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146 MetrieSpaeeSEGhapg
b. Show that the following set of metrics for the set of ntuples of real
numbers are equivalent:
ptxaw=[(X1y1)2+~+tx,ypz]1/2
p*tx,y)=lx1hl+'+lxnH
p+(x,y)=maxcfw_x1y1,.,xny,.
c. Fiudametricforthesetofutupleethauisawtequiva
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Seci 2] Imp!" In IllIII I  119
satisfy the last requirement, for from M f H = 0 we can only conclude
that f= 0 a. e. We shall, however, consider two measurable functions
to be equivalent if they are equal almost everywhere, and, 1T we do
no01stmgu1sh
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152 Metric Spaces EChap. 7
13. Proposition: Every subSpace of a separable metric space is sep
arable.
PraaLetXbeax   _ 
Wle collection cfw_0 of Open sets in
X such that each open set in X is a union of some subcollection of
cfw_0. By PrOposit
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Sec. 5 Convex Functions H5
Since v has a local maximum at y, we have D+1/(y) S 0. But D+lll
was nondecreasing, and so D+ S 0 on [0, 3)]. Consequently, (p is
nonincreasing on [0, y], and hence 111(3) S (M0) = 0. Thus the
maximumoftvonm, 1]isnonpositive,a
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132 The Classical Baaach Spaces EChap. 6
For the case p=1, let E: cfw_x: g(x)>M+e, and set f=
(Sgn 9)(12 Then f HI = mE, and
MmE= Will 2 ijgl 2(M +)mE.
Thus mE = 0, and llglloo S M. I
WearenowinaposWwerthefollownrgrhaW" ' ' ' '
the b
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5.3] E'EE . E I I 105
7. Lemma: If f is integrable on [a, b], then the function F dened
by
Fm Lfg) d:
is a continuous function of bounded variation on [a, b].
PraaIhecontimiithomemJZmposionuToshow
thatFisofbounefvariation,ieta=xo<xl<
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Sec4j '[heGeneLaliabesgueleegLaJ 89
Problems
3. Letfbe a nonnegative measurable function. Show that If: 0 implies
f 0 ac.
4. Letfbe a nonnegative measurable function.
a. Show that there is an increasing sequence (p" > of nonnegative
Wenieasd
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126 "liheLGlassical Banach Spaces EChap. 6
by Fatous Lemma. Hence 9" is integrable, and g(x) is nite for
almost all x.
For each x such that g(x) is nite the series 2 fk(x) is an abso
k = l
lutely summable series of real numbers and so must be sum
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L44 MetFiCSpaces EChap.7
c. Is the set in (b) always the closure of the ball
cfw_x2 p(x, y) < 6?
7. Wh1ch ofthe spaces W, C, If. E are separable?
3 Continuous Functions and
Hemeemerphisms
A functionfon a metric space (X, p) into a metric
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154 Metric Spaces EChap. 7
17. Lemma: A metric space X has the BolzanoWeierstrass pro
perty ifand only ifX is sequentially compact.
ProoSmceesLerylimitofasubsequencemLxnmisaclusteLpoim
WWW
property. Conversely, if (x, > has x for a cluster point, the
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124 the Classical Banach Spaces EChap. 6
M f, fm H < 6. It is easily veried that each convergent sequence is a
Cauchy sequence.
Denition;AnormedlinearspaceiscalledcampletLifeveryCauchy
WhDequ n in e space con erges, thta is, if for each Cauc
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Sec4]AbsohneContinuhy 109
each of total length less than 5, where K is the largest integer less
than 1 + (b a)/5. Thus for any subdivision we have t s K, and so
TsK.
o y. is a so u ' , iva ive
almost everywhere.
13. Lemma: If f is absolutely
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94 MWWWA
1). Under the same hypothesis there is a step function W such that
JlfdIKe.
ECorrrb'rnepartfajwith Proposition 3.22.]
c. Under the same hypothesis there is a continuous function 9 van
ishing outside a nite interval such that
Jalfgl<e.
16. Es
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130 The Classical Banach Spaces EChap. 6
Problems
19. Show that um pg mp.
20. Prove 6mm.
5Bounded4:'rneareFm=retlonalsonthe
LLipaces
We dene a linear functional on a normed linear space X to be a
mapping F of the space X into the set of
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Sec. 3] Differentiation of an Integral I07
the bounded convergence theorem implies that
1 C
J: F(x) dx lim fcfn(x) dx lim b (F(x + h) F(x) dx
. 1 c+h 1 +h
=11mlj F(x)dxEJ:J F(x)dx]
=Fcc9Fca;=lcfmx,
since F is continuous. Hence
[two 4m
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7 Metric Spaces
1 Introduction
The system of real numbers has two types of properties. The rst
type consists of the algebraic, dealing with addition, multiplication,
etc. The other type consists of properties having to do with the
Wmdw
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I06 Wion fChap. 5
Thus for some n we have
#01
ansoeither
fsO
or
bn
f0.
Inanycaseweseethatiffispositiveorkwepofrposivameasum;
then for some x a [a, b] we have
rhea
Similarly for f negative on a set of positive measure, and the l
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Sec. 53 Convex Functions H3
Then f0 = f a.e. (See Problem 17.b). Hence
= folg(X)g(X) aC
FF ( y) dy = [A F (9(X)g(x) dx
d.Isthe1xpressioninpartfc)stthrueifgisnotasW
monotone? Assume that g is bounded and that [is integrable on an interval
containi
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$813.44 MW 93
a special case of the Lebesgue Convergence Theorem, but the advan
tage of Fatous Lemma and the Monotone Convergence Theorem is
that they are applicable even 1f f 1s not integrable and are often a
good way of showmgthatmmtegrableliatou
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City, State or Zip Code
PRIVACY POLICY  CALIFORNIA TRANSPARENCY IN SUPPLY CHAINS ACT I TERMS OF USE  SITE MAP
2017 The Rockport Group. All rights reserved.
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Sec. 7] Compact Metric Spaces 153
denition, we say that a collection IL of open sets in a metric space
is an Open covering for a set K if K is contained in the union of the
sets in TL. A metric space X is said to be compact 1T every Open
coveringuso
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S .211 VI I. HHII l. 121
3. Lemma: Let 1 g p < oo. Thenfor a, b, t nonnegative we have
(a + tb)" > a + ptbap 1.
Proof: Set
(p(t) = (a + tb) a" ptba" 1.
TmnmQmm
t) = We + Ha)? a] 20
for p 2 1 and a, 1), t2 0. Thus cp(t) is increasing and hence
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Sec.2] OpenandClesedSets MB
Let cfw_0, consist of those balls SM for which x is in D and 6 is
rational. Then cfw_0; is a countable collection of open sets. If 0 is any
Open set and y a 0, then we want to show that for some 0, we have
yLeQC0.Since0
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116 DifferentiationarndJrntegizahen EChap.5
exp x = e". The inequality then becomes a generalization of the
inequality between the arithmetic and geometric mean:
21. Corollary: Letfbe an integrablefunction an IQ, l I. Then
[exp (f(t) dt 2 exp U f
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5"] DH . EH E . L01
since f is increasing. Thus
N M
Z xn) _f(xn ha) 2 2 Ni + k1) yl'h
n=l i=1
and so
v(s + e) > u(s 26).
Sincethisistrueforeachpositiyee,wehaeu52us.Butu>u,and
so 3 must be zero.
This shows that
f(x + h) f(X)
9959=11
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SecHDierentlati91oLMonotone Functions 99
lim (1,) = 0, the interval I must meet at least one of the intervals I,.
Let n be the smallest integer such that I meets I,. We have n > N,
and [(7) < k,_1 < 27(7,.) )Since x is in 1, and I has a point in commo
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198 BMW EGhap.5
Since n is arbitrary,
F'(x)>f(x) ae.
Consequently,
JV) F(x) dx 2 Jbx) dx = F(b) F(a).
ThusbyTheorem3wehave
Jb F(x) dx = F(b) F(a) = Jbx) dx
J (F '(x) f(X) dx = 0
Since F (x) f (x) 2 0, this implies that F (x)
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120 The Classical Banach 8136065 EChap. 6
1. Minkowski Inequality: Iffand g are in B with 1 S p S 00, then
so isf+ g and
H+ 911, S Hpr + Haip
If 1 < p < 00, then eqmziityTnnWdle if there are nonnegative
constants a and such that (if: 059. /