Hassani S. Mathematical Physics: A Modern Introduction to Its Foundations
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16.4
COMPACT
SETS
463
to
continue
his
research,
andthenext yearhe was
appointed
to theUniversity of Berlin,
where
he
remained
therestof his life.
Weierstrass's
great
creative
talents
wereevenly dividedbetweenhis
thinking
andhis
teaching. The
student
notesof his lectures,andcopies of thesenotes,andcopies of copies,
were
passed
from
hand
to
hand
throughout
Europe
andeven
America.
Like
Gauss
hewas
indifferent to fame, butunlike Gauss he endeared himself to
generations
of
students
by
thegenerosity withwhichhe
encouraged
themto developand
publish,
andreceive
credit
for,
ideas
and
theorems
that
heessentially
originated
himself.
Among
Weierstrass's
students
andfollowerswere
Cantor,
Schwarz,
Holder,
Mittag-Leffler,
Sonja
Kovalevskaya
(Weierstrass's
favoritestudent),
Hilbert,
Max
Planck,
Willard
Gibbs,
andmany
others.
In1885he
published
the
famous
theorem
nowcalledtheWeierstrassapproximation
theorem (see
Theorems
5.2.3 and8.1.1), whichwas givena
far-reaching
generalization,
withmany
applications,
bythemodem
American
mathematician
M.
H.
Stone.
The
quality
thatcametobeknownas
"Weierstrassian
rigor"
was
particularly
visiblein
his
contributions
to the
foundations
of
real analysis. He
refused
toacceptany
statement
as
"intuitively
obvious,"
but
instead
demanded
ironclad
proofbasedonexplicit
properties
of
thereal
numbers.
The
careful
reasoning
required
fortheseproofswas
founded
ona
crucial
property
of thereal
numbers
nowknownastheBWHB theorem.
We shall need the following proposition in our study of compact operators:
16.4.13. Proposition.
Let
W be.aclosed
proper
subspace
of
X and 8an
arbitrary
nonnegative number
with
0 :0 8 < 1. Then there exists a
unit
vector Ivo} E X
such that
IIx-
voll2:8
v [x) E W.
Proof
Choose a vector Iv} in X but not in W and let
d =
minl]»
-
xIII
[x) E W}.
We claim that d >
O.
To show this, assume otherwise. Then for each (large) n and
(sufficiently small)
f,
we could find distinct vectors (Ix
n
))
whose distance from
[u) would be
fin
and for which the sequence
(Ix
n
))
would have Iv} as a limit.
Closure
ofW
would then imply that Iv} is in W, a contradiction. So, d > 0,
Now, for any Ixo} E W, let
eW
lu}
sa
[x) _ Iv} - Ixo} = '(IIv -
xolllx)
+Ixo})
-Iv}
IIv
-xoll
IIv
-xoll
and note that by the definition of d, the norm of the numerator is larger than d.
Therefore,
lIuli
2:
dlllv
- xoll for every Ix},
Ixo}
E W.
If
we choose Ixo} snch
that
II
v -
Xo
II
<
as:
t, which is possible because
d8-
1
> d, then
lIu
II
2:8 for all
Ix} E W. Now let Ivo} = (Iv) -
Ixo})/lIv
- xoll. 0
464
16.
AN
INTROOUCTION
TO
OPERATOR
THEORY
16.5 Compact Operators
It
is straightforward to showthat if X is a compactset in 9ii and f :9i1 ..... 9i
2
is continuous, then
fiX)
(the image of X) is compact in 9i2. Since all bounded
operators are continuous, we conclude that
all bounded operators
map
compact
subsetsontocompactsubsets.
Thereis aspecialsubsetof
21
(9i1,9i2)thatdeserves
particularattention.
compact
operator
16.5.1. Definition. An operator K E 2l(9it, 9i2) is called a
compact
operator
if
it maps a boundedsubset
of9il
onto a subset Of9i2 with compact closure.
Since we will be dealing with function spaces, and since it is easier to deal
withsequencesoffunctionsthanwith subsetsof thespaceof functions,wefindit
moreusefultohaveadefinition ofcompactoperatorsin termsof sequencesrather
thansubsets.Thus,insteadof aboundedsubset,wetake a subsetof it consistingof
a (necessatily)boundedsequence.The image of this sequencewillbe a sequence
in a compact set, which,by defutition, must have a convergentsubsequence. We
thereforehavethe following:
16.5.2. Theorem.
An
operator KE 2l(9it, 9i2) is compact
ifand
only
iffor
any
bounded sequence (Ix
n
)} in
9it,
the sequence (K Ix
n
)} hasa convergent subse-
quence in 9i2.
product
of
two
compact
operators
Is
compact
16.5.3. Example.
Consider
2l(:J{),
thesetof
bounded
operators
onthe
Hilbert
space
X,
If
Kis a compactoperatorand T a bounded operator,then
KT
and TK arecompact. This is
because
(T
IXn)
==
IYn))is a
bounded
sequence
if
(Ix
n))
is,
and
(K
IYn)
= KTIXn))hasa
convergentsubsequence,because Kis compact. For the second part, use the firstdefinition
of the compact operator and note
that K maps bounded sets onto compact sets, which T
(beingcontinuous)maps onto acompact set.As aspecial case ofthis propertywe note that
the product of two compact operators is compact. Similarly,one can show that any linear
combination ofcompactoperators iscompact.Thus,anypolynomial ofacompactoperator
is compact.
In
particular,
~
n'
.
~
n'
.
(1-K)n
=
L....,
..
,(-K)]
=1+
L....,
'.
,(-K)]
==1-Kn,
j=O
J .(n - J).
j=l
] .(n - J).
whereK
n
is acompact
operator.
III
finite
rank
operators
16.5.4. Definition.
An
operatorT E
.c
(9it, 9i2) is called a
finite
rank
operator
if
its range isfinite-dimensional.
The following is clearfromTheorem 16.4.12.
16.5.5.
Proposition. A finite rank operator is
compact
In
particular,everylinear transformationof a finite-dimensional vector space
is compact.
linear
transformations
of
finite-dimensional
vector
spaces
are
compact
16.5
COMPACT
OPERATORS
465
16.5.6.
Theorem.
IflK
n}
E
,c(Xj,
X2) are
compactandK
E
,c(Xj,
X2) is such
that
II
K - K
n
II
--+ 0, then K is compact.
Proof.
Let {lxm)J be a bounded sequence in
Xl.
Let {Kl Ixm1)Jbe the convergent
subsequence guaranteed by the compactness
of
Kj.
Now, {lxm1)J is a bounded
sequence in
Xl.
It
therefore has a subsequence
{Ix
m,)}
such that
{K2Ixm,)}
is
convergent. Note that
{Kj
Ix
m,)}
is also convergent. Continuing this process, we
constroct the sequence
of
sequences
where each sequence is a subsequence
of
all sequences preceding it. Furthermore,
all the sequences
{I<J
Ix
mk
} }
for I = I,
...
, k are convergent. In particular, if we
pick the
diagonal
sequence
{IYm)J
==
{lxmm)J, then for any I E
!'ii,
the sequence
{K1IYm)} converges in X2. To show that K is compact, we shall establish that
{IYm)} is the subsequence
of
{Ix
m)}
such that {KIYm)} is convergent. Since X2
is complete, it is sufficient to show that {K
IYm)} is Cauchy. We use the so-called
"E/3
trick."
Write
K IYm)- K IYn} = K
IYm)
-I<J
IYm)
+
I<J
IYm}- K1IYn) +
I<J
IYn) - K IYn}
and use the triangle inequality to obtain
By choosing
m, n, and I large enough, we can make each
of
the three terms on the
RHS smaller than
E/3;the
first and the third ones because Kl --+ K, the second
one because
{I<J
IYn)} is a convergent sequence. 0
Recall that given an orthonormal basis
{lei)
}~j'
any operator T on a Hilbert
space X can be written as
L:0~1
Cij lei) (ejl,where Cij = (eil TIe
j).
Now let K
be a compact operator and consider the
finite
rank
operators
n
K
n
sa
L Cij Ie,)
(ejl,
i,j=l
Clearly,
11K
- K
n
II
--+
O.
The hermitian adjoints
{Kk}
are also of finite rank (there-
fore, compact). Barring some convergence technicality, we see that Kt, which is
the limit
of
the sequence
of
these compactoperators, is also compact.
Kis
compact
iff
Kt
is 16.5.7.
Theorem.
Kis a compact operator
if
and
only
ifKt
is.
A particular type of operator occurs frequently in integral equation theory.
These are called Hilbert-Schmidt operators and defined as follows:
16.5.8. Definition.
Let:J{
be a
Hilbert
space,
and
{lei)
}~l
an
orthonormal
basis.
Hiibert-Schmidt
An
operator
T E
,c(X)
is called
Hilberl-Schmidt
if
operators
466 16.
AN
INTROOUCTION
TO
OPERATOR
THEORY
00 00 00
tr(TtT)
sa L
(edTtT
lei) = L
(Ted
Tei)
= L IITedl
2
<
00.
i=l
i=l i=l
Hilbert-Schmidt
operators
are
compact.
Hilbert-Schmidt
kernel
16.5.9.
Theorem.
Hilbert-Schmidt operators are compact.
For
a
proof,
see
[Rich
78,
pp. 242-246].
16.5.10.
Example.
It is time to give a concrete example of a compact (Hilbert-8chmidt)
operator. For this, we retorn to Equation (16.2) with
w(y)
= I, and assume that Iu) E
r}(a,
b). Suppose
further
thatthefunction
K(x,
y) is continuous on the closedrectangle
[a, b] X [a, b] in the xy-plane (orIR
2
). Under such conditions,
K(x,
y) is called a
Hilhert-
Schmidt kernel. Wenow show
that
Kis compact. Firstnote thatdue to the continuity of
K(x,
y),
J: J:
IK(x,
y)I
2dx
dy < 00. Next, we calculate the trace of KtK. Let {lei)
)~1
be any orthonormal basis of ,e2(a, b). Then
trKtK
= f:(etl Kt
K
lei) = 't
fff
(eil
x) (xl K
t
Iy) (yl
Klz)
(zl ei)
dxdydz
l=l
/=1
=
fff
(YIKIX)*(YIKlz)'t(zlei)(eilx)dXdydZ
1=1
~o(x-z)
=
fff
(yl Klx)* (yl KIz)'(zl
(~Iei)
(eil)
Ix)
dxdydz
----
=1
III
Bernard
Balzano(1781-1848) was a Czechphilosopher, math-
ematician,
and
theologian
who
made
significant
contributions
to
both
mathematics
andthetheoryof knowledge.Heenteredthe
Philosophy Faculty.oftheUniversity of Prague in 1796, stodying
philosophy
and
mathematics.
He
wrote
"Myspecial
pleasure
in
mathematics
rested
therefore
particularly
onits purelyspecula-
tive
parts,
inotherwordsI prizedonly that
part
of
mathematics
whichwasatthe sametime
philosophy."
.10 the automo of 1800 he began three years of theological
studywhilehe was
preparing
a
doctoral
thesison
geometry.
He
received
his
doctorate
in 1804for a thesisin whichhe gave his view of
mathematics
and
whatconstitutes a
correct
mathematical
proof.In theprefacehe
wrote:
I couldnot besatisfiedwith a completelystrictproofif it werenot
derived
fromconceptswhichthe thesis to beprovedcontained, but
rather
madeuse
of some
fortuitous,
alien,
intermediate
concept,whichisalwaysan
erroneous
transition
to
another
kind.
16.6
SPECTRUM
OF
COMPACT
OPERATORS
467
Two days after receiving his doctorate
Balzano
was ordained a
Roman
Catholic priest.
However, he came to realize that teaching and not ministering defined his true vocation.
In the same year, Balzano was appointed to the chair of philosophy and religion at the
University
of
Prague. Because
of
his pacifist beliefs
and
his concern for economic justice,
he was suspended from his position
in 1819 after pressure from the Austrian government.
Balzano
had
not
givenup without a fight but
once
he was suspended on a charge
of
heresy
he was put under house arrest and forbidden to publish.
Althoughsome
ofhis
books
had
to bepublishedoutsideAustriabecause
of
government
censorship, he continued to write
and
to
play
an importantrole in the intellectual life of his
country.
Balzano
intended to write a series
of
papers
on
the
foundations
of
mathematics.
He wrote two, the first
of
which
was published. Instead
of
publishing the second one he
decided
to"
.
..
make
myself
better
known
to
the
learnedworldby publishing somepapers
which, by their titles, would be
more
suited to arouse attention."
Pursuing this strategy he published
Der binomische Lehrsatz
...
(1816)
and
Rein
analytischerBeweis
...
(1817),
which
containan attemptto free calculusfrom the concept
of
the infinitesimal. He is clear in his intention stating in the preface
of
the first that the
work
is
"a
sample
of
a new way
of
developing analysis."
The
paper
gives a
proof
of
the
intermediate value theorem
with
Bolzano's
new approach
and
in the
work
he defined what
is now called a Cauchy sequence.
The
concept appears in Cauchy's
work
four years later
but
it is unlikely that Cauchy
had
read
Balzano's
work.
After1817, Bolzanopublishedno further mathematicalworksfor
many
years. Between
the
late I820s and the 1840s, he workedon a
major
work
Grossentehre.
This attemptto
put
the whole
of
mathematics on a logical foundation was
published
in parts, while
Balzano
hopedthathis
students
would
finish
andpoblishthecomplete work.
His
work
Paradoxien des Unendlichen, a
study
of
paradoxes
of
the
infinite, was pub-
lished
in 1851, three years after his death, by
one
of
his students.
The
word
"set"
appears
herefor the first time.
In this
work
Balzano
gives examples
of
1-1
correspondencesbetween
the
elements
of
an infinite set
and
the elements
of
a
proper
subset.
Bolzano's
theories
of
mathematicalinfinity anticipatedGeorg Cantor'stheory of infinite
sets.
It
is also remarkable
that
he gave a function
which
is nowhere differentiable
yet
everywhere continuous.
16.6 Spectrum of Compact Operators
Our next task is to investigate the spectrum 0" (K) of a compact operator K on
a Hilbert space J£. We are particularly interested in the set
of
eigenvalues and
eigenvectors of compact operators. Recall that every eigenvalue of an operator on
a vector space
of
finite dimension is in its spectrum, and that every point
of
the
spectrum is an eigenvalue (see page 457). In general, the second statement is not
true.
In fact, we saw that the right-shift operator had no eigenvalue at all, yet its
spectrum was the entire unit disk of the complex plane.
We first observe that 0 E
O"(K), because otherwise 0 E
p(K),
which implies
that
K = K - 01 is invertible with inverse
K-
t. The product of two compact
operators (in fact, theproductof a compactand a boundedoperator)is compact(see
468
16.
AN
INTRODUCTION
TO
OPERATOR
THEORY
oE
cr(K)
ifKis
compact
Example 16.5.3). This yields a contradiction? because the unit operator cannot be
compact:
It
maps aboundedsequenceto itself, not to a sequence with a couvergent
subseqnence.
16.6.1. Proposition. For any compact operator KE
~(:Ji)
on an infinite dimen-
sionalHilbert space, we have
0 E u(K).
To proceed, we note that eigenvectors of K corresponding to the eigenvalue ).
belong to the null space
of
K-
),1.
So, let
lO
NJ.
= ker(K - ),1),
Nt =ker(K
t
-),*1),
:RJ.
es Range(K - ),1),
:Rt
=Range(Kt
-),*1).
generalized
eigenvector
16.6.2.
Theorem.
NJ.
and
Nt are finite-dimensional subspaces
of
X. Further-
-n.L _
Nt
more,
J\}..
- )..0
Proof. We use Theorem 16.4.12.
Let
(Ix.)} be a bounded sequence in
NJ..
Since
K is compact, (K Ix.) = ).Ix.)} has a convergent subsequence. So
(Ix.)}
has a
convergent subsequence. This subsequence will converge to a vector in
NJ.
if the
latter is closed. But this follows from Proposition 16.4.6, continuity
of
K- ),1, the
fact that
NJ.
is the inverse image of the zero vector, and the fact that any single
point of a space, such as the zero vector, is a closed subset. Finite-dimensionality
of
Nt follows from the compactness of Kt and a similar argument as above.
To show the second statement, we observe that for any bounded operator T,
wehavelllu)
ET(X).Liff(ulv)
= ofor all Iv)
ET(X)iff(uITx)
= ofor all
[x) E X
iff
(Ttul x) = 0 for all [x)
EX
iffTt
lu) = 0
iff
lu) E
kerTt.
This
shows that T(X).L
= ker
rt,
The desired result is obtained by letting T = K - ),1
and noting that
(W.L).L
= W for any subspace W of a Hilbert space. 0
We note that
NJ.
is the eigenspace of K corresponding to the eigenvalue )..
However, it may well happen that zero is the ouly number in o(K). In the finite-
dimensional case, this corresponds to the case where the matrix representation
of
the operatoris not diagonalizable. In such a case, the standardprocedure is to look
at generalized eigenvectors. We do the same in the case
of
compact operators.
16.6.3.Definition. A vector Iu) is a generaTked eigenvector
of
K
of
order m if
(K -
)'1)m-
l
l
u
)
#
Obut(K-
)'1)m lu) =
O.
The
set
of
such vectors, i.e., the null
space
of(K
- )'1)m, will be denoted by
Nt)
It
is clear that
{
O}
=
N(O)
C N esN(ll C
N(2)
C
...
c
Jlfm)
c
N(m+l)
...
A-A
),.-).-
_A_A
(16.6)
l~Our
conclusion is valid only in infinite dimensions. In finitedimensions, all operators, including 1, are compact.
In
what follows,we assumethat X
#=
O.
11Recall that T(9{) is the range
of
the operatorT.
16.6
SPECTRUM
OF
COMPACT
OPERATORS
469
and each
Nt') is a subspace of 1C In general, a subspace with higher index is
larger than those with lower index.
If
there happens to be an equalityat one link of
the above chain, then the equality continnes all the way to the right
adinfinitum.
To see this, let
p be the first integer for which the equality occurs, and let n > p
be arbitrary. Suppose lu) E Nin+l). Then (K - A1)p+1[(K - A
l)n-
p
lu)] =
(K-
A1)n+1 lu) =
O.
It
follows that (K - A
l)n-
p
lu) is in NjP+1).
But
NjP) =
NjPH).
So
(K - At)" lu) = (K - A
l)P[(K
- A
l)n-
p
lu)] =
o.
Thus every vector in
Ni
n
+
1
)
is also in
Ni
n).
This fact and the above chain imply
that
Ni
n
)
=
Ni
n
H)
for all n > p,
16.6.4.
Theorem.
Thesubspace
ll
Ni
n
)
isfinite-dimensionalforeach n. Moreover,
there is an integer p such that
for n
= 0, 1,2,
...
, p - 1
butNi
n
)
=
Ni
n
+
1
)
for all n
::::
p,
Proof
For the firstpart, use the resultof Exarnple 16.5.3to show that
(K-A
t)"
=
K
n
-
A
n1
where
Kn
is compact. Now repeat the proofof Tbeorem 16.6.2for
Kn.
If
the integer p exists, the secondpart
of
the theorem follows from the discus-
sion preceding the theorem. To show the existence of p, suppose, to the contrary,
that
Ni
n
)
oft
Ni
n
+
1
)
for every positive integer n. This means that for every n, we
can find a (unit) vector Iv
n)
E Nin+l) that is not in
Ni
n
)
and that by Proposition
16.4.13has the property
We thus obtain a bounded sequence {Iv
n)].
Let
us apply K to this sequence. If
j > I, then
(Ivj)
-
IVi»)
E
N~+I)
by the construction of IVj) and the fact that
N(l+l)
c N
U
+
1
)
Furthermore
A
-}..
,
but
11
For
infinite
dimensions, thefactthatlinear
combinations
of asubsetbelongtothesubsetisnotsufficient tomake
that
subset
intoa subspace. Thesubsetmustalsobe closed. We
normally
leaveoutthe
rather
technical proofof closure.
470 16. AN INTRODUCTION
TO
OPERATOR
THEORY
by the definition
ofN~+I).
Therefore, (K - A1)(lvj)
-Iv[))
E
N~).
Now note
that
ENf) ENf)
,-'-,
,-"-,
Klvj) - Klvl) =
AH<K-
A1)IVj) -
*(K-A1)
IVI)
+Ivj)
-IVI)}.
It
follows from Proposition 16.4.13 that the
nann
of
the vector in curly brackets
is larger than
~.
Hence,
II
KIvj) - KIVI)
II
:::
IA
I
/2,
i.e., since j and I are arbitrary,
the sequence (K
Iv
n
)}
does not have a convergent subsequence. This contradicts
the fact that Kis compact. So, there must exist a
p such that Nip) = Nl."+!). D
Wealso needtherange of various powers of K- A1.Thus, let
~in)
=
Range(K-
A
t)".
One can show that
1C=
~(O)
::::>
~(l)
::::>
•••
::::>
~(n)
::::>
~(n+l)
::::>
•••
),.-),,-
-J..-)"
-
16.6.5.
Theorem.
Each
~in)
is a subspace
of
1C.
Moreover, there is an integer q
h
ha
men)
m(n+l)fi
- 0
lib
men)
-
m(n+l)fi
II
sue t t
J\},
f=..AJ,.
orn
- , ,
...
, q - ,
ut.}\,J.
-..AJ..
ora
n::::
q.
Proof
The proofis similar to that of Theorem 16.6.4. The only extra step needed
is to show that
~in)
is closed. We shall not reproduce this step. D
16.6.6.
Theorem.
Let
q be the integer
of
Theorem 16.6.5. Then
1. 1C=
Ni
q
)
Ell
~iq)
2.
Ni
q
)
and
~iq)
are invariant subspaces
ofK.
3. The only vector in
~iq)
that K -
A1
maps to zero is the zero vector.
Infact,
when restricted to
~iq),
the operator K - A1 is invertible.
Proof
(l)
Recall that 1C =
Ni
q
)
Ell
~iq)
means that every vector
of
1Ccan be
written as the sum of a vector in
Ni
q
) and a vector in
~iq),
and the only vector
common to both subspaces in zero. We show the latter property first.
In
fact, we
show that
Ni
m
)
n
~iq)
= 0 for any integer m. Suppose [x) is in this intersection.
For each
n ::: q, there must be a vector Ix
n)
in 1Csuch that [x) = (K - At)" Ix
n)
because~in)
=
~iq)
forn:::
q.
If
[x)
i'
0, then Ix
n)
<t
Nr')
foreachn.
Nowletr
be the larger of the two integers
(p,
q) where p is the integer
of
Theorem 16.6.4.
Then
(16.7)
16.6
SPECTRUM
OF
COMPACT
OPERATORS
471
From
0=
(K-A1)'"
[x) =
(K-A1)'"+'lx
r)
and
it follows that
Ix
r)
E Ni,"+r).
But
Ni,"+r) =
Nf)
=
Nj"),
contradictingEquation
(16.7). We conclude that [x) mustbe zero,By the definition
of:Ri
q),
for any vector
lz)in :Ji, we have that (K - A1)q lz) E :Ri
q)
. Since :Ri
q)
= :Rfq), there must he a
vector
Iy)
E:Ji
such that
(K-A1)q
[z) =
(K-
A1)2q
Iy) or
(K-
A1)qllz) -
(K-
A1)q Iy)] =
O.
This shows that Iz) -
(K-
A1)q Iy) is in
Ni
q).
On the otherhand,
[z)
= liz) -
(K-A1)q
ly)J+(K-A1)q
Iy)..
and the first part of the theorem is done.
(2)
For
the second part, we simply note that (K - A1
)Ni
k
)
C;
Ni
k
-
1
)
C;
Ni
k
),
and that
K(Nl
q)
= (K - A1 +A
1)(Ni
q)
=(K - A
1)(Ni
q
)+ A1
(Ni
q)
c
Ni
q).
c:Ni
q
)
cNiq)
Similarly,
K(:Ri
q)
= (K - A1 +A1)(:Ri
q)
= (K - A1)(:Ri
q)
+A1(:Ri
q)
C :Ri
q).
~
c:Ri
q
+1)
c1l.i
Q
)
c:Ri
q
)
(3) Suppose [z) E :Ri
q)
and (K - A1) [z) =
O.
Then
[z) =(K - A1)q Iy) for some
Iy) in
Jf,
and 0 = (K - A1)[z) = (K - A1)q+lly), or Iy) E
Ni
q+
1
).
From part
(l)-withm
=
q+
I-we
conclude that lz) =
O.
It
follows that
K-A1
is injective
(or
1-1).
We also have
(K-A1):Ri
q)
=
(K-A1)(K-A1)q(:Ji)
= (K- A1)q+l(:Ji) = :Ri
q+
1)
= :Ri
q).
Therefore, when restricted to :Ri
q),
the operator K- A1 is surjective (or onto) as
well. Thus, (K- A1) : :Ri
q)
-->:Ri
q)
is bijective, and therefore has an inverse. D
16.6.7. Corollary. The two integers p
and
q introduced in Theorems 16.6.4 and
16.6.5 are equal.
472 16. ANINTROOUCTION
TO
OPERATOR
THEORY
Proof
The proof is left as a problem for the reader (see Problem 16.5). D
Thenext theorem characterizes the spectrum
of
acompact operator completely.
In
order to prove it, we need the following lemma.
16.6.8. Lemma.
Let
K, :
9<Iq)
-->
9<Iq)
be the restriction
ofK
to
9<Iq).
Then:
1. Each nonzeropoint
ofa(K)
is an eigenvalue
ofK
whose eigenspace isfinite-
dimensional.
2.
a(K,)
"" a(K).
3. Every infinite sequence in
a(K)
converges to zero.
Proof
(I)
If
A
oF
0 is not an eigenvalue of K, the null space
of
K - A1 is zero.
This says that
{O}
""
NIO)
"" Nil) ""
..
" i.e., p "" q ""
O.
From Theorem 16.6.6,
we conclude that:Ji ""
NIO)
EIl9<IO)
""
9<II).
Therefore, K - A1 is onto. Part (3)
of
Theorem 16.6.6 shows that K- A1 is one-to-one. Thus, K- A1 is invertible, and
A E p(K). So, A ¢ a(K).
(2) Clearly
a(K,)
<;a(K). To show the reverse inclusion, first note that
9<Iq)
is infinite-dimensional because
NIq)
has finite dimension. Thus by Proposition
16.6.1,0
E
a(K,).
Now let u-s-nonzero and distinct from
A-be
in a(K). By part
(I)
/Lis an eigenvalue
of
K,so there is a vector lu) E :Ji such that Klu) "" /Llu).
We also have (K - A1) lu} "" (/L -
A)
lu), or (K - A1)q lu) "" (/L -
A)q
lu).
Thus, (/L -
A)q
lu) (and, therefore lu) is in
9<Iq).
Therefore, we can restrict K to
9<Iq),
i.e., we can write
Klu)
"" /Llu) as K, lu} "" /Llu}, or
(K,
- /L1) lu) ""
O.
Hence, /L E
a(K,).
We conclude that every point of
a(K)
is a point
of
a(K,)
and
a(K) <;
a(K,).
(3) Let A be the limit of an infinite sequence in a(K) ""
a(K,).
If
A
oF
0,
K, -
A1
will be invertible (Theorem 16'6.6 part 3), indicating that A E
p(K,).
Since
p(K,)
is open, we can find an open round neighborhood of A entirely in
p(K,). This contradicts the property of alimit of an infinite sequence whereby any
neighborhood of the limit contains (infinitely many) other points of the sequence.
Therefore, we must conclude that no nonzero A can be the limit of an infinite
sequence in
a(K). D
16.6.9. Theorem. Let Kbe a compactoperator on an infinite-dimensionalHilbert
space
:Ji. Then
1. 0
E a(K).
2. Each nonzeropoint
of
a (K) is an eigenvalue
ofK
whose eigenspace isfinite-
dimensional.
3. a (K) is either afinite
set
or it is a sequence that converges to zero.