Hassani S. Mathematical Physics: A Modern Introduction to Its Foundations
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right-shift
operator
16.2
BOUNDED
OPERATORS
IN HILBERT
SPACES
453
discrete case, u; can only
"hop"
from i to i+I
and
thenbackto i.
Snch
a difference
(as opposedto differential) process is
notlocal;
it involves
not
only i
but
also i+1.
The
"point"
i does
not
have an (infinitesimally close) neighbor.
This essential difference betweendiscrete
and
continuous operators makes the
latterfar richerin possibilities for applications.
In particular,
if
L(x)
is considered
a differential operator, the equation
L(x
)u(x)
=
f(x)
leads directly to the fruitful
area
of
differential equation theory.
16.2 Bounded Operators in Hilbert Spaces
The
concept
of
an operatoron a Hilbertspaceis extremelysubtle. Eventhe elemen-
tary characteristics
of
operators, such as the operation
of
hermitian conjugation,
cannot generally be defined on the whole Hilbert space.
In
finite-dimensional vector spaces thereis a one-to-one correspondence be-
tween
operators
andmatrices. So, in some sense, the studyof
operators
reduces
to a stody
of
matrices, which are collections
of
real or complex numbers. Al-
though we have already noted an analogy between matrices
and
kernels, a whole
new realm
of
questions arises
when
Aij
is replaced by
K(x,
y)-questions
about
the continuity
of
K(x,
y) in both its arguments, about the limit
of
K(x,
y) as x
and/or y approach the
"end
points"
of
the interval on which K is defined, about
the boundedness and "compactness"
of
K, and so on.
Such
subtleties are
not
unex-
pected. Afterall, when we triedto generalize concepts
of
finite-dimensional vector
spaces to infinite dimensions in Chapter 5, we encountered difficulties. There we
were concerned aboutvectorsonly; the generalization
of
operators
is even more
complicated.
16.2.1.
Example.
Recallthat
<coo
is theset of sequencesla) = l"'i
J;';;l'
or of co-tuples
(al.
a2,
...
), thatsatisfythe convergence
requirement
L~l
laj 1
2
< 00 (see
Example
t.1.2). It is a Hilbert space with inner product defined by
(al
b) =
L:~I
"'jfJj.
The
standard
(orthonormal)
basisforCoois
{Iei}}~l'
wherelei} hasall
components
equal
to
zeroexcept the ith
one,
which is 1.Thenone has la} =
L~l
aj
[ej}.
Onecan
introduce
an
operator
Tr, calledtheright-shiftoperator,by
T,
la) =
T,
(f>j
le
j
)) =
f>j
Iej+I).
j=l j=l
In
otherwords,T
r
transforms("'10"'2
•...
) to (0,"'I. "'2,
...
). It is straightforward to show
that
Tr is
indeed
a
linear
operator.
..
The first step in
our
stody
of
vector spaces
of
infinite dimensions was getting
a handle on the convergence
of
infinite sums. This entailed defining a norm for
vectors and a distance
between
them. In addition, we noted
that
the
set
of
linear
transformations
L (V, W) was a vector space in its
own
right. Since operators are
"vectors"
in this space, the study of
operators
requires
constructing a norm in
L(V, W)
when
V and W are infinite-dimensional.
454 16.
AN
INTROOUCTION
TO
OPERATOR
THEORY
16.2.2. Definition.
Let
1f
l
and
1fz
be two Hilbert spaces with norms
II
.
111
and
II·
liz.
ForanyT
EL(1fl,1fz),thenumber
max {
II
TXllzl
lx}
'i-
o}
IIxlll
operator
norm
(if
it exists) is called
1
the operator
norm
ofT
and is denoted by
IITII.
A linear
transformation whose norm isfinite is calleda
bounded
lineartransformation. A
bounded linear transformationfrom a Hilbert space to
itself
is called a
bounded
bounded
operator
operator. The collection
of
all boundedlinear transformations, which is a subset
of
L(1f"
1fz),
will be denoted by 13(1fl,
1fz),
and
if1fl
=
1fz
es
1f,
it will be
denoted by
13(1f).
Note that
II
.
111
and
II
. liz are the norms induced by the inner product
of
1fl
and
1fz.
Also note that by dividing by
IIx
II
t we eliminate the possibility
of
dilating
the
norm
of
IITII
by choosing a "long" vector. By restricting the length
of
[x),
one can eliminate the necessity for dividing by the length.
In fact, the norm can
equivalently be defined as
IITII
=
max
IIlTxllzl IIxlll = I} =
max
IIlTxllzl
IIxliJ
="
I}.
(16.3)
It is straightforwardto show thatthe threedefinitions are equivalentand they indeed
define a norm.
16.2.3.
Proposition.
An
operator T is bounded
if
and only
if
it maps vectors
of
finite norm to vectors
of
finite norm.
Proof
Clearly,
ifT
is bounded, then
IITxll
has finite norm. Conversely, if
IITxllz
is
finitefor
all [x)
(ofunitlength),maxIIlTxllzlllxllt
= I} is also finite,
andTis
bounded. D
An innnediate consequence
of
the definition is
IITxllz
="
IITII
IIxllJ
'of
Ix) E
1f,.
(16.4)
Ifwechoose
Ix}-Iy}
instead
of
[x), it willfollowfrom (16.4)thatas [x) approaches
Iy},T [x) approachesT Iy}. Thisis the property thatcharacterizes continuousfunc-
tions:
bounded
operators
16.2.4.
Proposition.
The boundedoperatorT E 13(1fj,
1fz)
is a continuousfunc-
are
continuous
tionfrom
1f,
to 1fz.
Anotherconsequence
of
the definition is that
1The precisedefinitionuses "supremum"instead of"maxiroum." Rather
than
spending a lot
of
effortexplainingthe difference
between the two concepts, we use the less precise, but more intuitively familiar, concept
of
"maximum."
16.2
BOUNDEO
OPERATORS
IN HILBERT
SPACES
455
16.2.5. Box.
:B(:J-Cj,
:J-Cz)
is a vector subspace
of
!'(:J-Cj,
:J-Cz),
andfar:J-C, =
:J-Cz
=
:J-C,
we have 1 E
13(:J-C)
and
11111
= I.
16.2.6. Example. Wehaveseenthatinaninner
product
space,onecanassociate a
linear
operator
(linear
functional) to every
vector.
Thus,associatedwiththevectorIx} ina
Hilbert
space
Xts
thelinearoperatorIx :
:J£
-e- C
defined
byIx
(Iy»
'"
(x] y). Wewantto
compare
the
operator
norm
off
x
withthenormof
Ix}.
Firstnotethatbyusingthe
Schwarz
inequality,
we get
IIl
xll
=max[
Ilxl~/)IIIY)
"'0)
=max[
I(~~~)IIIY)
"'0)
s 114
Ontheotherhand,from
IIxli
Z
= Ix (Ix», weobtain
IIxll
=
IXI~~I)
~
max[
Ilxl~I»IIIY)
'" 0) =lllxll·
Thesetwoinequalities implythat
IIl
x
II
=
IIxli.
III
derivative
operator
is
unbounded
16.2.7.
Example.
The
derivative
operator
D dJdx is not a bounded
operator
on
the Hilbert space-
£}(a,
b) of sqnare-integrable fnoctions. Witha functionlike
f(x)
=
Jx
- a, onegets
b-a
""
Ilfll = .Ji '
norm
ofa
product
is
less
than
the
product
of
norms.
while
df
[dx
= 1/(2.,fx - a) gives 1I0fll
Z
= !
J%
dx/(x
- a) =
00.
Weconclude that
11011
=
00.
III
16.2.8. Example. Since£,O{) is an
algebra
aswell asavectorspace,onemaybe
inter-
estedin the
relation
betweenthe
product
of
operators
and
their
norms.
Morespecifically,
onemaywanttoknowbow
IISTII
is relatedto
IISII
and
IITII.
Iu this
example
we sbowthat
IISTII
~
IISII
II
TII.
Todo so, weuse the
definition
of
operator
nonnforthe
product
ST:
IISTII
= max[
II~:~II
I Ix)
'"
0)
=max[
IISTxll
IITxll
I Ix)
"'O",Tlx))
IITxll
IIxll
< max[
II
S
CT
IX)1I1
Tlx)
",o)max[
II
T
xlll
lx)
"'0).
- .
IITxll
~---,I,-Ix,-II
~_~
=IITII
(16.5)
2Herethetwo Hilbertspacescoincide,so thatthe
derivative
operator
actson asingle Hilbertspace.
456
16.
AN
INTROOUCTION
TO
OPERATOR
THEORY
Now note that the first term on the RHS does
not
scan all the vectors for maximality:
It
scans only the vectors in the image of T.
If
we include all vectors, we may obtain a larger
number. Therefore,
max{
11
5(Tlx))111
T1x);"0)
< max{
115.<
1
11
Ix)
;"0)
=
11511
IITxll
-
IIxll
'
andthedesiredinequalityisestablished.
A usefulconsequenceofthisresultis
IIT
n
II
::;
IITll
n,
which we shall use frequently.
IlIIII
We
can
put
Equation (16.5) to immediate good use.
16.2.9.
Proposition.
Let
1f
be a Hilbert space and T E 13(1f).
If
[T
II
< I, then
1 - T is invertible and (1 -
T)-t
=
L~o
Tn.
Proof. First note thatthe series converges, because
Il
f Tnl1s f
II
Tn
II
::::
f
IITli
n
=
I_I
T
n=O
n~O
n=O
II II
and the sum has a finite norm. Furthermore,
(1 - n
fT
n
= (1 - T)
(lim
tTn)
= lim (1 - T)
trn
n=O k->;oon=O
k--+oo
n=Q
= lim
(trn
-
trn+t)
= lim (1 - Tk+l) = 1,
k-+oo n=O n=O
k-HXi
because 0 <
Iim,
....co
IITk+
1
11
::;
limr
....oo
IITIlk+l
= 0 for
IITII
< I, and the
vanishing
of
the normimplies the vanishing
of
the operatoritself. One can similarly
show that
(L~o
T
n
)(l - T) = 1. D
A corollary
of
this proposition is that operators that are "close enough" to an
invertible operator are invertible (see Problem
16.1). Another corollary, whose
proof
is left as a straightforward exercise, is the following:
16.2.10.
Corollary.
Let
T E 13(1f) and Aa complex number such that
IITII
<
IAI.
Then T - Al is an invertible operator,
and
lOOT
n
(T -
Al)-t
=
--
L
(-)
A
n~O
A
Adjoints play an important role in the stody
of
operators. We recall that the
adjoint
ofT
is definedas (yl T Ix)* = (r]
rt
Iy) or (Txl y) = (x ITt
y).Inthefinite-
dimensional case, we could calculate the matrix representation
of
the adjoint in
a particular basis using this definition and generalize to all bases by similarity
transformations. That is why we never raised the question
of
the existence
of
the
adjoint
of
an operator. In the infinite-dimensional case, one must prove such an
existence. We state the following theorem without proof:
16.3
SPECTRA
OF
LINEAR
OPERATORS
457
16.2.11.Theorem. Let T E 1I(:Ji).Then the adjoint
ofT,
defined by
(Txly) = (x ITt
y),
T
and
Tt
have
equal
exists. Furthermore,
IITII
=
IITt
II.
norms
Anotheruseful theoremthat weshalluse lateris thefollowing.
16.2.12.Theorem.
Let N(T) and
~(T)
denote the null space (kernel) and the
range
ofT
E 1I(:Ji).
We
have
Proof
Ix}
is in N(Tt) iff
rt
[x) = 0 iff (y ITt
x)
= 0 for all
Iy}
E :Ji. This
holds
if andonlyif
(Ty
Ix) = 0 for all Iy) E :Ji. This is equivalentto thestatement that
Ix}
is in
~(T)-L.
This chain of argumentprovesthat N(Tt) =
~(T)-L.
The
second
partof thetheoremfollowsfrom the factthat (Tt)t = T. D
16.3 Spectra
of
Lineal" Operators
regular
point
of
an
operator
resolvent
set
and
spectrum
of
an
operator
every
eigenvalue
of
an
operator
on
a
vector
space
of
finite
dimension
is
in
its
spectrum
and
vice
versa
One of the most important results of the theory of finite-dimensional
vector
spacesis thespectraldecomposition theoremdevelopedin Chapter4.The
infinite-
dimensional analogueof that theoremis far more encompassingand
difficult
to
prove.
It
is beyondthe scopeof this bookto developall the machinery
needed
for
athoroughdiscussionof theinfinite-dimensional spectraltheory.Instead,weshall
presentthecentralresults, andoccasionallyintroducethe readerto the
peripheral
arguments when they seemto have theirownmerits.
16.3.1.Definition.
Let T E .t:,(:Ji). A complex number Ais called a regular point
ofT
if
the operator T - A1is boundedand invertible.
3
The set
of
all regularpoints
ofT
is called the resolvent set
ofT,
and is denoted by p(T). The complement
of
p(T) in the complex plane is called the spectrum
ofT
and is denoted by a(T).
Corollary16.2.10implies"thatifT isbounded,then p(T)isnot
empty,
andthat
thespectrumof a boundedlinearoperatoron a Hilbertspace is a boundedset.In
fact,an immediateconsequenceof the corollaryis that A::;
IITII
for all AE
aCT).
It
is instructive tocontrastthe finite-dimensional case againstthe
implications
of the above definition. Recall that because of the dimension theorem, a
linear
operator on a finite-dimensional vector space V is invertibleif and only if it is
eitherontoorone-to-one.
Now,
AE
a(T)
if andonlyifT - A1isnot
invertible.
For
finitedimensions, this impliesthat
S
ker(T- A1)
01
O.
Thus, in
finite
dimensions,
3IfTis
bounded,
then
T- ).1is
automatically
bounded.
40ne can
simply
choosea xwhose
absolute
value
is
greater
than
IITII.
SNotehowcritical
finite-dimensionality
is forthis
implication.
In
infinite
dimensions,
an
operator
canbeone-to-one
(thus
having
azero
kernel)
without
beingonto.
458 16.
AN
INTRODUCTION
TO
OPERATOR
THEORY
not
all
points
of
aCT)
are
eigenvalues
A E
,,(T)
if and only if there is a vector la) in V such that (T - A1) la) = O. This
is the combined definition
of
eigenvalue and eigenvector, and is the definition we
will have to use to define eigenvalues in infinite dimensions.
It
follows that in the
finite-dimensional case, ,,(T) coincides with
theset
of
all eigenvalues of T. This
is not true for infinite dimensions, as the following example shows.
16.3.2. Example.
Consider
the
right-shift
operator
actingon Coo.It is easy to see that
IIT
rall
=
lIall
for all ]«). This yields
IIT
rll
= I, so that any Athatbelongsto
a(T,)
must
besuch
that
1)..1
::::
1.Wenowshow
that
the
converse
is also
true,
i.e.,
that
if III:5 1,then
)" E o(T,.).Itis
sufficient
to show
that
if 0 <
1),,1
:::
I, thenTr - A1 is not
invertible.
To
establish
this,we shallshow
that
Tr - A1is notonto.
Supposethat T; -),,1 is onto.Thenthere must be avector la)suchthat(Tr-),,1) la) =
let}
where
leI}is
the
first
standard
basis
vector
ofC
co.
Equating
components
onbothsides
yieldsthe
recursion
relations
al
= -1/)", anda
j-l
= )"aj forall j
::::
2. Onecan
readily
solvethis
recursion
relation
to
obtain
aj =
-1/).)
forall
j.
Thisisa
contradiction,
because
00
00
1
I:
lajl2
=
I:
IAI
2j
J=l
J=l
will not converge ifO <
IAI
~
1,i.e., la}
f/.
Coo,
and
therefore T
r
-
A1
is not onto.
we
conclude thet c
rr-)
= {A E
Cj
O <
IAI:::
I). 1fwe couldgeneralize the result
of the
finite-dimensional
casetoCoo,we wouldconclude
that
allcomplex
numbers
whose
magnitude
isatmostI
are
eigenvalues
ofTr- Quitetoour
surprise,
thefollowing
argument
shows
that
T
r
hasno
eigenvalues
atall!
Suppose
that
Ais an
eigenvalue
of T
r.
Let la} be any
eigenvector
for
A.
Since T
r
preserves the tengthof a
vector,
we have (al a) =
(Tral
Tra)
= (Aal Aa) = IAI
2
(al
a).
It
follows
that IAI =
I.
Nowwrite la) =
{aj
)1=1 andletam be the firstnonzeroterm of
this
sequence.
Then 0 =
(Tral
em) =
(Aul
em) = Au
m
•
The firstequalitycomesabout
because
T; la} hasits
first
nonzero
term
inthe (m + l)st
position.
SinceA
=1=
0, we must
have
am = 0, which
contradicts
thechoiceof this
number.
II!I
16.4 CompactSets
This section deals with some technical concepts, and as such will be ratherformal.
The central concept
of
this section is compactoess. Although we shall be using
compactness sparingly in the sequel, the notionhas sufficientapplication in higher
analysis and algebra that it warrants an introductory exposure.
Let us start with the familiar case of the real line, and the intuitive notion
of
"compactness."
Clearly,
we
do
not wantto call the entirereal line
"compact,"
because intuitively, it is not. The next candidate seems to be a "finite" interval.
So, first consider the
open interval (a, b).
Can
we call it compact? Intuition says
"yes," but the following argument shows that it would not be appropriate to call
the open interval compact.
Considerthemap e :
IR
--->
(a, b) givenby
etr)
=
b"2
a
tanh t+bi
a
•
The reader
may check that this map is continuous and bijective. Thus, we
can
continuously
16.4
COMPACT
SETS
459
map all
of
R in a one-to-one manner onto (a, b). This makes (a, b) "look" very
much" like
lR.
How can we modify the interval to make it compact? We do not
want to alter its finiteness. So, the obvious thing to do is to add the end points.
Thus, the interval
[a,
b]
seems to be a good candidate; and indeed it is.
The next step is to generalize the notion of a closed, finite interval and even-
tually come up with a definition that can be applied to all spaces. First we need
some terminology.
open
ball
16.4.1. Definition. An open ball B;(x)
of
radius r and center Ix) in a normed
vector space
Vis the set
of
all vectors in Vwhose distance from Ix} is strictly less
than r:
open
round
neighborhood
bounded
subset
open
subset
boundary
point
closed
subsetand
closure
Br(x)
==
lly)
EVilly
-xII
<
r].
We call Br(x) an open
round
neighborhood
of
[x),
This is a generalization
of
open interval because
(a, b) =
{y
E lR
Ily
_
a:
b I<
b;
a}
.
16.4.2.
Example.
A prototype of finite-dimensional nonned spacesis
JRn.
An openball
of radius r centered at x is
Br(x) = {y E
JR
I(Yt -
Xt)2
+(Y2 - X2)2 +...+(Yn -
xn)2
< r
2
}.
Thus, all points inside a circle form an open ball
in
the xy-plane, and all interior points of
a solid sphereform an openbaIlin space. II
16.4.3. Definition. A bounded subset
of
a normed vector space is a subset that
can be enclosed in an open ball
of
finite radius.
For example, any region drawn on a piece of paperis a bounded subset of lR
2
,
and any "visible" part
of
our environment is a bounded subset
of
lR
3
because we
can always find a big enough circle or sphere to enclose these subsets.
16.4.4. Definition.
A subset (')
of
a normedvectorspace Vis called openifeach
of
itspoints(vectors) hasanopen round neighborhoodlying entirely in
(').
A boundary
pointof(')
is a point(vector) in Vall
of
whose open round neighborhoods contain
points inside
and
outside
(').
A closed subset e
of
V is a subset that contains all
of
its boundary points. The closure
of
a subset S is the union
of
S
and
all
of
its
boundary points, and
is denoted by
S.
For
example, the boundary
of
aregiondrawn on paperconsists of allits bound-
ary points. A curve drawn on paper has nothing but boundary points. Every point
is also its own boundary. A boundary is always a closed set.
In
particular, a point
is a closed set. In general, an open set cannot contain any boundary points. A
frequently used property of a closed set
eis that a convergent sequence of points
of
econverges to a point in
e.
6
In
mathematicaljargon one says that (a, b) and R are homeomorphic.
460
16.
AN
INTROOUCTION
TO
OPERATOR
THEORY
dense
subset
rational
numbers
are
dense
inthe
real
numbers
p(T)
is
open,
and
u(T)
is
closed
and
bounded
inC.
16.4.5.Definition. A subset W 01a normed vector space V is dense in V ifthe
closure
olW
is the entire space
V.
Equivalently, W is dense ifeach vector in W is
infinitesimally close to at least one vector in
V.
In otherwords, given any !u) E V
and any E > 0, there is a !w) E W such that
lIu
-
wll
< E, i.e., any vector in V
can be approximated, with arbitrary accuracy, by a vector in W.
A paradigm of dense spaces is the set of rational numbers io the normed
vector space of real numbers. It is a well-known fact that any real number can
be approximated by a rational number with arbitrary accuracy: The decimal (or
bioary) representation
of
real numbers is precisely such an approximation. An
iotuitive way ofimagioiogdensenessis that the (necessarily)infinite subsetis equal
to
almost all of the set, and its members are scattered "densely" everywhere io the
set. The embeddiogofthe rational numbers in the set ofreal numbers, and how they
densely populate that set, is a good mental picture of all dense subsets. A useful
property iovolviog the concept
of
closure and openness has to do with contiouous
maps betweennormedvectorspaces. Let
I:
1t:
1
-->
1t:2be a continuousmap. Let
(92
be an open setin1t:2. Let
1-
1
((92) denotethe ioverse imageof (92,i.e., allpoiots
of1t:1 that are mapped to
(92.
Let !XI)be a vector io
I-I
((92),
!X2)
=
I(!XI)),
and
let
B.(X2) be a ball contained entirely io
(92.
Then
I-I
(B.
(X2)) contaios !XI) and
lies entirely io
1-
1
((92). Because of the continuity of
I, one can now construct
an open ball centered at
IXI)
lyiog entirely io
I-I
(B. (X2)), and by inclusion, in
I-I
((92).This shows that every poiot of
I-I
((92)has a round openneighborhood
lyiog entirely io
1-
1
((92). Thus,
1-
1
((92) is an open subset. One can similarly
show the correspondiog property for closed subsets. We can summarize this io the
followiog:
16.4.6.Proposition.
Let
I : 1t:
1
-->1t:2be continuous.Then the inverse image 01
an open (closed) subset
011t:2
is an open (closed) subset ol1t:J.
Consider the resolvent set of a bounded operator T. We claim that this set is
open io C. To see this, note that if )"
E P(T), then T -
)"
1 is iovertible. On the
other hand, Problem 16.1 shows that operators close to an invertible operator are
iovertible. Thus, if we choose a sufficiently smallpositive number
E and consider
all complex numbers
I-' withio a distance E from )",then all operators of the form
T -
1-'1
are iovertible, i.e., I-' E P(T). Therefore, any)" E peT) has an open round
neighborhood io the complex plane all poiots of which are io the resolvent. This
shows that the resolvent set is open. In particular, it cannot contain any boundary
poiots. However,
peT) and O'(T) have to be separated by a common boundary?
Sioce
p(T) cannotcontaioany boundarypoiot,
0'
(T)mustcarry the entire boundary.
This shows that
0'
(T) is a closed subset of
Co
Recalliog that
0'
(T) is also bounded,
we have the following result.
7The spectrum
of
a bounded operatorneed not occupy any "area" in the complex plane. It may consist
of
isolated points or
line segments, etc., in which case the spectrum will constitute the entire boundary.
16.4
COMPACT
SETS
461
16.4.7. Proposition. For any T E 1l(9i) the set peT) is an open subset
ofC
and
IT (T) is a closed, boundedsubset
of
Co
Let us go back to the notion of compactness.
It
tums out that the feature
of
the closed interval [a, b] most appropriate for generalization is the behavior of
infinite sequences of numbers lying in the interval. More specifically, let la,
1;';;1
be a sequence of infinitely many real numbers all lying in the interval [a, b].
It
is
intuitively clear that since there is not enough room for these points to stay away
from each other, they will have to crowd around a number
of
points in the interval.
For example, the sequence
in the interval [-1, +1] crowds around the two points
-t
and
+t.ln
fact, the
points with even n accumnlate around +tand those with odd n crowd around -
t.
It
tums out that all closed intervals of ~ have this property, namely, all sequences
crowd around some points. To see that open intervals do not share this property
consider the open interval (0, 1). The sequence
12n~tl:l
=
Il,~,
... I clearly
crowds only around zero, whichis not a point
of
the interval. But we alreadyknow
that open intervals are not compact.
16.4.8. Definition. (Bolzano-Weierstrass property) A subset X
of
a normed vec-
compact
subset tor space is called compact if every (infinite) sequence in X has a convergent
subsequence.
The reason for the introduction
of
a subsequence in the definition is that a
sequence may have many points to which it converges. But no matter how many
of
these points there may exist, one can always obtain a convergent subsequence
by choosing from among the points in the sequence. For instance, in the example
above,one canchoosethe subsequenceconsisting
of
elementsfor whichnis even.
This subsequence converges to the single point +
t.
An important theorem in real analysis characterizes all compact sets in
~n:8
16.4.9.
Theorem.
(BWHB theorem) A subset
of~n
is compact if
and
only ifit is
closed and bounded.
We showed earlierthat the spectrum of a boundedlioearoperatoris closed and
aCT)
is
compact
bounded. Identifying C with
~2,
the BWHB theorem implies that
BBWHB stands for Balzano, Weierstrass, Heine, and Borel. Balzano and Weierstrass proved that any closed and bounded
subset of R has the Balzano-Weierstrass property. Heine and Borel abstracted the notion of compactness
in terms of open sets,
and showed that a closed bounded subset of R is compact. The BWHB theorem as applied to R is usually called the
Heine-
Borel theorem (although some authors call it the Balzano-Weierstrass theorem). Since the Balzano-Weierstrass property and
compactness are equivalent, we have decided to choose BWHB as the name of our theorem.
462 16.
AN
INTROOUCTION
TO
OPERATOR
THEORY
16.4.10. Box. The spectrum
of
a boundedlinearoperator is a compactsub-
set
of
C.
criterion
for
finite-dimensionality
An immediate consequence
of
the
BWHB
Theorem is that every bounded
subsetofJRnhas a compactclosure. Since
JR"
is a prototype
of
allfinite-dimensional
(normed) vectorspaces, the same statementis
true
for all suchvector spaces.
What
is
interesting
is
that
the
statement
indeedcharacterizes the
Donned
space:
16.4.11.
Theorem.
A normedvectorspace isfinite-dimensionalif
and
only ifevery
bounded subset has a compact closure.
This result can also be applied to subspaces
of
a normed vector space: A
subspace W
of
a normed vector space Vis finite-dimensional
if
and ouly
if
every
boundedsubset
ofW
has a compactclosure in W. A usefulversion
of
this property
is stated in terms
of
sequences
of
points (vectors):
16.4.12.
Theorem.
A subspaceW
of
a normedvectorspace Visfinite dimensional
if
and
only ifevery boundedsequence in W has a convergent subsequence in W.
Karl Theodor WDhelm Weierstrass (1815-1897) was
boththe
greatest
analyst
andthe
world's
foremost
teacher
of advanced mathematics of the last third of the nine-
teenth
century.
His
career
was also
remarkable
in
another
way-and
aconsolation toall
"late
starters"-forhebegan
thesolid
part
of his professional life at theage of
almost
40, whenmost
mathematicians
are
longpasttheir
creative
years.
His
father
senthimtotheUniversity
of
Bonntoqualify
forthe
higherranks
ofthe
Prussian
civilserviceby
studying
law and
commerce.
But
Karl
hadno
interest
in thesesub-
jects.He infuriatedhis fatherby rarelyattendinglectnres,gettingpoor grades,andinstead,
becominga
champion
beer
drinker.
Hedid
manage
tobecomea
superb
fencer,
butwhenhe
returned
home, he hadno
degree.
In
order
to earn his living, he madea fresh
start
by
teaching
mathematics,
physics,
botany,
German,
penmanship.
andgymnastics to the
children
of
several
small
Prussian
towns
during
the
day.
During
the
nights,
however,
he mingledwith the
intellectuals
of
the
past,
particularly
the
great
Norwegian
mathematician
Abel.
His
remarkable
research
on
Abelian
functions
was
carried
onforyears
without
theknowledge of
another
livingsoul;he
didn'tdiscuss
itwith
anyone
atall,or
submit
itforpublication inthe
mathernaticaljoumals
of theday.
All this
changed
in 1854when
Weierstrass
atlastpublished an
account
of his
research
on
Abelian
functions.
This
paper
caught
the
attention
ofan
alert
professor
atthe
University
of
Konigsberg
who
persuaded
his
university
to
award
Weierstrass
an
honorary
doctor's
degree.
The
Ministry
of
Education
granted
Weierstrass
a year'sleaveof
absence
withpay