4Hilbert spaces
II Linear Analysis
4.5 Operators
We are going to look at operators on Hilbert spaces. For example, we would like
to see how differential operators behave on spaces of differentiable functions.
In this section, we will at least require the space to be Banach. So let
X
be a Banach space over
C
. We will consider
B
(
X
) =
B
(
X, X
), the vector space
of bounded linear maps from
X
to itself. We have seen in the example sheets
that
B
(
X
) is a unital Banach algebra, i.e. it forms a complete algebra with
composition as multiplication. Our goal is to generalize some considerations in
finite dimensions such as eigenvectors and eigenvalues.
Definition
(Spectrum and resolvent set)
.
Let
X
be a Banach space and
T ∈
B(X), we define the spectrum of T , denoted by σ(T ) by
σ(t) = {λ ∈ C : T − λI is not invertible}.
The resolvent set, denoted by ρ(T ), is
ρ(t) = C \ σ(T ).
Note that if
T −λ
is is bijective, then by the inverse mapping theorem, we
know it has a bounded inverse. So if
λ ∈ σ
(
T
), then either
T −λI
is not injective,
or it is not surjective. In other words,
ker
(
T − λI
)
6
=
{
0
}
or
im
(
T − λI
)
6
=
X
.
In finite dimensions, these are equivalent by, say, the rank-nullity theorem, but
in general, they are not.
Example. Consider the shift operator s : `
∞
→ `
∞
defined by
(a
1
, a
2
, a
3
, ···) 7→ (0, a
1
, a
2
, ···).
Then this is injective but not surjective.
Now if
λ ∈ ρ
(
T
), i.e.
T −λI
is invertible, then (
T −λI
)
−1
is automatically
bounded by the inverse mapping theorem. This is why we want to work with
Banach spaces.
Definition
(Resolvent)
.
Let
X
be a Banach space. The resolvent is the map
R : ρ(T ) → B(X) given by
λ 7→ (T − λI)
−1
.
Definition
(Eigenvalue)
.
We say
λ
is an eigenvalue of
T
if
ker
(
T −λI
)
6
=
{
0
}
.
Definition
(Point spectrum)
.
Let
X
be a Banach space. The point spectrum is
σ
p
(T ) = {λ ∈ C : λ is an eigenvalue of T }.
Obviously, σ
p
(T ) ⊆ σ(T ), but they are in general not equal.
Definition
(Approximate point spectrum)
.
Let
X
be a Banach space. The
approximate point spectrum is defined as
σ
ap
(X) = {λ ∈ C : ∃{x
n
} ⊆ X : kx
n
k
X
= 1 and k(T − λI)x
n
k
X
→ 0}.
Again, we have
σ
p
(T ) ⊆ σ
ap
(T ) ⊆ σ(T ).
The last inclusion follows from the fact that if an inverse exists, then the inverse
is bounded.
An important characterization of the spectrum is the following theorem:
Theorem.
Let
X
be a Banach space,
T ∈ B
(
X
). Then
σ
(
T
) is a non-empty,
closed subset of
{λ ∈ C : |λ| ≤ kT k
B(X)
}.
In finite dimensions, this in particular implies the existence of eigenvalues,
since the spectrum is equal to the point spectrum. Notice this is only true for
vector spaces over C, as we know from linear algebra.
To prove this theorem, we will first prove two lemmas.
Lemma.
Let
X
be a Banach space,
T ∈ B
(
X
) and
kT k
B(X)
<
1. Then
I − T
is invertible.
Proof.
To prove it is invertible, we construct an explicit inverse. We want to
show
(I − T )
−1
=
∞
X
i=0
T
i
.
First, we check the right hand side is absolutely convergent. This is since
∞
X
i=0
kT
i
k
B(X)
≤
∞
X
i=0
kT k
i
B(X)
≤
1
1 − kT k
B(X)
< ∞.
Since
X
is Banach, and hence
B
(
X
) is Banach, the limit is well-defined. Now it
is easy to check that
(I − T )
∞
X
i=1
T
i
= (I − T )(I + T + T
2
+ ···)
= I + (T −T ) + (T
2
− T
2
) + ···
= I.
Similarly, we have
∞
X
i=1
T
i
!
(I − T ) = I.
Lemma.
Let
X
be a Banach space,
S
1
∈ B
(
X
) be invertible. Then for all
S
2
∈ B(X) such that
kS
−1
1
k
B(X)
kS
1
− S
2
k
B(X)
< 1,
S
2
is invertible.
This is some sort of an “openness” statement for the invertible bounded
linear maps, since if
S
1
is invertible, then any “nearby” bounded linear map is
also invertible.
Proof. We can write
S
2
= S
1
(I − S
−1
1
(S
1
− S
2
)).
Since
kS
−1
1
(S
1
− S
2
)k
B(X)
≤ kS
−1
1
k
B(X)
kS
1
− S
2
k
B(X)
< 1
by assumption, by the previous lemma, (
I − S
−1
1
(
S
1
− S
2
))
−1
exists. Therefore
the inverse of S
2
is
S
−1
2
= (I − S
−1
1
(S
1
− S
2
))
−1
S
−1
1
.
We can now return to prove our original theorem.
Theorem.
Let
X
be a Banach space,
T ∈ B
(
X
). Then
σ
(
T
) is a non-empty,
closed subset of
{λ ∈ C : |λ| ≤ kT k
B(X)
}.
Note that it is not hard to prove that it is closed and a subset of
{λ ∈ C
:
|λ| ≤ kT k
B(X)
}. The hard part is to prove it is non-empty.
Proof.
We first prove the closedness of the spectrum. It suffices to prove that
the resolvent set ρ(T ) = C \ σ(T ) is open, by the definition of closedness.
Let
λ ∈ ρ
(
T
). By definition,
S
1
=
T −λI
is invertible. Define
S
2
=
T −µI
.
Then
kS
1
− S
2
k
B(X)
= k(T −λI) − (T − µI)k
B(X)
= |λ − µ|.
Hence if
|λ − µ|
is sufficiently small, then
T − µI
is invertible by the above
lemma. Hence µ ∈ ρ(T ). So ρ(T ) is open.
To show σ(T ) ⊆ {λ ∈ C : |λ| ≤ kT k
B(X)
} is equivalent to showing
{λ ∈ C : |λ| > kT k
B(X)
} ⊆ C \ σ(T ) = ρ(T ).
Suppose |λ| > kT k. Then I − λ
−1
T is invertible since
kλ
−1
T k
B(X)
= λ
−1
kT k
B(X)
< 1.
Therefore, (I − λ
−1
T )
−1
exists, and hence
(λI − T )
−1
= λ
−1
(I − λT )
−1
is well-defined. Therefore λI − T , and hence T − λI is invertible. So λ ∈ ρ(T ).
Finally, we need to show it is non-empty. How did we prove it in the case
of finite-dimensional vector spaces? In that case, it ultimately boiled down to
the fundamental theorem of algebra. And how did we prove the fundamental
theorem of algebra? We said that if
p
(
x
) is a polynomial with no roots, then
1
p(x)
is bounded and entire, hence constant.
We are going to do the same proof. We look at
1
T −λI
as a function of
λ
. If
σ
(
T
) =
∅
, then this is an everywhere well-defined function. We show that this is
entire and bounded, and hence by “Liouville’s theorem”, it must be constant,
which is impossible (in the finite-dimensional case, we would have inserted a
det
there).
So suppose
σ
(
T
) =
∅
, and consider the function
R
:
C → B
(
X
), given by
R(λ) = (T − λI)
−1
.
We first show this is entire. This, by definition, means
R
is given by a power
series near any point
λ
0
∈ C
. Fix such a point. Then as before, we can expand
T −λI = (T − λ
0
I)
h
I − (T −λ
0
I)
−1
(T −λ
0
I) − (T − λI)
i
= (T −λ
0
I)
h
I − (λ − λ
0
)(T −λ
0
I)
−1
i
.
Then for (λ − λ
0
) small, we have
(T −λI)
−1
=
∞
X
i=0
(λ − λ
0
)
i
(T −λ
0
I)
−i
!
(T −λ
0
I)
−1
=
∞
X
i=0
(λ − λ
0
)
i
(T −λ
0
I)
−i−1
.
So this is indeed given by an absolutely convergent power series near λ
0
.
Next, we show R is bounded, i.e.
sup
λ∈C
kR(λ)k
B(X)
< ∞.
It suffices to prove this for λ large. Note that we have
(T −λI)
−1
= λ
−1
(λ
−1
T −I)
−1
= −λ
−1
∞
X
i=0
λ
−i
T
i
.
Hence we get
k(λI − T )
−1
k
B(X)
≤ |λ|
−1
∞
X
i=0
|λ|
−i
kT
i
k
B(X)
≤ |λ|
−1
∞
X
i=0
|λ|
−1
kT k
B(X)
i
≤
1
|λ| − kT k
B(X)
,
which tends to 0 as |λ| → ∞. So it is bounded.
By “Liouville’s theorem”,
R
(
λ
) is constant, which is clearly a contradiction
since R(λ) 6= R(µ) for λ 6= µ.
Of course, to do this properly, we need a version of Liouville’s theorem for
Banach-space valued functions as opposed to complex-valued functions. So let’s
prove this.
Proposition
(Liouville’s theorem for Banach space-valued analytic function)
.
Let
X
be a Banach space, and
F
:
C → X
be entire (in the sense that
F
is given
by an absolutely convergent power series in some neighbourhood of any point)
and norm bounded, i.e.
sup
z∈C
kF (z)k
X
< ∞.
Then F is constant.
This is a generalization of Liouville’s theorem to the case where the target
of the map is a Banach space. To prove this, we reduce this to the case of
complex-valued functions. To do so, we compose this F with a map X → C.
Proof.
Let
f ∈ X
∗
. Then we show
f ◦ F
:
C → C
is bounded and entire. To see
it is bounded, just note that f is a bounded linear map. So
sup
z∈C
|f ◦ F (z)| ≤ sup
z∈C
kfk
X
∗
kF (z)k
X
< ∞.
Analyticity can be shown in a similar fashion, exploiting the fact that
f
∗
is
linear.
Hence Liouville’s theorem implies
f ◦F
is constant, i.e. (
f ◦F
)(
z
) = (
f ◦F
)(0).
In particular, this implies
f
(
F
(
z
)
− F
(0)) = 0. Moreover, this is true for all
f ∈ X
∗
. Hence by (corollary of) Hahn-Banach theorem, we know
F
(
z
)
−F
(0) = 0
for all z ∈ C. Therefore F is constant.
We have thus completed our proof that
σ
(
T
) is non-empty, closed and a
subset of {λ ∈ C : |λ| ≤ kT k
B(X)
}.
However, we want to know more. Apart from the spectrum itself, we also
had the point spectrum
σ
p
(
T
) and the approximate point spectrum
σ
ap
(
T
), and
we had the inclusions
σ
p
(T ) ⊆ σ
ap
(T ) ⊆ σ(T ).
We know that the largest set
σ
(
T
) is non-empty, but we want the smaller ones
to be non-empty as well. We have the following theorem:
Theorem. We have
σ
ap
(T ) ⊇ ∂σ(T ),
where
∂σ
(
T
) is the boundary of
σ
(
T
) in the topology of
C
. In particular,
σ
ap
(T ) 6= ∅.
On the other hand, it is possible for
σ
p
(
T
) to be empty (in infinite dimensional
cases).
Proof.
Let
λ ∈ ∂σ
(
T
). Pick sequence
{λ
n
}
∞
n=1
⊆ ρ
(
T
) =
C \ σ
(
T
) such that
λ
n
→ λ. We claim that R(λ
n
) = (T −λ
n
I)
−1
satisfies
kR(λ
n
)k
B(X)
→ ∞.
If this were the case, then we can pick
y
n
∈ X
such that
ky
n
k →
0 and
kR(λ
n
)(y
n
)k = 1. Setting x
n
= R(λ
n
)(y
n
), we have
k(T −λI)x
n
k ≤ k(T − λ
n
I)x
n
k
X
+ k(λ − λ
n
)x
n
k
X
= k(T −λ
n
I)(T − λ
n
I)
−1
y
n
k
X
+ k(λ − λ
n
)x
n
k
= ky
n
k
X
+ |λ − λ
n
|
→ 0.
So λ ∈ σ
ap
(T ).
Thus, it remains to prove that
kR
(
λ
n
)
k
B(X)
→ ∞
. Recall from last time if
S
1
is invertible, and
kS
−1
1
k
B(X)
kS
1
− S
2
k
B(X)
≤ 1, (∗)
then S
2
is invertible. Thus, for any µ ∈ σ(T ), we have
kR(λ
n
)k
B(X)
|µ − λ
n
| = kR(λ
n
)k
B(X)
k(T −λ
n
I) − (T − µI)k
B(X)
≥ 1.
Thus, it follows that
kR(λ
n
)k
B(X)
≥
1
inf{|µ − λ
n
| : µ ∈ σ(T )}
→ ∞.
So we are done.
Having proven so many theorems, we now look at an specific example.
Example. Consider the shift operator S : `
∞
→ `
∞
defined by
(a
1
, a
2
, a
3
, ···) 7→ (0, a
1
, a
2
, ···).
Then
S
is a bounded linear operator with norm
kSk
B(`
∞
)
= 1. The theorem
then tells σ(S) is a non-empty closed subset of {λ ∈ C : kλk ≤ 1}.
First, we want to understand what the point spectrum is. In fact, it is empty.
To show this, suppose
S(a
1
, a
2
, a
3
, ···) = λ(a
1
, a
2
, a
3
, ···)
for some λ ∈ C. In other words,
(0, a
1
, a
2
, ···) = λ(a
1
, a
2
, a
3
, ···).
First consider the possibility that
λ
= 0. This would imply that the left is zero.
So a
i
= 0 for all i.
If
λ 6
= 0, then for the first coordinate to match, we must have
a
1
= 0. Then
for the second coordinate to match, we also need
a
2
= 0. By induction, we need
all a
i
= 0. So ker(S − λI) = {0} for all λ ∈ C.
To find the spectrum, we will in fact show that
σ(S) = D = {λ ∈ C : |λ| ≤ 1}.
To prove this, we need to show that for any
λ ∈ D
,
S − λI
is not surjective.
The
λ
= 0 case is obvious. For the other cases, we first have a look at what the
image of S − λI looks like. We take
(b
1
, b
2
, b
3
, ···) ∈ `
∞
.
Suppose for some λ ∈ D, there exists (a
1
, a
2
, ···) such that we have
(S − λI)(a
1
, a
2
, ···) = (b
1
, b
2
, ···)
In other words, we have
(0, a
1
, a
2
, ···) − (λa
1
, λa
2
, λa
3
, ···) = (b
1
, b
2
, b
3
, ···).
So −λa
1
= b
1
. Hence we have
a
1
= −λ
−1
b
1
.
The next line then gives
a
1
− λa
2
= b
2
.
Hence
a
2
= −λ
−1
(b
2
− a
1
) = −λ
−1
(b
2
+ λ
−1
b
1
).
Inductively, we can show that
a
n
= λ
−1
(b
n
+ λ
−1
b
n−1
+ λ
−2
b
n−2
+ ··· + λ
−n+1
b
1
).
Now if
|λ| ≤
1, we pick
b
n
such that
|b
n
|
= 1 and
λ
−i
b
n−i
=
|λ|
−i
. Then we must
have
|a
n
| → ∞
. Such a sequence (
a
n
)
6∈ `
∞
. So (
b
n
)
6∈ im
(
S − λI
). Therefore
for |λ| ≤ 1, S − λI is not surjective.
Hence we have
σ
(
S
)
⊇ D
. By the theorem, we also know
σ
(
S
)
⊆ D
. So in
fact σ(S) = D.
Finally, we show that
σ
ap
(S) = ∂D = {λ ∈ C : |λ| = 1}.
Our theorem tells us that
∂D ⊆ σ
ap
(
S
)
⊆ D
. To show that indeed
∂D
=
σ
ap
(
S
),
note that if |λ| < 1, then for all x ∈ `
∞
.
k(S − λI)xk
`
∞
≥ kS
x
k
`∞
− |λ|kxk
`
∞
= kxk
`
∞
− |λ|kxk
`
∞
= (1 − λ)kxk
`
∞
.
So if
|λ| <
1, then there exists no sequence
x
n
with
kx
n
k
`
∞
= 1 and
k
(
S −
λI)x
n
k
`
∞
→ 0. So λ is not in the approximate point spectrum.
We see that these results are rather unpleasant and the spectrum behaves
rather unlike the finite dimensional cases. We saw last time that the spectrum
σ
(
T
) can in general be complicated. In fact, any non-empty compact subset of
C
can be realized as the spectrum of some operator
T
on some Hilbert space
H
.
This is an exercise on the Example sheet.
We are now going to introduce a class of “nice” operators whose spectrum
behaves in a way more similar to the finite-dimensional case. In particular, most
of
σ
(
T
) consists of
σ
p
(
T
) and is discrete. This includes, at least, all finite rank
operators (i.e. operators T such that dim(im T )) < ∞) and their limits.
Definition
(Compact operator)
.
Let
X, Y
be Banach spaces. We say
T ∈
L
(
X, Y
) is compact if for every bounded subset
E
of
X
,
T
(
E
) is totally bounded.
We write B
0
(X) for the set of all compact operators T ∈ B(X).
Note that in the definition of
T
, we only required
T
to be a linear map, not
a bounded linear map. However, boundedness comes from the definition for free
because a totally bounded set is bounded.
There is a nice alternative characterization of compact operators:
Proposition.
Let
X, Y
be Banach spaces. Then
T ∈ L
(
X, Y
) is compact if and
only if T (B(1)) is totally bounded if and only if T (B(1)) is compact.
The first equivalence is obvious, since
B
(1) is a bounded set, and given any
bounded set
E
, we can rescale it to be contained in
B
(1). The second equivalence
comes from the fact that a space is compact if and only if it is totally bounded
and complete.
The last characterization is what we will use most of the time, and this is
where the name “compact operator” came from.
Proposition.
Let
X
be a Banach space. Then
B
0
(
X
) is a closed subspace of
B(X). Moreover, if T ∈ B
0
(X) and S ∈ B(X), then T S, ST ∈ B
0
(X).
In a more algebraic language, this means
B
0
(
X
) is a closed ideal of the
algebra B(X).
Proof.
There are three things to prove. First, it is obvious that
B
0
(
X
) is a
subspace. To check it is closed, suppose
{T
n
}
∞
n=1
⊆ B
0
(
X
) and
kT
n
−T k
B(X)
→
0.
We need to show T ∈ B
0
(X), i.e. T (B(1)) is totally bounded.
Let ε > 0. Then there exists N such that
kT −T
n
k
B(X)
< ε
whenever
n ≥ N
. Take such an
n
. Then
T
n
(
B
(1)) is totally bounded. So there
exists
x
1
, ··· , x
k
∈ B
(1) such that
{T
n
x
i
}
k
i=1
is an
ε
-net for
T
n
(
B
(1)). We now
claim that {T x
i
}
k
i=1
is an 3ε-net for T (B(1)).
This is easy to show. Let
x ∈ X
be such that
kxk ≤
1. Then by the triangle
inequality,
kT x − T x
i
k
X
≤ kT x − T
n
xk + kT
n
x − T
n
x
i
k + kT
n
x
i
− T x
i
k
≤ ε + kT
n
x − T
n
x
i
k
X
+ ε
= 2ε + kT
n
x − T
n
x
i
k
X
Now since
{T
n
x
i
}
is an
ε
-net for
T
n
(
B
(1)), there is some
i
such that
kT
n
x −
T
n
x
i
k < ε. So this gives
kT x − T x
i
k
X
≤ 3ε.
Finally, let
T ∈ B
0
(
X
) and
S ∈ B
(
X
). Let
{x
n
} ⊆ X
such that
kx
n
k
X
≤
1. Since
T
is compact, i.e.
T (B(1))
is compact, there exists a convergence subsequence
of {T x
i
}.
Since
S
is bounded, it maps a convergent sequence to a convergent sequence.
So
{ST x
n
}
also has a convergent subsequence. So
ST (B(1))
is compact. So
ST
is compact.
We also have to show that
T S
(
B
(1)) is totally bounded. Since
S
is bounded,
S
(
B
(1)) is bounded. Since
T
sends a bounded set to a totally bounded set, it
follows that T S(B(1)) is totally bounded. So T S is compact.
At this point, it helps to look at some examples at actual compact operators.
Example.
(i)
Let
X, Y
by Banach spaces. Then any finite rank operator in
B
(
X, Y
) is
compact. This is since
T
(
B
(1)) is a bounded subset of a finite-dimensional
space (since
T
is bounded), and any bounded subset of a finite-dimensional
space is totally bounded.
In particular, any
f ∈ X
∗
is compact. Moreover, by the previous propo-
sition, limits of finite-rank operators are also compact (since
B
0
(
X
) is
closed).
(ii)
Let
X
be a Banach space. Then
I
:
X → X
is compact if and only if
X
is
finite dimensional. This is since
B(1)
is compact if and only if the space is
finite-dimensional.
(iii)
Let
K
:
R → R
be a smooth function. Define
T
:
C
([0
,
1])
→ C
([0
,
1]) by
the convolution
(T f)(x) =
Z
1
0
K(x − y)f(y) dy.
We first show T is bounded. This is since
sup
x
|T f(x)| ≤ sup
z∈[−1,1]
K(z) sup
y
|f(y)|.
Since
K
is smooth, it is bounded on [
−
1
,
1]. So
T f
is bounded. In fact,
T
is compact. To see this, notice
sup
x
d(T f)
dx
(x)
≤ sup
z∈[−1,1]
|K
0
(z)|sup
y
|f(y)|.
Therefore if we have a sequence
{f
n
} ⊆ C
([0
,
1]) with
kf
n
k
C([0,1])
≤
1,
then
{T f
n
}
∞
n=1
is uniformly bounded with uniformly bounded derivative,
and hence equicontinuous. By Arzel`a-Ascoli theorem,
{T f
n
}
∞
n=1
has a
convergent subsequence. Therefore T (B(1)) is compact.
There is much more we can say about the last example. For example, requiring
K
to be smooth is clearly overkill. Requiring, say, differentiable with bounded
derivative is already sufficient. Alternatively, we can ask what we can get if we
work on, say, L
2
instead of C([0, 1]).
However, we don’t have enough time for that, and instead we should return
to developing some general theory. An important result is the following theorem,
characterizing the point spectrum and spectrum of a compact operator.
Theorem.
Let
X
be an infinite-dimensional Banach space, and
T ∈ B
(
X
) be a
compact operator. Then
σ
p
(
T
) =
{λ
i
}
is at most countable. If
σ
p
(
T
) is infinite,
then λ
i
→ 0.
The spectrum is given by
σ
(
T
) =
σ
p
(
T
)
∪ {
0
}
. Moreover, for every non-zero
λ
i
∈ σ
p
(T ), the eigenspace has finite dimensions.
Note that it is still possible for a compact operator to have empty point
spectrum. In that case, the spectrum is just
{
0
}
. An example of this is found
on the example sheet.
We will only prove this in the case where
X
=
H
is a Hilbert space. In a
lot of the proofs, we will have a closed subspace
V ≤ X
, and we often want to
pick an element in
X \ V
that is “far away” from
V
in some sense. If we have a
Hilbert space, then we can pick this element to be an element in the complement
of
V
. If we are not in a Hilbert space, then we need to invoke Riesz’s lemma,
which we shall not go into.
We will break the full proof of the theory into many pieces.
Proposition.
Let
H
be a Hilbert space, and
T ∈ B
0
(
H
) a compact operator.
Let
a >
0. Then there are only finitely many linearly independent eigenvectors
whose eigenvalue have magnitude ≥ a.
This already gives most of the theorem, since this mandates
σ
p
(
T
) is at most
countable, and if
σ
p
(
T
) is infinite, we must have
λ
i
→
0. Since there are only
finitely many linearly independent eigenvectors, the eigenspaces are also finite.
Proof.
Suppose not. There there are infinitely many independent
x
1
, x
2
, x
3
, ···
such that T x
i
= λ
i
x
i
with |λ
i
| ≥ a.
Define
X
n
=
span{x
1
, ··· , x
n
}
. Since the
x
i
’s are linearly independent, there
exists y
n
∈ X
n
∩ X
⊥
n−1
with ky
n
k
H
= 1.
Now let
z
n
=
y
n
λ
n
.
Note that
kz
n
k
H
≤
1
a
.
Since
X
n
is spanned by the eigenvectors, we know that
T
maps
X
n
into itself.
So we have
T z
n
∈ X
n
.
Moreover, we claim that T z
n
− y
n
∈ X
n−1
. We can check this directly. Let
y
n
=
n
X
k=1
c
k
x
k
.
Then we have
T z
n
− y
n
=
1
λ
n
T
n
X
k=1
c
k
x
k
!
−
n
X
k=1
c
k
x
k
=
n
X
k=1
c
k
λ
k
λ
n
− 1
x
k
=
n−1
X
k=1
c
k
λ
k
λ
n
− 1
x
k
∈ X
n−1
.
We next claim that
kT z
n
− T z
m
k
H
≥
1 whenever
n > m
. If this holds, then
T
is not compact, since T z
n
does not have a convergent subsequence.
To show this, wlog, assume n > m. We have
kT z
n
− T z
m
k
2
H
= k(T z
n
− y
n
) − (T z
m
− y
n
)k
2
H
Note that
T z
n
− y
n
∈ X
n−1
, and since
m < n
, we also have
T z
m
∈ X
n−1
. By
construction, y
n
⊥ X
n−1
. So by Pythagorean theorem, we have
= kT z
n
− y
n
− T z
m
k
2
H
+ ky
n
k
2
H
≥ ky
n
k
2
= 1
So done.
To prove the previous theorem, the only remaining thing to prove is that
σ
(
T
) =
σ
p
(
T
)
∪{
0
}
. In order to prove this, we need a lemma, which might seem
a bit unmotivated at first, but will soon prove itself useful.
Lemma.
Let
H
be a Hilbert space, and
T ∈ B
(
H
) compact. Then
im
(
I −T
) is
closed.
Proof.
We let
S
be the orthogonal complement of
ker
(
I − T
), which is a closed
subspace, hence a Hilbert space. We shall consider the restriction (
I − T
)
|
S
,
which has the same image as I − T .
To show that
im
(
I −T
) is closed, it suffices to show that (
I −T
)
|
S
is bounded
below, i.e. there is some C > 0 such that
kxk
H
≤ Ck(I − T )xk
H
for all x ∈ S. If this were the case, then if (I − T )x
n
→ y in H, then
kx
n
− x
m
k ≤ Ck(I − T )(x
n
− x
m
)k → 0,
and so
{x
n
}
is a Cauchy sequence. Write
x
n
→ x
. Then by continuity, (
I−T
)
x
=
y, and so y ∈ im(I − T ).
Thus, suppose (
I − T
) is not bounded below. Pick
x
n
such that
kx
n
k
H
= 1,
but (
I − T
)
x
n
→
0. Since
T
is compact, we know
T x
n
has a convergent
subsequence. We may wlog
T x
n
→ y
. Then since
kT x
n
− x
n
k
H
→
0, it follows
that we also have x
n
→ y. In particular, kyk = 1 6= 0, and y ∈ S.
But
x
n
→ y
also implies
T x
n
→ T y
. So this means we must have
T y
=
y
.
But this is a contradiction, since y does not lie in ker(I − T ).
Proposition.
Let
H
be a Hilbert space,
T ∈ B
(
H
) compact. If
λ 6
= 0 and
λ ∈ σ(T ), then λ ∈ σ
p
(T ).
Proof.
We will prove if
λ 6
= 0 and
λ 6∈ σ
p
(
T
), then
λ 6∈ σ
(
T
). In other words,
let
λ 6
= 0 and
ker
(
T − λI
) =
{
0
}
. We will show that
T − λI
is surjective, i.e.
im(T −λI) = H.
Suppose this is not the case. Denote
H
0
=
H
and
H
1
=
im
(
T − λI
). We
know that
H
1
is closed and is hence a Hilbert space. Moreover,
H
1
( H
0
by
assumption.
We now define the sequence {H
n
} recursively by
H
n
= (T −λI)H
n−1
.
We claim that
H
n
( H
n−1
. This must be the case, because the map (
T − λI
)
n
:
H
0
→ H
n
is an isomorphism (it is injective and surjective). So the inclusion
H
n
⊆ H
n−1
is isomorphic to the inclusion H
1
⊆ H
0
, which is strict.
Thus we have a strictly decreasing sequence
H
0
) H
1
) H
2
) ···
Let
y
n
be such that
y
n
∈ H
n
,
y
n
⊥ H
n+1
and
ky
n
k
H
= 1. We now claim
kT y
n
− T y
m
k ≥ |λ|
if
n 6
=
m
. This then contradicts the compactness of
T
. To
show this, again wlog we can assume that n > m. Then we have
kT y
n
− T y
m
k
2
H
= k(T y
n
− λy
n
) − (T y
m
− λy
m
) − λy
m
+ λy
n
k
2
= k(T −λI)y
n
− (T − λI)y
m
− λy
m
+ λy
n
k
2
H
Now note that (
T −λI
)
y
n
∈ H
n+1
⊆ H
m+1
, while (
T −λI
)
y
m
and
λy
n
are both
in
H
m+1
. So
λy
m
is perpendicular to all of them, and Pythagorean theorem
tells
= |λ|
2
ky
m
k
2
+ k(T − λI)y
m
− (T − λI)y
m
− λy
m
k
2
≥ |λ|
2
ky
m
k
2
= |λ|
2
.
This contradicts the compactness of T . Therefore im(T − λI) = H.
Finally, we can prove the initial theorem.
Theorem.
Let
H
be an infinite-dimensional Hilbert space, and
T ∈ B
(
H
) be a
compact operator. Then
σ
p
(
T
) =
{λ
i
}
is at most countable. If
σ
p
(
T
) is infinite,
then λ
i
→ 0.
The spectrum is given by
σ
(
T
) =
σ
p
(
T
)
∪
0. Moreover, for every non-zero
λ
i
∈ σ
p
(T ), the eigenspace has finite dimensions.
Proof.
As mentioned, it remains to show that
σ
(
T
) =
σ
p
(
T
)
∪{
0
}
. The previous
proposition tells us
σ
(
T
)
\{
0
} ⊆ σ
p
(
T
). So it only remains to show that 0
∈ σ
(
T
).
There are two possible cases. The first is if
{λ
i
}
is infinite. We have already
shown that λ
i
→ 0. So 0 ∈ σ(T ) by the closedness of the spectrum.
Otherwise, if {λ
i
} is finite, let E
λ
1
, ··· , E
λ
n
be the eigenspaces. Define
H
0
= span{E
λ
1
, ··· , E
λ
n
}
⊥
.
This is non-empty, since each
E
λ
i
is finite-dimensional, but
H
is infinite dimen-
sional. Then T restricts to T |
H
0
: H
0
→ H
0
.
Now
T |
H
0
has no non-zero eigenvalues. By the previous discussion, we know
σ(T |
H
0
) ⊆ {0}. By non-emptiness of σ(T |
H
0
), we know 0 ∈ σ(T |
H
0
) ⊆ σ(T ).
So done.