence and therefore
l f(t )  A I < lf(t)  f,(t ) I + Jf,(t)  A,. I + l A ,.  A I .
( 1 9)
We first choose n such that
ff( t)  J,( t) I < e
(20)
3
for all
t e E (this
is possible by the uniform convergence), and such that e
l A ,.  A I < ·
(2 1 )
3
Then, for this
n, we choose a neighborhood
V
of x such that
(22)
if t e V n E, t #= x.
Substituting the inequalities (20) to (22) into ( 1 9), we see that
provided
t e V n £, t:#: x .
l f(t)  A I � e, This is equivalent to ( 1 6).
150
PRINCIPLES OF MATHEMATICAL ANALYSIS
If {/,} is a sequence of continuous functions on E, and if/, + 1· uniformly on E, then f is continuous on E.
7.12
Theorem
This very important result is an im mediate corollary of Theorem 7. 1 1 . The converse is not true ; that is, a sequence of continuous functions m ay converge to a continuous function, although the convergence is not uniform . Example 7.6 is of this kind (to see this, apply Theorem 7.9). But there is a case in which we can assert the converse .
Suppose K is compact, and (a) {/"} is a sequen ce of continuous functions on K, (b) {/n } conv erg es pointwise to a continuous function f on K, (c) f,(x) > /, + 1 (x) for all x e K, n = I , 2, 3, . . . . Then /, + f unifo rm ly on K. Proof Put g" = fn  f Then g" is continuous, g" + 0 poi ntwise, and
7. 1 3 Theorem
g" > g n + t · We have to prove that g" + 0 uniformly on K. Let e > 0 be given . Let K" be the set of all x E K with g"(x) > e . Since g" is conti nuous, K" is closed (Theorem 4. 8), hence com pact (Theorem 2. 35). Since g" > gn + t ' we have K" => Kn + t · Fix x e K. Since gn(x) + 0, we see that X ¢ K, i f n is sufficiently large. Thus X ¢ n K" . I n other words, n Kn is empty. Hence KN is empty for some N (Theorem 2. 36). It fol lows that 0 < g"(x) < e for a l l x e K and for a l l n > N. This proves the theorem. Let us note that compactness is really needed here. For i nstance, i f f,(x)
= nx
(0 < x < 1 ; n
I
+ I
= 1 , 2,
3, . . . )
then f"(x) + 0 monotonically in (0, 1 ), but the convergence is not u n iform . 7. 14 Definition I f X is a metric space, <e'( X) will denote the set of al l complex valued, cont inuous, bounded functions wi t h domai n X.
[Note that boundedness is redundant if X is compact (Theorem 4. 1 5) . Thus �( X) consists of all complex continuous functions on X if X is compact. ] We associate with each f E �(X) its supren1 um norm sup lf(x) 1 . 11 ! 11
=
xe X
=
Since f is assumed to be bounded , 11 ! 11 < oo . It is obvious that 11 / 11 f(x) 0 fo r every x e X, that is, only iff = 0. If h f + g, then l h(x) I < Jf(x) I + lg(x) I < llfll + llg ll for all x e X; hence
=
I ll + g il � 1 1/ 11
+
llg ll 
= 0 only if
SEQUENCES AND SERIES OF FUNCTIONS
If we defi ne the distance between f E �( X) and it follows that A xioms 2. 1 5 for a metric are satisfied . We have thus 1nade �( X) into a metric space. Theorem 7. 9 can be reph rased as follows :
g
E �(X) to be
151
11/  g il ,
A sequence {fn} converges to f with respect to the metric of �( X) if an d only if In + f uniformly on X.
Accord i ngly, closed subsets of �(X) are sometimes called uniformly closed, the clos ure of a set d c �(X) i s called its uniform closure, and so on. 7. 15
Theorem
The above metric makes �( X) into a complete metric space.
Proof
Let {.fn} be a Cauchy sequence i n �(X). This means that to each e > 0 corresponds an N such that 11/n  / il < e i f n > N and 1n > N. It fol lows (by Theorem 7.8) that there is a function f with domain X to which {/n} converges u niformly. By Theorem 7 . 1 2, f is continuous. Moreover, f is bou nded , si nce there is an n such that lf(x)  fn(x) I < 1 for all x E X, and In is bou nded. Th us f E rc(X), and since In + f u niformly on X, we have 11/  .fn ll + 0 as n + oo . m
U N I FO R M C O N V E R G E N C E A N D I NTEGRATIO N 7.16 Theorem Let r:x. be monotonically increasing on [a, b ]. Suppose /, e f!l (r:x.) on [a, b ], for n = I , 2, 3, . . . , and suppose f, + f uniformly on [a, b ]. Then f e f!l(r:x.) on [a, b ], and
..J b f dr:x.
(23)
a
= lim
n . oo
I
bfn dr:x..
a
(The existence of the limit is part of the conclusion.) Proof It suffices to prove this for real f, . Put
en = sup l .fn (x)  f(x) I ,
(24)
the supremum bei ng taken over a < x < b. Then In  en < f (x)
=
lx
I
{ l < x < l)
and extend the definition of (x
+ 2) =
q> (x).
Then, for all s and t ,
(36) In particular,
q>
I < Is  t 1 . is continuous on R 1 • Define l q>(s)  q>( t)
(37)
f(x)
00
= I (i )" q> (4" x) . n=O
Si nce 0 < q> < 1 , Theorem 7. 1 0 shows that the series (37) converges uniformly on R 1 • By Theorem 7. 1 2, f is contin uous on R 1 • N ow fix a real number x and a positive integer m. Put
+ .1 . 4  m bm = 
(38)
2
where the sign is so chosen that no i nteger lies between 4mx and 4m(x + This can be done , since 4m I bm I = ! . Define
(39)
Yn =
q>(4"( x
+ b m)) 
bm ) .
q>(4"x)
When n > m, then 4"bm is an even i nteger, so that Yn = 0. When O < n < m , (36) i mplies that 1 r" I < 4" . Since I Ym I 4m, we conclude that :;::
f(x
+
bm)  f(x)
> 3 m m2:l 3" =O n = ! (3 m + 1 ). As m �
oo ,
b m � 0. It follows that f is not differentiable at x .
EQUI C O NTIN U O US FAM ILIES O F F U N CTIONS In Theorem 3.6 we saw that every bounded sequence of complex num bers contains a convergent subsequence, and the question arises whether something similar i s true for sequences of functions. To m ake the question more precise, we shall define two kinds of boundedness.
SEQUENCES AND SERIES OF FUNCTIONS
155
Definition Let {/, } be a sequence of functions defined on a set E. We say that {/,} is pointwise bounded on E if the sequence {f,(x)} is bounded for every x E E, that is, i f there exi sts a finitevalued function l/J defi ned on E such that
7. 19
lfn (x) I < l/J (x)
(x E E, n = I , 2, 3, . . . ) .
We say that {/,} is uniformly bounded on E if there exists a number M s uch that (x e E, n = I , 2, 3, . . . ) .
lfn (x) I < M
N ow i f {/, } is pointwise bounded on E and £ 1 i s a countable subset of E, it is always possi ble to find a subsequence {f, k } such that {f,; k (x)} converges for every x E £1 • This can be done by the diagonal process which is used i n the proof of Theorem 7 .23. H owever, even if {/, } is a uniformly bounded sequence of contin uous functions on a compact set E, there need not exist a subsequence which con verges pointwise on E. I n the following exam ple, this would be quite trouble some to prove wi th the equip ment which we have at hand so far, but the proof is quite simple if we appeal to a theorem from Chap. I I .
7.20
Example
Let
fn (x) = sin nx
(0 < x
< 2rr, n = I , 2, 3 , . . . ) .
S uppose there exists a sequence {nk} such that {sin nkx} converges, for every x E [0, 2rr] . I n that case we m u st have l i m (sin nkx  si n n k + 1 x) = 0 k + oo
(0 < x
< 2rr) ;
hence
(40)
l i m (sin nkx  si n nk + 1 x) 2 = 0
k +
(0 < x < 2rr).
00
By Lebesgue's theorem concerni ng integration of boundedly convergent sequences {Theorem I I .32), (40) i mp lies
(4 1 )
lim k + 00
J
2 1t
0
(sin nkx  sin nk + 1 x) 2 dx = 0.
But a si m ple calculation shows that
Jo
2 1t
which contradicts (4 I ) .
(sin nkx  sin nk + 1 x) 2 dx = 2rr�
156
PRINCI PLES OF MAT H E MATI CAL ANALYSIS
A nother q uestion is whether every co nvergent sequence conta i n s a uniformly convergent subseq uence. Our next exam ple will show that th i s need not be so, even if the sequence i s uniformly bounded o n a com pact set . (Exan1ple 7.6 shows th at a sequence of bou nded functions may converge without being uniformly bou nded ; but it is trivial to see that uniform conver gence of a sequence of bounded functions implies uniform bounded ness. )
7.21
Let
Example
/,(
x)
= x2
x2
(0
+ ( I  n x) 2
< x < 1, n
=
1 , 2, 3, . . . ) .
Then l.fn (x) I < I , so that {fn } is uniformly bounded on [0, 1 ]. A l so lim fn (x)
II + 00
=
(0 < X < 1 ),
0
but
(n
=
1 , 2 , 3 , . . . ),
so that no subseq uence can converge uniform l y on [0, 1 ] . The concept which is needed i n this connection is that of cqu iconti nuity ; it is given in the follo\\'·ing defi nition .
7.22 Definition A family !F of complex functi ons f defi ned on a set E i n a m etric space X i s said to be equicontinuous on E. if for every c > 0 there exi sts a {) > 0 such that
lf(x)  f(y) I < c;
whenever d(x, y) < b, x E £, y E £, and f E !F . H e re d denotes the metric of X. I t is clear that every member of a n equicontinuous family i s u n i formly con tinuous. The sequence of Exam ple 7.2 1 is not equicontinuous. Theorems 7.24 and 7.25 will show that there is a very close relation between equicontinuity, on the one hand, and uniforn1 convergence of sequences of conti n uou� fu nctions, on the other. But first we descri be a selecti on process which has nothing to do with continuity.
7.23 Theore m Jf {fn} is a pointwise bounded sequence oj� co1nplex functions on a coun table set E, then {f,} has a subsequence {j� k } such that {fnk ( x)} converg es for every x E £.
SEQUENCES AND SERIES OF FUNCTIONS
157
Proof Let { x ; } , i = 1 , 2, 3, . . . , be the points of 1:, arranged in a sequence. Since {f,,(x 1 ) } is bounded, there exists a subsequence, which we shall denote by {ft ,k}, such that {/1 ,k(x1)} converges as k + oo . Let us now consider sequences Sh S2 , S3 , , which we represent by the array •
sl : It , 1 h , 2 h , 3 s2 : /2 , 1 /2 ,2 /2 , 3 s3 : 13 , 1 13 ,2 13 , 3 . . . . . •
•
•
•
•
•
•
.
•
1. ,4 /2 , 4 . . . /3 ,4 . . . . . . .
•
•
and which have the followi ng properties :
( a) sn is a subseq uence of sn  1 ' for n = 2, 3, 4, . . . .
(b ) {f,. ," (xn)} converges, as k + oo (the boundedness of {/n (xn ) } makes it possible to choose sn i n this way) ; (c) The order i n which the functions appear is the same in each se quence ; i.e. , if one function precedes another i n S , they are in the same 1 relation in every S" , until one or the other is deleted . Hence, when goi ng from one row in the above array to the next below, functions may move to the left but never to the right. We now go down the diagonal of the array ; i .e . , we con sider the sequence s .· 1f1 , 1 f2 ,2 f3 , 3 1r4 ,4 . . . . By (c), the sequence S (except possi bly its first n  l terms) is a sub sequence of S" , for n = I , 2, 3, . . . . Hence (b ) impl ies that { fn , n(x i) } converges, as n + oo , for every X ; E £.
7.24 Theorem If K is a conzpact nzetric space, if/,. e �(K)for n = I , 2, 3, . . . , an d if{f,.} converges unijorn1ly on K, then { f,} is equicontinuous on K.
(42)
(4 3 )
Proof Let e > 0 be given. Since {fn} converges uniformly, there is an i nteger N such that I I /,.  IN I I < e
(n > N).
(See Definition 7. 1 4.) Since contin uous functions are uniformly con tinuous on compact sets, there i s a {) > 0 such that
l/;(x)  [;( y) I < e if 1
N and d(x, y) < �, i t follows that
lfn(x)
 f,.(y) I < lf, (x) fN(x) I + lfN(x)  fN(Y) I + lf"(y)  /,,(y) I < 3e . 
In conjunction with (4 3), this proves the theorem.
158
PRINCIPLES OF MATHE MATICAL ANALYSIS
7.25 Theorem If K is con1pact, if. In E f(}( K) for n pointwise bounded and equicontinuous on K, then
=
I,
2, 3 , . . . , and if {.fn} is
(a) {j�} is unifonnly bounded on K.. (b) {fn} contains a uniform�r convergent subsequence. Proof
(a) Let e > 0 be given and choose b > 0, i n accordance with Definition
7.22, so that (44)
l fn(x)  f,.(y) I
0, and pick 0 as i n the begi nning of this proof. Let V(x, �) be the set of all y E K with d(x, y) < b . Since E is dense in K, and K is com p act, there are finitely many points xh . . . , xm i n E s uch that •
•
•
n
(4 5) that
Si nce {g ; (x)} converges for every x e
£,
there is an integer N such
(46) whenever i > N, j > N, I < s < m. If x E K, (4 5 ) shows that x E V(x5 , b) for some s, so that
for every i. If i > N and j > N, it follows from (46) that
l g i(x)  gi(x) I < l g ;(x)  g ; (Xs) I + l g ;(xs)  gi(xs) I + l gi(xs)  gi(x) I < 3e. This completes the proof.
SEQU ENCES AND SERI ES OF FUNCTIONS
159
TH E ST O N EWEI E R STRASS TH E O R EM 7.26 Theorem If f is a continuous con1plex function on [a, b ], there exists a sequence of polynoJnials Pn such that lim Pn(x) = f(x)
n +
:�:;
unijorn1ly on [a, b ] . Iff is real, the Pn may be taken real. This i s the form in which the theo rem was origi nal ly discovered by Weierstrass.
Proof We may assume, without loss of general i ty, that [a, b] = [0, 1 ] . We may also assume that /(0) = /( I ) = 0. For if the theoren1 i s proved for this case, consider
(0 < X < 1 ) .
g (x) = f(x)  /(0)  x [f( I )  /(0)]
Here g(O) = g( I ) = 0 , and if g can be obtai ned as the limit of a unifo rmly convergent seq uence of polynomial s , i t is clear that the same i s true for J, si nce f  g is a polynom ial . Furthermore , we defi ne j'(x) to be zero for x outside [0, 1 ] . Then f i s uniformly conti nuo u s on the whole li ne. We put
( 47 )
(n = I , 2, 3, . . . ), where cn is chosen so that
J
(48)
1
1
Q n (x) dx = I
( n = 1 , 2,
3, . . . ) .
We need some information abo ut the order of magnitude of en . Si nce
f
 1
( I  x2)" dx = 2
( (I  x2 )" dx 2 f1.r, ( I  x2 )n dx 0
>
0
>2 >
i t follows from (48) that
(49)
f1 0
J·ln
4
3J � 1
, 'V
r;,
( l  nx2) dx
160
PRINCIPLES OF MATHEMATICAL ANALYSIS
Th e i n equa l ity ( I  x 2 )" > I  nx 2 which we used above is easily shown to be tru e by consi d e ring the function
(1 
x
2 " )
 1 + nx2
which i s zero at x = 0 and whose derivative is positi ve i n (0, 1 ) . For any {) > 0, (49) implies (50)
( < l x I < 1 . N o\v set (5 1 )
J
P.(x) =
1
 1
f(x + t) Q.( t) dt
(0 < X < 1).
Our assumptions a b out / show, b y a simpl e change of va ri abl e , that
P.(x) =
J
1 x 
x
f(x + t) Q. ( t) dt =
t f(t) Q.(t  x) dt , 1
and the last i ntegra l is clearly a po l y n om i al i n x. Thus {Pn } is a sequence of polynomials, which are real if f is real. Given e > 0, we choose b > such that IY  x I < {> i m p lies
0
l f(y)  f(x) I
e
< 2·
Let M = sup J f( x) 1 . Using (48), (50), and th e fact that Q n (x) > 0, we see that for 0 < x < I ,
I Pn(x)  f( x) I ' =
0 be given. Ct , , Cn SUCh that n •
(5 3)
•
163
By Corollary 7.27 there exist real numbers
•
(  a < y < a) .
L c ;y i  I Y I < e
i= 1
Si nce
14
is an algebra, the function
g
n
=L
i= 1
C ;/ i
is a member of a . By ( 5 2) and (53), we have
(x e K) .
J g(x)  Jf(x) I I < e
Since STEP
2
14
is uniformly closed , this shows that 1 / 1 e !!1.
Iff e 14
and
g
e
then
14,
max (J, g) e 14
By max (J, g) we mean the function
h(x) and min (J,
g)
=
{f(x) g(x)
h
and min (J, g)
e !!1 .
defined by
iff(x) > g(x) , i ff(x) < g(x) ,
is defined likewise.
Proof Step 2 follows from step I and the identities g + I (J, max , g ) = 2
min (J, g) =
f
+
If  g I , 2
; g  If ; g 1 .
By iteration, the result can of cou rse be extended to any finite set of functions : If It , . . . , In e 14, then max (ft , . . . , fn) e 14, and min (ft , STEP
. . , /,) e 14 . .
Given a real function J, continuous on K, a point x e K, and e > 0, there exists a function g x e 14 such that gx(x) = f(x) and (54) (t e K). 3
Proof Since d c a and d satisfies the hypotheses of Theorem 7.3 1 so does 14 . Hence, for every y e K, we can find a function h, e 14 such that
(55)
h,(x) = f(x),
h,(y) = f(y) .
164
PRINCIPLES OF MATHEMATICAL ANALYSIS
By the continuity of hy there exists an open set such that
hy(t) > f(t)
(56)
(t E Jy)
 8
JY ,
containing
y,
.
Si nce K i s compact, there is a finite set of poi nts y 1 ,
•
•
•
, Yn such that
(57) Put By step 2, g e 81, and the relations (55) to (57) show that g x has the other requ i red properties. STEP
4
h e 81
Given a realfunction /, continuous on K, and E > 0, there exists a function such th a t l h(x)  f(x) I
0) ,
if {xn} is a sequence of distinct point s of (a, b), and if "'L I Cn I converges, prove that the series 00
f(x) = L Cn l(x  Xn) n= l
9.
converges uniformly , and that f is continuous for every x i= Xn . Let {/,} be a sequence of continuous functions which converges uniformly to a function f on a set E. Prove that lim f,(xn) = f(x) for every sequence of points Xn E E such that Xn � x, and this true ?
x
E £. Is the converse of
SEQUENCES AND SERIES OF FUNCTIONS
10.
Letting (x) denote the fractional part of the real number x (see Exercise 1 6, Chap. 4, for the definition), consider the function f(x) =
11.
·
(x real).
n= l
·
3.42.
Suppose g and f,(n 1 , 2, 3, . . . ) are defined on (0, oo ) , are Riemannintegrable on [ t , T] whenever 0 < t < T < oo , I f, I < g, ln 7 f uniformly on every compact sub set of (0, oo ) , and =
I
00
0
Prove that lim n + oo
13.
f (n�) n
Find all discont inuities of /, and show that they form a countable dense set. Show that f is nevertheless Riemannintegrable on every bounded interval. Suppose {f,}, {gn} are defined on E, and (a) l.: In has uniformly bounded partial sums ; (b) On 7 0 un i formly on E; (c) g 1 (x) > g 2 (x) > g 3 (x) > for every x E E. Prove that :E fngn converges uniformly on E. Hint : Compare with Theorem ·
12.
167
I
00
g(x) dx < oo .
J,(x) dx =
0
I
00
l(x) dx.
0
(See Exercises 7 and 8 of Chap. 6 for the relevant definitions.) This is a rather weak form of Lebesgue's dominated convergence theorem (Theorem 1 1 .32). Even in the context of the Riemann integral, uniform conver gence can be replaced by pointwise convergence if it is assumed that f E !11 . (See the articles by F. Cunningham in Math. Mag., vol. 40, 1 967, pp. 1 791 86, and by H . Kestelman in Amer. Math. Monthly, vol. 77, 1 970, pp. 1 B21 87.) Assume that {f,} is a sequence of monotonically increasing functions on R1 with 0 is an isometry (a distancepreserving mapping) of X onto c'l>(X) c �(X). Let Y be the closure of c'I>( X) in �(X). Show that Y is complete. Conclusion: X is isometric to a dense subset of a complete metric space Y. (Exercise 24, Chap. 3 contains a different proof of this.) Suppose 4> is a continuous bounded real function in the strip defined by 0 < x < 1 ,  oo < y < oo . Prove that the initialvalue problem y' = cp (x, y),
y(O) = c
has a solution. (Note that the hypotheses of this existence theorem are less stringent than those of t he corresponding uniqueness theorem ; see Exercise 27, Chap. 5 . ) Hint: Fix n. For i = 0, . . . , n, pu t x, = i/n. Let f,. be a continuous function on [0, 1 ] such that f,.(O) = c, [',(t ) = cp(x, , J,(x , ))
if x, < t < Xt � , +
and put
�,.(t ) = /�(!)  cp(t, f,.(t)), except at the points
x, ,
where �,.(!) = 0. Then
/.(x) = c +
Choose M
I < M. Verify the following assertions.
(a) l /� 1 < M, l � n l < 2M, �,. e rH, and I fn i < l e i + M = M� , say, on [0, 1 ], for all n. (b) {/,.} is equ i c on ti nuous on [0, 1 ], since I /� I < M. (c) Some {f,,J converges to some /, uniformly on [0:. 1 ] . (d) Since cp is uniformly continuous on the rectangle 0 < x < 1 , I Y I < Mh
cp(t, f,k(t)) � (t, f(t )) uniformly on [0, 1 ]. (e) �,.( t) � 0 uniformly on in (x, ,
)
x, + . .
[0, 1 ], since
�,.(!) = cp(x , , f,.(x,))  (t, f,.(t))
SEQUENCES AND SERIES OF FUNCTIONS
171
(/) Hence f(x) = c +
J ( t. f(t )) dt. %
0
26.
This f is a solution of the given problem. Prove an analogous existence theorem for the initialvalue problem y ' = cl» (x , y),
y (O) = c,
where now c E Rk , y E Rk, and 4» is a continuous bounded mapping of the part of Rk defined by 0 < x < 1 , y E Rk into Rk. (Compare Exercise 28, Chap. 5.) Hint: Use the vectorvalued version of Theorem 7.25. + 1
8 SOME SPECIAL FUNCTIONS
POWER S E R I E S
In this sect ion we sha 1 1 derive some propert ies of funct ions wh ich a re represe nted by power series, i .e. , fu nct ions of the form (1)
f(x) = L cn x" oc
n=O
or, more genera lly, (2)
00
f(x) = I cn(x  a) ". ·1 = 0
These are ca l led analy tic functions. We sha l 1 restrict ourse1 ves to rea l val ues of x . I nstead of c i rcles of con vergence (see Theorem 3 . 39) we shal l therefore encounter i nterva ls of conver gence. If ( 1 ) converges for a l l x i n (  R, R), for some R > 0 (R may be + oo ) , we say that f i s expanded i n a power series abou t the poi nt x = 0. Simi larly, i f (2) converges for I x  a I < R, f is said to be expanded i n a power series about the poin t x == a. As a matter of convenience, we sha l l often take a = 0 wit hout any l oss of genera l ity.
SOME SPECIAL FUNCTIONS
8.1
Theorem
173
Suppose the series CX)
L cn x n
(3 )
n=O
converges for I x I < R, and define CX)
f( x) = L cn xn n=O
(4)
( l x l < R).
Then (3 ) conve rg es uniformly on [  R + e, R  e ] , no matter which e is chosen . The function f is continuous and differentiable in (  R, R), and
(5)
>
0
CX)
f ' (x) = I ncn xn  l ( I x l < R). = l n Proof Let e > 0 be gi ven. For l x l < R  e, we have I cn xn I < I cn( R  e)n I ; and since
LCn( R  e) n converges absol utely (every po\ver series con verges absol u tely i n the i n terior of its interval of con vergence, by the root test), Theorem 7. 1 0 shows the un i form convergence of (3 ) o n [  R + e, R  e]. Since .V n � I as n � oo , we have r
l i m sup y/n I c: f = l i m s u p y/ I en I , n + oo
so that t he series (4) and (5) have the same i n terva l of co nvergence. S i nce (5) is a power series, it converges un iformly in [  R + e, R  e], for every e > 0, and we can appl y Theorem 7. 1 7 (for series in stead of sequences). It fol lows that (5) holds if I x l < R  e. But, given any x s uch that I x I < R , we ca n fi nd an c: > 0 s uch that I x I < R  e. This shows that (5) holds for I x I < R. Cont i n uity off follows from the existence off ' (Theorem 5.2). Under the hypotheses of Theorem 8 . 1 , f has derivatives of all orders in (  R, R), ·which are given by Corollary
(6) (7)
CX)
J (x) = I n (n  1 ) n=k In particular,
· · ·
(n 
k + 1 )en xn  k .
(k = 0 , 1 , 2 , . . . ) .
(Here J means J, and J is the kth derivati ve off, fo r
k
= 1 , 2, 3 , . . . ) .
174
PRI NCIPLES
OF MATHE MATICAL AN ALYSIS
Proof Equat ion (6) fol lows i f we apply Theorem 8 . 1 successively to f, f ', f ", . . . . Putti n g x = 0 i n (6), we obtain (7).
Formula ( 7 ) i s very i n teresting. It shows, on the one hand, that the coefficien ts of the power series developmen t of f are determ i ned by the va l ues off and of its deri vati ves at a si ngle point. On the other hand, i f the coefficients are given , the val ues of the derivatives of f at the center of the i n terva l of con vergence can be .. read off i mmed iately from the power series . Note, however, t hat although a function f may have derivat i ves of a l l n orders, t h e ser ies �en x , where en i s computed b y (7), need not co nverge t o f(x) for any x i= 0. I n th i s case, / can not be expan ded i n a power ser ies about x = 0. n For i f we had f(x) = �a n x , we shou ld have n !an = / ( )(O) ;
n
hence an = en . A n exa m ple of thi s situation i s given i n Exerc ise I . If the series (3) converges at an end point, say at x = R , then / i s contin uous not only in (  R, R), but also at x = R. This follows from A bel ' s theorem (for s i mpl ic ity of notation, we take R = I ) : 8.2
Theorem
Suppose �en converges . Put
f(x) = I en xn n=O 00
(

1 < X < 1 ).
Then 00
l i m f(x) = L en . n=O x + 1
(8)
Proof Let sn = Co + . . . + en , s  1 = 0. Then m
m
1
m n n I en x = nI (sn  Sn  l )x" = ( I  x) I Sn X + n=O =O n=O
For l x l < I , we let
m � oo
sm xm .
and obtain
f(x) = ( I  X) I sn xn . n= O 00
(9) Suppose s = l i m i m pl ies
n+0
s" .
Let e > 0 be given . Choose N so that n > N
1 75
SOME SPECIAL FUNCTIONS
Then, since 00
=
( I  x) L x" n=O
( I X I < I ),
I
we obtai n from (9)
if x > I

b, for some suitably chosen b
>
0. Th is i mpl ies (8).
As an appl ication, let us prove Theorem 3 . 5 I , which asserts : If' 'La" , 'Lb" , 'Lc" , converge to A , B, C, and if c" = a0 b" + · · · + a" b0 , then C = A B. We let f(x)
=
00
I a" x" ,
g (x)
n=O
00
= L bn x" , n=O
h(x)
=
00
I en x" ,
n=O
for 0 < x < I . For x < I , these series converge absolutely and hence may be m u ltiplied according to Definition 3 . 48 ; when the multiplication i s ca rried out, we see that f(x) · g(x)
( I O)
=
h(x)
( 0 < X < I ).
By Theorem 8 . 2, f(x) � A ,
(I I)
g(x) � B,
h(x) � C
as x + I . Equat ions ( I O) and ( I I ) i mply A B = C . We now require a theorem co ncerning an i nversion i n the order of sum mation. (See Exercises 2 and 3. ) 8.3 Theorem suppose that
( I 2)
Given a double sequence { a ii}, i 00
L l a ii l = h i =l
j
=
I,
(i = I , 2, 3,
2,
3 , . . . , j = I , 2, 3, . .
. '
. . .)
an d 'Lb i converges. Then
( I 3)
00
00
00
00
I I a ij = I I a ij . i= l j= l j= l i= l
Proof We could establish ( I 3) by a direct procedure s i milar to (although more i nvolved t ha n) the one used in Theorem 3 . 55. However, t he fol l owi ng method seems more interesti ng.
176
PRINCIPLES OF
MATHEMATICAL ANALYSIS
Let E be a countable set , consisti ng of the p oint s x0 , xb x2 , • • • , and suppose x" + x0 a s n + oo . Defi n e ( 1 4)
fi (x o )
=
( 1 5)
fi ( x" )
=
00
I a ii
(i = 1 ' 2, 3 , . . . ) ,
L a ii
(i, n
j= l " j= l 00
g (x) = I flx)
( 1 6)
=
1 , 2, 3 , . . .),
( x e £).
i= 1
Now, ( 1 4) and ( 1 5), together with ( 1 2) , show that each fi is con ti n u ou s at x0 • Since l fi (x) I < b i for x E £, ( 1 6) converges u n i formly, so t h a t g is contin uous at x0 (Theorem 7 . 1 1 ) . I t fol lows that 00
00
00
I L a ii = I fi (xo )
i= l j= l
i= l
=
g (x0)
oo
l i m L .fi (xn )
=
=
=
Theorem
n+ oc, ct:;
n
l i m L L a ii
n+ oo i = l n + oo n oo oo = lim ' . . = ' � ' � a lj � n + oo j = l i = l j= l
8.4
l i m g (x " )
i= l j= l oo ' � a l). i= l •
Suppose f(x)
=
I en x" , 00
n=O
the series converging in I x I < R. If  R < a < R, then f can be expanded in a power series about the point x = a �vhich converges in I x  a I < R  I a I , and
( 1 7)
f ( n ) (a ) , (x f(x) = � n0 n . oo

a)
"
(lx
·
a l < R  l a l ).
Th is is an extens i on of Theorem 5 . 1 5 and i s also known as Taylor 's theorem . Proof We have f(x )
=
L c" [(x  a ) + a] " 00
n=O
SOME S PECIAL FUNCTIONS
177
Th is i s the desired expansion about the point x = a. To prove its validi t y , we have to justify the c hange which was made i n the order of summat ion. Theorem 8 . 3 shows that this is permissible i f ( 1 8) converges . But ( 1 8) is the same as ( ] 9)
CX)
I cn I · ( I X nL =O

a I + I a I )n ,
and ( 1 9) converges i f l x  a l + l a l < R . F i nal ly the forn1 of the coefficient� in ( 1 7) fol lows from (7). ,
It shou l d be noted that ( 1 7) may ac t ually converge in a larger interval than t he one given by l x  a l < R  l a l . I f two power series converge to the same function in ( R, R), (7) shows that the t wo series must be i dentical , i.e., t hey m ust have the same coefficients. It is i nterest ing that the �a1ne concl usion can be ded uced from much weaker hypotheses : 
8.5 S
=
n Suppose th e series Lan xn and f.}Jn x conL'erge in the segment (  R. R ) . Let E he the set of all x E S at which
Theorem
rr = 0
(20)
if E has a lin1it point in S, then a/1 X E S.
Proof Put
(2 1 )
2: a, x" CX)
en
=
an
=
=
I hn x" . CX)
n=O
bn for n
=
0. 1 , 2, . . . . Hence (20) holds for
an  bn and f(x)
=
CX)
L
,. = 0
cn x
n
(x E S).
Then f(x) = 0 on E. Let A be the set or a l l l i n1 i t poi nts of E i n S, and let B consis t of all o ther points of S. It i� c l e a r from the defi n i tio n of H l i m i t poi nt" that B is open. Suppose \\t e ca n prove that A is open . Then A and B are disjoi nt open sets. Hence they are separated ( De fi n i t ion 2.45). Since S = A u B, and S is con nected, one of A and B m ust be empty . By hypot hes is, A is not em pty. Hence B is empty, and A = S. Since .f i s contin u ous in S, A c E. Thus E = S, and (7) shows t hat en = 0 for n = 0, I , 2, . . . , which is the desi red conclusion.
178
PRINCIPLES OF MATHEMATICAL ANALYSIS
tha t
Thus we have to prove that A i s open . I f x0 E A , Theorem 8. 4 shows
f(x)
(22)
=
L dn (X  Xo) n 00
( l x  xo l < R  l xo l ) .
n=O
We claim t hat dn = 0 fo r a l l n. Otherwise, let k be the sma1 1est non negative  i nteger. s u c h that dk i= 0. Then
( l x  x0 1 < R  l x0 ! ) ,
(23) where
x) = I dk + m (x  Xo )m. m=O i s cont i nuous at x0 and 00
g(
(24) Si nce
g
=
g(x0)
dk i= 0, there exi sts a b > 0 such that g(x) i= 0 i f ! x  x0 I < b. It follo\vs from (2 3) that f(x) i= 0 i f 0 < I x  x0 I < b . But th i s contrad ict s the fact t h a t x0 i s a l i m i t point of E. Thus dn = 0 for a l l n, so that f (x ) = 0 for a l l x for which (22) holds, i.e. , in a neighborhood of x0 • This shows that A is open, and co m p letes t he proof.
THE EXPON ENTIAL A N D LOGARITHMIC FU N CTI O N S
W e define E(z)
(25)
=
zn L n = O n'. 00
The rat io test shows that this series converges for every complex z. A pplying Theorem 3 . 50 on m u ltiplica t ion of absol utely convergent ser ies, we obtain z k wn  k E(z)E(w) = I  L = L I m n = O n ! m = O ! n = O k = O k ! (n  k) !  � 1 �  k  � (z + w) n k n zw  L  L  L zn
00
oc
li;'m
n
oc
(11)

n=O n ! k=O k
n=O
n!
'
which gives u s the i mporta nt addition formula (26)
E(z
+
w)
=
E(z)E(»')
w
(z,
complex).
One consequence i s that (27)
E(z)E(  z)
=
E(z  z) = £(0)
=
I
(z complex).
SOME SPECIAL FUNCTIONS
179
fhis shows that E(z) i: 0 for all z. By (25), E(x) > 0 if x > 0 ; hence (27) shows that E(x) > 0 for a l l real x. By (25), E(x) + + oo as x + + oo ; hence (2 7 ) shows t hat E(x) + 0 as x +  oo along the real axis. By (25), 0 < x < y i mpl ies that E(x) < E(y) ; by (27), i t fol lows that E(  y) < E(  x) ; hence E is strictly in creasing on the whole real axis. Th� addition formula also shows that 1. E(h) . E z + h)  E(z) = E(z ) 1 m I1m (
(28)
h
h=O
h
h=O
I
=
E(z) '·
the last equal i ty fol lows directly from (25). I terat ion of (26) gives ( 29) Let us take z1 = · · · = z" = I . Since E( I ) i n Defin ition 3 . 30, we obtai n (30) If p
=
E(n) njn1 , where n ,
(3 1 )
m
= e
"
= e,
(n =
where
e
is the number defined
I , 2, 3 , . . . ) .
are positive i n tegers, then [E(p) ] m
=
E( mp )
=
E(n)
" = e ,
so that (32)
E(p)
= e
P
(p > 0, p rational).
It fol lows fro m (27) that £( p) = e  P if p is positive and rational . Thus (32) holds for a l l rati onal p. In Exercise 6, Chap. I , we suggested the defin ition (3 3) where the sup is taken over a l l rational p such that p x > 1 . If we t h us defi ne, for any rea l x , (3 4 )
< y,
for any real y, and
( p < x, p rational),
the conti nui ty a nd monotonici ty properties of E, toget her with (32), show that ( 3 5)
E(x)
= e
x
for all real x. Equation (35) explains why E i s called the exponential function . The notation exp (x) is often used i n place of eX, expecially when x is a compl icated expression. Actually one may very wel l use (35) i nstead of (34) as the definition of ex ; (35) is a much more convenient starting poi nt for the investigation of t he x properties of e . We sha l l see presently that (33) may also be replaced by a m ore conven ient defini tion [see (43)].
180
PRINCIPLES OF MATHEMATICAL ANALYSIS
We now revert to the customary n otat io n , eX, in place of E(x), and sum marize what we have proved so far.
8.6 Theorem Let ex be defined on R 1 by (35) an d (25). Then
ex is continuous and differentiable for all x; (ex)' = ex ; ex is a strictly increasing function of x, and ex > 0 ; ex + )' = exey ; ex � + 00 as X � + 00 , ex � 0 as X �  00 ; (f) lim x .... + 00x"e x = 0, for every n. Proof We have a lready proved (a) to (e) ; (25) shows that
(a) (b) (c) (d) (e)
x" + t ex > (n + 1 ) ! for
x
> 0, so that _
x" e x
0),
= x
(x rea l) .
=
Differen tiating (37), we get (compare T he orem 5. 5) Writing y = (38)
L'(E(x)) · E(x)
E(x) , this g i ves us
1 L'(y) = y
L(y)
I.
(y > 0).
Taking x = 0 i n (37), we see that L( l ) (39)
=
=
=
0. Hence (38) impl ies
Y
Jt
dx
.
X
SOME SPECIAL FUNCTIONS
181
Quite frequent ly, (39) i s taken as the start i ng poi n t of the theory of the logarithm and the exponent i a l funct ion . Writing u = E(x), v = E(y) , (26) gives L ( uv)
so that
=
L ( E(x) E (y )) ·
=
L(E(x + y))
=
x + y,
=
L( u ) + L(v) ( u > 0, v > 0). This shows that L has the fa mil iar property which makes logarithms useful tools for computation. The customary notation for L(x) i s of cou rse log x. As to the behavior of log x as x � + oo a n d as x � 0, Theorem 8 . 6(e) s hows that as x � + oo , log x � + oo
L(uv)
(40)

oo
x"
=
log x � It i s eas i ly seen t hat (4 1 ) if x > 0 and
(4 2)
n
as x � o. E(n L(x))
i s a n i nteger. S i m i larly, i f n1 i s a positive i nteger, we have x l fm
=
E
(,� )
L (x) ,
s in ce each term of ( 42), when raised to the 1nth power, yields t he correspond i ng term of (37). Combin i ng (4 1 ) and (42), we obta i n (4 3)
x�
=
E(�L( x))
= e(I log
x
for any rational a . We now defi ne xa, for any rea l � and any x > 0, by (43). The con t i n u it y and monot on icity of E and L sho\v t hat this defin it ion leads to the same res ult as the prev iously s uggested one. The facts stated in Exerc ise 6 of Cha p. I , are trivial conseq uences of ( 43). If we different iate (43), we obta i n , by Theorem 5. 5, (44)
Note that we have previously used (44) only for i ntegral val ues of ct , in wh ich case (44) fol lows easi ly from Theorem 5. 3(b). To prove (44) d i rect ly from the defin ition of the derivat ive, i f xa i s defined by (33) and a is i rrat iona l , i s qu i te troublesome. The wel lknown i ntegrat ion formula for x(l fol l ows fron1 (44) i f ':I. #  I , and from (38) i f a =  I . We wish to demonstrate one more property of log x, namely, l i m x  a l og x = O (45) x +
+ oo
182
PRINCIPLES
OF M ATH EMATICAL ANALYSIS
for every a > 0. That i s, log x � + oo "s lower" than any pos i t ive power of x, as x � + oo . For i f O < e < a , and x > 1 , then
x  « log x = x  « = xa
I1 t X
1
I1 t• X
dt < x  «
eX .
1
0 for all x > 0, hence S'(x) > 0, by (49), hence S i s strictly i ncreas ing ; and si nce S(O) = 0, we have S(x) > 0 if x > 0. Hence if 0 < x < y , we have (50)
S(x)(y  x)
0, ( 5 0 ) ca nnot be
SOME SPECIAL FUNCTIONS
183
Let x0 be t he smal lest positive n umber such t hat C(x0 ) == 0. This exists, s i nce the set of zeros of a continuous fu nction is closed, and C(O) i= 0. We define the n u n1ber rr by n
(5 1 )
=
2x0 .
Then C(rr/2) = 0, and (48) shows that S(n/2) = + I . Since C(x) > 0 i n (0, rr/2), S i s i ncreasi ng i n (0 , n/2) ; hence S(rr/2) = I . Thus i G £ )
=
i,
and the add i tion formula gives E(rri)
(52 )
=
 I,
E(2ni)
=
I;
hence E(z + 2rri)
(5 3 ) 8.7
=
(z complex).
E(z)
Theorem (a) The function E is periodic, ·with period 2rri. (b) The functions C and S are periodic, with period 2n. (c) lf O < t < 2rr , then E(it ) i= I . (d) If z is a complex number with I z I = I , there is a unique t in [0, 2n) such that E(it ) = z.
By (53} , (a) holds ; and (b) fol lows from (a) and (46). Suppose 0 < t < rr/2 and E( it) = x + iy, with x, y rea l . Our preceding work shows t hat 0 < x < I , 0 < y < I . Note that Proof
E(4it )
=
(x + iy)4
=
x 4  6x2y2 + y 4 + 4ixy (x 2  y2) .
I f E(4 it ) is real , i t follows t hat .x 2  y 2 = 0 ; s i nce x2 + y2 we have x2 = y 2 = ! , hence £(4it) =  1 . This proves (c). I f 0 < t 1 < t 2 < 2rr, then £{ it 2 )[£(it1 )]  1
=
E(it2  it1 ) i= I ,
=
I , by (48),
by (c ) . Th is establ ishes the u n iqueness assert ion in (d). To prove t he existence assert ion i n (d), fix z so that I z I = I . Write z = x + iy, with x and y rea l . Suppose first t hat x > 0 and y > 0. On [0, rr/2], C decreases from I to 0. He nce C(t ) = x for so me t E [0, n/2]. Si nce C 2 + S 2 = 1 and S > 0 on [0 , rr/2], it fol lows that z = E(it ). If x < 0 and y > 0. the preced i ng condit ions are satisfied by  iz. Hence  iz = E(it ) for some t E [0, n/2], and si nce i = E (rri/2), we obtain z ;=: E(i(t + n/2)). Final ly, i f y < 0, the preceding two cases show that
184
PRINCIPLES OF MATHEMATICAL ANALYSIS
 z = E(it ) for some t e (0, n) . Hence z =  E(it ) = E(i(t + n) ) .
Thi"s proves (d), and hence the theorem .
It fol lows from (d) and (48) that the curve y defined by ( 5 4) y(t) = E(it) (0 < t < 2rr) is a simple closed curve whose ra nge is the u n i t circle i n the plane. Since y' (t) = iE(it) , the length of y is
f0
2 1t
1 y'(t) 1 dt = 2n,
by Theorem 6.27. This is of course the expected result for the circumference of a circle of radius I . It shows that ::, defined by (5 1 ), has the usual geometric significance. I n the same way we see that the point y(t) descri bes a ci rcular arc of length 1 0 as t i ncreases from 0 to t 0 • Considerat ion of the triangle whose vertices are z 1 = 0, z 3 = C (t0) z 2 = y(t 0 ) , shows that C(t) and S(t ) are i ndeed identical with cos t and sin t, if the latter are defi ned i n the usua l way as rat ios of the sides of a right triangle. I t should be stressed that we derived the basic propert ies of the trigono metric functions from ( 46) and (25), without any appeal to the geometric notion of angle. There a re other nongeometric approaches to these funct ions. The papers by W. F. Eberle i n (A n1er. Math. Monthly, vol . 74, 1 967, pp. 1 223 1 225) and by G. B. Robison ( 1\lath. Mag . , vol . 4 1 , 1 968, pp. 6670) dea l with these topics .
T H E ALGEBRA IC C O M PLETE N E SS O F TH E C OM P LEX FIELD
We are now in a position to give a simple proof of the fact that the complex field is algebraica l ly complete, that i s to say, that every n onconsta nt polynomial with complex coefficients has a complex root. Suppose a0 , • . • , an are complex numbers, n > I , an i= 0, n P(z) = I ak zk . 0 Then P(z) = 0 for som e complex number z. Proof Without loss of general ity, assume an = I . Put
8.8
Theorem
(55)
I f l z l = R , then (56)
J1
= inf I P(z) I
(z complex)
n 1 J P(z) l > Rn [l  l an  t i R   · · ·  l a o f R  ].
SOME SPECIAL F UNCTIONS
185
The right side of (56) tends to oo as R + oo . Hence there exi sts R0 such that I P(z) I > J1 if I z I > R0 • S i nce I P I is conti n uous on the c losed disc with cen ter at 0 and rad i us R0 , Theorem 4. I 6 shows that I P(z0) I = J1 for some z0 • We claim that J1 = 0. If not , put Q(z) = P(z + z0)/P(z0). Then Q is a nonconstant poly nomial, Q(O) = I , a n d I Q(z) I > I for a l l z. There is a sma1 Iest i nteger k, I < k < n , such that (57) (58)
Q(z)
=
1 + bk zk +
·· ·
+ bn z" ,
b i= 0. k
By Theorem 8. 7(d) there i s a rea l 0 such that e ikobk
=

k If r > 0 and r l bk l < I , (58) i mpl ies
I I + bk rke iko l
=
I bk 1 . k I  r I bk I ,
so that For sufficiently sma l l r, the expression i n braces i s positive ; hence I Q(re ;8) I < I , a contradiction . Thus JL = 0, that i s, P(z0) = 0. Exercise 27 conta i n s a more genera l result. FOURIER SER IES 8.9
Definition
A trigonon1etric po/yno1nia/ is a fi n i te sum of t he form
f(x)
(59)
=
a0 +
L (an cos nx + bn s i n nx) N
n= l
where a0 , , aN , b 1 , . . . , bN are co mplex n u mbers. (46), (59) ca n also be wri tten i n the for m •
.
•
f(x)
(60)
=
N
LN en e inx

(x rea l), On
account of the i dentities
(x real),
which i s more conven ient for most purposes. It i s clear that every trigonomet ric polynomial is periodic, with period 2rr. If n is a nonzero i nteger, ein x i s the der ivat ive of einx/in , which also has period 2rr. Hence (6 I )
I 2n
n e inx dx  { I J n 0 _
(if n (i f n
= =
0 ),
+ I , + 2, . ) . .
.
186
PRINCIPLES OF MATHEMATICAL ANALYSIS
Let us mult iply (60) by e  im x , where m is an integer ; if we i ntegrate t he product, (6 1 ) shows that (6 2) for I m I < N. I f I m I > N, the i ntegral i n (62) i s 0. The fol lowing observat ion ca n be read off from (60) a n d (62) : The trigonometric polynomial f, given by (60), i s real if and on ly if c = c" for n = 0, . . . , N. In agreement with (60), we define a trigonometric series to be a ser ies of the form n
(6 3 )

(x real) ;
oo
the Nth part ial sum of (63) i s defined to be the right side of (60). Iff is an i ntegrable functi on on [  n , n], the n umbers em d�fi ned by (62) for a l l integers m are ca l led the Fourier coefficien ts off, and t he series (63) formed with these coefficients is ca l led the Fourier series off The natural question which now arises is whether the Fourier series of .f converges to f, or, more generally, whether / i s determi ned by i ts Fourier ser ies. That is to say, i f we know t he Fourier coefficients 0f 3 function , ca n we fi n d the function, a n d if so, how ? The study of s uch series, and, i n part icular, the problem of re pre�c n t i n g a given function by a trigonometric series, originated i n phys ica l problems such as the theory of osc i 1 lat ions and the theory of heat conduct i on ( Fourier's "Theorie analytique de Ia cha leu r ' ' was pub l ished i n 1 822). The many difficu lt and del icate problems which arose duri ng t h i s st udy caused a t horough revi�ion and reformulation of the whole theory of funct i on s of a rea l va riable. Among many prominent na mes, those of R iemann, Ca ntor, and Lebesgue are i n t i mately con nected with this field, which nowadays, with a l l its general i zations and ra m i ficat ions, may wel l be sai d to occu py a centra l pos i t ion i n the whole of analysis. We sha l l be content to der ive some basic t heorems which are easi ly accessible by the methods developed in the preceding chapters. For mor e thorough i nvestigations, the Lebesgue in tegral i s a natural and i n d ispensab le tool. We sha l l fi rst study more general systems of functions which share a property analogous to (6 1 ). 8.10 Definition Let { cPn } (n on [a, b ], such that
(64)
=
1 , 2, 3, . . . ) be a sequence of complex functi ons
J fi>n(x)fi>m(x) dx b
a
=
0
(n
#
m) .
SOME SPECIAL FU NCTIONS
187
Then { c/J n} is sai d to be an orthogonal system offunctions on [a, b ]. If, i n add i tion,
J l cf>n(x) l 2 dx b
(6 5)
1
=
a
for all n , { c/Jn} is said to be orthonormal. n For example, the fu nct ions (2rr)  iei x form an orthonorma l system [  rr, rr]. So do the real fu nct ions
on
cos x sin x cos 2x si n 2x
I
J2rr ' J rr ' J n ' J rc ' J rc '
If { cPn} is orthonormal on [a, b] and if
(66)
Cn
=
J f(t ) cPn( t) dt b
(n
=
a
l , 2, 3 , . . . ) ,
\Ve ca l l en t he nth Fourier coefficient of / relative to { c/Jn } . We write CX)
f(x) "' L en cPn(x)
(67)
1
and ca l l this �eries t he Fourier series of f (rela tive to { c/J n} ). Note that the sy mbol .. used i n (67) implies nothing about t he conver gence of the serie� ; it merely sa ys that t he coefficients are given by (66) . The fol lowing t heorems show that t he partial sums of the Fourier series off' have a certa in m i n i n1 um property. We sha l l assume here and in the rest of t h is cha pter t hat f E .!1 , although this hypothesis ca n be weakened. 8. 1 1
Theorem
Let { ,.} he orthonor1nal on [a, h ] . Let s,.(x)
(68 )
=
L em m (x) ,.
m= l
the nth partial s un1 of the Fourier series of(, and suppose n ( 69) tn(X) = L Ym c/J , (x) .
he
m= l
Then
(70)
J l f  sn l 2 dx < J l f  tn l 2 dx, b
b
a
a
and equality holds if and on ly if (7 1 )
Y m = em
(m
=
1,
.
.
. , n ).
Tha � is to say, among all fu nctions tn , sn gives the best possible mea n square approximation to f
188
PRI NCI PLE S OF MATH EMATICAL ANALYSIS
Proof
l:
Let J de n ote the i ntegra l over [a, b ] ,
I/111 IJI Ymm Ym 1/J,.
I "i'k 0. 8. 12
Theorem
L 1
I
em
12
0 be given. Si nce f e .11 and /( rr ) = f(  n), the construction described in Exercise 1 2 of Chap. 6 yields a cont i n uous 2nperiodic func tion h wi th (87)
11 /  h l! 2 < e . By Theorem 8 . 1 5, there is a trigonometric polynomial P such that l h (x)  P(x) / < e for all x. Hence ll h  P II 2 < e . If P has degree N0 , Theorem 8. 1 I shows that
(88)
l l h  sN (h) ll 2 < ll h  P ll 2 < e
for a l l N > N0 • By (72), with h  f i n place off, ll sN ( h)  s N(/) 11 2 = ll sN (h  /) 11 2 < ll h  /11 2 < e.
( 89)
Now the tria ngle inequal ity (Exercise I I , Chap. 6), combined wi th (87), (88), and ( 89 ) , shows that (N > N0).
II /  s N(/) 11 2 < 3 e
(90)
This proves (83). Next , (9 I )
I 2n
J
n
 n
N
I c dx = L s N(f)g "  N 2 7r
J
and the Schwarz inequal ity shows that
n
 n
.
N
e' "x g(x) dx = L en Yn , N
192
PRINCIPLES OF MATHEMATICAL ANALYSIS
which tends to 0, as N + oo , by (83). Comparison of (9 I ) and (92) gives (84). Finally, (85) is the special case g = f of (84). A more general versi on of Theorem 8 . I 6 appears in Chap. 1 1 .
TH E GAMMA :FUNCTI O N This function i s closely related to factorials and crops up i n many unexpected places in analysi s. J ts origin, history, and development are very wel l described in an interesting article by P. J. Davis (Amer. Math. Monthly, vol . 66, I 959, pp. 849869). Art in's book (cited in the Bibliography) is anofher good elemen tary introduct ion . Our presentation \Vi i i be very condensed , with only a few comments after each theorem. This section may thus be regarded as a large exercise, and as an opport un ity to apply some of the material that has been presented so far.
8.17
Definition
(93)
For 0 < x < oo , r(x)
=
J
.. oc
0
The in tegral converges for these be looked at.)
t"  l e  t dt.
x.
(When x < I , both 0 and oo have to
8.18 Theorem (a)
The functional equation r(x + I ) == xr(x)
holds if O < x < oo . (b) r(n + I ) = n ! for n = I .. 2, 3, . . . (c) log r is convex on (0, oo ).
.
Proof An i ntegrat ion by parts proves (a) . Since r{ I ) = I , (a) i mpl ies (b), by i nduct ion . If 1 < p < oo and ( 1 /p) + ( 1 /q) = I , apply Holder's inequa l i ty ( Exercise I 0, Chap. 6) to (93), a n d obta in
r(; �) +
0 and x > I , then 0 < t/l x(s) < t/1 1 (s). (e) So t/1 1 (s) ds < oo .
(a)
The convergence theorem stated i n Exercise 1 2 of Chap. 7 can therefore be app l ied to the i ntegral ( I 07), and shows that this in tegral converges to Jn as x + oo , by ( I 0 I ). Thi s proves ( I 03) . A more deta iled version of this proof may be found in R. C. Buck's "Advanced Calculus," pp. 2 1 62 1 8. For two other, entirely different, proofs, see W. Feller's article in Amer . Math. Monthly, vol. 74, 1 9 67, pp. 1 223 1 225 (with a correction in vol . 7 5, 1 9 68, p. 5 I 8 ) and pp. 2024 of Artin's book. Exercise 20 gives a si mpler proof of a less precise result.
196
PRINCIPLES OF MATHEMATICAL ANALYSIS
EXERCISES 1 . Define
{e
f(x) =
(x # 0), (x = 0) .
 1 tx 2
0
Prove that f has derivatives of all orders at x = 0, and that j 0).
log b
log ( 1 + x)
lim ( 1 +
x o O
=
x " + = ex. n
.
oo
=
+
oo
may occur).
SOME SPECIAL FUNCTIONS
S. Find
the following l imits
.
(a) I 1m
e
 (1 + x)
x + 0
n +
.
lx
.
n [n 1 1n  1 ] . log n
oc
lffi x + 0
l
X
(b) lim
( c) I
197
tan x  x . x( l  cos x)
.
 Sin X • (d) lim + 0 tan X X x X
6.
Suppose f(x)f(y) = f(x + y) fo r all real x and y. (a) Assuming that f is differentiable and not zero, prove that
f(x) h ere
= ec x
is a constant . (b) Prove the same thing, assuming only that f is continuous. w
7.
c
If 0 < x
 oc
=
co
:L
r(n + �) ' 11 .
n=O
r< > x" �
if 1 < x < 1 a n d ex > 0. 23. Let y be a continuously different i able closed curve in the complex plane, with parameter interval [a, b] , an d assume that y( t ) # 0 for every t e [a, b] . Define the index of y to be 
Ind (y) = 2 . 1T /
1
fb a
y (t ) dt. Y( t ) '
Prove that I nd (y) is always an i n teger . y' fy, cp(a) = 0. Hence y exp(  cp) Hint: There exists cp on [a, b] with cp' is constant. Since y( a) y( h) it fol l ows t hat exp cp(h) exp cp(a) l . Note that cp(b) = 21Ti l nd (y) . · Compute l nd (y) when y(t ) = e 1 " ' , a = 0, b = 2 1T . E xpla i n why l nd (y) is often cal led the windin,q numher of y around 0. 24. Let y be as in E xerc i se 23, and assume i n addition that t he range of y does not intersect the negative real axis. Prove t h a t l nd ( y) 0. Hint : For 0 < c < oo , lnd (y + c) i s a cont inuous i n t ege r va lu e d fu nction of c. Also, l nd (y + c) � o as c � oo . =
=
=
=

=
202
PRINCIPLES OF MATHEM ATICAL AN ALYSIS
25. Suppose
/' t and /'2 are curves as i n Exercise 23, and (a < t < b). I /' t (t )  ')' 2(t ) I < I Y t ( t ) I Prove that l nd (y t ) = I nd (y2). Hint: Put y = y2I Y 1 · Then I I  Y l < I , hence lnd (y) = 0, by Exercise 24 .
Also,
y' Y
26. Let
I
Y2
Yt /' t
y be a closed curve in the complex plane (not necessari ly different iable) wit h parameter in terval [0, 2 7T ] , such that y( t ) =!= 0 for every t E [0, 27T ] . Choose S > 0 so that I y ( t ) l > S for all t E [0, 2 7T ] . I f P1 and P 2 are trigo
nometric polynomials such that I Pi(t)  y(t ) l < S/4 for all tence is assured by Theorem 8. 1 5), prove t hat I nd
t E [0, 2 7T ]
(their ex is
(P. ) = Ind (P2)
by applying Exercise 25. Define this common value to be l nd (y). Prove that the statements of Exercises 24 and 25 hold wi thout any differenti abil ity assumption. 27. Let f be a continuous complex function defi ned in the complex plane. Suppose there is a positive integer 11 and a complex number c =1= 0 such that l im z  "f(z) = c .
J z J + oo
Prove that f(z) 0 for at least one complex number z. Note that this is a generalization of Theorem 8 . 8 . Hint: Assume /(z) =1= 0 for a l l z, define =
y,(t) = f(re1')
for 0 < r < oo , 0 < t < 27T, and prove the following statements about the curves
y, :
28.
(a) Ind (yo) = 0. (b) I n d (y,) = n for all sufficiently large r. (c) Ind (y,) is a continuous function of r, on [0, XJ ) . [In (b) and (c), use the last part of Exercise 26. ] Show that ( a), (b) , and (c) are contradictory, since n > 0. Let fJ be the closed uni t disc in the complex plane. (Thus z E D if and only i f I z I < 1 .) Let g be a continuous mapping of D in to the un it ci rcle T. (Th us, l g(z) I = 1 for every z E lJ.) Prove that g(z) =  z for at least one z E T. Hin t : For 0 < r < 1 , 0 < t < 27T, put
y,( t )
=
g(re i' ) ,
and put t/J(t) = e  "yt (f ). I f g(z) =1=  z for every z E T, then t/J(t) =!=  1 for every t E [0, 27T ] . Hence Ind (t/J) 0, by Exercises 24 and 26. I t follows that lnd (y1 ) = 1 . But Ind (yo) = 0. Derive a con trad iction, as i n Exercise 27 . =
SOM E SPEt:IAL FUNCTIONS
29. Prove that every con t i nuous mapping f of
203
D i nto D has a fixed point i n D. (Th i s i s the 2d i mensional case of B r o u we r s fixedpoint theorem .) Hint: Assu me f(z) j. z for every z E D. Associate to each z E lJ the point '
g(z) E T wh ich lies o n the ray that starts at f(z) and passes through z. Then g maps lJ i nt o T, g(z) = z i f z E T, a n d g i s c onti nu o u s, because g( z)
=
z  s( z) [f(z)  z ] ,
where s(z) is the un ique n on negat ive root of a certain quadratic equation whose coefficients are cont i nuous fu nc tions of f and z. Apply Exercise 28 .
9 FU NCTION S OF SEVERAL VAR I ABLES
LINEA R T R A N S F ORM ATI O N S
We begin this chapter with a d i scu ssion of sets of vectors i n eucl idean nspace R". The algebraic facts presented he re extend without change to fi n ited i mensional vector spaces over any field of scal ars. However, for our purposes it is quite sufficient to stay with i n t he fam i l iar framework provided by the eucl idean spaces. 9. 1
Definitions
(a) A nonempty set X c: R" i s a vector space if x + y E X and for a l l x E X� y E X, and for a l l scalars c. (b) If x b , xk E R" and c 1 , , ck are scalars, the vector .
.
.
.
.
ex
E X
.
is cal led a linear combination of x 1 , . , xk . If S c: R" and if E is the set of a l l l i near combination s of elements of S, we say that S spans E, or that E is the span of S. Observe that every span is a vector space. .
.
FUNCTIONS OF SEVERAL
VARIABLES
205
( c ) A set consisting of vectors x b . . . , xk (we shal l use the notation {x 1 , • • • , x k } for such a set) is sa id to be independent i f the relation c1 x 1 + · · · + ckxk = 0 impl ies that c 1 = · · · = ck = 0 . Otherwise {x 1 , • • • , xk }
i s said to be dependent. Observe that no independent set �ontains the n u l l vector. ( d) I f a vector space X contains an in dependent set of r vectors but con ta ins no independent set of r + I vectors, we say that X has diln ension r, and write : d i m X = r. The set consisting of 0 alone i s a vector space ; i ts d i mension i s 0 . (e ) A n i ndependent subset of a vector space X which spans X is cal led a basis of X. Observe that if B = {x 1 . , xr} is a basis of X, then every x E X has a un ique representation of the form x = I.ci x i . S uch a representat ion ex ists si nce B spans X. and it is un ique since B is in dependent. The nun1 bers c1 , • . • , cr a re ca l led the coordinates of x with respect to the basis B. , en }, where The most fam i liar exan1 ple of a basis is the set { e 1 , ei is the vector in R n whose jth coord i nate is I and whose other coord inates are a l l 0. l f x E R n . x := (x1 xn ), then x = I.xiei . We shal l cal l ..
.
.
.
•
•
•
•
.
,
{e 1 '
the standard b as is of R n .
•
.
.
. ' en }
9.2 Theorem L e t r be a posit ire integer. If a l'ector spac e set of r z: ectors, then d i m X < r .
X
is spanned by a
Proof
If this is fa lse. there is a vector space X \vh ich contains an i nde pendent se t Q = {y 1 Yr + 1} and wh ich is span ned by a set 50 con sist i n g o f r vectors. S uppose 0 < i < r. a n d suppose a set· S; has been const ructed wh ich spans X and which consists of a l l yi w·ith I < j < i plus a certa i n col lect ion of r i members of S0 , say x 1 . , xr ; ( I n other words. S; is obtained from S0 by replacing i of its elements by members of Q, without altering the spa n . ) S ince S; spans .A'. y ; + 1 is in the span of S; ; hence there are scalars a b . . , a; + 1 , b 1 • • • • , b r  ; , with a; + 1 = I . such that •
•
•
•
�

..
.
i+
I
.
.
_
.
ri
0i Y i + kL b k x k = O. L = l
j
= l
I f all bk 's were 0, the independence of Q \vould force all a/ s to be 0 , a contradictio n . I t follows that some xk E S; i s a l i near combi nation of the other members of T; = S ; u {Y; + 1 }. Remove this xk from T; and cal l t he remaining set S; + •1 • Then S; + 1 spans the sa me set as T; , na mely X, so that si + 1 has the pro pe r t ies po s'tulated fo r si with i + 1 in place of i.
206
PRINCI PLES OF MATH EMATICAL ANALYSIS
Starting with S0 , we thus construct sets S 1 , S, . The last of these consists of y 1 , y, , and our construction shows that i t spans X. But Q is independent ; hence y, + 1 is not i n the span of S, . This cont ra dicticn establishes the theorem . •
• •
Corollar y
dim Rn
•
.
•
•
.
= n.
Proof Since { e 1 , , en } spans Rn, the theorem shows that din1 R n Since { e1 , , en} is independent, dim R n > n. •
•
9.3 Theorem
•
•
•
•
< n.
Suppose X is a vector space � and din1 X = 11 .
(a) A set E of n l'ectors in X spans X if and only if E is independent. (b) X has a basis, and every basis consists of n l'l'Ctors. ( c) If 1 < r < n and {y 1 y,} is an independent set in �Y. then X has a basis containing {y 1 , y,}. •
Proof Suppose
E =
•
{x 1 ,
•
•
•
•
•
•
•
•
•
•
x n}. Since din1 A' =
•
n. t he set {x 1
•
•
•
•
•
x" . y}
is dependent, for every y e X. I f E is independent. it fol l ows thdt y is in the span of £ ; hence E spans X. Conversely. if E is dependen t , one of i t s members ca n be removed \vithout chan ging t he �pan of lJ,. Hence �:· ca nnot span X. by Theorem 9.2 . This proves (a). Since d i m X = n. X con tains an independen t set of 11 vectors. and (a) sho\vs that every such set is a basis of X : (b) now foi i O\\'S fron1 9. I ( d ) and 9.2 . x n } be a basis of X. The set To prove (c). let {x 1 •
S
=
•
•
•
{y I
,
�
•
•
•
, y , X1 r
•
•
•
•
� Xn}
spans X and is dependent, since it contains more than n vectors. The argun1ent used in the proof of Theorem 9.2 shows t hat one of the X ; � s i s a linear combination of the other members of S. I f we remove this X ; fron1 S. the ren1aining set sti l l spans X. This process can be repea ted r tin1es and leads to a basis of X which contains {y 1 , y,}. by (a). •
9.4 Definitions A mapping A of a vector space to be a linear transforn7ation if
X
•
•
•
into a vector space
A ( c x)
=
Y
is said
cA x
for alJ x. x 1 , x 2 e X and all scalars c. Note that one often wri tes A x instead of A( x) if A is l inear. Observe that AO = 0 if A is linear. Observe also that a l i nea r tran sforn1a tion A of X into Y is completely determined by its act ion on any ba�i� : I f
207
FUNCTIONS OF SEVERAL VARIABLES
{x1 , form
• • •
, xn} is a basis of X, then every x e X has a u nique representation of the
and the linearity of A allows u s to compute Ax from the vectors Ax1 , Ax,. and the coordinates c 1 , , en by the formula n Ax = L ci A xi . i= 1 Linear transformations of X into X are often called lin ear operators on X. If A is a linear operator on X which (i) is onetoone and (ii) maps X onto X, we say that A is invertible . In this case we can define an operator A  t on X by requiring that A  1 (Ax) = x for all x e X. It is trivial to verify that we then also have A(A  1 x) = x, for all x e X, and that A  t is linear. An important fact about linear operators on finitedimensional vector spaces is that each of the above conditions (i) and (ii) implies the other : • • • ,
• • •
A linear operator A on a finitedimensional vector space X is onetoone if and only if the range of A is all of X.
9.5 Theorem
Proof Let { X t ' ' xn} be a basis of X. The linearity of A shows that , Axn}· We therefore its range fJl(A) is the span of the set Q = {Ax1 , infer from Theorem 9.3(a) that fJl(A) = X if and only if Q is independent. We have to prove that this happens if and only if A is onetoone. Suppose A is onetoone and :Eci Axi 0. Then A(:Ecixi) = 0, hence :Ecixi = 0 , hence c1 = · · · = en = 0, and we conclude that Q is independent. Conversely, suppose Q is independent and A(:Ec i X ;) 0. Then :Eci Axi = 0, hence c1 = · · · = en = 0, and we conclude : Ax = 0 only if x = 0. If now Ax = Ay, then A(x  y) = Ax  Ay 0, so that x  y 0 , and this says that A i s onetoone. 0
0
0
• • •
=
=
=
9.6
=
Definitions
(a) Let L( X, Y) be the set of all linear transformations of the vector space X into the vector space Y. Instead of L( X, X), we shall simply write L( X). If A 1 , A 2 e L( X, Y) and if cb c2 are scalars, define c1A 1 + c2 A 2 by (c1 A 1 + c2 A 2)x
=
c1 A 1 x
+
c2 A 2x
(x e X).
It is then clear that c1 A 1 + c2 A 2 e L( X, Y). (b) If X, Y, Z are vector spaces, and if A e L( X, Y) and B e L( Y, Z), we define their p roduct BA to be the composition of A and B : Then BA e L( X, Z).
(BA)x = B(Ax)
(x e X).
.
l08
PRINCIPLES OF MATHEMATICAL ANALYSIS
A B, e ve n if X = Y = Z. ( c) For A e L( R", Rm ), define the nornz II A II of A to be the s u p of n u mbe rs I Ax I , where x ra n ges over all vectors in R" w it h I x I < 1 . Observe that the i neq ual i ty BA
N ote that
ne ed not be the sa me as
all
I A x l < II A II I x l
holds for all x e R". A l so, if ;w i s such that I A x I then I I A ll < )
1 . If f is a real function \Vith dotnain (a, b) c R 1 a n d if x e ( a, b), t he n f' ( x) is u s u a l ly defi ned to be the real nun1ber (7)
. f(x + h)  f(x) , I1 m
h + 0
h
provided, of course, that t h i s l i mit exists. Thus
f(x + h)  f(x) = f' (x) h + r(h) where the "remainder" r ( h) is small, in the sense that
(8)
(9)
rtm r (h)  0 .
h + 0
h
212
PRINCIPLES OF MATHEMATICAL ANALYSIS
Note that (8) e x pre ss e s t he d i ffere n ce f(x + h)  f(x) as the sum of t h e linear function that takes h to f'(.>.:)h. pl u s a small re n1ai nder. We can therefore re g a r d the derivative of f at x, n o t as a rea l n u mber, but as the linear operator on R 1 that ta kes h t o f'(x)h. [Observe that every rea l n um be r t:�. gives r i se to a l i near o pe r a tor on R 1 ; the o perato r in question is simply m ult i pl i c at io n by a. C o n verse l y , every li near function that carries R 1 to R 1 is m u l t i p l i cation by some rea l n u n1 be r. I t is this n at ur a l 1  1 co rre s po nden ce between R 1 and L( R 1 ) which mo t i va t e s the pre ced i ng st a te me n ts ] 1 Let u s next con s ider a fu nct i o n f t h a t maps (a, b) c R i n to Rm. I n t h a t case, f '(x) w a s defi ned to be tha t ve c t o r y E Rm ( i f t here is one) fo r which .
(1
h + 0
can again rewrite t h i s
We
{
. f (x + h)  f (x) I 1m y
( 1 0)
h
=
0.
i n the fo rn1
f (x + h)  f (x )
1)
}
=
hy + r(h),
where r(h)/h � 0 as h � 0. The m a i n te rn1 on the r i gh t si de of ( I I ) is aga i n a linear function of h. Eve ry y E Rm i nd u ce s a l i near transformat i o n of R 1 i n t o 1 Rm, by a ss oc i a t i n g to each h E R t h e vector hy E Rm. T h i s iden t i ficat i o n of Rm with L( R 1 , Rm) allows us to rega rd f '(x) a s a m e m ber of L( R 1 , Rm). Thus, if f is a d i ffere n t i a ble ma ppi ng of (a, b) c R 1 i n to Rm, and if x E (a, b), 1 : he n f '(x) is the l i near transformation of R i n.to Rm that sati sfies . f (x + h)  f (x)  f '(x) h hm h h + 0
( 1 2) or,
0,
=
equivalen tly,
.
( 1 3) We are
9. 1 1
now
Definition
!��
I f (X + h)  f (X)  f (X )h I ,
ready fo r t h e S u ppose
E is
lhl
>
case n
A
t hen we say that
f is differentiable at x,
O.
of
i n R", f m a p s E i n to Rm, and x E £. R" into Rm such tha t
. j f (x + h )  f (x)  A h l lim IhI h + 0
( 1 4)
=
I.
a n o pe n se t
If t here e x i s ts a l i near tra n sfo r m a t i o n
=
o,
a n d we w r i te
f '(x) = A .
( 1 5) If f is

d i ffere n t i a ble a t every
x E £, we
say th at
f is differentiable in E.
F U N CTIONS OF SEVERAL VARIABLES
213
I t is of cou rse u nderstood i n ( 1 4) that h E Rn . If I h I is small enough, then x + h E £, si nce E is open. Thus f (x + h) is defi ned , f (x + h) E Rm , and since A E L(R" , Rm ), A h E Rm. Thus f ( x + h)  f (x)  Ah E Rm . The norm in the n u rnerator of ( 1 4) is that of Rm . In the den ominator we have the R"  norn1 of h. There is an obvious un iqueness problem which has to be settled before we go a ny further. 9. 1 2 Theorem Suppose E and f are as in Definition 9. 1 1 , x with A = A 1 and with A = A 2 . Then A 1 = A 2 • Proof
E
£, and ( 1 4) holds
If B = A 1  A 2 , the i nequality I Bh l < l f(x + h )  f ( x)  A 1 h l + l f ( x + h)  f (x)  A 2 h l
shows that I Bh I I I h I
�
0 as h � 0. For fixed h i= 0, it fo llows that
I B( t h) �0 th I l
( 1 6)
as
t � 0.
The l i nearity of B shows that the left side of ( 1 6) is i ndependent of t. Th us Bh = {) for every h E R" . Hence B = 0. 9. 13
Remar k s
(a )
( 1 7)
The relation ( 1 4) ca n be re\vritten i n the form f( x + h)  f (x) = f '( x) h + r(h )
where the rema i nder r(h) satisfies ( 1 8)
I r( h) I = 0. Iim h .. O lhl
\Ve may interpret ( 1 7 ). as in Sec. 9. 1 0, by saying that for fi xed x and stna ll h, the left side of ( 1 7) is approx imately equa l to f '(x)h. that is. to the val ue of a l i near tra nsformation appl ied to h. (b) Suppose f and E are as i n Defi nition 9. 1 1 , and f is differen tiable in E. For every x E £. f '(x) is then a function , natnely, a linear transformation of R" i nto Rm. But f ' is also a function : f ' maps E into L( R" , Rm ). (c ) A glance at ( 1 7) shows that f is contin uous at any point at which f is differen tiable. (d) The derivative defined by ( 1 4) or ( 1 7) is often called the differential of f at x, or the total derivative of f at x, to d istinguish i t from the partial derivatives that will occur later.
214
PRINCIPLES OF MATHEMATICAL ANA LYSIS
9. 14 Example We have defined derivati ves of fu nctions ca rryi ng R" to Rm to be l inear transformatio ns of Rn i n to Rm. What is the derivative of such a linear transformation? The answer is very simple . If A e L(R" , Rm) and if x e Rn , then A '(x) = A .
( 1 9)
Note that x appears on the left side of ( 1 9), but not on the right. Both sides of ( 1 9) are members of L(R", Rm), whereas A x e Rm. The proof of ( 1 9) is a t riviality, since A(x + h)  A x = A h, (20) by the l inearity of A . With f (x) = A x, the numerator i n ( 1 4) is thus 0 for every h e R". I n ( 1 7), r(h) = 0. We now extend t he chain rule (Theorem 5 . 5) to the present situation . 9. 15 T'heorem Suppose E is an open set in R", f maps E into Rm, f is differentiable at x0 e £, g maps an open set containing f (E ) into R k , and g is differentiable at f (x0 ) . Then the mapping F of E into Rk defined by F(x) = g(f (x))
is differentiable at x0 , and F' (x0) = g'(f(x0 )) f ' (x 0 ) .
(2 1 )
On the right side of (2 1 ), we have the prod uct of two linear transforma tions, as defi ned i n Sec. 9.6. Proof
Put Yo = f (x0), A = f ' (x0), B = g' (y0 ) , and define u(h) = f (x0 + ll)  f (x0)  A h, v(k) = g(y0
+
k)  g ( y0 )  Bk ,
for all h e R" and k e Rm for which f (x0 + h) and g(y 0 + k) a re defined . Then (22) l u(h) l = e (h) l h l , 1 v(k ) 1 = 17 (k ) k 1 , where e(h) + 0 as h + 0 and 17 ( k ) + 0 as k + 0. Given h, put k = f (x0 + h)  f (x0). Then l k l = I Ah
(23)
and
+ u(h) j < [ II A II
I
+ e (h) ] I h i ,
F(x0 + h )  F(x0 )  BAh = g(y0 + k)  g(y0)  BA h = B(k  A h) + v( k ) = Bu(h)
+ v(k).
FUNCTIONS OF SEVERAL VARIABLES
215
Hence (22 ) and (23) imply� for h =F 0, that I F( x o + h) It
�
� (x o )  BA hI < II B II s(h) + [II A ll +. t:(h) J'7 (k) .
Let h 0. Then e (h) 0. Also, k + 0, by (23), so that 17 (k) + follows that F '{x0) = BA , which i s what (2 1 ) asserts. +
+
0.
9. 16 Partial derivatives We again consider a function f that maps an open set E c R" into Rm. Let {e b . . . , e"} and {u . , . . , um } be the standard bases of R" and Rm. The components of f are the real functions /1 , , fm defined by .
•
( 24)
f (x)
m i= 1
= L h(x)u i
•
•
(x e £),
or, equivalen tly, by /,(x) = f (x) · u ; , 1 < i < m. For x E £, 1 < i < m, 1 < j < n, we define (25) provided the l i mit exists. Writing /;(x 1 , , xn) in place of /,(x), we see that Difi is the derivative of[; with respect to xi , keeping the other variables fixed. The notation •
( 26)
•
•
o!"i OX)·
is therefore often used i n place of Difi , and Difi is called a partial derivative. I n many cases where the existence of a derivative is sufficient w hen dealing with functions of one va riable, continuity or at least boundedness of the partial derivat ives is needed for fu nctions of several variables. For example, the functions f and g described in Exercise 7 , Chap. 4, are not continuous, although their partial derivatives exist at every point of R2• Even for continuous functions. the existence of a l l partial derivatives does not i m ply differentiability in the sense of Defi nition 9. 1 1 ; see Exercises 6 and 1 4, and Theorem 9.2 1 . However, i f f is known to be d ifferentiable at a point x, then its partial derivatives exist at x, and they determine the linear transformation f '(x) completely : 9. 17 Theorem Suppose f maps an open set E c R" into Rm , and f is differentiable at a point x E £. Then the partial derivatives ( Di/;)(x) exist, and
(27)
m f '(x)ei = L ( Di/;)(x)u i i= 1
( 1 < j < n).
216
PRINCIPLES OF MATHEMATICAL ANALYSIS
Here, as in Sec. 9 1 6 { e 1 , of R" and Rm . .
,
.
.
•
, e,} and {u b . . , u m } are the standard bases .
Proof Fix j. Since f is differentiable at x, f(x + tei)  f(x)
= f '(x)(tei) + r(tei)
where I r(tei) 1 /t + 0 as t + 0. The linearity of f '(x) shows therefore that . I tm
(28)
t + 0
f(x + tei)  f(x) t
= f '(x)ei .
If we now represent f in terms of its components, as in (24), then (28) becomes {29)
. I lffi � t + 0 i = l
� [;(x
+ tei)  /;(x)
t
U;
= f '(X)ej .
It follows that each quotient in this sum has a limit as t � 0 (see Theorem 4. 1 0), so that each ( Di[;)(x) exists, and then (27) follows from (29). Here are some consequences of Theorem 9. 1 7 : Let [f '(x) ] be the matrix that represents f '(x) with respect to our standard bases, as in Sec. 9.9. Then f '( x )ei is the jth co lu m n vector of [f '(x) ], and (27) shows therefore that the number ( Difi )(x) occupies the spot in the ith row and jth column of [f '(x) ]. Thus [f '(x)]
If h
=
= �hi ei is any vector in
.......................... ( D t fm ){ x) · · · ( D,./m)(x) R", then (27) implies that
(3 0)
Example Let y be a differentiable mapping of the segment (a, b) c R 1 into an open set E c R", in other words, '}' is a differentiable curve in E. Let f be a realvalued d ifferentiable function with domain £. Thus f is a differentiable mapping of E into R 1 • Define
9.18
(3 1 )
g( t)
= f(y(t))
.(a < t < b).
The chain rule asserts then that (3 2 )
g '(t)
= f'(y(t))y'(t)
(a < t < b).
FUNCTIONS OF SEVERAL VARIABLES
217
Since y'( t ) E L( R 1 � Rn ) and f'(y(t )) E L( Rn � R 1 )� (32) defi nes g'(t ) as a l i nea r operator on R 1 • This agrees with the fact that g maps (a. b) into R 1 • HoweveL g'(t ) can also be regarded as a real nun1 ber. (This was discussed i n Sec. 9. 1 0. ) This number ca n be co mpu ted i n tern1s o f the part ial derivat i':es of I and the derivatives of the components of y. as we shall now see. , en} of R", [ y ' ( t ) ] is the n by I With respect to the standard basis {e 1 � �, matrix (a "colu m n matrix ) which has y � (t) i n t he ith row� where Y t � . . . . }' n are the components of y. For every x E E. [l ' (x) ] is the 1 by n n1at rix (a " row matrix") which has ( Di/)(x) i n the jth col umn. Hence [g'(t ) ] is the 1 by 1 n1atrix whose only entry is the real number n g'(1) = L ( D;/)(y(t ) h'; (1 ). (33) •
•
•
i= l
This is a frequently encountered special case of the chain rule. I t ca n be rephrased in the following n1anner. Associate with each x E E a vector, the socalled "grad ien f " of I at x, defined by (34)
( V/)(x)
=
n
L ( D;/)(x)e; . I
' =
Since (3 5)
y'(t )
=
,,
L y� (t )e, ,
' =
I
(33) can be wri tten i n the fonn g ' (t)
(36)
=
(Vf)(y(1 )) · y ' (1).
the scalar prod uct of the vectors (VI)(y(t)) and }' ' ( 1 ). Let us now fi x an x E £� let u E R" be a unit vector (that is. I u ! specialize y so that
(37 )
y(t )
Then y'(t )
=
u for every t .
=
(
x + tu
00
< 1 < 00 ).
Hence (36) shows that
(38)
g'(O}
=
(V/)( x) ·
u.
On the other hand, (3 7) shows that g( t )  g(O)
= f(x
+ t u)  l(x) .
Hence (38) gives
(3 9 )
lim
t + 0
f(x
+ tu )  /(x) = ( V/) (x) t
·
u.
=
I ). and
218
PRINCI PLES
OF MATH EMATICAL ANALYSIS
The l i m i t in (39) is usually cal led the directional derivative off at x. i n the direction of the u n it vector u , and may be denoted by ( D0/)(x). I ff and x are fi xed but u varies , then (39) shows that ( D0/)(x ) attains its maximum when u is a posit ive scalar m ultiple of ( V/)(x}. [The case ( V/)(x) = 0 should be excluded here. ] If u = 'Lu; e; , then (39) shows that ( Duf)(x) ca n be expressed in terms of the partial derivatives of f at x by the formula ,
n
( Duf)(x) = L ( D;/)(x)u ; .
(40)
i
= I
Some of these ideas will play a role in the following theorem. 9. 19 Theorem Suppose f n1aps a convex open set E c R" into R m , f is differen tiable in £, and there is a real number M such that
l l f (x ) l l < M '
x E
for every
£. Then l f(b)  f(a) l < M l b  a !
for all a
E
£, b e E.
Proof Fix a
E
£, b E £. Define y(t) = ( 1  t)a + tb
for all t E R 1 such that y(t ) E £. Since E is convex, y(t ) E E i f 0 < t < 1 . Put g(t) = f(y(t)).
Then
g'(t) = f '(y(t))y'(t) = f '(y(t))(b  a),
so that I g' ( t) I < II f ' (y( t)) I l l b  a I < M I b  a I
for all t E [0, 1 ]. By Theorem 5. 1 9, l g( l )  g(O) I < M l b  a l .
But g(O) = f(a) and g( l ) = f (b). This completes the proof. Corollary
If, in addition, f '( x ) = 0 for all x e £, then f is constan t.
Proof To prove this, note that the hypotheses of the theore m hold now with M = 0.
FUNCTIONS Of SEVERAL VARIABLES
219
9.20 Definition A differentiable mapping f of an open set E c R" i nto Rm is sa id to be continuousfr differentiable in E if f ' is a continuous mapping of E into L( R", Rm ). M ore explicitly. it is required that to every x e E and to every e > 0 co rresponds a � > 0 such that ll f '(y)  f '(x) ll
13. Suppose f is a different iable mappi ng of
+
a cos t ) 2 •
R 1 into R 3 such that I f(t ) I
Prove that f'(t) · f(t ) = 0. Interpret this result geometrical ly.
14. Define /(0, 0)
=
==
1 for every t.
0 and .
f (x , y ) ==
x3
X 2 i )' 2
i f (x, y) # (0, 0) .
(a) Prove that D f a n d D2J are bounded functions in R 2 • ( Hence f is con t i n uous.) (b) Let u be any u n i t vector i n R 2 • Show t hat t he d i rect i onal derivat i \ C ( Duf)(O, 0) exists, a n d that its absol ute value i s at most 1 . ( c ) Let y be a d ifferent i able mapping of R 1 I n t o R 2 ( i n ot her words, y is a d i ffer entiable curve in R 2 ), with y(O) = (0, 0) and I y'(O) j > 0. Put g( t ) = f(y( t )) and prove t h a t g is different iable for every t E R 1 • If y E CC', prove that g E (x, t) =
X
x + 0
2V t
and put q>(x, t) =  q>(x, I t J ) if t < 0.
(O � x �vt ) (vt � x � 2Vt) (otherwise),
FUNCTIONS OF SEVERAL VARIABLES
243
Show that cp is continuous on R2, and ( D 2 cp)(x, 0) = 0 for all x. Define f( t ) = Show that /(!) = t if I t l < 1. Hence f'(O) =f.
r. cp(x, t) dx.
r / D cp)(x, 0) dx. .
29. Let E be an open set in R". The classes ct'(E) and ctn(E) are defined in the text. By induction, ct(£) means that the partial derivatives D1/, . . . , Dnfbelong to ct< "  1 )(£). Assume f e ct 3, then D t 2 t 3/ 30.
=
D3u 2 f
for every f e ct(£), where E is an open subset of R". Fi x a e E, and suppose x e R" is so close to 0 that the points
p(t) = a + tx lie in E whenever 0 < t :::;;: 1 . Define
h(t) = f(p(t)) for all t e R1 for which p(t) e E. (a) For 1 :::;;: k � m, show (by repeated application of the chain rule) that
h < "> (t ) = L ( Dt t
•••
'" f)(p(t )) Xt t
•••
Xr" .
The sum extends over all ordered ktuples (i� , . . . , i") in which each i1 is one of the integers 1 , . . . , n . (b) By Taylor's theorem (5. 1 5 ) ,
for some t e (0, 1 ). Use this to prove Taylor's theorem in that the formula
n
variables by showing
244
PRINCIPLES OF MATHEMATICAL ANALYSIS
/(a t x ) =
m 1 1
"� k ! L ( D, I . . . ' " /)(a)x; 1
• • •
x ,k ; 
r (x)
represents /(a + x ) as the sum of its socaJ ied "Taylor polynom ial of degree m  1 ," pl us a remainder t hat satisfies I .• m
x + 0
r (x )
IXI
m 
•
__
0
.
Each of t he in ner sums extends over a l l ordered ktu ples (ih . . . , i,.), as in part (a) ; as usua l , t he zeroorder derivative of I is sim ply f, so t hat the constant term of t he Taylor polynomial of f at a is /(a). (c) Exercise 29 shows that repet ition occurs in the Taylor polynom ial as wri tten in parl (b). For i nstance, D 1 1 J occurs three t i mes, as D t t J , D 1 J 1 , D J J I · The sum of the corresponding three terms can be wri tten in the form 3(D� DJI)(a)x� X 3 . Prove ( by calculati ng how often each deriva tive occurs) that the Tay l or polynomia l i n (b) can be written in the form
Here t he summat ion extends over a l l ordered ntuples (s 1 , sn) such that each s; is a nonnegative i n teger, and s . + · · · + sn < m  I . 3 1 . Suppose I E �( 3 > i n some neigh borhood of a poi nt a E R 2 , the grad ien t of I is 0 at a, but not a l l secondorder deri vat ives of f are 0 at a . Show how one can then determ ine from the Taylor pol ynomial of I at a ( o f degree 2) whether f has a local maximum, or a local minimum, or neit her, a t t he point a . Extend this to R" in pl ace of R 2 • .
•
.
.
•
10 I NTEG RATION OF D IFFERENTIAL FORMS
I ntcgrati on can be st ud ied on many levels. I n Chap. 6, the theory was devel oped for reasonably wellbehaved functions on subi ntervals of the real line. I n Chap. I I �·e shall enco unter a very highly developed theory of i ntegration that can be applied to much large r cla sses of funct i ons, \\'hose domai ns are more or less arbitrary sets 't not necessarily subsets of Rn . The present chapter i s devoted t o those aspects of integrati on theory t hat are closel y related to the geometry of euclidean spaces, such as the change of variables formula, l i ne integrals, and the machinery of d i fferential forms that i s used i n the statement and proof of the nd i mensional analogue of the fundamental theorem of calculus, namely Stokes' theorem . INTEGRATIO N 1 0.1
Definition
such that (1 )
Suppose
a l.
Jk
is a kcell i n Rk , consisting of all x =
< X· < b .

l

l
(x 1 ,
•
•
•
,
xk)
( i = I , . . , k) , .
246
PRINCIPLES OF MATHEMATICAL ANALYSIS
is the jce11 in Ri defined by the first j inequalities ( I ), and f is a real con tinuous function on Jk . Put f = fi , and define fi _ 1 on Jk  l by
Ji
bk
= I .fi(x l , xk  1 xk ) dxk The uniform continuity of .h on Jk shows that h _ 1 is continuous on l k Hence we can repeat this process and obtain functions .fj , continuous on Ji, such that fj_ 1 is the integral offj , with respect to xi , over [ai , bi]. After k steps we arrive at a number .fo , which we call the integral off J k � we write it in the form f f(x) dx or I f (2) h  1 (x 1 ,
·
·
·
xk  1 )
,
·
.
·
,
•
·
Qk
 t .
o ver
6
Jk
Jk
priori, this definition of the integral depends on the order in which the k integrations are carried out. However, this dependence is only apparent . To prove this, let us introduce the temporary notation L(f) for the integral (2) and L' (f) for the result obtained by carrying out the k integrations in some other order. A
10.2
f e rt(Jk ) , L(f) = L' (f) . If h(x) = h 1 (x ) hk (xk ) , where hi e �( [ai , bi]), then
Theorem Proof
For every
1
• • •
L(h)
k
= n1 I !=
b;
h ; (X; ) dx; = L'(h).
a;
If .91 is the set of all finite sums of such functions h, it follows that L(g) = L' (g ) for all g e d. Also, .91 is an algebra of functions on Jk to which the StoneWeierstrass theorem applies. k Put v = n (b ;  ) Iff E r.t(J k) and e > 0, there exists g E .91 such that 11 /  gil < e/ V, where 11! 11 is defined as max lf(x) I (x e lk ). Then I L(f  g ) I < e , I L '(f  g) I < e , and since L(f)  L ' (f) = L(f  g ) + L'( g  f) , we conclude that I L(f)  L '(f) I < 2 e . In this connection, Exercise 2 is relevant. 1
a;
.
The support of a (real or complex) function f on R k is the closure of the set of all points x e Rk at which f(x) # 0. I f f is a continuous 1 0.3
Definition
INTEGRATION OF DIFFERENTIAL FORMS
247
funct ion with com pact su pport, let I" be any kcell which contains the support off, and define (3)
The i ntegral so defi ned is evidently independent of the choice of J k, pro vide d only that I " conta ins the su pport off I t is no\\' tempti ng to extend the defi nition of the integral over R" to fu nct ions which are li mits (in son1e sense) of conti nuous functions with compact support. We do not want to discuss the cond itions under which this can be done : the proper setting for this question is the Lebesgue integral . We shall merely descri be one very si mple exa mple which will be used i n the proof of Stokes ' theorem. 10.4 Example let Q " be the ksi mp1ex which cons i sts of all points x = (x 1 , x1 J i n R" for which x 1 + + x" < 1 and X ; > 0 for i = I , . . . , k. If k = 3. for exa mple, Q " is a tetrahed ron, with vertices at 0, e 1 , e2 , e 3 • I ffe �( Q"), extend f to a fu nction on I " by setting f(x) = 0 off Q" , and define ·
·
·
·
�
I"
·
f
(4) Here
·
Qk
is the · 4 unit cube 't ' defined by 0
< X;
. Poi nts of D will be denoted by u (u 1 , , uk ). We shall confine ours e l ves to the s i m ple sit uation in which D is either a kcel l or the ksi mplex Q k descri bed i n Exam ple I 0.4. The reason for this i s that we shal l have t o integrate over D , and w e have not yet discussed i ntegr a ti o n over more compli c ated subsets of R k . I t wi ll be seen that this restriction on D (wh ich will be tacitly made from now on) entai ls no s i gn i ficant loss of generality in the resulti ng theory of differential forn1s. We stress that ksurfaces i n E are defi ned to he rnappin,qs i nto £, n ot subsets of E. This agrees with our earlier defi nition of curves ( Defi nition 6. 26). In fact , ] surfaces are precisely the same as con t i nuously differentiable curves. •
=
•
•
10.1 1 Definition Suppose E is an open set i n R " . A differential fornz o.f order k > I in E (briefly, a kform in E) is a function w , symbolically represented hy the sum
(34) (t h e indices i1 , , ik range i ndependently from I to n), \vhich assign s to each ksurface i n E a nu n1ber oJ ( ) J w , according to the rule •
•
•
=
J
(35)
w
=
J L a; , D
.
. ;. (( u) ) O(x; , .
•
0( U 1 ,
.
.
.
.
.
.
, x;.) du , , Uk)
where D is the parameter do mai n of . The fun ctions a i • • • ;,._ are assumed to be real and co n t i n u ous i n }�. I f 4>n are the com ponents of , the J acob i an i n (35 ) i s the o ne dete rmined ¢1 , by the mapping • . .
1
,
(u1 ,
.
.
. , uk ) + (¢ i 1 ( u) , . , ¢;k(u)) . .
.
N ote that the right side of (35) is an i ntegr a l over D , as defi ned in Defi ni t i on I 0. 1 (or Example I 0.4) and that (35) is the de_finition of the sym bol J w . A kform w is said to be of class rJ ' or f6 ' if the functions a i • • • ; k in ( 3 4) are all of class fC' or n. •
•
In other words,
•
The anticommutativity expressed by (42) i s the reason for the inordi na te amount of attention that has to be paid to minus signs when studying different ial forms.
INTEGRATION OF DIFFERENTIAL FORMS
Basic kforms
10.14
< ik < n,
kindex,
and if
I is
If
i1 ,
0 o 0 ,
are integers such that
ik
the ordered ktuple
{ib . . . , ik} ,
I < i1 < i2
0 such that b1(x) > 0 for satisfy I x1  v1 I � h. Let D be the kcell only if I u, l � h for r 1 , . . . , k. Define •
•
that b1(v) > 0 for some v e E , jk} · Since b 1 is continuous, all x e Rn whose coordinates in Rk such that u e D if and
•
=
(49)
(u) = v + L u, eir t k
(u e D).
r=
Then is a ksurface in E, with parameter domain D, and b1{(u)) > 0 for every u e D. We claim that
( 5 0)
t
co
=
t bicl>(u)) du.
Since the right side of (50) is positive, it follows that w() i= 0 . Hence (50) gives our contradiction. To prove (50), apply (35) to the presentation (48). More specifically, compute the Jacobians that occur in (35). By (49), o(Xi t ' . . . ' Xjk)
o(ut ' . . . ' uk)
=
1.
For any other increasing kindex I :1: J, the Jacobian is 0, since it is the determinant of a matrix with at least one row of zeros. 10.16
(5 1 )
Products of basic kforms
Suppose
J = {jt ,
I = {it , . . . , i } , p
. . . , jq}
where 1 � it < · · · < iP < n and 1 < jt < · · · < jq � n. The product of the co r responding basic forms dx1 and dx1 in Rn is a (p + q)form in Rn , denoted by the symbol dx1 A dx1 , and de fi ned by (52)
dx1 A dx1 = dxit A
•••
A
dx1p A dxi1 A
•••
A
dxi, ·
INTEG RATION OF DIFFERENTIAL FORMS
259
I f I and J have an element i n common, then the discussion in Sec. 1 0. 1 3 shows that dx 1 1\ dx1 = 0. I f I and J have no element i n com mon, let us write [I, J] for the i ncreasing ( p + q)index wh ich is obtai ned by arrangi ng the members of I u J in increasing order. Then dxu . 1 1 is a basic (p + q)form . We clai m that
2
dx1 A dx1 = (  1 ) dxu , 1 1
(53)
where a is the n u mber of di fferences j,  is that are negative. (The number of positive differences is thus pq  a .) To prove (53), perform the fol lowi ng operations on the numbers (54)
M ove iP to the right, step by step, until i ts left neighbor is less than iP . The n u mber of steps is the n u mber of subscripts t s uch that iP < }, . (Note that 0 steps are a distinct possibil i ty.) Then do the same for ip  1 , , i1 The total n u m ber of steps taken is t:�. . The fi nal arrangement reached is [/, J]. Each step, when appl ied to the right side of (52), mul tiplies dx 1 1\ dx1 by  I . Hence (53) holds. Note that the right side of ( 53) is the standard presentati on of dx1 A dx1 • Next, let K = (k 1 , , k,) be an increasing ri ndex i n { I , . . . , n}. We shall use (53) to prove that •
•
•
•
•
•
•
(5 5 )
I f a ny two of the sets I, J, K have an element in co mmon , then each side of (55) is 0, hence they are equal . So let us assume that I, J, K are pairwise disjoint. Let [I, J, K] denote the i ncreasing (p + q + r)index obtained from t hei r union. Associate P with t he ordered pai r (J, K) and y with t he ordered pai r (I, K) i n the way that a: was associated with (I, J) in (53). The left si de of (55) is then (  I )2 dx u . 11 A dxK = (  1 Y'(  I )P + Y dx u . 1 • K J by two applications of (53), and the right side of (55) is (  t )P
dx1
1\
dx [ J , K ] = (  1 )P(  Iyr + y dx[ I, J , K ] ·
Hence ( 5 5) is correct. 10. 17 Multiplication Suppose w and A. are p and qforms, respectively, i n some open set E c R" , with standard presentations (56)
w
=
L b1(x) dx1 , I
A. = L
J
( )
c1 x
dx1
where I and J range over all i ncreasing pindices and over all increasing qindices taken from the set { 1 , . . . , n} .
260
PRINCIPLES
OF MATH EMATICAL ANALYSIS
Their product, denoted by the symbol w A
w
(57)
)� =
A
L b i ( x) c1 (x ) dx 1
I,
J
A. , is defined to be A
dx1 .
In this sum, I and J range independently over their possible values, and dx 1 is as in Sec. I 0. 1 6. Thus w A A. is a (p + q)form i n E .
A
dx1
I t is quite easy to see (we leave the details as an exercise) that the distribu tive laws
and w
A
( l 1 + A2) = (w
A
)� 1 ) + (ro
A
l 2)
hold, with respect to the addition defined i n Sec. I 0. 1 3. I f these distributive laws are combi ned with (55), we obtain the associative law (ro
(58)
A
A)
(J = W
A
A
( ). A (J)
for arbitrary forms oJ, A., (J in £. In this discussion it was tacitly assumed that p > I and q � I . The product of a 0form f with the pform ro given by ( 5 6) is simply defined to be the pform fw = wf = L f(x ) b1 (x) dx1 • I
I t is customa ry to write fw rather than j' A w, when f is a 0form. ,
� 0.18 Differentiation We shall now define a differentiation operator d which associates a (k + I )form dw to each kform ro of class �' in some open set E c R".
A 0form of class �' i n E is just a real function f e �'(£), and we define df = L (Di.l)(x) dx i . i= 1 n
(59)
I f w = Ib 1(x) dx 1 is the standard presentation of a kform ro, and b 1 e Cl'(£) for each i ncreasing kindex /, then we define dw =
(60)
L (db 1) 1
A dx 1 .
10. 19 Ex ampl e Suppose £ is open in R", f E r�' ( £), and y is a continuously differentiable cu rve in £. with don1ain [0. I ]. By (59) and (35), (6 1 )
f
f
1
( DJ)( y(t ))y;(t ) dt. L df = 1 0 i "' n
=
INTEGRATION OF DIFFERENTIAL FORMS
261
B y the chain rule, the last i ntegrand is (f y)' (t). Hence o
f df = f( y( l ))  f( y(O)),
(6 2)
y
and we see that J Y dj is the same for all y with the same initial point and the same end point, as in ( a) of Example 1 0 . 1 2 . Comparison with Example I O. I 2(b ) shows therefore that the 1 form x dy is not the derivative of any 0form f This could also be deduced from part ( b) of the following theorem , since
d(x dy) = dx A dy i= 0. 10.20
Theorem
(a) If w and A are k and mforn1s, respectively, of class d(w A A ) = (dw ) A 1 + (  l )k w A dA .
(63 )
(b) If w is of class
r.t"
rc'
in E, then
in E, then d2w = 0 .
Here d 2 w means, of cou rse, d(dw) . Proof Because of ( 57) and (60), (a) follows if (63) special case w = f dx 1,
(64)
ts
proved for the
)� = g dx1
where f, g E rove (b) first for a 0form f E �" :
d 2f = d
Ct< Di/)(x) dxi)
n
= L d(Dif) j= l
A dxi
n
= L (D iif)(x) dx ; i, j = 1
A dxi
.
Since D iif = Diif (Theorem 9.4 1 ) and dx ; A dx1 =  dx1 A dx; , we see that d 2f = 0. If w = f dx 1 , as in (64), then dw = (df) A dx 1 . By ( 60) d(dx I ) = 0. Hence (63) shows that ,
10.21 Change of variables Suppose E is an open set in R", T i s a �' ma ppi ng m of E into an open set V c R , and w is a kform in V, \vhose standard presenta tion is
(6 5 ) (We use y for points of V, x for points of E. ) Let t 1 , , tm be the components of T: I f •
then (66)
Y;
•
•
Y = (y 1
= t ;(x). As in (59),
. . . ' Ym)
"
dt; = I (Di t;) (x) dxi j= l
Thus each dt ; is a 1 form in £. The mapping T transforms (67)
'
= T {X) (l
< i < m) .
into a kform wT i n £, w hose definition is
(O
w r = L b 1(T (x)) dt;, I
A •·· A
dt;k .
I n each summand of (67), I = {i 1 , , ik} is an increasing kindex . Our next theorem shows that addition, multiplication, and differenti a ti o n of forms are defined in such a way that they com mute with changes of variables. •
•
•
INTEGRATION OF DIFFERENTIAL FORMS
263
With E and T as in Sec. 1 0.2 1 , let w and A. be k and mforms in V, respectively. Then 10.22
Theorem
(a) ((u + A.) r = W r + A. r if k = m ; (h ) (w A A. ) r = W r A A r ; (c) d( w r) = (dw) r if w is of class �' and T is of class �". Proof Part (a) follows i mmediatel y from the definitions. Part (b) i s al most as obvious, once we realize that (6 8)
(dy ' 1 A
• • •
·
A dy ,,.) T ·
=
dt ' A ·
1
• • •
A dt ,,. ·
, i,} is i ncreasing or not ; (68) hol d s because regardless of whether {i 1 , the same number of minus signs are needed on each side of (68) to produce increasing rearrangements. We turn to the proof of (c) . I f f is a 0form of class �' in V, then • • •
fT( x)
=
�f( T (x) ),
By the chain ru le, it foilo\VS that d(fr )
(69)
=
L ( Difr)(x) dxi
=
L Ii ( D if)(T(x))( Di t i)(x) dxi
j
j
=
L ( D ;/)( T (x)) dt i i
=
I f dy 1
=
dy ; 1 A
• • •
(df) r ·
A dy ;" then '
1 0.20 shows that
(dy1) r = dti1 A
d((dy 1) r)
(70)
=
• • •
A dt;" ' and Theorem
0.
(This is where the assumption T e �" is used.) Assume now that w = f dy 1 . Then Wy
=
fr(x) (dy r) r
and the preceding calculations lead to d(w r)
= =
d(fr) A (dy 1) r = ( df) r A (dy 1) r ( (df) A dy 1)r = (dw) y .
The first equality holds by (63) and (70), the second by (69), the third by pa rt (b), and the last by the definition of dw. The general case of ( c) follows from the special case just proved, if w� apply (a). This co m pletes the proof.
264
PRINCIPLES OF MATHEMATICAL ANALYSIS
Our next objective is Theorem I 0.25. Thi s will follow d irectly from two other important transformation properties of differential forms, which we state first.
Suppose T is a � ' mapping of an open set E c R" into an open set V c Rm , S is a �'mapping of V into an open set W c RP, and w is a kform in W, so that w5 is a kform in V and both (ws) r and Wsr are kforms in E, where ST is defined by (ST )(x) = S(T(x)). Then 10.23
Theorem
(7 1 )
= Wsr .
( Ws) r
Proof I f w and l are forms i n
((w
and
=
" )..)s) r
W,
Theorem 1 0.22 shows that
(ws " ls) r = (ws) r A (A.s) r
= Wsr A
(w " A.)s r
Asr ·
Thus if (7 1 ) holds for w and for A, it follows that (7 1 ) also holds for w A A. Since every form can be built up from 0forms and 1 forms by addition and multiplication, and since (7 1 ) is trivial for 0forms, it is enough to prove ( 7 1 ) in the case w = dzq , q = 1 , . . , p. (We denote the points of E, V, W by x, y, z, respectively.) , sP be the com po Let 1 1 , , tm be the components of T, let s 1 , nents of S, and let r 1 , r P be the components of ST. I f w = dzq , then .
•
•
•
•
,
•
•
•
•
•
dsq = I ( Di sq)(y) dyi ,
w5 =
j
so that the chain Tule implies (ws) r = I (Di sq)(T (x)) dti j = I ( Di sq)( T (x)) I ( D ; ti)(x ) dx ; i j = I ( D ; rq)(x) dx ; = drq = Ws r · i
10.24 Theorem Suppose w is a kform in an open set E c Rn, is a ksurface in E, with parameter don1ain D c Rk , and � is the ksurface in Rk , with parameter domain D, defined by �(u) = u (u E D). Then
f
cz,
W
=
J
Wcz, .
11
Proof We need only consider the case w=
a{x) dx .
II
"
···
"
dx .
II
(u))J(u) du t a((u))J(u) du
Let [A ] be the
D
k
by
k
1\
1
duk =
• • • 1\
tnat rix with entries
t
w"' .
(p, q = 1 , . . . , k).
Then d¢ ;p =
I a( p , q) duq q
so that d¢ ; 1
A · · · A
d¢ ;k = L a(I, q1 )
• • •
a (k, qk ) duq 1
A · · · A
duq k .
I n this last sum , q 1 , , qk range i ndependently over I , . . . , k . The anti com mutative relat i on ( 42) i mplies that •
du q 1
•
•
A · · · A
du qk = s (q1 ,
•
•
•
, qk) du 1
A · • · A
duk ,
where s is as i n Defi nition 9 . 3 3 ; applying this defi nition, we see that d¢ ; 1
A · • • A
d¢ ;k = det (A ] du 1
and since J(u) = det [A ], (72) is proved.
A · · · A
duk ;
The final resu l t of this secti on combines the two preced i ng theorems. 10.25 Theorem Suppose T is a ct'n1apping of an open set E c R" into an open set V c Rm, is a ksur.face in £, an d ltJ is a kjorn1 in V. Then
266
PRINCIPLES OF MATHE MATICAL ANALYSIS
Proof Le t D be the para meter domai n of define � as i n Theorem I 0.24.
�
(hence also of T or
r,
(88) The boundary o of the oriented ksimplex = T c a is defi ned to be t he (k  I ) chain o = T(iJu).
(89)
I n j ustification of (89), observe that if T is affine, then = 1· u i s an oriented affine ksi mplex. in which case ( 89) is not a matter of defini tion , but is seen to be a consequence of (85). Thus (89) ge n e ral i ze s this spec i a l c a se . It is in1med iate that o i� of class fC" i f this is true of . Finally, we define the boundary o'Y of the kchain \f' = 1: ; to be the (k  1 ) chain �.
(90)
o'Y =
I c ; .
10.31 Positively oriented boundaries So far we have associated bo unda ries to chains, not to subsets of R". T h is notion of bou n d a ry is ex act l y the one t h a t is most suitable for the statement and proof of Stokes' theorem. H owe\: er, i n applications, especially i n R 2 o r R 3 , i t i s customary and convenient t o talk about "oriented boun<Ja ries" of certain sets as well . We shal l now describe this briefly. Let Q" be the standa rd simplex i n R" , let u0 be the identity mappi ng with domain Q" . As we saw in Sec. 1 0. 26, u0 may be regarded as a positively oriented nsimplex in R". I ts boundary cu0 is an affi ne (n  1 )chai n. This c h a in is called the positively oriented boundary of the set Q " . For example, the positively oriented boundary of Q 3 is
[e 1 , e 2 , e 3 ]  [0 , e 2 , e 3 ] + [0, e 1 , e 3 ] [0, e 1 , e 2 ] . N ow let T be a 1  1 mapping of Q" into R", of class 0, and define (I l l)
Then y we have
y( t )
= (r cos t,
sin t )
(0 < t < 2 rr) . is a curve (an H oriented 1 si mplex " ) i n R 2  {0} . Si nce y(O) r
iJ y
( I I 2)
=
=
y(2rr),
0.
D i rect computation sho\vs that
J
( 1 1 3)
'I
17
=
2rr
#
0.
The d iscussi on i n Remarks 1 0. 3 5( b ) and (c) shows that we can draw two concl usi ons from ( 1 1 3) :
First, '1 is not exact in R 2  {0 } , for otherwise ( 1 1 2 ) would force the integral ( 1 1 3) to be 0. Second ly, y is not the boundary of any 2chain in R2 {0} (of class (6 ) , for ot herwi se the fact that 11 i s closed wou ld force t he i ntegral ( 1 1 3) to be 0. 
10.37 ( 1 1 4)
Example
Let E
=
"
R 3  {0} 3space with the origin re moved . Defi ne
x dy A dz + y dz A dx + z dx "' � = (x2 + y 2 + z 2 ) 3 / 2 ,
1\
dy
where we have written (x, y, z) i n place of (x 1 , x2 , x3 ). Di fferentiation shows that d( = 0, so that ( is a cl osed 2form i n R 3  {0} . Let I: be the 2chai n i n R 3  {0} that was constructed i n Exa mple I 0.32; recall that I: is a paramet rization of the unit sphere i n R 3 . Using the rectangle D of Example 1 0. 32 as pa ra meter domain, it is easy to com pute that ( 1 1 5)
J ( = J sin u du dv I
D
=
4rr
#
0.
A s in the preceding ex::tmple, we can now concl ude that ( is not exact in RJ  {0 } (since o i: = 0, as was shown i n Example 1 0. 32) and that the sphere I: i s not the boundary of any 3chain i n R 3  {0} (of class �"), although oi: = 0. 'fhe fol lowing result wil l be used in the proof of Theorem 1 0. 3 9 .
178
PRINCIPLES OF MATHEMATICAL ANALYSIS
10.38 1 '(p0), i n accordance with Defi nit ion 9. 1 1 . Let us n ow assume that the rank of T is 2. ( I f it is 1 or 0, then N = 0, a nd the ta n gent plane menti oned bel ow degenerates t o a l i ne or to a point .) The range of the affi ne ma pping
(u, r) � (p0) + T(u, v)
is then a plane n , cal led the tangent plane t o a t p0 . [One would like t o cal l n the tangent plane at (p0), rat her tha n at p0 ; i f is n ot onetoone, this runs into difficulties . ] I f w e use ( 1 33) i n ( 1 29), w e obtai n
( I 3 5)
N
=
( et. 1 fJ 3  a 3 fJ 1 )e 1
+ (a 3 P 1  a
1
fJ 3 )e 2 + ( (1. t fi 2  a 1 fi 1 )e 3 ,
and ( 1 34) shows that ( 1 36)
Te 1
=
' L 3
i=
1
Te 1
. a I. e ,
=
3
L /J; e; . i= J
A st r �ightforward com putat ion n ow leads to ( 1 37)
lIenee N i s perpendicular to n . I t is therefore ca lled the nor1na/ to at P o . A second property of N , also verified by a di rect computat ion ba sed o n ( 1 35) and ( 1 36). i s that the determinant of the linear transformation o f R 3 t h a t takes {e 1 , e 2 , e 3) t o { Te1 , Te 1 , N 1 is I N 1 2 > 0 ( Exercise 30). The 3si m plex
[0 , Te 1 , Te 1 , N]
( 1 38)
is thus posit ice/y oriented. The third property of N that we shall use is a conseq uence o f t h e fi rst t w o : The above mentioned determinant , whose va l ue is I N 1 2 , i s the volutne of the pa ra llelepiped with ed ges [0, T� 1 ], [0, Te 2 ], [0, N ] . By ( 1 37), [0, N] is pe rpen dicular to the other two edges . The area o.l the paral/elogran1 with L�ertices ( I 39)
is there,{ore I N 1 .
This pa rallelogram i s the i mage u nder T of the unit square i n R 1 • I f E 1 i s any recta ngle i n R , i t fol l ows (by the linearity of T) that the a rea o f the para l lelogra m T(E) is ( 1 40)
A ( T(E) )
=
I N I A ( £)
=
J I N( 0 , v0 ) ! du dv. u
E
INTEGRATION OF DIFFERENTIAL FORMS
285
We conclude that (1 32) i s correct when in each rectangle by the cor res p onding tangent plane. The sum of the areas of the resulting parallelogram s , obtained via (140), is then an approximation to A(Cl>). Fi nally , one can j usti fy ( 1 3 1 ) from ( 1 32) by approxi mating f by step functions. 10.47
Example
Let 0 < a
0 when t = a. For example, take u = v = n/2, t = a. This gives the la rgest val ue o f z on 'I'(K), and N = a(b + a)e 3 poi nts " ' upward " ' for this choice of (u, v).
3 10.48 Integrals of 1forms in R 3 Let �'/ be a r6''curve in an open set E c R , with pa rameter i nterval [0, I ], let F be a vector field i n £, as in Sec . 1 0. 42. and defi ne ;. F by ( 1 24). The i ntegral o f .A F over y can be rewritten in a ce rt a i n \Nay which we now descri he. For any u E [0, 1 ], I
y (u)
=
y ; ( u)e 1 +
}'; ( u )e 2 + }' � ( u )e 3
is called the tangent z.:cctor to }' at u . We defi ne t the direction of y1(u) . Th us
y (ll )
=
I
[I f yl(u) By (35),
( 1 42)
=
0 for some u. put t(u)
J/ F
= =
=
t(u) to be the unit vector
I
I y ( u) I t( u).
e 1 ; any other choice would do ju�t
=
in
a�
\Vei l . ]
i�t fo Fi ( i• (u) ) y i (u) du 1
3
J F( }'(u)) · y '(u) du 1 f F( }'(u)) · t(u) I }' '(u) I du. 1
0
=
• 0
Theorem 6.27 makes it reasonable to cal l I y1(u) I du the e/en1ent t�( arc length along }'· A customa ry n otation for it is ds, and ( 1 42 ) is rewritten in the form ( 1 43)
J AF J (F =
y
•
t) ds.
y
Since t is a unit tangent vector to y, F t is cal led the tangential con1ponent of F a long y. ·
INTEGRATION OF DIFFERENTIAL FORMS
'1.87
The right side of ( 1 43) should be regarded as just an abbreviation for the last integral in ( 1 42). The point is that F is defined on the range of y, but t is defined on [0, 1 ] ; thus F t has to be properly interpreted. Of course, when y is onetoone, then t(u) can be replaced by t(y(u)), and this difficulty disappears. •
3 Let (y, x) of R 2 onto R2 is not the con1pos it ion of any two prilnit ive n1appi ngs, in any neighbo rhood of the origi n . (Th is shows that the fl i ps B, cannot be on1it ted from the statemen t of Theorc1n 1 0. 7 . ) For (x, y) e R 2 , defi ne F(x, y) = (ex cos y  1 , ex sin y).
Prove that F
==
G2
o
G � , w here
G t (X, y) = ( ex G z(u, v)
COS y 
= (u, ( I
I , y)
t u) tan v)
are pri rn i t i ve in sorne nei g h borhood of (0, 0).
Cornpute the Jacobians of G � , G z , F a t (0, 0). Defi ne H2(x, y)
and find
=
(x,
ex sin y)
H , (u, t•) = (h ( u, v), v)
so that F = H . o H 2 is some neighborhood of (0, 0). S. For rn ul a te and prove an analogue of Theoren1 I 0.8, in wh ich K is a com pa ct subset of an arbi trary n1et ric space. ( Replace the funct ions rp; that occur in the proof of Thc ore rn I 0.8 by fu nct ions of t he type constructed i n Exercise 22 of Chap. 4 . ) 6 . S treng t hen the concl usion of Theorern 1 0.8 by s h o w in g t hat the fu nct ions � ; can be made d i fferent i able , and even infinitely differentiable. ( Use Exercise I of Chap. 8 in the const ruct ion of t h e au x i l i a ry funct i ons cp; . ) 7. (a ) Show that the simplex Q" is the sn1al lest convex subset of R k t h at contains 0,
e• ·
. . . , e, . (b) Show that a ffi ne n1 a ppi ngs ta k e convex set s t o con , ·c x s e t s . 8 . Let H be the parallelogram in R 2 w·hose vert ices are ( 1 , 1 ), ( 3 , 2 ), (4, 5 ), (2, 4). Find the affine n1ap T wh ich sends (0, 0) t o ( I , I ), ( I , 0) to ( 3, 2). ( 0, I ) to ( 2, 4). Show that J r
=
5 . Use
T to convert the integral
to an in tegral over 1 2 and thus compute
oc .
290
PRINCIPLES OF MATHEMATICAL ANALYSIS
9. Define (x, y)
= T(r, 8) on the rectangle 0 < r < a,
by the equat ions
x = r cos 8 ,
y = r sin 8 .
Show that T maps this rectangle onto the closed disc D with center a t (0, 0) and radius a , that T is onetoone in the interior of the rectangle, and that J T( r , 0) r. I f I e (C( D), prove the formu l a for i ntegration in polar coord inates : ==
t
/(x, y) dx dy
= I: r·
f< T(r, 8 ))r dr d8 .
Hint: Let Do be the interior of D, m i nus the interval from (0, 0) to (0, a).
As it stands, Theorem 1 0.9 appl ies to continuous funct ions I whose support l ies in Do . To remove this rest rict ion, proceed as in Exarnple I 0.4.
10. Let
a > oo in Exercise
I
R2
9 and prove that
f(x, y) dx dy =
r r· 0
f< T(r, 8 ) )r dr dB,
0
for con tinuous funct ions I that decrease suffil:iently rapidly as I x I � I y I '). x . (find a more precise fonnulation . ) A pply this to
l(x, y)
==
exp (

x
2

.v 2 )
to derive formula ( 1 0 1 ) of Chap . 8.
1 1 . Define (u, v) = T(s, f) on the strip 0 <s
0. Let ( x, y, z) define a 2surface , with parameter domain / 2 , by
=
Cl>(s, t )
x = g(t )h . (s),
Prove that
I ' = o. oD
directly from (35). Note the shape of t he range of : For fi x e d s , (s, t ) runs over a n i n t e r v a l on a line through 0. The range of ci> t h u s lies i n a co ne with vertex at the origi n . "
"'
(d) Let E be a closed rectangle in D, \V i t h edges paral lel t o those of D. Su ppose f E is a set fu nct ion defined on 91 if 4> assigns to every A e .JII a. n u m ber l/>(A ) of the ex tended real n umber system. 4> is additive if A n B = 0 impl ies l/>(A u B) = l/>(A ) + l/>(B),
(3)
and 4> is coun t ably additire if A ;
n
A i = 0 (i :/: }) im pl ies
(4) We shall al ways assume that the ra nge of 4> does not conta i n both + oo a nd  oo ; for if it did, the right side of (3) could become meaningless. Also, we exclude set fu nct ions whose only value is + oo or  oo . I t is i nterest ing to note t hat t he left side of ( 4) i s independent of the order i n wh ich t he An , s are arra nged . Hence the rearrangement theorem shows that the right side of (4) converges a bsol utely if it converges at all ; if it does not converge , t he pa rt ial sums tend to + oo , or to  oo . If 4> is additive, t he fol lowing propert ies are easily verified : ¢(0) = 0.
(5 )
¢(A 1 u · · · u A n) = l/>(A 1 ) + · · · + ¢( A ")
(6) if A i
n
A i = 0 whenever i :/: }.
302
PRINCI PLES OF MATH EMATICAL ANALYSIS
cf> (A 1
(7)
u
A 2 ) + c/> (A 1
If cf> (A ) > 0 for a l l A , and A 1
c
n
A 2 ) = l/> (A 1 ) + (A 2 ) .
A 2 , then
c/> (A 1 ) < c/> (A 2 ) .
(8)
B e cause of (8) , non negat i ve ad d i t ive set fu nct i ons a re often ca l1cd m onoton ic.
( A  B) = l/J(A )  cf>( B)
(9) if B e A , and j ( ¢B ) I < +
oo .
1 1 .3 Theorem Suppose cf> is countably addit ire on a ring (n = 1 , 2 , 3 , . . . ), A 1 c A 2 c A 3 c · · · , A E :f/1, and
Jf .
Suppose A n
E
:Jf
00
A = U A,. . 1
n=
Then, as n + Cf.J ,
cf> ( A,. ) + cf> (A ) . (n =
Then B;
n
u · · · u
Bj = 0 fo r i =I= }, A ,. = B1
2 , 3, . . . ) .
B,. , and A = U B,. . He nce
n
cf> (A ,. ) = L cf> ( B ; ) j
=
1
and 00
cf>( A ) = L ( B ; ). 1 i
=
C O N ST R U CTI O N O F T H E L E BE S G U E M E A SU R E 1 1 .4 Definition Let RP denote pd i mens ional eucl idea n space . By a n interral i n R P we mea n the set of points x (x 1 , , xp) such that =
( 1 0)
Q I·
< X· 
< b I.
I 
•
.
.
(i = 1 ' . . . ' p),
or the set of po i nts wh ich is characterized by ( I 0) w i th any or a l l of the < signs replaced by < . The possibi l i ty that a ; = h i for a ny val ue of i i s not ru led out ; i n part icular , t he empty set i s i ncluded a mong the i ntervals.
THE LEBESG UE THEORY
303
If A is t he un ion of a fi n ite number of i ntervals, A is said to be an e/emen la r r set. If I i s a n interval , we defi ne "
nz(J) = n (b i  a j) , p
i
=
1
no matter whether eq ual i ty is incl uded or excluded i n any of the i nequalities ( 1 0). If A = /1 u · · · u In , and if these intervals are pai rwise disjoint, we set (I I )
nz(A ) = rn(/1 ) +
We
· · ·
+ nz(ln ).
let 6 denote the fa rnily of all eletnentary subsets of RP. At t his point, t he following properties should be verified :
( 1 2 ) o is a ring. but not a erring. ( 1 3 ) If A E 6' , t hen A is t he un ion of a finite nu mber of disjoint interval s. ( 1 4) If A e o�, tn(A ) is well defined by ( I I ) ; that i s . if two different decomposit ions of A into disjoint intervals are used, t!ach gives rise to the same val ue of nz(A ). ( 1 5 ) nz is addit ive on o. N ote that if p = I , 2. 3 , t hen
nz
i � length, area , and vol ume, respectively.
A nonnegat ive addit ive set fu nct ion 4> defi ned on G is sa id to be regular if the follo\ving is true : To every A E & and to eve ry e > 0 there cxi�t se ts F e 6 . C E 6 s uch t hat F i s closed , G is open , F c A c G, and 1 1 .5
D efinition
¢( G)  e < c/> ( A) < cf>( F ) +
( 1 6) 1 1 .6
c.
Examples
(a)
The set function nz is regular. I f A is an interval, it is trivial that t he requi rements of Definition
1 1 . 5 are sat isfied . The general case follows from ( 1 3). (b) Take RP = R 1 • and let rx be a monoton ical ly increasing func t ion , defined fo r a l l real x. Put Jl( [a, h)) = cx(b  )

�(a  ),
cx(b + )  ex (a  ), Jl((a, b]) = cx(b + )  �(a + ), Jl((a, b)) = cx ( b  )  ct(a + ).
Jl ([a , b] ) =
1Iere [a, h) is the set a < x < b, etc. Because of t he possible discon tinuities of ct, these cases have to be dist i nguished . If Jl is defi ned for
304
PRINCIPLES OF MATHEMATICAL ANALYSIS
elementary sets as in ( 1 1 ), of (a).
J1.
i s regular on
�.
The proof is just l i ke that
Our next objective is to show that every regular set fu nction on 6 can be extended to a countably additive set function on a aring w hich contai ns d. 11.7 Definition Let J1. be additive, regular, nonnegat ive, and finite on & . Consider countable coveri ngs of any set E c RP by open elementary sets A n :
Define JJ,*(£)
( 1 7)
=
00
inf L Jl(A n ), n= 1
the i nf being taken over all cou ntable coveri ngs of E by open elementary sets. p,*(E) is cal led the outer measure of £, corresponding to Jl. It is clear that Jl *(E ) > 0 for all E and that ( 1 8)
1 1.8 Theorem (a)
For every A
(b) If E = U 00
E
e ' J1 *(A )
=
J1(A ).
En , then
1
J1 * (E ) < L Jl*(E"). n= l oc
( 1 9)
Note that (a) asserts that 11* is an extension of J1 from all subsets of R P . The property ( 1 9) is called subadditivity. Proof Choose A
e iff
o
to the fa m i ly of
and e > 0.
The regulari ty of J1 shows that A is contai ned in an open elementary set G such that J1(G) < J1 ( A ) + e . Si nce J1 *(A) < Jl(G) and si nce e was arbitrary, we have p*(A ) < p( A ) .
(20)
The definition of p* shows that there is a sequence {A " } of open elementary se ts whose union contains A , such that
L p(An) < p * ( A ) + 00
n= l
e.
THE LE BESG UE THEORY
The regu l a rity of Jl shows t hat that Jl(F)
> Jl ( A)
A
 c ; and s i nce F is compac t , we have
Fc
N.
for some Jl( A )
co ntai n s a closed e l e me n t a ry set
AJ
u
. . . u
u ···
u
F such
AN
He nce
< Jl ( F) + c; < Jl(A 1
305
N
A , J + c; < L Jl ( An ) + e < Jl*(A ) + 2e. 1
(20), t h i s proves (a). Next, suppose E = U En . and assu me that Jl * (E") G i ve n r, > 0, t h e re a re coveri ngs {Ank) , k = I . 2. 3, . . . I n conj u nct ion w i t h
< +
oo
fo r all n .
, of 1:..n by open
elementa ry se t s such that
(2 1 )
L1 Jl(A n k) < Jl * (£,.) + 00
k
=
2  " f. .
Then Jl * ( /:..' ) < a nd
( 1 9) 1 1 .9
( 1 9)
fo l l ov.' s .
L L Jl ( A ni J < L 00
00
00
rr = l k = l
rr = t
J l * ( f:�" ) +
f. ,
I n t he excl u ded case, i . e . . i f JL * ( £,.) =
+ oo for some n,
is of cou rse t r i v i a l .
Definition
RP,
B
S( A . B ) =
(A
d(A .
Jl * ( S(A .
For a ny A
(22)
(23)
c
W e �· ri te A n __.. A i f
B) =
c 
RP,
\\'e defi ne
B) u ( 8 
A ).
B) ) .
I i In d( A . A n )
= 0.
I f t here i s a seq u e nce { A n1 of e l e rnen tary set s �uch t h a t A ,. + A . we say
t h a t A is jinite(l' Jl lneasurahle a n d wri te
A
E �.ll r(Jl ) .
I f A i s t h e u n i o n o f a cou n t a ble col lect i o n o f fi n i te l y Jl nl ea s u ra ble sets , w e s a y t hat A i s JlIHeasurahle a nd w r i t e A
E �.ll ( Jl ) .
S( A , t h a t d( A .
B ) i s t h e 50ca l l ed · · sym n1et r i c d i ffere n ce· · B ) i s esse n t i a l l y a d i stance fu nct ion .
of A a n d
B.
W e sha l l see
The fo l l owi ng t h eore n1 w i l l enable u s to obta i n t h e des i red e x te n s i o n of Jl .
1 1.10
Theorem
�.ll( Jl) is a aring, and Jl *
is countahly
additire
on
�lJl ( J L ) .
Be fore v.·e t u rn t o t h e proof o f t h i s t heoren1 . we deve l o p some of t he propert ies of S( A ,
B)
a nd d( A ,
B).
We h ave
306
PRiNCI PLES OF MATHEMATICA L A N A LYSIS
S( A . B) = S(B , A),
(24)
S(A , B)
(25) (26)
c
S( A 1 u A 2 , B1 \..) 82 ) S(A 1 n A 2 , B1 n B2 ) S(A 1  A 2 , B1  B2)
S(A , A ) = 0.
S( A , C ) u S( C, B) . c
S( A 1 , B1 ) u S(A 2 , B 2 ) .
(24) is clear, and (25) follo\\'S from (A  B)
c
(A

C)
u
( C  B),
(B  A )
c
u
(C  A )
( 8  C) .
The first formula of (26) is obtained from (A 1 u A 2 )  (B1 u B2 )
c
(A 1  B1 )
u
(A 2  B2 ).
Next , writing Ec for the com plement of E, v.'e have S(A 1 n A 2 , B 1 n B2) = S(A� u A 2 , Bf u B2) c
S( A � , B f)
u
S( A � . B2 ) = S( A 1 • B1 )
u
S( A 2 , B2 ) :
and the last formula of ( 26) is obtai ned if \\·e note that A 1  A 2 = A 1 n A2 .
By (23), ( 1 9), and ( 1 8 ) , these propert ies of S(A , B) im ply ( 27 )
(28 ) (29 )
d(A , B) = d( B, A ) .
d( A , A ) = 0,
d(A , B) < d( A , C) + d(C. B), d( A 1 u A 2 , B 1 u B 2) d( A 1 n A 2 • B 1 n B 2 ) d( A 1  A 2 , B 1  B 2 )
< d( A 1 , B 1 ) + d( A 2 B 2 ) . •
The relat ions (27) and (28) show t hat d(A , B ) sat isfies t he requirernent� of Defi n i t i on 2. 1 5, except t hat d( A , B) = 0 does not i m ply A = B. For instance, if Jl = n 1 . A is countable. and B is em pty. \Ve have d(A , B) = 1n*(A ) = 0 ;
to see this, cover the nth poi n t of A by an interval /" such that 111(/n ) < 2  "t: . But if we define two �et � A and B t o be eq uivalent , provided d(A , B) = 0.
we divide the subsets of RP i n to equi valence classes , and d(A , B) ma kes the set of these equ ivalence classes into a met ric space. �.ll1 (ll ) is then obt a i ned a � t h e closure of o' . This i n terpretat ion i " not essential for t he proof, but i t expl a i n s t he underlying idea . ·
,
T H E LEBESG UE TH EORY
307
We need one more property of d(A , B), name1y, ( 30)
I JL*(A)  Jl*(B) I < d(A , B),
For s u pp o se 0 < JL*(B) < JL*(A).
if at least one of Jt*(A), JL*(B) is fin i te. Then ( 2 8 ) shows that
d(A ,O ) < d( A , B) + d (B ,
0),
that is, Jt* (A ) < d( A ,
B) + JL * ( B) .
Since Jt * ( B) i s fin ite, it fol lows that Jt * ( A )  JL *( B) < d(A , B).
(3 1 )
Proof of Theorem 1 1.10 Suppose A e 9JlF(JL), B e 9JlF(JL). Choose {A n}, { Bn} such that An e o. Bn e o', A n + A , Bn + B. Then (29) and ( 30) s h o w t hat An u
n
B,: + A
u
B,
( 32)
An
(33 )
A n  Bn + A  B '
( 34)
JL *(A n) + Jt *(A ) ,
a nd Jt *(A ) < + By (7),
oo
Bn + A n B,
s i nce d(A n A ) + 0. By ( 3 I ) and (3 3 ) , 9Jl F(Jl) i s a ri ng. �
Jl(An ) + JL ( Bn ) = Jl( A n
Letting n +
oo ,
U
Bn) + Jl (An
fl
Bn} .
we obtai n . by ( 3 4 ) and Theorem I 1 . 8(a), Jl * ( A ) + JL * ( B) = Jl �: (A u B) + Jt * (A n B).
If A n B = 0. t h e n Jl *(A n B) = 0. I t follows that J l * is add i t i ve on �Jl,..(JL ). Now let A E �Jl(Jl) . Th en A ca n be represented a s t he un ion of a cou ntable col lect ion of disjoint sets of 9Jlr(JL). For if A = U A ;, wit h A ;, e �lJl,.(JL), write A 1 = A ; . and A n = (A ; u . . · u A ;, )  (A ;, u · . . u A ;, _ I )
Then ( 35 )
A =
U An 00
II =
I
is the required representat ion . By ( 1 9 ) (3 6 )
JL * (A )
A1
u
· · · u A n ; and by the add itivity of
11 *(A) > p*(A t u · · · u An ) = J1 * ( A t ) + · · · + J1 * (A n ) ·
(37)
Equations (36) and ( 3 7) i mply (38)
Jl*(A )
that
A
E
I Jt*(A n).
n= J
Suppose J1 *(A ) is finite. Put Bn = A 1 u · · · u A n . Then (38) shows d(A , Bn )
as
=
00
n + oo .
9Jl F(J1 ).
=
00
00
U A ;) = I Jl *(A )
J1 * (
;
i=n+ 1
i=n+ l
...!.+
0
Hence Bn + A ; and since Bn e 9JlF(J1), it i s ea sil y see n tha t
We have thus sh own that A e 9JlF(J1) if A e 9Jl(Jl ) and Jl *(A ) < + I t i s now clear tha t p* i s cou ntably add it ive o n �Jl(Jl). For if
oo .
where {A n} is a sequence of disj o i nt sets of 9Jl(Jl), we have shown that (38) holds i f 11*(An) < + oo fo r every n , and i n the other ca se (38) i s t'r i v i a l . Finally, we have to show that 9Jl(J1) is a aring. I f A n E 9Jl (J1 ) , n = I , 2, 3 , . . . , i t is clear t hat U A n e 9Jl(J1) (Theorem 2. 1 2). Su ppose A e 9Jl(Jl), B e 9R(J1), and
where A n , Bn e 9JlF(J1). Then the i de n tity An n B =
00
i
U (A n n = 1
B;)
shows that A n n B e 9Jl(J1) ; and si nce 11*(A n
A n n B E 9JlF(J1). Hence A  B = U:= 1 (A n  B).
n
B) < J1*(A n) < +
An  B
E
9JlF(J1),
oo ,
and
A  B e 9Jl(J1)
. s t n ce
We now replace Jl*(A ) by J1(A ) if A e 9Jl(JL). Th us J1, ori g i nally only de fined on C, i s extended to a cou ntably addit ive set fu nct ion on the ari ng !Dl(JL). Th i s extended s e t fu nct ion is ca l led a measure. The s pec i a l case J1 = m i s called the Lebesgue measure on R P .
THE
1 1 .1 1
LEBESG UE
TH EORY
309
Remarks
If A is open, then A e 9Jl(Jl) . For every open set i n RP is the uni on of a countable col lect ion of open i ntervals. To see this. i � is sufficient to construct a cou ntable base whose members are open i nterva ls. By taking complements, it follows that every cl osed set is in 9Jl(Jl) . (b) If A E �Jl(J1) and e > 0, there exist sets F and G such that (a)
Fe A
c
G,
F is closed, G is open , and
(3 9)
J1 ( G  A )
0, {sn} 1 1.20
may be chosen to be a monotonically increasing sequence. Proof If/ > 0, define
En i
=
{
X
i
I
i
< f(x) < 2" n 2

}
'
Fn
=
{x l f(x) > n}
314
PRINCIPLES OF MATHEMATICAL ANALYSIS
for n
=
I,
2, 3 ,
(50)
. . . , i = 1 , 2, . . . , n2n . Put n 2" i  I Sn =
i
�
2n
l
KEni + n KFn .
In the general case, let f = j+  f , and apply the preceding construction to f + and to f  . It may be noted that the sequence {sn} given by ( 50) converges uniformly to f iff is bounded .
INTEGRATI O N We shall defi ne integration on a measurable space X, i n which 9Jl i s the aring of measurable sets, and Jl is the measure. The reader who wishes to visual ize a more concrete situation may think of X as the real l i ne, or an i nterval , and of Jl as the Lebesgue measure m.
1 1.21
Definition
(5 1 )
Suppose s(x) =
L c i KE; (x) n
i
=
(x e X,
1
ci
> 0)
i s measurable, and suppose E e 9Jl . We defi ne n
(52)
JE(s) =
i
L =
1
c;
p(E n E;).
Iff is measurable and nonnegative, we define
JE f dJL = sup I,ls),
(53)
where the sup is taken over all measurable simple funct ions s such that 0 < s < f The left member of (53) i s called the Lebesgue integral of f, with respect to the measure Jl, over the set E. It should be noted that the integral may have the value + oo . It is easi ly verified that ( 5 4) for every non negative simple measurable function
1 1.22
Definition
s.
Let / be measurable, and consider the two integrals
(55) where / + and /  are defined as i n (47).
THE LEBESG UE THEORY
315
If at least one of the i ntegrals (55) is finite, we define (56) If both i ntegrals in (55) a re finite, then (56) is finite, and we say that f is integrable (or summable) on E in the Lebesgue sense, with respect to p ; we write f e !l'(p) on E. If Jl = m, the usual notation is : f e !l' on E. Th is term i nology may be a little confusing : If (56) is + oo or oo , then the i ntegral of 1· over E is defined, although f is not integrable in the above sense of the word ; .f is integrable on E only if its integral over E is finite. We shall be main ly i nterested in i ntegrable functions, although in some cases it is desi rable to deal with the more general situ a tion . 
1 1 .23
Remarks
The fol lowi ng properties are evident :
(a) If f is measurable and bounded on E, and if p(E) < + oo , then f e !l'(p) on E. ( b) If a < l(x) < b for x e £, and p(E) < + oo , then
OJL(E)
(A )
n). Then 9n is measurable on E, and 0 < g 1 (x) < g2 (x)
0, Fatou's theorem sho\vs t hat
I (f E
t
g) dJt < lim i nf n  oo
E
I fc/11 < limn  ooinf I f. d11. E
:t}(Jl) and 1· e !t'(Jl)
I (.f. + g) dJt .
or (85)
e
E
322
PRINCIPLES OF MATHEMATICAL ANALYSIS
Si nce g  In > 0, we see similarly that
so that
I (g  f) dp, < lim inf J (g
which is the same as (86)
n  oo
E
E
 J . f dJ1 < lim inf [  J n  oo
E
E

fn) dp, ,
]
J, dJ1 ,
J f dJ1 > lim sup J f dJ1. n  oo
E
E
The existence of the limit i n (84) and the equal ity asserted by (84) now fo1 1ow from (85) and (86). Corollary If p,(E) < + oo , {/n} is uniformly bounded on then (84) holds .
E,
andfn( x)
+
f(x ) on E,
A u niformly bounded convergent seq uence is often sa id to be boundedly
convergent .
COMPARI S O N WITH THE RIEMANN INTEGRAL
Our next theorem will show that every function which is Rieman nintegrable on an interval is also Lebesgueintegrable, and that Riemanni ntegrable fu nc tions are subject to rather stringent continuity conditions. Quite apart from the fact that the Lebesgue theory therefore enables us to i ntegrate a much la rger class of fu nctions, its greatest advantage lies perhaps in the ease with which many li mit operations can be handled ; from th is point of view, Lebesgue ' s convergence theorems may well be regarded as the core of the Lebesgue theory. One of the difficulties which is encountered i n the Riemann theory is that limits of Riemannintegrable functions (or even conti nuous functions) may fail to be Rieman nintegrable. This difficulty is now al most eliminated , si nce l i mits of measurable functions a re always measu rable. Let the measure space X be the interval [a, b ] of the real l i ne, with p, = m (the Lebesgue measure), and 9Jl the family of Lebesguemeasu rable subsets of [a, b ]. Instead of
I f dm X
it is customary to use the familiar notation
I fdx b
G
THE LEBESGUE THEORY
323
for the Lebesgue integral of f over [a, b]. To distinguish Riemann i ntegrals from Lebesgue integrals, we shall now denote the former .by 9t
J f dx. b
a
1 1.33 Theorem
(a) Iff e 9l on [a, b ], then f e !£ 011 [a, b ], and
J f dx = J f dx. 9t
b
(87)
a
b
a
(b) Suppose f is bounded on [a, b ]. Then f e 9l on [a, b] if and only iff is continuous almost everywhere on [a , b ]. Proof Suppose f i s bounded . By Definition 6. 1 and Theorem 6.4 there is a sequence {Pk} of partitions of [a, b], such that Pk + l is a refinement of Pk , such that the distance between adjacent points of Pk is less than 1 /k, and such that
(88 )
lim L(Pk , f ) =
k +
00
9t
J fdx,
lim U(Pk , f)
k +

=
00
9t
J fdx. 
(In this proof, all i ntegral s are taken over [a, b ]. ) If pk = {xo ' xb . . . ' xn}, with Xo = a, Xn = b, defi ne
Uk(a)
=
L1c,(a) = f(a ) ;
put Uk(x) = M ; and Lk(x) = mi for X;  1 < x < X ; , I < notation introduced in Definition 6. 1 . Then
i < n , using the
J
(89)
U(Pk , f) = Uk dx, and
(90) for all x
[a, b ], since Pk
+
1
refines Pk
L(x) = lim Lk(x) ,
(9 1 )
k+
that (92)
e
00
•
By (90), there exist
U(x) = lim Uk(x). k + oo
Observe that L and U are bounded measurable functions on L(x) < f(x) < U(x)
(a < x < b),
[a, b ],
3Z4
PRINCIPLES OF MATHEMATICAL ANALYSIS
and that
JL dx
(93)
=
9f
fldx, Ju dx = :Jl fldx,
by (88), (90), and the monotone convergence theorem . So far, noth ing has been assu med about / except that / i s a bounded real funct iCJn on [a, b ]. To complete the proof, note· that . f e 31 if and only if its upper and lower Riemann integrals are equal , hence if and only if
JL dx = Ju dx;
(94)
x
since L < U, (94) happens if and only if e [a, b] (Exercise I ) . In that case, (92) implies that
L(x) U(x) for al most · all =
L(x) f(x) U(x) =
=
(95)
f
almost everywhere on [a, b] , so that is measurable, and (87) fol lows from (9 3 ) and (95). Furthermore, if belo ngs to no Pk . it is qu ite easy to see t hat = ( ) if and only if/is continuous at Si nce the union of the sets Pk is cou nt able, its measure is 0, and we concl ude that i s cont inuous almost every where on [a, b] if and only if = U(x) al most everywhere, hence (as we saw a bove) if and only iff e PA. This completes the proof.
x
Lx
x. L(x)
U(x)
f
The fam i liar connection between integration and differentiation is to a large degree carried over into the Lebesgue theory. Jf f e .!£ on [a, b] . and
x J fdt
F( )
(96)
F'(x)
=
X
(a
0, there is a fu nction g, continuous on [a, b ], such that II!  gi l
=
{ ( (f  g)2 dxr12
< e.
Proof We shal l say that .f i s approxi mated i n 22 by a sequence { g" } i f 0 as n + 00 . ll f'  9n 11 Let A be a closed su bset of [a, b ], and KA i ts characteristic funct ion . Put +
t(x)
=
( y E A)
inf l x  Y l
and g" (x )
=
1
(n
I + nt(x)
Then Bn is continuous on [a, b ], 9n(x) where B = [a, b]  A . Hence ll g" 
KA II
=
=
=
I , 2, 3 , . . . ) .
1 on A, and g"(x) + 0 on B,
{ fsu: }" dx
2
+ 0
by Theorem 1 1 . 32. Thus characteristic fu nctions of closed sets ca n be approxi ma ted i n 22 by continuous functions. By (39) the same is true for the characteristic fu nct ion of any measurable set , and hence also for si mple measurable fu nct ions. If I > 0 and f e 2 2 , let {s"} be a monotonically i ncreasi ng sequence of simple nonnegative measu rable funct ions such that s"(x) +l(x). Si nce 1/  sn l 2 < / 2 , Theorem 1 1 .32 shows that l lf'  sn ll + 0 The general case follows. .
Definition We say that a sequence of complex functions { ¢n} is an orthonormal set of functions on a measu rable space X i f 1 1 .39
(n (n
:/=
m), = m).
I n pa rt icular. we m ust have n E 2 2 (J1). I f I E 22(/1 ) and if
( we write
as i n Defi n ition 8 . 1 0.
n
=
I , 2, 3, . . . ) ,
328
PRINCIPLES OF MATHEMATICAL ANALYSIS
The definition of a trigonometric Fourier series is extended i n the same way to !£2 (or even to !l') on [  n , n]. Theorems 8 . 1 1 and 8. 1 2 (the Bessel inequal ity) hold for any f e !l' 2 (p). The proofs are the same, word for word. We can now prove the Parseval theorem .
Suppose
1 1.40 Theorem
f(x)
(99)
L c,ein x ' 00
"'

oo
where f e !l' 2 on [  n, n]. Let s, be the nth partial sum of ( 99). Then l i m II !  s, ll = 0, ( 1 00) , ..... 00
f
1 (101) / 1 2 dx . 1 2n Proof Let e > 0 be given . By Theorem 1 1 . 38 , there is a continuo us function g such that 00
I I c, 1 2 =
 00

•
n
e 2
II !  g i l < 
·
Moreover, it is easy to see that we can arra nge it so that g(n) = g(  n). Then g can be extended to a periodic continuous fu nc tion. By T h e o rem 8. 1 6, there is a trigonometric polynomial T, of degree N, say, such that e 11 9  T i l <  · 2
Hence, by Theorem 8. 1 1 (e x tended to !l' 2 ),
n > N
implies
li s,  !I I < I I T /1 1 < e , and ( 1 00) follows. Equation ( 1 0 1 ) is deduced from ( 1 00) as in the proof of Theorem 8 . 1 6. Corollary
1/.f e !l'2
J
.n
then llfll = 0 .
on [  n, n ], and if f(x)e  i n x dx = 0
(n = 0, + I , + 2, . . . ) ,
 n
Th us if two functions i n !l'2 have the same Fou rier series, they differ at mos t on a set of measure zero .
THE LEBESGUE THEORY
329
11.41 Definition Let I and fn e !l' 2 (p) (n = 1 , 2, 3, . . . ) We say that {f,.} converges to I in !l'2 (p) if l lf,.  Il l + 0. We say that {f,.} is a Cauchy sequence in !l' 2 (p) if for every 8 > 0 there is an integer N such that n � N, m � N implies .
II J,. lm II � 8 .
Theorem If {f,.} is a Cauchy sequence in !l' 2 (p), .function f e !l' 2 (p) such that {f,.} converges to .f in !l' 2 (p) .
11.42
This says, in other words, that !l'2(J.J.) is a
then there exists a
complete metric space.
Proof Since {/,} is a Cauchy sequence, we can find a sequence {nk} , k = 1 , 2, 3 , . . . , such that
(k Choose a function
=
1 ' 2 , 3, . . . ) .
g e !l'2(J.J.) . By the Schwarz inequality,
Hence
J)x i o(J...  !... + , ) 1 dp < ll u l l . 00
( 1 02)
By Theorem 1 1 . 30, we may interchange the summation and integration in (1 02) It follows that .
( 103)
00
l g(x) l L lfnk(x)  f,k + l (x) l < + k= l
oo
almost everywhere on X. Therefore ( 1 04)
00
link + 1 (x)  .f,. k(x) I < + L k= l
oo
almost everywhere on X. For if the series in ( 1 04) were divergent on a set E of positive measure, we could take g(x) to be nonzero on a subset of E of positive measure, thus obtaining a contradiction to (1 03). Since the kth pa rtial sum of the series 00
L (f,.k + 1 (x)  f,.k(x)) ,
k= l
which converges almost everywhere on X, is
Ink+ 1 (x) /,.l (x),
330
PRINCIPLES OF MATHEMATICAL ANALYSIS
we see that the equation
f(x) = lim /,k(x) k  oo
defines f(x) for almost all x e X, and it does not matter how we define f(x) at the remaining points of X. We shall now show that this function f has the desired properties. Let e > 0 bt given, and choose N as indicated in Definition 1 1 .4 1 . If nk > N, Fatou ' s theorem shows that
I I /  f,.k ll < li m in f 1 1 /,i  /,k ll < e . i .... oo
Thus f  /,k E 2 2 (JL), and si nce f = (f  Ink) + /,k , we see that f E 22(J1). Also, si nce e is arbitrary, l i m 1 1/  /,.k ll = 0. k .... oo Fi nally, the inequal ity ( 1 05) shows that {/,} converges to f in 2 2 (JL) ; for if we take n and n k large enough, each of the two terms on the right of ( I 05) can be made arbi trari ly small. The RieszFischer theorem Let { c/>" } be orthonormal on X. Suppose l: I c, 1 2 converges, and put s, = c1 c/>1 + · · · + c,c/>n . Then there exists a function 1 1.43
f e !£ 2 (JL) such that {s,} converges to f in 2 2 (JL) , and such that
Proof For n > m,
li s,  sm l1 2 = l cm + 1 1 2 + . . . + l c, l 2 , so that {s,} is a Cauchy sequence in 2 2 (JL) . By Theorem 1 1 .42, there is a function f e .f£ 2 (JL) such that lim I I/  s, II = o . Now, for n > k,
, ..... <X>
THE LEBESG UE THEORY
331
so that
Letting n
.
oo ,
we see that ( k = 1 ' 2 , 3, . . . )
,
and the proof is complete. 1 1 .44 Definition An orthonormal set f e !t' 2 (J1), the equations
{ c/>n } is said to be complete if, for (n = 1 , 2,
3,
. .
.)
i mply th at 11! 11 = 0. In the Coro l lary to Theore m 1 1 .40 we deduced the completeness of the trigonometric system from the Parsevat equation ( 1 0 1 ). Conversely, the Parseval equation holds for every complete orthonormal set : 1 1 .45
Theorem
Let {c/>n } be a complete orthonornzal set. Iff E !t' 2 {J1) and if
( 1 06)
then ( 1 07)
r I en 1 2 converges . Putting C1 c/> 1 + . + Cn c/> n '
Proof By the Bessel i nequality,
Sn
=
. .
the RieszFischer theorem shows that there is a function g that ( 1 08) and such that ll g  sn ll __. 0. Hence ll sn I I __. ll u ll . Since 2 II s" 11 2 = I c 1 1 + · · · + I c" 1 2 , we have ( 1 09)
e
!t' 2 (J1) such
332
PRINCIPLES OF MATHEMATICAL ANALYSIS
Now ( 1 06) , ( 1 08), and the completeness of {n} show that II /  g i l so that ( 1 09) i mpl ies ( 1 07).
=
0,
Combin ing Theorems 1 1 .43 and 1 1 .45, we arrive at the very interesting conclusion that every complete orthonormal set induces a 1  1 correspondence between the fu nctions f e 22(J1) (identifying those which are equal al most everywhere) on the one hand and the sequences {en} for which 1: I en 1 2 converges , on the other. The representation
together with the Parseval equation, shows that 22 (J1) may be regarded as an infinitedimensional euclidean space (the socalled " H ilbert space"), in wh ich the point / has coordinates en , and the functions n a re the coo rdinate vectors .
EXERCISES 1.
If I> 0 and f E I d!L = 0, prove that /(x) 0 almost everywhere on E. Hint : Let En be t he su bset of E on whichl(x) > 1 /n. Write A U En . Then �L(A ) 0 if and only =
=
=
if �L(En ) = 0 for every n.
f
I d!L = 0 for every measurable subset A of a measurable set E, then /(x) almost everywhere on E.
2. If 3.
..
0
If {/,} is a sequence of measurable functions, prove that the set of points x at which {/n(x)} converges is measurable.
4. If I E 2(�L) on E and g is bounded and measurable on E, then lu 5.
= =
Put
g (x)
=
{�
l2k( x) = g(x) f2k + . {x) = g ( l  x)
(O < x < } ), (� < X < 1 ), (0 < X < 1 ), (0 < X < 1 ).
Show that l im inf llx) = 0 but
[Compare with (77). ]
Jo'
(0 < X < 1 ),
[.(x) dx = ! .
E
2(�L)
on E.
THE LEBESG UE THEORY
6.
333
Let f,(x)
1
Then f,(x) .) 0 u n iform ly on R 1 , but
( I X I < 11),
n 0
=
( l x l > n).
r., /. dx = 2
(n
= 1 , 2, 3, . . ) . .
(We write J�oo in place of J, u .) Thus uniform convergence does not imply domi
nated convergence in the sense of Theorem 1 1 .32. However, on sets of finite measure, uniformly convergent sequences of bounded functions do satisfy Theo rem 1 1 .32. 7. Find a necessary and sufficient condition that af( a) on [a, b] . Hint: Consider Example 1 1 .6(b) and Theorem 1 1 . 33. 8 . I f f � on [a, b] and if F(x) J: f(t ) dt, prove that F'(x) f(x) almost every where on [a, b). 9. Prove that the function F given by (96) is continuous on [a, b). 10. If p.( X) < + oo and f e: !£ 2(p.) on X, prove that f e !l'(p.) on X. If
=
E
this
is
/E
p.(X) =
false. For instance, if
f(x )
+
=
cc ,
= I 1 I xI , +
then f e !£ 2 on R ' , but / ¢ !£ on R 1 • 1 1 . If /, g .,!ll ( ,._, ) on X, define the di s t ance be t wee n / a nd g by
E
f
.. X
I f  u l d,._, .
Prove t ha t !t'(/L) is a cornplete n1etric space. 12. Suppose ( a) /( x , y) < 1 if 0 < x < 1 , 0 < y < 1 , (b) for fixed x, f(x, y) is a conti nuous function of y, (c) for fixed y, f(x, y) is a contin uous funct ion of x.
I
I
Put
g (x ) 1 3.
Is g con t i nuo u s ?
=
J
1
0
/( x , y) dy
(0
< X