weak and norm convergence of sequences in .../67531/metadc500521/...chapter 1 introduction the...
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WEAK AND NORM CONVERGENCE
OF SEQUENCES IN
BANACH SPACES
THESIS
Presented to the Graduate Council of the
University of North Texas in Partial
Fulfillment of the Requirements
For the Degree of
MASTER OF ARTS
By
Arthur J. Hymel, B.S.
Denton, Texas
December, 1993
Hymel, Arthur J., Weak and Norm Convergence of Sequences in Banach Spaces, Mas-
ter of Arts (Mathematics), December, 1993, 71 pp., bibliography, 5 titles.
We study weak convergence of sequences in Banach spaces. In particular, we compare
the notions of weak and norm convergence. Although these modes of convergence usually
differ, we show that in V they coincide. We then show a theorem of Rosenthal's which
states that if {x,} is a bounded sequence in a Banach space, then {x} has a subsequence
{X'} satisfying one of the following two mutually exclusive alternatives; (i){x'} is weakly
Cauchy, or (ii) {'} is equivalent to the unit vector basis of fl.
TABLE OF CONTENTS
Chapter
1 INTRODUCTION ..............
2 PRELIMINARIES ..............
2.1 Normed Linear Spaces . . . . . .
2.2 Banach Spaces . . . . . . . . . .
2.3 Linear Transformations . . . . .
2.4 Spaces of Linear Transformations
2.5 Dual Spaces . . . . . . . . . . . .
3 THREE FUNDAMENTAL THEOREMS
3.1 Baire Category Theorem . . . . .
3.2 Uniform Boundedness . . . . . .
3.3 Hahn-Banach Theorem.....
3.4 Finite Dimensional Normed Linear Spaces . . . . . . -.. 4........
4 CONVERGENCE OF SEQUENCES IN BANACH SPACES.........
4.1 Weak and Norm Convergence of Sequences .................
4.2 Weak Convergence in J2 and V ......................
5 ROSENTHAL'S THEOREM.. .............. ............
5.1 Equivalence to 1-Basis . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 M ain Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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CHAPTER 1
INTRODUCTION
The purpose of this thesis is to study and compare weak and norm convergence of sequences
in Banach spaces. Although these modes of convergence usually differ, we show that in V1
they coincide.
We begin by recalling the definition of a Banach space and some theorems regarding
these spaces. We notice that we can define an isometry between many dual spaces and
other Banach spaces.
In chapter 3 we discuss three fundamental theorems of functional analysis. We note the
significance of these theorems and use there results later in the thesis. These fundamental
theorems include the Baire Category Theorem, the Uniform Boundedness Theorem, and
the Hahn-Banach Theorem.
In chapter 4 we define the notions of norm convergent, weakly convergent, and weakly
Cauchy. We compare these modes of convergence and find that most of the time they are
not equivalent. We observe, however, that in 1 these modes of convergence are the same.
This relationship in il is known as the Schur Property.
We recall what it means for a sequence of a Banach space to be equivalent to the
unit vector basis of P. A theorem of Rosenthal's is discussed and his proof is presented.
This theorem states that if {x,,} is a bounded sequence in a Banach space, then {x,} has a
subsequence {x' } satisfying one of the following two mutually exclusive alternatives; (i){'4}
is weakly Cauchy, or (ii) {x'} is equivalent to the unit vector basis of '.
1
CHAPTER 2
PRELIMINARIES
2.1 Normed Linear Spaces
In this chapter we will discuss some basic definitions and some fundamental results. We will
define and talk about normed linear spaces, Banach spaces, linear transformations, spaces
of linear transformations, and dual spaces. The results in these sections are important
in developing the appropriate foundations from which we will be able to work with later
theorems.
We should note that much of the material found in this first chapter was taken from
sections 69, 80, 81, and 82 of Foundations of Mathematical Analysis, by Richard Johnson-
baugh and W.E. Pfaffenberger [2]. Any terms or results not defined or stated in this chapter
may be found in this text.
We begin by first defining what a linear space (vector space) is. Notice that for linear
spaces the field F will just be the set of real numbers, denoted R.
Definition 2.1.1 (Vector Space) Given the ordered pair (V, F), V is said to be a vector
space over the field F with operations + : V x V -* V and. F x V -+ V satisfying:
(1) x+y= y+x for all x,yE V
(2) x + (y+z) (x + y) + z for all x,y,z E V
(3) There exists90 E V such that x + 0 = x for all x EV
2
3
(4) for allix E V, there exists yEC V such that x + y = q
(5) c.(d-x)=(c.d)-x for all c,dE F, andx c V
(6) 1-x=x for allxEI V
(7) c-(x+y)=c.x+c.y for allcEF, and x,yEV
(8) (c+d)-x=c-x+d-x for all c,dEF, andxEV
Definition 2.1.2 (Norm) A function * V -+ [0, oc) is said to be a norm on the vector
space (V, R) provided:
(1) |lxii - 0 if and only if x = 0
(2) Icl -xI = Ic|-||lxll for all c E R and x E V
(3) i|x + y| jj ixil+ Ilyl| for all x, y E V
Next we define some linear spaces which will be referred to often in this thesis.
Definition 2.1.3 (1) E1 - {{an} :f{an} is a sequence in R and , anj E R}.
(2) 2 - {{an} :{an} is a sequence in R and .'(a,) 2 E R}.
(3) to {{an} :f{an} is a bounded sequence in R}.
(4) co = {{an} : {an} is a sequence in R and lim. oo an = 0}.
Theorem 2.1.4 The spaces 0', 2, oo, and c0 are vector spaces over R.
Proof. This is easy to see if we define scalar multiplication and addition coordinate wise.
a
4
Definition 2.1.5 (Normed Linear Space) A Normed Linear Space is an ordered pair
(V,11 *1) where V is a vector space and |11*1 is some norm on V.
Theorem 2.1.6 (Cauchy-Schwartz for Rn) If (a,,..., an) and (b 1,.. ., bn) E R, then
Eakbk < E a 1: b 2.
k=1 (k=1 ) k=1
Proof. If bk = 0 for all k E {1,...,n}, then we are done. Thus suppose that there is
some k E .1,...,n} such that bk $ 0. Therefore
n
Eb? > 0.i=1
Clearly for any real number x we have
n
0 Z (ai - xbi) 2.i=1
After expanding we get
n n- 2x aibi +x 2 Zb.
i:=1 i=1
Therefore if we let
we get
- 2b? aibi +
This, however, reduces to
n (n ( 2 n
b? - ( a-b .
Some simple algebra gives us what we wanted. I
n
E=a?
i alaib b l .
n aibi-En lb?
5
Theorem 2.1.7 (Cauchy-Schwartz for 2) If a and b E 2, then
E anbn < I: an E b2.
n=1 n=1 n=1)
Proof. Let {an} and {bn} E 2. By theorem 2.1.6 we have that for every n E N,
n n n
1:|ak| -|bkj :5 E ak12 E bk 2.
k=1 k=1 k=1
Notice that this is the same as saying,
1: |ak| -bk| E ak E b k.k=1(k=1 k=1
Thus, taking the limit of n on the right we get
n 00 2)
E |ak|-|bk| < Eak ( bkk=1 k=1 k=1
Therefore we have that the sequence of partial sums of the left hand side is bounded for
every positive integer n. Thus the limit on the left must exist. Hence, we have
CO -l000a )('b2
jak bk|< (7aE ) bE .k=1
From this point we can conclude that
>Iakbk0 a2)(0b2k=1 k=1 k=1
We will use the Cauchy-Schwartz inequality to help prove many theorems throughout
the course of this thesis.
Next we will discuss two similar notions of equivalence among normed linear spaces.
6
Definition 2.1.8 (Isomorphic) Given V and W normed linear spaces, we say that V is
isomorphic to W, denoted V ^4 W, if and only if there exists a continuous linear transfor-
mation, T : V -> W, which is one to one and onto and whose inverse is also continuous.
If T is such a transformation then T is called an isomorphism.
Definition 2.1.9 (Isometrically Isomorphic) Given V and W normed linear spaces,
we say that V is isometrically isomorphic to W, denoted V =W, if and only if there exists
an isomorphism T: V -+ W such that JT(x)j| = jjx|| for all x E V. T is referred to as an
isometry.
The next theorem gives us the usual norms for R, l,f12, too, and some other common
vector spaces.
Theorem 2.1.10 (1) The equation jj(xi,..., xn)jj_= - (x+... + x) defines a norm on
R".
(2) The equation |j{an}jj = E janj defines a norm on f1.
(3) The equation ||{a}j| = a defines a norm on f2.
(4) The equation |j{an}1| = lub{janj|: n N } defines a norm on f .
(5) The equation j1f|1 = lub{jf(x)j : x E [a,b]} defines a norm on B[a,b], the set of bounded
real functions on [a, b], and hence on C[a, b], the set of continuous real functions on
[a, b], and R,[a, b], the set of Riemann-Stieltjes integrable real functions, where a is
of bounded variation on [a, b].
Proof. The first step is to show that |(x1 ,...1, xn)j = /(x + .... + x2) defines a norm
on R'. We begin by showing that 1|(x1, .. ., nX)j = 0 if and only if (x 1,-.. ., xn) = 0. Clearly,
7
I + n.+X2 = 0 if and only if x = 0 for all i E {1,...,n},and for all nE N. Thus we
have that (x1,...,xnjj= 0 ifandonly if(x 1 ,...,Xn)=
Next we will show that I{c(x1,- ...,X)I = Ic .f(xI,- . .. , Xn)jj for all (x1 ,. .. ,x) E R4.
This follows from the following equalities
I c(x, . .. , Xn)|(cXi, . . . , cX,)
(cx 1)2 + ... + (cxn) 2
V(2( ?+ . + X2)
C- I (XI+...+x )
Finally, we need to show the triangle inequality. We want to show that IIx + Vll I
lxii +|Ijy|| for all x,y E R n. We have by our definition that lix + y|l= Zl-(xk +yk) 2 .
By theorem 2.1.6 we have that
n n n
ZxkYk ( xk (Yk)k=1 k=1 (k=1
This is true if and only if
n n n
2 E Xk k<2 E xk yk=1 (k= ) (k=1
Therefore,
n n n n n + n
EX2+ 2Exkyk +Ey x +2 (jxk(yk jy2.k=1 k=1 k=1 k=1 k=1 k=1 /k=1
By moving things around we can write this as
n ) 5n 2 E (n + n2
Z(xk + 2xkyk+ yk ) E Zx4 +2 _xk y + yk.kk=1 k=1
8
Thus, factoring both sides and taking the square root we get
Z(xk+yk)2 72 + 2
k=l k 1 kk=1
Which is
jx +\ yI \ail If + Yf
and we are done. Thus, fI(x1 ,..X.,xn)II= /(x + . .. + x2) defines a norm on Rn.
Next we will show that |f{an}ff = Z0 1 Ianj defines a norm on il. Notice that for all
{an} E illE', 01 an is a real number, making ff * ff well defined.
We start by showing that |f{an7 }f= 0 if and only if {an} = 9. Suppose that |f{an}fI = 0.
This is true if and only if El 1 Ian1= 0. This is necessary and sufficient to give us that for
all n E N, fa72 = 0, and this is true if and only if {an)} = .
Now we will show that IIc{an}ff-= ci- f{a 2n}f for all c E R,and for all {an} E fl. Notice
that
ffc{an}ff=I{can}11 = >3canf.n=1
Now, let k E N. Clearly, EI 1 f Ican nc1 fa72. This is true for all k E N. Thus,
taking the limit on k we get
E3 can4 = c 3 janf = jcl -jj{an}fj.nr=1 n=1
Lastly we wish to show that |f{an7 } + {bn}|f < ff{an}jj|+ jf{bn}f for all {an}, {bbn} E fl.
We have
ff{an} + {bn}|f = {an + bn}II00
= E |an + b2n=1
00
< >(IanI+ Ibn|).n=1
9
Letting k E N we getk k k
Z(IanI+Ibn)= k Y an|+ jbnj.n=1 n=1 n=1
This is true for all k E N. Thus, taking the limit on k we get
0000 00
5(Ian-+ jbn| 1n+ Ib=I) - '+ E jbnj = II{an}II + II{bn}I
n=1 n=1 n=1
and finally,
II{an}+ {bn}I < I{an}II|+ II{bnJI
for all {an},f{bn} E E.
Next we will show that II{an}1I Z 1 (an)2 defines a norm on f2. First notice that
Z'E=1 (an) 2 is a real number, making 1*11 well defined.
We start out by showing 1f{an}II = 0 if and only if {an} = 0. Suppose II{an}II = 0.
Therefore Z/ L1 (a) 2 = 0. This is true if and only if Z' 1 (an) 2 = 0. Giving us that
{an} = 0.
The next thing to show is that flc{an}|j= cl I-- {a1|I. Clearly,
Ic{an}i = |i{can}ii00
E(can) 2
n=1
00
c2(an)2.n=1
Now letting k E N, we have _ c2(anc)2 =cI - 1 (an)2.
This is true for all k E N. Therefore by taking the limit on k we get
(){00
E c2(an )2 - ZQI E an) -Ic-11a}-n=1 n=1
10
Lastly, we wish to show that II{an} + {bn}I 1{a}fI + I{bj}II for all {an}, {bn} E 2.
From part (1) we have that for every k E N,
kZk k
(an+ b)a+ Zbn.n=1 n=1 n=1
Taking the limit on the right we get,
k 00 00
(an+ bn)2 Sa+ Zb .n=1 n=1 n=1
This gives us that the left hand side is bounded for every k E N and thus the limit exists.
Therefore {an} + {bn} E 2 and II{fa}+ {b}II II{an}II + jf{bn}jj. Hence, taking the limit
on the left we get our conclusion and thus, we are done. Therefore this defines a norm on
2.
Next we will show that the equation II{a }1 = lub{janj : n E N} defines a norm on to.
We start by supposing that I|{a }II = 0. This is true if and only if lub{Ian|I: n E N} = 0.
But that is true if and only if for all n E N, janj 0. This is true if and only if for all
n E N, jan|j= 0. Which is true if and only if {an} = 9.
Now let {an} E P" and c E R. Then we get the following equations
1c{an}|I = j{can}jj
= lub{Ican: n E N}
= lub{cl -I.xn : n E N}
= i lub{Iaj : n E N}
= Icl -I{a }1I.
11
Let {an},{b,} E f'. Then our conclusion follows from the following equations,
ll{an} + {bn}ll = ll{an + bn}ll
= lub{an + bI : n E N}
lub{jan+ IbnI: n E N}
lub{|anj : n E N} + lub{lbnl: n E N}
ll{an}|l|+l|{bn}ll.
Hence this defines a norm on t .
Now we will show that the equation IjfII = lub{If(x)I: x E [a, b]} defines a norm on
B[a, b]. First notice that this definition makes is well defined since f E B[a, b] implies that
lub{If(x) I: x E [a, b]} exists.
First we show that If II = 0 if and only if f = 9. Suppose f = 0. This is true if and only
if for all x E [a, b]f(x) = 0, which is true if and only if If(x)I = 0 for all x E [a, b]. And this
is the necessary and sufficient condition for lub{If(x)I: x E [a, b]} = If II= 0.
Next we show that jjcf1 = =cj I-|f 11, for all f E B[a, b]. We get this from the following
set of equations.
|IcfIl = lub{cf(x) : x E [a, b]}
= lub{IcI -If(x)I : x E [a, b]}
= Jc lub{if(x)I : x E [a, b]}
= IdC h-|f I.
Finally we will show that If + gI < lfi11+ ugh|, for all f,g E B[a, b]. This result comes
12
from the following inequalities.
Ilf + g = lub{|(f + g)(x)f : E [a, b]}
= lub{jf(x) + g(x)j: x E [a, b]}
< lub{lif(x)J + jg(x)j : x E [a, b]}
< lub{jf(x)1 : x E [a, b]} + lub{Ig(x)j : x E [a, b]}
11f 1| + 11g1|.
I
The next theorem introduces the idea of the metric induced by a norm.
Theorem 2.1.11 If (V, 1|*|1) is a normed linear space, then the equation
IIx - y11 = d(x, y)
defines a metric on V.
Proof. Let x, y, z E V.
Certainly d(x, y) = 0 if and only if x = y.
Next we show d(x, y) = d(y, x). Notice that
d(x, y) = x - yll
-l(-1)(y - x)Il
-l 1-I y - X11
=Ily -X)
=d(y, x).
13
And so we are done.
Lastly we show d(x, z) d(x, y) + d(y, z). We get this by the following
d(x, z) = lix - zil = lix - y + y - zl lix - yll + ly - zjj= d(x, y) + d(y, z).
Therefore d defines a metric on V. I
Proposition 2.1.12 Let V be a vector space and let d be a metric for V which satisfies:
(a) d(cx,cy)= lcId(x,y), for all c E R and x, y E V.
(b) d(x, y) = d(x + z, y + z), for all x, y, z E V.
If for x E V, we define lix|| = d(x, 9). Then (V,||1 * jj) is a normed linear space.
Proof. First we show llxii = 0 if and only if x = 9. Certainly |lxii = 0 if and only if
d(x, 9) = 0, and this is true if and only if x = 9.
Second we show iicxii =tei- llxl, for all c E R and x E V. We get this from the equations:
lcxii d(cx,90) =-d(cx, c9) = jcjd(x,90) = jcl -jlxii.
Finally we show jlx + yll |lxii+ flyll, for all x, y E V. Noticing that
ix+yll = d(x+y,9)
= d(x+y-y,9-y)
= d(x, -y)
K d(x,0)+ d(9,-y)
= d(x,) + d(-y,90)
- ilxi+ll-y
= 112I + llYll,
14
we get this without difficulty. 1
2.2 Banach Spaces
Definition 2.2.1 (Banach Space) A Banach space is a complete normed linear space.
It can be shown that R,fl , 2, , and B[a, b] with norms defined as in theorem 2.1.10
are Banach spaces. We will only show the following.
Proposition 2.2.2 ( 1, |* J) is a Banach space.
Proof. We want to show that V? with 11 * fl defined by II{a}= ZE', asf is a Banach
Space. Let {an} be a Cauchy sequence in 0 and e> 0. Thus there exists N E N such that
for all m, n > N, Jan - a l < c. Notice that for all i E N, {a } is a Cauchy sequence in
R. Thus {ag} converges in R. for all i E N, since R is complete. Thus for all i E N define
limn.... a = a;. Notice that for all n > N
00 00
Etan_ N i aIk=1 k a +1
k=1 k=1
00
k=1
Thus, we have
00 00
k=1 k=1
for all n>2 N. Letting p be any positive integer and taking the limit on n, we have
k=1 k=1
15
Hence the sequence of partial sums of Zi, akI is bounded. Therefore {ak} E 1 If
pEN, we havethat _aM- a j <(c for all m, n > N. Taking the limit on m, we get
E_ jak - anj < e, for all n > N. Since this is true for all p E N we have E' aj -- a =
lak - aflj <; E. Therefore, an -+ {ak} in 4?. Thus, V? is complete. U
Proposition 2.2.3 Let (V, 11 * 11) and (V 2, 11 * 112) be normed linear spaces.
(a) If we define addition and scalar multiplication by the equations
(x1,,y1) + (X2, Y2) = (x1 + x2,Y1 + y2) and c(x, y) = (cx, cy)
and the norm by the equation
11(x, y)11= 11x1 +|1Y112,
then V, x V2 is a normed linear space
(b) If V1 and V2 are Banach spaces, then V, x V2 is a Banach space.
Proof (a) Clearly V, x V2 with addition and scalar multiplication defined coordinatewise
is a vector space.
Now, show |* defines a norm on V, x V2. Show I(x, y)11= 0 if and only if (x, y) = 0.
We have that 11(x,y)1l = 0 if and only if |lxi1i + IIYI12 = 0. This is true if and only if
lixi1 = -|11y12. Which is true if and only if both lixil, and 1Y112 = 0. Hence we have that
i(x, y)1i 0 if and only if (x, y) = (01, 02) =90.
Next we want to show ic(x,y)ii =tcl - l(x,y)1l. We have,
jjc(x, y)jl =- l (cxcy)ll
16
= I11 ii+ 1CY112
= ICf - 111 1+ I1- Y12
= IcI(1XII1 + 1Y112)
= ICI -I(xy)1.
Giving us the second condition for a norm.
Finally we need 1(x1,,y1) + (X2 , Y2 )11 5 II(X,,y)f + 1(x2, Y2)11. This comes from the
following inequalities
ff(x1, Y1) + (X 2 , Y2)ff = ff(xI + X 2 , Y + Y2)f
= Iix1+ x2111+11Y1+ Y2112
< 5 I1111i+ lIx2111 +f11112 + fY2112
- lX11 + fly12112 If2 11 + ffY2112
= ff(x,, Y1)f1+ fl(x2, Y2)11.
And we are done with the proof of part (a).
(b) Let {(as, b)} be a Cauchy sequence in V, x V 2. Thus for all e > 0, there exists
N E N such that for all m, n> N, l(anbn) - (am, bm)11 < c. However, this is true if and
only if 1(an - am)ffi + f(b - bm)11 2 < E. Hence 1|(an - am)1 < c and fl(b - bm)11 2 < <.
Therefore {an} and {b} are Cauchy in V, and V2, respectively. Therefore there exists
a E V, and b E V2 such that a -> a and b -+ b.
Next we will show that the above implies that (an, b) -> (a, b). Let c > 0. Since an -+ a
17
there exists N 1 E N such that for all n > N1 , Ilan - all1 < . Similarly there exists
N2 E N such that for all n > N2 , llb - b11 2 < <. Let N = lub{N1, N2}, thus for all
n N, Ian - all 1 + lb - b11 2 < - + = E. Therefore, jf(a - a), (b - b)I < E implying
(an, bn) - (a, b)I < e, and thus, (a, bn) - (a, b).
Hence, V x V2 is complete. Proving that (V 1 x V2,|*|I) is a Banach space. U
2.3 Linear Transformations
We will begin this section by first recalling the definition of a linear transformation.
Definition 2.3.1 (Linear Transformation) A function, T, from a linear space, V, into
a linear space, W, is called a linear transformation provided
1. for allv and w E V, T(v+w) =T(v)+T(w), and
2. for all v E V and any scalar r, T(rv)= rT(v).
The following theorem gives a list of equivalences which prove very useful in dealing
with continuous linear transformations.
Theorem 2.3.2 When V, W are normed linear spaces and T : V -+ W is a linear trans-
formation, the following are equivalent:
(1) T is continuous on V.
(2) T is continuous at some point in V.
(3) T is uniformly continuous on V.
(4) there exists k such that IT(x)ll < kjjxf, for all x E V.
18
(5) A ={ fT(x)||: x E V and |fxj| < 1} is bounded.
(6) B = {f|T(x)| : x E V and jjxfj|= 1} is bounded.
Proof. Clearly (1) implies (2). Next we will show (2) implies (3). Let T be continuous
at a E V and c> 0. Thus there exists 6 > 0 such that if fix - ajf <6 then jT(x)- T(a)fj < C.
Let y E B and suppose fx -y| <.6. Then jx-a+a-yf = Ilx -yII, thus lx-a+a-yjiy <6.
Therefore we have ff(x + a - y) - al < 6 implying that jJT(x + a - y) - T(a)ll < e. Hence
ffT(x) + T(a) - T(y) - T(a)jj < c. Which gives us, finally, jIT(x) - T(y)fj < C.
Next we will show (3) implies (4). Assuming (3) we have that for all E > 0, there
exists 6 > 0 such that if jjx - yfj < 6 then j|T(x) - T(y)|j < 1. In particular if f|xjf < 6
then |IT(x)lf < 1. Now let x E V such that x # 0. Thus, 1 x = < 6. Therefor,
T( 6x) < 1, but
6 6T( x) 1 T(x)
2||x l2||x||
6=21x1 j- T(x)jf < 1
thus,
I|T(x)1I < j<
forallxEVx#0.
Let K = }.
Next we will show (4) implies (5). Supposing (4), let x E V such that jlxjj < 1. Thus,
jjT(x)jj < Kffxjj < K. And hence A is bounded by K.
19
Showing (5) implies (6) is clear since B C A.
Next we will show (6) implies (1). Suppose B is bounded by K. Let c > 0. Let 6 =
(K+1). Suppose fix - yf < 6. If x 0 y, then jI = -land thus, T( ~_))11<K.
Hence ||T(x - y)iI Kfjx - yiI. And so, fIT(x) - T(y)fj < K6 = K' < 6. If x = y then
ffx - y11 = 0 < 6, and we are done. U
The next theorem gives us a natural way of identifying elements in l with continuous
linear transformations from t' into R.
Theorem 2.3.3 Let {an} E V'. Define T : f'-+ R by T({x}) = Z1 (anxn). T is a
continuous linear transformation on f'.
Proof. Clearly T is linear. We let A = {fT({x,})| :{X,}fj 1} and show A is
bounded, implying that T is continuous. Let {x} E o such that |f{x}jf < 1. Thus,
lT({xn}) = Z 1(anxn)1 < IE _, an|, since for all n C N, x K 1. Therefore, A is
bounded by IE ,an I. Thus, we have T continuous. I
The next corollary gives us a natural way of identifying elements in fl with continuous
linear transformations from c0 into R.
Corollary 2.3.4 Let {an} E fl. Define T : c0 -+ R by T(x) = .|. anX,. Then T is a
continuous linear transformation on c,.
Proof. Clearly T is linear. We first notice that the norm for f' also defines a norm for
cO. Thus since c is a subspace of tE we know that T is continuous. I
The next theorem gives us a natural way of identifying elements in R with continuous
linear transformations from R' into R.
20
Theorem 2.3.5 Let {ak} E R'. Define T: Rnh -+ R by T(x) =E' akxk. Then T is a
continuous linear transformation on Rn.
Proof. Clearly T is linear. We begin by letting a = (a1,..., a n ) E R. We want to show
that
|T(x|| :5Kl|x||,
where K E R. This follows from the following inequalities
n
IT(x)l = Zakxkk=1
< ( a) ( nx ) (by theorem 2.1.6)(k 1 k=1
Thus, we have that T is continuous on R'.
The next theorem gives us a natural way of identifying elements in f' with continuous
linear transformations from Vl into R.
Theorem 2.3.6 Let {an} E Ef. Define T :1 -+ R by T(x) = a Then T is a
continuous linear transformation on f'.
Proof. Clearly T is linear. We begin by letting {an} E Ef. We want to show that
|T(x)j< Kjjxjf,
where K E R. This follows from the following inequalities
00
IT(x)I =E>anXn
n=1
21
00
E Ziani in|n=100
n=100
n=1
- halfl jxii.
Thus, we have that T is continuous on '.1I
The next theorem gives us a natural way of identifying elements in 2 with continuous
linear transformations from 2 into R.
Theorem 2.3.7 Let {an} E 2. Define T : J2 -+ R by T(x) anxm. Then T is.a
continuous linear transfromation on 2.
Proof. Clearly T is linear. We begin by letting {an} E 2. We want to show that
IT(x)i KfjzxI,
where K E R. This follows from the following inequalities
00
iT(x)i = anXnn=1
< ( a2 ) ("fX2) (by theorem 2.1.7)
in=l n=1
-- 1 al| - ||xj |.
Thus, we have that T is continuous on 2. 1
22
Theorem 2.3.8 Let c be the set of all convergent real sequences. If {xn} E c, let fI{x}II
lub{jxn|f: n E N}. Then,
(a) c is a normed linear space, and
(b) the equation T({x, }) = lim_. x,-n, for all {xj} E c, defines a continuous linear trans-
formation on c.
Proof. (a) Notice that V with * f, as defined above, is a normed linear space, and
since c C f, it suffices to show that c is a subspace of f'. To do this we need only show that
if {xn},{yn} E c and r E R then r{xn} + {y} E c. First suppose that {x},{yn} E c and
r E R. Thus lim +o x, and lim,.o y exist. Therefore, by using the algebraic properties
of limits we get that
r lim Xn + lim y = lim (rxn + y,).n-+oo n-+oo n-+oo
Since r lim a_.o Xn + lim oo y E R then lim_.O(rxn + yn) E R, which means the limit
exists, and therefore r{xn} + {y)} E c.
(b) Clearly T is a function since the definition makes sense for all {xn} E c, and limits
of real sequences are unique.
Now, we will show that T is linear. First let {xn},f{yn} E c and r E R. Consider
T({x,} + {yn}),
T({xn} + {yn}) = lim (xn + yn) = lim Xn + lim y= T({xn}) + T({y}).n-*oo n-+oo n-.oo
Now consider T(r{X}),
T(r{Xn}) = T({rXn}) = lim rXn = r Him Xn = rT({xn}).n-.+oo n-.+oo
23
Finally, we will show that T is continuous. Let {xn} E c such that |f{x,}| 5 1. Hence,for
all n E N, lxI K 1. Therefore, lim .oo xn 1, but this is true if and only if IT({x})I 1,
implying that {IT({xn})I: j1{x1} 1 1} is bounded. Therefore we have that T is continuous.
I
2.4 Spaces of Linear Transformations
In this section we will observe that spaces of linear transformations are themselves linear
spaces. This fact will be useful in characterizing properties of these spaces.
Definition 2.4.1 Let L'(V, W) denote the set of all linear transformations from V into W,
where V and W are vector spaces.
Definition 2.4.2 Let L(V, W) denote the set of all continuous linear transformations from
V into W, where V and W are normed linear spaces.
Notice that the set L'(V, W) is a vector space if we define addition and scalar multipli-
cation as follows:
+ : L'(V, W) x L'(V, W) -+ L'(V, W) by, (T + U)(x) = T(x) + U(x)
R x L'(V, W) -+ L'(V, W) by, (cT)(x) = cT(x)
Theorem 2.4.3 Let V and W be normed linear spaces. Then L(V, W) is a vector space
where the operations are defined pointwise.
Proof. Since L(V, W) C L'(V, W) then it suffices to show that L(V, W) is a subspace. Let
T, U E L(V, W) and c E R. By theorem 2.3.2 there exists KT and KU such that IIT(x)II
24
KTjiXi| and |iU(x)ii Kujixii. Thus, c ii-T(x)ii iciKTiixii. Implying that iicT(x)iI
iciKTixiI. Therefore, we have licT(x)ii +IiU(x)ii icljiKixl + Kulixfj. Giving us,
I|cT(x) + U(x)II jcT(x)lI|+|I|U(x)l|
< |c|KT||x|| + Ku||x||
= (lcIK2 + Ku)iixii.
Therefore cT + U is a continuous linear operator, and hence CT + U E L(V, W). Thus,
L(V, W) is a vector space. I
Observe that we can define a norm on L(V, W) and L'(V, W), making them normed
linear spaces.
Theorem 2.4.4 Let V and W be normed linear spaces. Then the equation|1T|| = lub{|IT(x)h1
x E V and lIxII = 1} defines a norm on L(V,W).
Proof. If T E L(VW), then let AT = {iiT(x)Ii : x E V and ljx|i 1}. Since T is
continuous then AT is bounded and hence lubAT exists.
If T = 0, then AT = {0}, and thus lubAT - 0 implying that iT| = 0. If IT11 = 0,
then lubAT = 0, and since AT ; [0, oo) then AT = {0}. Therefore,for all x E V, T(x) = 0,
implying T = 0.
Let T E L(V, W) and c E R. If c = 0 then iicTii = 0 = OliTti = cliTli and we are done.
Suppose c #4 0. If x E V and jixj = 1, then
ll(cT)(x)ii = tcl - iT(x)li < Icl -|ITil.
25
Hence iicTii Icl -1|Th]. Also,
Icl - iT(x)hII= I(cT)(x)hI |IcTiI,
for all x E V such that lixj! = 1. Therefore, |IT(x)hI IICTi implying that jc - T11< lIcTil
and therefore I|cTlI = Icl - 1Th
Let T, U E L(VW). If x E V such that lixIl = 1, then |IT(x)l :|T11 and JIU(x)hl K fjUfl.
Therefore,
I(T + U)(x)1 = IT(x) + U(x)I1 IIT(x)hi + IIU(x)ll< |T11|+|Ul.
Thus, it follows that j|T + U< |T11+ l|Ull. Hence, we have that this defines a norm on
L(V,W). U
This next theorem gives us other equivalent ways of viewing the norm on L(V, W), where
V and W are normed linear spaces.
Theorem 2.4.5 Let V and W be normed linear spaces and let T E L(V, W). Then 11Th =
lub{IT(x)il : x E V and lxii 1} =_glb{K: IT(x)ii Kfjxjj, for all x E V}.
Proof. Let A= {liT(x)i: x E V and ljxii 1},B = {11T(x)11 : x E V and l1xii = 1},
and C = {K : iiT(x)ii Kflxjf, for all x E V}. Since B C A, then iT!!-= lubB < lubA.
Let K E C. If x E V and l1xii 1, then iiT(x)ii Kflxjj K K. This is true for all K E C
and for all x E V such that IxIl K 1. Hence for all K E C, K is an upper bound for A.
Therefore lubA K K, for all K c C. Thus lubA is a lower bound for C and hence we have
lubA < glbC.
Suppose that x E V such that x#54O, then =1, and hence, <T(x)IT11|.
This reduces to JiT(x)ii liTi l- xii, for all x E V such that x 54 0. If x = 0 then T(x) = 0
26
implying that IIT(x)ii = 0 < |T11h -lxl = 0. Thus, IiT(x)hi < T11- -lxii, for all x E V.
Therefore 11Th| E C implying glbC < fjTj1.
Therefore we have 11Th = lubB < mbA < glbC < f|T|f. Hence they are all equal. 0
Corollary 2.4.6 Let V and W be normed linear spaces and let T E L(V, W). ThenIiT(x) I|
|T11- -ix|l, for all x E V.
Proof. The proof follows easily from the previous theorem. 1
2.5 Dual Spaces
In this section we will the discuss the idea of a dual space. We will recognize that these
spaces are Banach spaces and that for many common Banach spaces there is a natural
correspondence between its dual space and another Banach spaces. This information will
be very useful in that it will simplify the way we look at many of our examples of dual
spaces.
Definition 2.5.1 For V a normed linear space, L(V, R), the space of all continuous linear
transformations from V into R, is called the dual of V, denoted V*. The norm on V* is
given by 11T|| = lub{iT(x)i|: x E V and l|xii = 1}, for all T E V*.
The following theorems in conjunction with earlier theorems gives us some of these
equivalences between dual spaces and other Banach spaces. For example, we will see in this
next theorem that elements in (Rf)* can be identified with elements in R.
Theorem 2.5.2 Let T E (Rn)*. Then there exists a = (a1, ... , an) ER' such that T(x) =
Z ( 1(akxk), for all x E R n, and 11T|| = ||all.
27
Proof. Let T E (Rf)* and let 6 k = (0,...,0,1,0,... ,0), where 1 is in the kth position.
Set ak = T(dk), for all k E {1,...,n} and let a = (a1,...,an). If x = (x 1,...,x,) E R,
then T(x) = T(Z-l(kok)) = 1T(xk6k) -=EE(T(6 )) - Z I(xkak). Now,
lT(x)l = lE"i_.(Xkak)l < Zy - x~ a2 - l|xii - ll. By previous theorem 11ThJ
flail. If a = 0 then 11ThI = 0 = t1all. If a / 0, then T(a) = h|al2 and thus T( a) < 1Th.
Therefore 11Th = hall. U
Combining the previous theorem with theorem 2.3.5 we can, in fact, conclude the fol-
lowing corollary.
Corollary 2.5.3 (Rn = (Rf)*) The linear transformation T: Rnh -+ (Rn)* defined by
T(a) = a* where a* is defined by a*(x) = J_' axxk is an isometry.
This next theorem lets us identify elements in c* with elements in fl.
Theorem 2.5.4 Let T E c*. Then there exists a E f such that T(x) = 1 (xnan), for
all x E c, and ITII = |all.
Proof. Let T E c*. Let
1 if n=m
= 0 otherwise.
Set an = T(6b). Let x = {x} E c0 and let y = Zl(xkk). Then
T(yn) = T(Z(xk k))
k=1
- (xkT( bk))k=1n
- >(xkak).k=1
28
Now, x - yn = (0,... .,Q,0x + 1,x, + 2, ... ) and thus lix - ynI| = lub{xkI : k > n}.
Hence llm i+o jjx - yn= lim Iub{xIkl k > n} = lima-o lubjx"f = 0, since {xn} E
co. Thus, lim.o y = x E co and since T is continuous then T(x) = limn-.o T(yn) =
limn,-oo, Z=I(xkak) = En =(Xnan).
Next we show a E 0. To do this let
I if m < n and an > 0
7 = -I if m< n and an < 0
0 ifm>n.
Then T(7 n) = El (Qn} am) = EnZ lakl. Notice that -y E c. Also notice that |/Ill = 1;
so by previous theorem <T")j K jTl. Thus E' ladj 1Th|, for all n E N, and hence
a E L. Now taking the limit we get 1al= fan|j 1Th. On the other hand, if x E c and
jxI = 1, then IT(x)] = |Z =j(xnan)h S= 1(hxnl - janj) Z5E , lanj = hlall. Therefore,
by previous theorem 11Th| fjajj and hence they are equal. I
Combining the previous theorem with corollary 2.3.4 we can conclude the following.
Corollary 2.5.5 (il'= c*) The linear transformation T : f -+ (co)* defined by T(a) = a*
where a* is defined by a*(x) = E', anx, is an isometry.
This next theorem lets us identify elements in (l')* with elements in L0.
Theorem 2.5.6 Let T E (l)*. Then there exists a E L' such that T(x) = (xna,),
for all x E Li, and 11Th =l =|all.
Proof. First let T E (tL)*. Let
1 if n= m
0 otherwise,
29
and let a, = T(6). Now let x = {x} E fl. Let y, = Z ,l(k6k). Thus T(yn) =
T(k E( k)) =Z (kT())=Z (xkak). Now, x - yn = (0,...,,xn+1,xn+2,..)
Therefore fix - yi = E3+ jnlxki, but lim,..+0 ffx - y2ff = lim-..oo Z =n+ 1 fxkf = 0.
Therefore limn,., yn = x in fl. Therefore, since T is continuous, T(x) = limw T(yn) =
lim7 1. Zn i(xkak) = E" 1(xna,).
Next we show a E '. Recall, T(67 ) = an. Notice that 6b E ', for all n E N and
jfbnj5f= 1, for all n E N. Thus fT(6b)f ilTil. But fT(6b)i = 1ani JITff, for all n E N, and
hence {an} is bounded by ffTll. Thus a = {an } E 00, and flal < 11Th.
Now consider IT(x)l. We have that
Z(xnan) 0(0X -lan)n=1 n=1
00
< 1 (IXnl -||a||)
n=100
=ia|iEixni
n=1
Therefore we have iT(x)l <hall - lix , implying by previous theorem that 11 T< 1 |a|l. Hence
they are equal. U
Combining the previous theorem with theorem 2.3.6 we can conclude the following.
Corollary 2.5.7 (f =1)*) The linear transformation T : 00- (1)* defined by T(a) =
a* where a* is defined by a*(x) = 1 x is an isometry.
This next theorem lets us identify elements in ( 2)* with elements in 2.
30
Theorem 2.5.8 Let T E ( 2)*. Then there exists a = {aj} 2 2 such that T(x) = E'_(xna),
for all x E 2, and T11| = |ail.
Proof. First let T E (2)*. Let
1 if n=m
0 otherwise,
and let an = T(6b). Now let x = {xn} E 2. Let y ="l(xkok). Thus
T(yn) = T(Z(xkk))k=1
n
3(xkT(bk))k=1
Z(xkak).k=1
Nowx-yn (0,. .. ,0, xn+1,xn+2, ... ). Therefore Ix-yn1 = |-n+ xbutlima 0 jx-
YnjI = limn_+oo L|'n+1 xl = 0. Therefore lim,, yn x in 2. Therefore, since T is con-
tinuous,
T(x) = lim T(y)n-,oo
- lim (xkak)
n-oo00
- (xnan).n=1
Next we show a E 2. Let n E N. We have that
k (T( )ak)k=1 k=1
31
n
=- T (6ak)k=1
T( (6ak)k=1
- -11-1Ia(kkak)k=1
=JT 1- {akB k111
= IT 1 a2.
k=1
Hence,
n a T
implying that VE"_ af IT11, for all n E N. ThusE a T11, and hence a E
with h1al 11.
Now let x E 2.
00 00
IT(x) - >(xnan) (n -ant)n=1 n=1
00 00
n=1 n=1
jx j - ||1a .
Therefore we have hT(x)t hiatt l- xii, implying by theorem 2.4.5 that 11Th| hiatt. Hence
they are equal. U
Combining the previous theorem with theorem 2.3.7 we can conclude the following.
32
Corollary 2.5.9 (P2 = (f 2 )*) The linear transformation T : f2 (_ J2 )* defined by T(a)
a* where a* is defined by a*(x) = ZE anx is an isometry.
The next theorem shows us that every dual space is in fact a Banach space.
Theorem 2.5.10 Let V and W be normed linear spaces with W complete. Then L(V, W)
is complete.
Proof. Let T be a Cauchy sequence in L(V, W). Define T V -+ W by T(x) =
lim_.oo T(x), for all X E V.
Note that T(x) is Cauchy in W, for all x E V, and W is complete. Thus T(x) is defined
for all x E V.
Next we show that T is a linear function. Letting x, y E V and a E R
T(ax + y) = lim T(ax + y)n-+oo
=lim (aT(X) + T(y))n-,oo
= a lim Tn(x)+ lim T(y)n-.oo n-.-oo
= aT(x)+T(y).
Hence T is linear.
Finally we show that T is continuous. Since {T} is a Cauchy sequence of real valued
bounded linear transformations then {ITll} is a Cauchy sequence of real values. This im-
plies that {jjTjj) is a bounded sequence in R and thus there exists K E R such that IITII
K, for all n E N. Hence, for all x E V, and for all n c N, IT(x)I Klxjj. Thus taking the
limit on n we get that IT(x)i KjIxII. And so T is continuous. I
Corollary 2.5.11 If V is a normed linear space, then V* is a Banach space.
CHAPTER 3
THREE FUNDAMENTAL THEOREMS
3.1 Baire Category Theorem
Much of the material discussed in this chapter was taken from chapter 7 sections 8, 9, and
10 of Real Analysis, third ed., by H.L. Royden [5].
In this chapter we discuss three fundamental theorems for the study of functional analy-
sis. The first of these is the Baire Category Theorem. We first recall the following definitions.
Note that in each of the following definitions we will let (X, d) be a metric space.
Definition 3.1.1 (Nowhere Dense) E C X is said to be nowhere dense if E is dense
in X.
Definition 3.1.2 (First Category) E C X is said to be of first category (or meager) if
it is the countable union of nowhere dense subsets of X.
Definition 3.1.3 (Residual) E C X is residual if ~E is of first category in X.
Definition 3.1.4 (Second Category) A set is of second category if it is not of first cat-
egory.
Definition 3.1.5 (G-delta) A subset of X is a G6 if it can be written as the countable
intersection of open subsets of X.
Definition 3.1.6 (F-sigma) A subset of X is an F, if it can be written as the countable
union of closed subsets of X.
33
34
Theorem 3.1.7 (Baire) Let (X, p) be a complete metric space and O a countable collec-
tion of dense open subsets of X. Then f=1 On is dense in X.
Proof. Let U be open in X. Let x 1 E 01 U and S be a ball radius r 1 centered
at x1 such that 1 01 fl U. Since 02 is dense there exists x2 E 02n fl. Since 02 is
open, there exists S2 a ball radius r2 < (I)r1 and r2 < r1 - p(x1, x2). Thus S2 C S1.
Proceeding inductively, we obtain a sequence {Sn} of balls such that Sn C Sn_1 and Sn
On, for all n E N, and {rn} tends to zero. Let {xn} be the sequence of centers of these
balls. Then for n, m > N we have xn E SN and xm E SN. Therefore, p(xn, xm) <;2 rN-
Thus {xn} is a Cauchy sequence, since rn -* 0. By the completeness of X there is a
point x E X such that xn -+ x. Let N E N. Since Xn E SN+1, for all n > N, we have
x E SN+1 C SN 9 ON. Consequently, x E flO On, and x E U. Thus, for all open U
C X, U f(ol On) 0. Therefore fnl On is dense. U
Corollary 3.1.8 (Baire Category Theorem) Let X be a complete metric space. Then
no nonempty open subset of X is of first category.
Proof. Let U be open and suppose {En} is a countable collection of nowhere dense
subsets of X such that U= U' En. Let On= X - En. Thus {0 } is a countable
collection of dense open subsets of X. Thus U fl(fl= 1 On) 4 0. Hence, there exists x E
U such that x g' (}nLI En. This contradicts U = U0 En. I
Proposition 3.1.9 If 0 is open and F is closed, then the sets 0 - 0 and F - F' are
nowhere dense. If F is closed and of first category in a complete metric space, then F is
nowhere dense.
35
Proof. We want to show that X - (0 - 0) is dense in X. Notice that 0 - 0 = (0 - 0)
since 0-0 is closed. We have X - (0Y- 0) = (X -06)U O. Thus
X = (X-0)UO
=(X-0)UO
= [X -(P0-0)].
Hence X - (0 - 0) = X and 0 - 0 is nowhere dense.
Now we will show F -F' is nowhere dense. Let x E F - F, and e> 0. Now let B((x) be
the open ball radius e centered at x. Since x V F*, B(x) F - F. Therefore there exists
y E B(x) such that y F - F*. This is true for all e> 0. Hence, x is an accumulation
point of X - (F - F*) and thus X - (F - FO) = X.
Now suppose F is closed and of first category in a complete metric space X. Clearly,
then, every subset of F is of first category. Hence F* is of first category and is open. Thus,
by corollary (3.1.8), F' = 0. Hence, F - F' = F is nowhere dense. I
Proposition 3.1.10 A subset of a complete metric space is residual if and only if it con-
tains a dense G6. Hence a subset of a complete metric space is of first category if and only
if it is contained in an F, whose complement is dense.
Proof. (-) Let U C X, a complete metric space such that U is residual. Then X - U
is of first category. Thus X - U = U'-1 En, where for all n E N, E is nowhere dense.
Hence for every n E N, E, contains no nonempty open sets. But U= 1 E U 1 E= . This
36
implies that X - U C Unh En. Therefore X - (U,, E,) = nAnIj(X - En) g U. Clearly,
for all n E N, X - En is open. Thus, fl, 1 (X - En) is a dense G and is a subset of U.
(<=) Let U ( Xwhere X is a complete metric space. Suppose U contains 0, where
O is a dense G6 . Hence O = l 1 On, where the On's are open. Since 0 C On, for all
n E N, and 0 is dense, then for all n E N, O is dense in X. Notice that for all n E N,
O = X. Hence, by the previous proposition X - On is nowhere dense for all n E N. Hence
UnL1(X - On) is of first category. But X - U C U'(X - On), thus X - U is of first
category. Clearly, X - U = (U 1 (X - On)) A(X - U)= U l((X - On) f(X - U)), and
since (X - On) fl(X - U) C (X - On) is nowhere dense, then (X - O) l(X - U) is nowhere
dense. And hence we are done. I
Proposition 3.1.11 Let {Fn} be a countable collection of closed sets with X = 1 Fn.
Then 0 = U' 1(Fn)* is a residual open set. If X is a complete metric space, then 0 is
dense. In particular, there exists n E N such that Fn 0.
Proof. The set En = Fn - (Fn)* is nowhere dense for all n E N. Let E = U' 1 En. It is
a set of first category. However X - 0 C E, and so 0 is residual. This implies that X - 0
is of first category. Notice that since 0 is open then X - 0 is closed. Thus by a previous
proposition X - 0 is nowhere dense. Hence 0 is dense. I
3.2 Uniform Boundedness
In this section we prove many important boundedness principles on sequences of real valued
continuous functions. Recall that elements in dual spaces are continuous functions. At the
same time recall that we have proven that in many cases we can identify these continuous
37
linear transformations with elements of other Banach spaces in a natural way. Hence, much
of the results we find here concerning sequences of real valued continuous functions can be
used to tell us about sequences in respective Banach spaces.
Theorem 3.2.1 (A Uniform Boundedness Principle) Let ! be a family of real val-
ued continuous functions on a complete metric space X, and suppose that for all x E
X, there exists Mx E R such that If(x)I Mx, for all f E . Then there exists 0 (nonempty
open) X and M E R such that If(x)I I M, for all f E ! and for all x E 0.
Proof. For every m E N, let Em,f = {x : If(x)j m}, and set Em A=fE Em,f.
Since each f is continuous, Em,f is closed, and consequently Em is closed. For every
x E X, there exists m E N such that If(x)f < m, for all f E ; i.e. there exists m E
N such that x E Em. Hence, X = U=1 En.
Since X is a complete metric space, then by corollary 3.1.8 there exists m E N such that Em
is not nowhere dense. Since this Em is a closed set, it must contain some open ball 0. But
for all x E 0 we have If(x)j < m, for all f E a. I
Proposition 3.2.2 (The Uniform Boundedness Principle) Let X be a Banach space
and !a a family of bounded linear operators from X to a normed space Y. Suppose that for
all x E X, there exists Mx E R such that IT(x)I M., for all T E . Then the operators in
! are uniformly bounded; That is, there exists M E R such that JT|J M, for all T ECQ.
Proof. Let T E a. Define f(x) = f|T(x)j|. Since the family of these functions is bounded
at each x E X and X is complete, then by theorem 3.2.1, there exists open 0 C X on which
these functions are uniformly bounded. Thus there exists M' E R such that JIT(x)lI M',
38
for all x E 0 and for all T E !. Let y E 0. Since 0 is open, there exists a sphere
S = {x f:f|x - yfj < 6} for some radius 6 centered at y and contained in 0. Let T E 9
and z E X such that 0 < lizil b 6. Then T(z) = T(y + z) - T(y) with y + z E S C 0.
Hence, IIT(z)I| |iT(y + z)ff|+ IIT(y)II 2M'. If fjz|j = 0 then clearly we have T(z) < 2M'.
Therefore for every z E X such that lizil 5 6, we have ||T(z)II 5 2M'. Now Let w E X such
that |twit 5 1. Thus IIT(w)ii|= }IIT(bw)I|. Since 116w|| 6 we have that IIT(w)II < for
all W E X such that |twit < 1. Consequently, |T11 M, for all T E Q.
Theorem 3.2.3 Let {Tj} be a sequence of continuous linear operators on a Banach space
X to a normed linear space Y, and suppose that for all x E X the sequence {Tn(x)} converges
to a value T(x). Then T is a bounded linear operator.
Proof. Let x E X. Since lim_., Ta(x) exists, for all n E N, then they are bounded
in norm. Consequently there exists Mx E R such that |lT(x)I1 M., for all n E N.
Therefore by theorem 3.2.2, there exists M E R such that 11T1 M, for all n E N. Thus
IT(x)|I Mx, for all n E N and for all x E X. Suppose T(x) -+ T(x), for all x E X.
Thus, jjT(x)I= liMn-.oolTn(x)| M5jjx|, for all x E X. Therefore IT11< KM. Thus T is a
bounded linear operator. U
3.3 Hahn-Banach Theorem
Theorem 3.3.1 (Hahn-Banach) Let p be a real valued function defined on the linear
space X such that
(1) p(x + y) < p(x) + p(y), for all x and y E X, and
39
(2) p(ax) = ap(x), for all a > 0 and for all X E X.
Suppose that f is a linear functional defined on a subspace S and that f(s) < p(s) for
all s E S. Then there exists F a linear functional defined on X such that F(x) , p(x) for
all x E X, and F(s) = f(s) for all s E S.
Proof. Consider all linear functionals g defined on a subspace of X and satisfying
g(x) < p(x) whenever g(x) is defined. This set is partially ordered by setting g, 92 if 92
is an extension of g1.
By the Hausdorf Maximal Principle, there exists {g},I a maximal linearly ordered
subfamily that contains f (from above). Define F on the union of the domains of the g
by, F(x) = ga(x) if x is in the domain of g.
We claim that the domain of F is a subspace, and F is a linear functional. Let x and
y be in the domain of F. Then x is in the domain of g, and y is in the domain of g for
some a and 3. By the linear ordering either g, g or gl <g. Without loss of generality,
suppose that g, go. Therefore x and y are in the domain of g and so Ax + py is in the
domain of go. Thus Ax + py is in the domain of F, and
F(Ax + py) = gg(Ax + py)
= Ag(x)++ pgp(y)
- AF(x) + pF(y).
Consequently, F is an extension of f. Moreover, F is a maximal extension. For if G is any
extension of F, g, F < G. This implies that G E {gc}aED. For if G V' { g}aj then we
40
can create another chain, namely G U{g-}ED. However, this contradicts {g},jE being a
maximal chain. Hence we must have G E {ga}aED, which implies that G K F. Giving us
that G = F.
It remains only to show that F is defined for all x E X. Since F is maximal, this will
follow if we can show that each g that is defined on a proper subspace T of X and satisfies
g(t) < p(t) has a proper extension h.
Let y be an element in X - T. We shall show that 9 may be extended to the subspace
U spanned by T and y, that is, to the subspace consisting of elements of the form Ay + t
with t E T and A E R. Therefore, if h is an extension of g, we must have h(Ay + t)
Ah(y) + h(t) = Ah(y) + g(t). Then h is defined as soon as we specify h(y).
For L1 and t2 E T we have
g(t1 ) + g(t2 ) = g(t1 + t 2) < p(t1 + t2) < P(ti - y) + p(t2 + y).
Hence,
-p(tl - y) + 9( 1 ) < p(t2 + y) - g(t2 )
and so,
lubtET{-p(t - y) + g(t)} inf {p(t + y) - g(t)}.tET
Define h(y) = a, where a E R and
lubtET{-p(t - y) + g(t)} < a < inf{p(t + y) - g(t)}.tET
We must now show, h(Ay + t) = Aa + g(t) < p(Ay + t). Three cases:
(1) If A > 0, then
Aa +g(t) =A[a +g( )t
41
tK A[p(-+y) -g()+g(t)]
t= Ap(X + y)
- p(t + Ay)
(2) If A = -- < 0, then
-pa +g(t)t
= [-a +g(- ]
t t IS"[p(- -y) - g(-)+ g(-)]
t- pp(-y)
= p(t +(-u)y) =p(t + Ay).
(3) If A = 0 then it is clearly true.
Thus, h(Ay + t) p(Ay + t) for all A E R, and h is a proper extension of g. I
Proposition 3.3.2 Let Y be a subspace of the normed linear space X. Let y* E Y*. Then
ther exists x* E X* such that x*(y) = y*(y) for all y E Y and |Jx*jj =|y*|1.
Proof. Let p(x) = I1y*I| -ijjxil. Thus p is a real valued function defined on the linear
space X satisfying the conditions of p in theorem 3.3.1. Now define f on Y by f(y) = y*(y)
for all y E Y. Clearly f(y) < p(y) for all y E Y. Thus by theorem 3.3.1 there exists F
a linear functional defined on X such that F(x) < p(x) = |y* l- xii for all x E X, and
F(y) = f(y) = y*(y) for all y E Y. Clearly IIy*II < lF simply by the fact that F covers
42
a larger domain. Since F(x) < p(x) = I|y*I l-j|xil for every x E X, then we have that
|IFII < I|y*|I. This gives us that h|ll|= |ly*1I and that F is continuous. Hence F E X*. I
Proposition 3.3.3 Let x E X, where X is a normed linear space. Then there is a bounded
linear functional f on X such that f(x) =jjxfj and lf1l= 1.
Proof. Let S be a subspace consisting of all multiples of x and let BS denote the unit
ball of S. Define f on S by f(Ax)= AlixII. Set p(y) = jyll. Note that
1|fAl = sup {lf(y)}yEBs
1
-1.
Then by proposition 3.3.2 there is an extension f E X* such that lIf| = 1 andfi = j. I
Proposition 3.3.4 Let T be a linear subspace of a normed linear space X, and let y E
X such that d(y, T) 6 > 0. Then there is a bounded linear functional f on X such that |f ft|<
1, f(y) = 6, and f(t) = 0 for all t E T.
Proof. Let S be the subspace spanned by T and y. Define f(ay + t) = ab. Then f is
a linear functional on S, and since Ilay + tl = al - Ify + '1 1 >a6, we have f(s) f1sjf on
S. Therefore Ifff= 1. By proposition 3.3.2 we may find an extension f of j on all of X so
that I|f1| = 11fil. By the definition of j on S, we have f(t) = 0 for all t E T and f(y) = 6. 1
43
3.4 Finite Dimensional Normed Linear Spaces
Much of the material used in this section was found in by Joseph Diestel. Any terms or
results not defined or shown may be found in this text.
This next theorem gives us a lot of information on finite dimensional normed linear
spaces. In fact it tells us that all finite dimensional normed linear spaces are essentially the
same. Therefore we will conclude that they are all Banach spaces.
Theorem 3.4.1 If X and Y are finite dimensional normed linear spaces of the same di-
mension, then they are isomorphic.
Proof. Suppose the dimension of X is n. Then we will show that X ~ fl, where l is
defined to be all finite sequences of n terms with the usual V1 norm.
Define the norm on i to be1|(a1, a2 ,...,an)I = Ia+a2+. .. +Ia1, forall(a,a 2 ,...,an) E
fl. Let {X 1,x 2 , - -., x,} be a Hamel basis for X. Define 1:l -+X by
I(al, a2,...,)ai)= 1x 1 + a2 x 2 + ... + an n.
First we show I is linear. Let a and b E 4' and r E R. Therefore
I(ra+ b) = (ra +bi)xi + (ra2 + b 2 )X 2 + ... + (ran+ bn)Xn)
= r(aix + a2 x2 + ... +anzX)+ (bixi + b2 x2 + .. .+ bX,)
- rI(a) + I(b).
Thus I is linear.
44
Next we show that I is continuous. Let (a, a2 , ... ,an) E f. Thus,
II(ai,a2 ,... an) = ||aixi + a2 x2 + .... + anzn||
< |aix1|+11a2x211 + ... + jjannjj
= laillIxll|+|a211X211+ ... +anljjxnl|.
Let K = lub{xllxI: i E {1,, n}}. Hence,
III(a1,a2,..., an)I <; K(ja1|+|a 2 |+ ... .+|ani) = K|f(a, a2,..., an)f.
Next we show I is one to one. We will accomplish this by showing that the ker(I) = {}.
Suppose I(a, a2, ... , an) = 0. This is true if and only if a 1x 1 + a2 x2 + ... + anxn = 0,
but since {x 1 , x2 ,..., xn} is a basis then this implies that (a 1, a2 ,...I, an) = 0.
We now show that I is onto. This is clearly follows from the construction of I and by
the definition of a Hamel basis.
Now to prove 1-1 is continuous, we need only show that I is bounded below by some
m > 0 on the closed unit sphere S of f4; an easy normalization argument then shows that
I1 is bounded on the closed unit ball of X by .
We define the function f : S -+ R by f(a 1,a2 , ... ,an) = jjai1x + a 2x2 + ... + anznxj.
The axioms of a norm quickly show that f is continuous on the compact subset S of R .
Therefore, f attains a minimum value m> 0 at some (a?, a,... , a) E S. Let us assume
that m = 0. Then
I|axii + a 0x2 + ... + ao j =0
so that ai + ajx2 + . . + a nxn = 0. Since {x 1 , x2 , .-- , Xn} is a Hamel basis for X, then
45
(aa2,.. .,an) 0=. But 0 S. Thus m > 0, and jI(a1,a2 , ... ,a)ff > m > 0. Thus we
have that I-1 is 'continuous, and hence defines an isomorphism between X and 4n.
We can similarly prove that Y 2 'i, and thus have X e Y.E
Corollary 3.4.2 Finite dimensional normed linear spaces are complete.
Proof. Let Y be a finite dimensional normed linear space. Then Y is isometric to l,
which is complete. Hence Y is complete. U
Lemma 3.4.3 (Riesz's Lemma) Let Y be a proper closed linear subspace of the normed
linear space X and let Sx denote the unit sphere of X. Choose 0 < 0 < 1. Then there is
some Xo E Sx such that I|xo - yf| > 0 for all y E Y.
Proof. Let x E X - Y. Since Y is closed, the distance from x to Y is positive.
d0 < d = glb{jjx - z|: z E Y} < -;
Therefore there exists z E Y such that ix - z||< K#- Let xo = x . Clearly Xo E Sx since
llxoll = 1. Furthermore, if y E Y, then we have
X - zIXO - Y1-= Il- y
|x -zZ11x-11
llx z i x - z|lylx - Z11 ~ |X - Z11 ~ |X - Z11
- II -Ix - (z +IIx - zlly)|lIX - Z11
> -d=0.d
I
46
In the next theorem we will conclude that a space in which every closed bounded subset
is compact must be finite dimensional.
Theorem 3.4.4 In order for each closed bounded subset of a normed linear space X to be
compact, it is necessary and sufficient that X be finite dimensional.
Proof. If the dimension of X is finite then X I l for some n E N. Thus, the compactness
of closed bounded subsets of X follows from the Heine- Borel theorem.
If the dimension of X is infinite, then SX is not compact, although SX is closed and
bounded. We prove this by constructing a sequence in SX so that for any distinct m and
n, 1IXm - Xnfj > I12
Let x1 be an element of SX, thus < x 1 > is a closed linear subspace of X. Hence by
lemma 3.4.3 there exists x2 E Sx such that Jfx2 - axJ1 > , for all a E R. Thus < x1 , x2 >
is a closed linear subspace of X. Hence by lemma 3.4.3 there exists x3 E SX such that |1x3 -
ax2 -/x1iI> 3, for all a and # ER. Continue this process inductively. Thus we have {x,}
a sequence such that for every m and n distinct positive integers I|xm - x, 1 .
We can conclude thus conclude the following concerning infinite dimensional normed
linear spaces.
Corollary 3.4.5 If X is an infinite dimensional normed linear space then there are bounded
sequences with no norm convergent subsequences.
Proof. From the proof of theorem 3.4.4 we see that such a sequence in fact exists within
Sx. I
CHAPTER 4
CONVERGENCE OF SEQUENCES IN BANACH SPACES
4.1 Weak and Norm Convergence of Sequences
Definition 4.1.1 (Norm Convergence) Let X be a normed linear space. We say a se-
quence {xn} of X converges in norm to x E X if for all c > 0 there exists N E N such that
for all n> N, I|x, - xj| < c (denoted xn - x).
Definition 4.1.2 (Weak Convergence) Let X be a normed linear space. We say a se-
quence {xn} of X converges weakly to x E X if x*(xn) - x*(x) in norm, for all x* E X*.
Definition 4.1.3 (Weak Cauchy) Let X be a normed linear space. We say a sequence
{xn} of X is weakly Cauchy if {x*(xn)} is a Cauchy sequence for all x* E X*;
Theorem 4.1.4 If {xn} is weakly Cauchy, then {xn} is a norm bounded sequence.
Proof. Let X be a normed linear space and {x} a sequence in X such that {xn} is
weakly Cauchy.
For all n E N, think of xn : X* -+ R by x(x*) = x*(xn). Notice that X* is a Banach
space and for all n E N, x, is a bounded linear operator.
Let n E N, let x* and y* E X*, and let a E R. Thus
xn(x* + ay*) = (X* + ay*)(Xn)
- x*(xn) + ay*(xm)
47
48
= Xn(X*) + aXn(y*).
Hence, xn is linear for all n E N. We also have that
IXn(X*)I=IX*(Xn)I 5 x*1 I-kXnIj,
and so for all n E N, X is bounded.
Now, since {xn} is weakly Cauchy, we get that for all x* E X* there exists Mx* E R
such that Ixn(x*)I 5 Mx. for all n E N. Thus by 3.2.1 {xn} is uniformly bounded. That is,
there is some M E R so that jjx1 nj < M for all n E N. I
Theorem 4.1.5 Let X be a normed linear space and let {xn} be a sequence in X. If {xn}
converges in norm to some x E X, then {xn} converges weakly to x. Also, if {xn} converges
weakly to some x E X, then {x,} is weakly Cauchy.
Proof. First we will show that norm convergence implies weak convergence. Suppose
{xn} is a sequence in a normed linear space X such that {x} converges in norm to x E X.
Now let x* E X*. Since x* is continuous, then clearly x*(xn) converges to x*(x). Hence we
are done.
Suppose we have {xn} a sequence of X such that {xn} converges weakly. Hence x*(xn)
exists for every x* E X*. Thus {xn} is weakly Cauchy. I
Theorem 4.1.6 The reverse implications of the previous theorem do not hold.
Proof. In c0, let {60} be defined by
1 if n= k
0 otherwise.
49
Now we will show that k -+ 9 weakly but not in norm.
Let x* E c*. From previous theorem we know that there exists a E P such that for all
x E c,x*(x) = E' 1 anx,. Therefore x*(bk) = anbn = ak. Therefore x*(bk) + 0,
since ak -+ 9. Thus bk -+ 0 weakly in c.
Note that 114 - bjj = 1 if j # k. Thus { 6k} has no norm convergent subsequence.
In cO, let 7Ik be defined by
1 if n<k
0 otherwise.
Now we will show that {Yk} is weakly Cauchy but not weakly convergent.
Let x* E c*. From corollary 2.3.4 we know that there exists a E V such that for all
x E co, x*(x) = E*_, a na. Therefore x*($n) = Zn k = a1 +a2 +...+ak = 1 an.
Thus, {x*(7k)} is the sequence of partial sums of J:' 1 an. Since E',an is absolutely
converging in R, {x*(yk)} is Cauchy. Hence {Y} is weakly Cauchy in co.
Now we will show that {7} is not weakly convergent in co. Let bk be defined as before.
Thus 6k E E . Therefore for all p E N, we can define x* E c* by x*({y }) ="n, for
all {yn} E c. Therefore X*({yn}) = y for all {y} E c. Suppose {7k} weakly converges
to 7 E c. Thus for all x* C c*, {x*(7k)} converges to x*(y). Therefore for all p E N,
x*(7k) -+ x*(y). But
00
-(k = ,k = 7PXp(-Yk) p k kn=1
Therefore we have that for every p E N, x*(yk) = 7y. Thus limk_., xyk) = 1 for all
p E N. This implies that for all p E N, x*(7) = 1. Hence for all p C N, y' = 1. Implying
that {-} = (1, 1, 1,...) which is not an element of co. This is a contradiction. Therefore,
{ -Yk} does not converge weakly. U
50
We conclude this section with the following two theorems.
Theorem 4.1.7 Let Y be a subspace of X, a Banach space. Let {xn}, be a sequence in Y.
Then {x,} is weakly Cauchy in Y if and only if {xn} is weakly Cauchy in X.
Proof. Suppose {xn} of Y, is weakly Cauchy in Y. Let x* E X*. ThusX E Y*. There-
fore for all e> 0, there exists N E N such that for all m and n > N, I|x (xm)-x *(xn)II <
c. This implies that IIx*(xm) - x*(Xn)I| < c. Hence {x*(xn)} is Cauchy. Therefore, {xn} is
weakly Cauchy in X.
Now suppose {xn} of Y, is weakly Cauchy in X. Let y* E Y*. By proposition 3.3.2, there
exists x* E X* such that x* = y* and f|x*jj = |ly*II. Thus for all c> 0, there exists N E
N such that for all m and n > N, IIx*(xm) - x*(xn)II < c. But since for all n E N,
x*(xn) = y*(Xn), then IIy*(Xm) - y*(xn)II < e, for all m and n > N. Thus {y*(Xn)} is
Cauchy and hence {xn} is weakly Cauchy in Y. I
Theorem 4.1.8 Let Y be a closed subspace of X, a Banach space. Let {xn} be a sequence
in Y. Then {xn} is weakly convergent in Y if and only if {xn} is weakly convergent in X.
Proof. Suppose the sequence {xn} of Y, is weakly convergent in Y. Suppose that
y*(xn) -+ y*(x), for some x E Y, and for all y* E Y*. Let x* E X*. Thus x E Y* and fory
all E > 0, there exists N E N such that for all m and n > N, Ixj'(Xm) - xr(x)I1 < f.
This implies that IIX*(Xm) - x*(xn)ji <cE. Thus {x,} is weakly convergent in X.
Now suppose that {xn} is weakly convergent in X. Suppose that x*(xn) ->X*(X),
for some x E X, and for all x* E X*. Let y* E Y*. By proposition 3.3.2, there exists
x* E X* such that x* = y*. What we need to show now is that x E Y.
51
Suppose x Y. Hence d(x, Y) > 0 since Y is closed. Thus let a E R such that d(x, Y) =
a. Thus there exists f, a bounded linear functional on X, such that If II = 1, f(x) = a, and
f(y) = 0 for all y E Y. Thus, by proposition 3.3.4, f E X*, and f(x") = 0 for all n E N.
Hence f(x,) -+ 0. This is a contradiction. Hence, x E Y.
Therefore, xry(xn) = x*(xn) for all n E N, and x* (x) = x*(x). Therefore {xf(xn)}
converges to x* (x). Then y*(xn) converges to y*(x). Hence {xn} is weakly convergent in
Y.E
4.2 Weak Convergence in 2 and f'
We recall from corollary 3.4.5 that infinite dimensional normed linear spaces have bounded
sequences with out norm convergent subsequences. The behavior of weakly convergent
sequences in normed linear spaces will not be as predictable. For instance we show that
every bounded sequence has a weakly convergent subsequence in 2. We first prove the
following lemma.
Lemma 4.2.1 Let {xn} be a bounded sequence of elements in 2 and suppose that limnO xk =
0 for all k E N, then xn -+ 9 weakly.
Proof. We want to show that for all x* E (2)* and for all e> 0, there exists N E N
such that for all n > N, fx*(xn)l < c. Let x* E (2)* and e> 0. Thus, there exists {an} E 2
such that for all {y} E 2, x*(yn) = Elay
Since {xn} is bounded there exists M E R such that |xInj < M for all n E N. Since
{an} E , then E', a exists. Thus there exists N, E N such that for all m > No,
00 a2 - E",L alI < (A:)2 . This implies that IE'=N. a <(1< ) 2 . Also, since for all
52
k E N, Xkim =x 0, there exists N E N such that for all n > N and k = 1,...,N- 1,
jXkI < Cn 4(No - 1) ENO~' a2a=1 k
as long as ENO-" a2 $0. For n > N,
00 No-1 00Zakn< _ak Xk k
Ix*(xn)I= x <+ axx1 k =1 k=No
No-1 c0
kakakxn +I: akXxk=1 k=No
No-1 No-1 00 00
Z a2 E(Xk)2+ + :a{ (k)2k=1 k=1 k=No k=N,
No-1 No-1 C2 EM< (a 2 1 + --
k 4(No - 1) ZO1 a 22M
E E
2 2
I
Theorem 4.2.2 In 2 bounded sequences have weak convergent subsequences.
Proof. Let {xn} be a bounded sequence in 2. Thus for all k E N, {xk} is a bounded
sequence in R. Hence, for {x,} there is some subsequence {xn,} C {x,} such that xi - xi
where x1 E R.
Now, {xa,1 } is a subsequence of {xn}, thus it is bounded. Therefore there exists {xn2} C
{xn} so that x22 -+ X2 where x 2 E R.
Continuing this procedure inductively we get {x+1 } is a bounded sequence in R, and
so there is some subsequence {xnk+ 1 9 Xnk } such that xk+l - xk+1 where xk+l E R.
53
Now for all n E N we define {y,} = {x,}, and let x = {xk}g 1 . Thus we have
{y,} ; {x,} such that y -+ xk for all kE N, since for every k E N, {y} is eventually in
{Xn,}.
We claim next that x E 2. Suppose it is not. Suppose M is a norm bound for {x}. Since
x 2, then E',(x k) 2 = oo. Thus, E_1(Xk) 2 > M2 . Thus there is some N E N such
that for all m > N, Ek(Xk) 2 > M2 . This is true if and only if Ek"_1(im-0 y) 2 > M2 ,
and that is true if and only if lims, LE_1(y') 2 > M2 . Thus, there is an n E N such that
1 > M 2 , but Zki 1 (yn) 2 (yn) 2 > M2 , which is a contradiction. Thus
x E j2.
This gives us that for all k E N, y - xk -+ 0. This implies that y, - x -+ 9 weakly.
And hence, for all x* E ( 2)*, X*(yn - x) -+ 0. This is true if and only if x*(yn) - x*(x) -+ 0.
Thus x*(yn) -+ x*(x), and hence y -* x weakly. I
We contrast the results of the previous theorems by showing that the only weakly Cauchy
sequence in 01 are in fact norm convergent.
Lemma 4.2.3 Every weakly null sequence in 1 is norm null.
Proof. Suppose that {x4} is a sequence in 1 which is weakly null. Thus {x} is bounded
in norm. This implies that there exists an M E R such that lx,|j < M for all n E N. We
claim this implies that for all k E N, lim. x = 0.
To prove this claim, let k E N and define the sequence {6,} of f' as before. Thus x*
defined by x*(xr) Z x,6 is in (01)*. But, *(x,) = x4. Since lima+c x*(x,) = 0
for each k E N, we have lim_.oo x = 0 for each k E N.
54
Now, suppose {xn} is not norm null. Since {jjxIj} is bounded in R and does not
converge to zero, there must be a subsequence {y,} of {x,} such that 11y,11 > L > 0 for all
n E N and for some L E R where L > 0.
Now we will find a subsequence of {y} which is not weakly null. We will create this
subsequence inductively. Starting with n, = 1.
Since {yk' } is in fl, E'_1 yk, I exists. Thus, there is an N 1 E N such that Zgi=N, I I
y. But since ffyn,11 > L ZNIlyk If> 2L
Since y -+ 0 for all k E N, then for every j < N1, there exists a Pj E N such that for
all n > P,j< < . Let P = lub{P, : j < N1 }. Choose n2 E N so that n2 > ni and
n 2 > P. Thus Iy 21 < L for all j 5 N1 , and hence EN' yAl < L. As before, there exists
N2 E N such that N2 > N1 and E=N2 +1 k Lf2 < . Thus, EN2
Continue this process inductively. Since y -+ 0 for all k E N, then for every j Ni,
there exists a Pj E N such that for all n > Pj, yjI < L. Let P = lub{P, : j K Ni}.6N(
Choose n2+i E N so that n-+1 > ni >--> n 1 and n-+ 1 > P. Thus Iyi 1 < L for all
j<Ni,thusZENIJy4, I<L. As before, there exists Ni+1 E N such that Ni+1 > Ni and
Zk=N2 +1 IA+1 I < .L Thus, E I+A+1 I>
Now define {Mk} c P(N) by M 1 = {1,...,N 1}, and Mi = {N_1,...,Ni} for all
i E N - {1}. Thus the Ms are pairwise disjoint and
(1) ZjEM yJ| > , and
(2) Z J~ Y2?I <4L
Now define : - R by q(zn) = 1 a z for every {z} E EV and where a3 =
sgn(y3) if j E M . Thus {a} E E and therefore 0 E (0)*.
55
Notice that N jEyk a =y = E MyIk > -, and that Z k ajyV k - ZJMk I4
yT rore 4(y) = EjEyasyi + Ej, a y' > I L- - Hence, for every k E N,
(yk) > 4. Thus y, cannot converge to 0 weakly, which is a contradiction. Therefore {x,}
is norm null. U
Theorem 4.2.4 (The Schur Property) In V weakly Cauchy implies norm convergence.
Proof. Suppose {x,}, a weakly Cauchy sequence in l that does not converge in norm.
Then there exists a subsequence {xn,-} of {xn} and e> 0 such that jx,., - I f,, > 1 for
all j E N. Since {x,} is weakly Cauchy, it follows that {xn,} is weakly Cauchy and hence
the sequence {xn, - X ,,} is weakly null. By lemma 4.2.3, {x,, - X,,l } must be norm
null, which contradicts IIXT,- - Xni,1 >1c for all j E N. U
CHAPTER 5
ROSENTHAL'S THEOREM
5.1 Equivalence to l-Basis
Much of the material in this chapter was taken from A Characterization of Banach Spaces
Containing fl, by H.P. Rosenthal [4], and from Applications of Ramsey Theorems to Banach
Space Theory, by E. Odell [3].
Definition 5.1.1 (Equivalence to l-basis) A bounded sequence of {xn} in a Banach
space X is said to be equivalent to the usual 1 basis, denoted {x} ~ {e71 }, if there exists
6 > 0 such that for all n and choices of scalars, C 1, ... 7,c,
n n
6Zci<; Z cixi
i=1 i=1
Proposition 5.1.2 Let {xn} be a bounded sequence in X such that {x,,} is equivalent to
the usual 1' basis. Then the mapping T :f'1 -+ [xn] (the closed linear span of {xn}) defined
by T(e 1) = xn or T({an}) = E 1anx is an isomorphism onto [Xn].
Proof. Let 6 > 0 be such that
n 7n
6ZIci Zcixi
i=1 i=1
for all n E N and c1 ,...,c, scalars.
Next we wish to show that T is well defined. Let {an} E fl. We need to show
that E a7a x exists. To this end let k E N and consider E a71 7 . Observe that
56
57
EZi =1IanxnII < f1sup{xn} -. 1 klanj 1 1sup{xnj -EiI lanf. Thus the sequence of
partial sums of ZEn I|a~nxII is bounded. Hence J:'1 Ianx converges, which implies
that E*_1 anzx converges. Hence T is well defined.
Now we want to show that T is a linear transformation. Let {an},{bn} E l and c E R.
We get
00
T(an+-cbn) =Z(an + cbn)xnn=1
00
= Z(anxn+cbnxn)n=100 00
- Z(anxn) + >3(cbnxn)n=1 n=1
00
- T(an) + c Z(bnxn)n=1
- T(an) + cT(bn)
Giving us that T is linear.
Next we will show that T is one to one. We will do this by showing that ker(T) = {}.
Suppose {an} E V1 such that T(an) = 0. Hence, 'a 1 = 0. However, 6 E ja;f
I a xijf. Taking the limit on k and using the fact that * is continuous, we get
6 I l fanf II anznxf. But recall that jc anxn|j = IIT(an)II = = 0 (in X).
Thus, 6 ' lanj = 0. Implying that _'|anf = 0, and therefore {an} = 0.
Now we will show that T is continuous by show that T is bounded. Letting {an} E l
we have,
IT(an) =- Sanxnn=1
58
00
K Ejanxn||n=1
00
I:SlanL -IIXnIln=100
< (kanjlubIjXn|I)n=1
00
- lubI|xnIIE|ani.n=1
This is true for any {an} E fl, hence T is continuous.
Finally we wish to show that T is onto. Letting x E [xn] we have that x = lim7 2 ,m an
where {an} is a sequence of elements in < xn >, the linear span of {xn}. Thus, for every
n E N,kn
an = Zc xi.
i=1
Therefore,k,
x = lim3C~xi.n-+oo
i=1
Now define {b} a sequence of elements in 0l by,
cn if m <k,b n=
10 otherwise.
Notice that for every n E N,
00kn
T(b) = Zb~xi = L c = an.i1 i=1
Therefore, lim72 .00 T(bi) = im7 _,0 a7 = x.
Since {a,} converges in norm, {an } is Cauchy in norm. Hence for all c > 0 there exists
N E N so that for all p, q > N,
Hap - aqjj < .
59
But recall Ila, - afl ZPi cx; - EkA c= X . Thus by the appropriate substitution we
can write jfap-aqf11as|E , bx - E' b1xll. But this give us, lap-ag I1= lZlEg(b - b)xll.
Thus, for all n E N,
6 Kb - biq| < (bP - bq )xi < C.i=1
This gives us that for every E > 0 there exists an N E N such that for all p > q N,
oo kp
Elb -b=E lb- -b||=||bP - bq||< C.i=1 i=1
Implying that {bf} is a Cauchy sequence in V. Since 0i is a Banach space, we have
b = lim_. bn E . Therefore, since T is continuous we have,
T(b) = lim T(b h) = lim an = x.n-.oo n--oo
Hence T is onto. U
Proposition 5.1.3 If {xn} ~~ {e} then every subsequence of {xn} is equivalent to {en}.
Proof. Let {xn1 1 } be a subsequence of {xn}. Let 6 > 0 such that for all k E N and
C1 ,...,Ck E Rk k
6Zlcn >ICnxn.
n=1 n=1
Let k E N and let c1 ,..., cC E R. Consider the sum = 1cgzj. Define {a, . . ., aI,} by,
c if m=njam
{0 if m#n3
Thus, M= amxm = 1 CX, and " laml = = 1Icf. Therefore, {xI} f{e}.
I
For our next proposition we will need to first prove the following lemma.
60
Lemma 5.1.4 If T : X -+ Y is a continuous linear operator, then T* : Y* -+ X* by
< T*(y*),x >=< y*, T(x) > is a continuous linear operator.
ProoL We first need to show that T* is well defined. To this end we let y* E Y*. Now
consider T*(y*). By definition this is < T*(y*),x >=< y*,T(x) >= y*(T(x)). Since y* is
a continuous linear operator and T is a continuous linear operator then T*(y*) is just the
composition of continuous linear operators which in turn is a continuous linear operator.
Giving us what we wanted.
Now we will show that T* is linear. This fact follows from the following equations.
< T*(y* + cx*), x > = (y* + cx*)(T(x))
= y*(T(x)) + cx*(T(x))
= < T*(y*), x > +c < T*(x*), x >
Finally we wish to show that T* is continuous. Letting y* E Y* we get,
< T*(y*), x > 11= Ily*(T(x))I< I1y*II -IIT(x)I|.
Proposition 5.1.5 If {x} ~. {en}, then {x} has no weakly Cauchy subsequences.
Proof. First we will show that {en} has no weakly Cauchy subsequences. Since in
weakly Cauchy is the same as norm convergence, then it is sufficient to show that {en} has
no norm convergent subsequences. Let {enj} be a subsequence of {en} and let i f j E N.
Hence, Ieni- en, = 2. Thus {enj } is not norm Cauchy, hence it cannot be norm convergent.
61
Define T : [x,] -+ +1 by, T(x,) =-e Thus, T is a continuous linear operator. Now
define T* : (i1)* -+ [x,]* by < T*(y*),x >=< y*,T(x) > for all y* E ( 1)*. Thus, T* is a
continuous linear operator.
Now let {x,} be a subsequence of {x,}. Thus since {x,} ~ {e,} then {xn,} {en,}.
But since {ej, } is not weakly Cauchy, then there is some y* E (el)* so that {y*(en,)} is not
Cauchy. Since T*(y*) E X*, we have
< T*(y*),x,, >= y*(T(Xnj)) = y*(en )
which we know is not Cauchy on [Xn]. Hence, by theorem 4.1.8 {x,,,} is not weakly Cauchy
in X.E
5.2 Main Theorem
In the previous chapter, we noted that the only weakly Cauchy sequences in i' were norm
convergent. Thus, in fl, there are many bounded sequences without weakly Cauchy subse-
quences. We show as our main theorem that in some sense the only way a bounded sequence
in a Banach space X can not have a weakly Cauchy subsequence is if it behaves like the
unit vector basis of 0.
We will construct the following example to help develop an intuitive idea of some later
definitions and our main theorem.
We will need the following definition for this example.
Definition 5.2.1 (Rademacher Functions) Let A be the dyadic intervals in [0, 1].
62
That is to say for a fixed n,
A .- 1, , for i = 1,2,.. ,2 n; n = 0, 1,...
For example, for n = 0, i = 1; Ao, = (0,1]. For n = 1, i = 1, 2; A1 , = (0,1], A1 2 = (},1].
We define r(t) = 2 _l(-1)i+XA (t) for all t E [0,1] and n E N.
Example: Let rn(t) be the Rademacher functions on [0,1] in f ([0, 1]). Let {q}!U 1 be a
finite sequence of +1. Then there is a unique j such that I < j 2 and if t E An,, then
r(t) = e.
The claim is that {rn} {en} in f [0,1]. We can show this by letting n E N and
C,..., cn E R. Thus we have the following inequality.
n =1
>3cric< > -1cirloo
i==1n
= 1?Icil - IriIloon
> cI, since 1riH = 1.7=1
Hence, ffDn1 Cr;||o ZE IIc;I. Now choose qi so that ccZ = jcj for i = 1, ... , n. Thus,
there exists some j such that if I E A, then r(t) = ei for all i. Therefore,
n n n
cir >3er(t) = > c;i.
1 o i=1 i=1
Hence we have for all n E N,
n n7 T1 < 71 r
i=1 i=1
And we are done.
63
We will first consider the main theorem for the case X = f 0 (S), the space of bounded
functions on a set S. We show that if {ff} j g(S) does not have a weakly Cauchy
subsequence then it must have a subsequence that behaves like the Rademacher function.
Definition 5.2.2 (Independent) A sequence of subsets {An, B}0 1 of a fixed set S with
AnABn = 0 for all n E N, is said to be independent if for all {c;} 1 , with ei = 1,
nk k= $Ai 0, where
Ai if Ci = 1
B if i = -1.
Example: If n E N, An {t E [0,11: r (t) = 1} and B = {t E [0,1]: rn(t) = -1}. Then
{An, Bn} is clearly independent.
Lemma 5.2.3 Let {fn} be a uniformly bounded sequence of real valued functions on a set
S, and let r E R and 6 > 0. Let B, = {s E S: fn(s) < r} and An = {s E S : fn(s) > r+6}.
Assume that {An Bn} is independent. Then {f,} is equivalent to the unit vector basis of
f in supremum norm.
Proof. Let {a;}i= 1 be a real valued finite sequence. It is enough to show that
k ak
a f > 2 ail.
=1 00 i=1
Let F = {i : ai > 0}, G = {i : a < 0}. Let s E (fiEFAi)fnl(AiEG Bi) and t E
(AiEG Ai) fl (iEF B) . Then
k
aifi(s) >3 E ail(r + 6) - 3 lail(r)i=1 iEF iG
and
k
- > aifi(t) > - > ail(r)+ E laiI(r + 6).i=1 iEF iEG
64
Thus
k k
ai(ft(s) - f;(t)) b kja;l
i=1 i=1
Implying that
k (kk k
aif; > lub a-f;(s) , a fi(t) - ail.
Definition 5.2.4 (Convergence) Let {An, Bn} be a sequence of subsets of a set S with
A. f Bn = 0 for all n C N. {An, B} is said to converge on S if for all s E S, s belongs to
at most a finite number of A' s or a finite number of B' s.
Definition 5.2.5 (jM-Work) Let {An, Bn} be a sequence of subsets of a set with An f Bn =
0 for all n E N. We shall say that j E N and M (infinite) C N work if {An, BnInEM has
no subsequence convergent on either X fl A or X l Bj.
Without loss of generality, suppose X = S and {A, B}TnEN have no convergent subse-
quence on X. The following will be referred to as the Basic Algorithm and is constructed
as follows. Let ni E N. If n1 an N do not work, then let N 1 (infinite) C N such that
{ An, Bn}nEN1 converges on An 1 or Bn,. Suppose that k > 1 and the infinite subset Nk-1
of N and the element nk_1 of N have been defined. Choose nk E Nk-1 so that nk > nk-1.
If nk and Nk_1 do not work, let Nk (infinite) Q Nk-1 so that {A, B}nE-Nk converges on
Ank or Bnk.
Proposition 5.2.6 If n3 's and N1 's are selected by the Basic Algorithm, then there is some
k > 1 such that nk and Nk_1 work, where N = N.
65
Proof. Suppose that such a k does not exist. Then we obtain nk, Nk and n, = 1 for
each k such that nk > n_1 and nk E Nk-1, Nk g Nk-1, and {A, B,}nEN, converges on
en, Ank , where no = 0.
Now let M == {ni,n2 ,...}. Then for all k, {AniBn}nEM is a tail-end subsequence of
{An, Bn}nENk. Hence, {An, Bn}lEM converges on U 1 Enk Ank. Now we can choose an in-
finite subset M' of M so that cm = 1 for all m E M' or cm = -1 for all m E M'.
Suppose that the former is true. Since {An, Bn}nEM' is a subsequence of {An, Bn}nEM
and UnEM, An = UnEM'EnAn C U 1 efln Ank, then {An, Bn}nEM, converges on UnEM, An.
Since {An, Bn}n,=M, does not converge, then there exists and x so that {n E M' : x c An}
and {n E M': x E B,} are infinite. But then x E UnEM, An, and hence {An, Bn}nEM, does
not converge on UnEM'EnAn. This is a contradiction. The latter case is similar. I
Lemma 5.2.7 Let > 1, {An, Bn}nEN a sequence of pairs of subsets of a set S with
An f B, = 0 for all n E N, X 1 ,... , Xi disjoint subsets of S. Suppose for all i such that
1 i 1, {An, Bn}nGN has no subsequence convergent on Xi. Then there exists j and
M (infinite) C N so that for all i such that 1 < i 1, {An, Bn}nEM has no subsequence
convergent on Xif A and no subsequence convergent on X fl Bj.
Proof. Let 1 = 1. Suppose that {An, Bn}nEN are independent and let X be a subset of
S such that {An, Bn}nEN has no convergent subsequence on X. Without loss of generality,
suppose that X = S. Now, by applying the Basic Algorithm and the proposition, we have
the case for 1 = 1 of our lemma.
Now suppose that our lemma is true for I = r and let Xi's and {An, Bn}nEN satisfy the
hypothesis for the I = r+1 case. We shall say that j and M r-work if for every i c {1, ... , r}
66
and c =- 1, {Ar, B,}CM has no convergent subsequence on Xi cA. By the induction
hypothesis, we may choose ni and N (infinite) C N so that ni and Nf r-work. If ni and N
do not work, choose N 1 (infinite) G N1 so that {An, B,}nEN 1 converges on An, l Xr+ 1 or
on Bn, f X,+i. Suppose that k > 1, and that the subset Nk-1 of N and the element nk-1 of
N have been defined. Since {An, B}nEN9 {Ani, B}nEN, we may apply the induction
hypothesis to choose an nk E Nk-1 with nk > n-1 and an infinite subset N of Nk1 so
that nk and Nk r-work. Again, if nk and Nk do not work, choose Nk (infinite) g N such
that {An, Bn}nEkNk converges on Ank f Xr+i or on Bn, l Xr+ 1 . Now, this process cannot
be continued indefinitely, since the ink's and Nk's, constructed this way satisfy the criteria
of the Basic Algorithm and {An, B}nEN has no subsequence convergent on Xr+1. Thus,
there exists k > 1 such that nk and N' work. By construction, nk and N' r-work. Hence,
nk and N' satisfy the conclusion of this lemma. U
Theorem 5.2.8 Let {An, Bn}nEN be a sequence of pairs of subsets of a set S with An f Rn =
0 for every n E N, and suppose that {AnBn}nEN has no convergent subsequence. Then
there exists M (infinite) C N such that {An, Bn}nEM in independent.
Proof. Applying Lemma 5.2.7 for the case 1 = 1, choose ni and M 1 an infinite subset
of N so that {An, Bn}nEM, has no subsequence convergent on either An,, or Bni. Suppose
n, < n2 < ... < nk and Mk have been chosen, so that on each of the 2 k disjoint sets
n _ An,, {A , Bn}nEM, has no convergent subsequence, when == (1f,...B, ek) ranges over
all choices of signs 1 for all i. Now applying Lemma 5.2.7 to I = 2 k choose nk+1 E Mk
with nk+1 > nk, and Mk+1 (infinite) g Mk so that for every c, {An, Bn}nEMk+ has no
subsequence convergent on n _1(e An,) f Ank+, and also no subsequence convergent on
67
n k cAy) f B,+1. Thus, we construct the n's and M's inductively. Letting M
{In, n2 , ... } we get that M satisfies our conclusion. I
Theorem 5.2.9 Let {fn} be a uniformly bounded sequence of real valued functions on a
set S. Then there is a subsequence {f,} which is either pointwise convergent on S or which
is equivalent to the unit vector basis of 0 in the supremum norm.
Proof. Assume that {fn} has no pointwise convergent subseqence on S. For r E R and
6 > 0 let
A6 = {s : f(s) > r + 6} and
Br6= {s : fn(s) < r}.
The claim is that there exists r, 6 and M (infinite) C N such that {A'1, B }nEM has no
convergent subsequences.
Suppose the claim is not true. Let {ri, i}' be dense in R x R+. Choose an infinite
subset of M1 of N so that {A' 1, Bnr 1 }nEM converges on S. Inductively choose a sequence
{M} of infinite subsets of N so that Mi+i C Mi and {AE''5 , B"'6'I}nEM converges on S
for all i E N.
Let M be a diagonal sequence (i.e. M - Mi {ni , .... ,ni- 1 }, where n5 E M3 for all
j E {1,... , i - 1}). We claim that {fU}nEM is pointwise convergent on R.
To prove this claim we first let s C S. Since {f }EM is uniformly bounded then
{fn(s)}nEM is a bounded sequence in R. Thus there exists a subsequence {fn(s)}nEM'
which converges in R. Suppose that {fn(s)}nEM, converges to L E R. We claim that in
68
fact {ff(s)}IEM converges to L. Let e > 0. Thus there exists N E M' such that for all
n E M' with n > N0 , fn(s) - LI <C. -Now choose i E N so that r < L, IL - rf < ,
and 6b < .. This gives us that the open interval (ri,ri + 6;) C (L - E, L + c) and that
L E (ri,ri + 6?). Thus we can find an Ni E M' such that for all n E M' with n > N1 ,
ffn(s) - LI < min{IL - rj, IL - ri - 6;I}. Thus, fn(s) E (ri, ri + 6;) for all n E M' where
n > N1.
Since {A'n 8', Bn"' },EM, converges on S then for at most finitely many n E Mi, E An
or s E B. Without loss of generality suppose the former case. Thus there exists an
N2 E Mi such that for all n E Mi with n > N2 , fn(s) r + 6?. Hence for all n E M with
n > N2, fn(s) 5 r;+ bi.
Now choose j E N so that L -c < r < rj +63 <ri. Thus the open interval (r, rj+6) C
(L - c, L + c). Since {Ar'6'1,Br?"' }nEM, converges on S then for at most finitely many
n E M, s C Arn?' or s C Br''. The former case cannot be true since for all n E M' CMj
with n > N1 we have f,(s) > r? > r2-+ 6b. Therefore there exists an N3 E M, such that for
all n E M2 with n > N3 , fn(s) > rj. Hence for all n C M with n > N3, f,(s) > r-. Letting
N = max{N 1 , N 2 , N3} we have that for all n C M such that n > N, rj <ffn(s) +ri+6.
Thus fn(s) C [r, r; + bi] C (L - c, L + c). Hence, {fn}InEM is pointwise convergent on R.
This contradicts our original assumption. Hence, the claim must be true.
Thus by theorem 5.2.8 {A'1, B }nEM is independent. Which, by lemma 5.2.3 implies
that {f'} is equivalent in supremum norm to the usual V basis. I
Theorem 5.2.10 (Main Theorem) Let {fn} be a bounded sequence in a real Banach
space B. Then {fn} has a subsequence {f,'} satisfying one of the following two mutually
69
exclusive alternatives:
(1) {f,} is a weakly Cauchy sequence.
(2) {fj is equivalent to {e}, the unit vector basis of t'.
Proof. Let {f,} be a bounded sequence in B. Let S denote the unit ball of B*. Define
f,(s) = s(f,) for all s E S and for all n E N. Notice that the f,'s are uniformly bounded
on S. We have that for all s E S and for all n E N
|fMS)l = 18(fn)l < 113s1 - Jfnj| <-M,
where M is a bound for {fn}. Thus by theorem 5.2.9 there exists {fn} a subsequence of
{fI} such that either,
(1) {fn} is pointwise convergent on S, or
(2) {f,} is equivalent in supremum norm to the unit vector basis of t1.
Suppose the first case happens. Then for every s E S, {ff(s)} converges. That is to
say, lim. f,(s) exists for all s E S. This is true if and only if lim_.x s(f,) exists for
all s E S. Now let b E B*. Also since 9 E S, suppose b 4 9. Therefore b E S. Thus
lim.,( 1 )b(fl) exists, implying that lim_. b(f,) exists for all b E B*. This gives us
that {f } is weakly Cauchy.
Now suppose that the second case happens. Therefor there exists a b6> 0 such that for
all n E N and choices of scalars, C,... , c
n n
6ZIciI ZcifI
i=1 i= 00
70
However, from the following inequalities we get {fe} ,
n n
6ZIciI|< ctfi=1 i=1
= lubSs { cifj(s)}n
lubses cis(f)}n
lubSEs is (c
< lubses{sI - c f}
n
< Z cil- fii=1
Thus, we have our conclusion. U
BIBLIOGRAPHY
[1] Joseph Diestel, Sequences and Series in Banach Spaces, Springer-Verlag, New York
(1984).
[2] Richard Johnsonbaugh and W.E. Pfaffenberger, Foundations of Mathematical Analysis,
Marcel Dekkar, Inc., New York (1981).
[3] E. Odell, Applications of Ramsey Theorems to Banach Space Theory,in " Notes in Ba-
nach Spaces " (H.E. Lacey, ed.), University of Texas Press, Austin, Texas (1980), 379-
404.
[4] H.P. Rosenthal, A Characterization of Banach Spaces Containing ', Proc. Nat. Acad.
Sci. (USA) 71 (1974), 2411-2413.
[5] H.L. Royden, Real Analysis, 3rd ed. Macmillan Publishing Company, New York (1988).
71