Arnold Lecture notes on functional analysis [sharethefiles com]

FUNCTIONAL ANALYSIS

Douglas N. Arnold

References:
John B. Conway,

A Course in Functional Analysis

, 2nd Edition, Springer-Verlag, 1990.

Gert K. Pedersen,

Analysis Now

, Springer-Verlag, 1989.

Walter Rudin,

Functional Analysis

, 2nd Edition, McGraw Hill, 1991.

Robert J. Zimmer,

Essential Results of Functional Analysis

, University of Chicago Press,

1990.

CONTENTS

I. Vector spaces and their topology

:::::::::::::::::::::::::::::::::::::::::::::::

Subspaces and quotient spaces

::::::::::::::::::::::::::::::::::::::::::::

Basic properties of Hilbert spaces

:::::::::::::::::::::::::::::::::::::::::

II. Linear Operators and Functionals

::::::::::::::::::::::::::::::::::::::::::::::

The Hahn{Banach Theorem

::::::::::::::::::::::::::::::::::::::::::::::

Duality

::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::

III. Fundamental Theorems

:::::::::::::::::::::::::::::::::::::::::::::::::::::::

The Open Mapping Theorem

:::::::::::::::::::::::::::::::::::::::::::::

The Uniform Boundedness Principle

::::::::::::::::::::::::::::::::::::::

The Closed Range Theorem

::::::::::::::::::::::::::::::::::::::::::::::

IV. Weak Topologies

:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::

The weak topology

:::::::::::::::::::::::::::::::::::::::::::::::::::::::

The weak* topology

::::::::::::::::::::::::::::::::::::::::::::::::::::::

V. Compact Operators and their Spectra

::::::::::::::::::::::::::::::::::::::::

Hilbert{Schmidt operators

:::::::::::::::::::::::::::::::::::::::::::::::

Compact operators

:::::::::::::::::::::::::::::::::::::::::::::::::::::::

Spectral Theorem for compact self-adjoint operators

::::::::::::::::::::::

The spectrum of a general compact operator

:::::::::::::::::::::::::::::

VI. Introduction to General Spectral Theory

::::::::::::::::::::::::::::::::::::::

The spectrum and resolvent in a Banach algebra

:::::::::::::::::::::::::

Spectral Theorem for bounded self-adjoint operators

::::::::::::::::::::::

These lecture notes were prepared for the instructor's personal use in teaching a half-semester course

on functional analysis at the beginning graduate level at Penn State, in Spring 1997. They are certainly

not meant to replace a good text on the subject, such as those listed on this page.

Department of Mathematics, Penn State University, University Park, PA 16802.

Email: dna@math.psu.edu. Web: http://www.math.psu.edu/dna/.

Typeset by

-TEX

I. Vector spaces and their topology

Basic denitions: (1) Norm and seminorm on vector spaces (real or complex). A norm

denes a Hausdor topology on a vector space in which the algebraic operations are con-

tinuous, resulting in a

normed linear space

. If it is complete it is called a Banach space.

(2) Inner product and semi-inner-product. In the real case an inner product is a positive

denite, symmetric bilinear form on

X X

. In the complex case it is positive denite,

Hermitian symmetric, sesquilinear form

X X

. An (semi) inner product gives rise

to a (semi)norm. An inner product space is thus a special case of a normed linear space.

A complete inner product space is a Hilbert space, a special case of a Banach space.

The polarization identity expresses the norm of an inner product space in terms of the

inner product. For real inner product spaces it is

(

) = 14(

;

)

For complex spaces it is

(

) = 14(

;

)

In inner product spaces we also have the parallelogram law:

;

= 2(

)

This gives a criterion for a normed space to be an inner product space. Any norm coming

from an inner product satises the parallelogram law and, conversely, if a norm satises the

parallelogram law, we can show (but not so easily) that the polarization identity denes

an inner product, which gives rise to the norm.

(3) A

topological vector space

is a vector space endowed with a Hausdor topology such

that the algebraic operations are continuous. Note that we can extend the notion of Cauchy

sequence, and therefore of completeness, to a TVS: a sequence

in a TVS is Cauchy if

for every neighborhood

of 0 there exists

such that

;

for all

A normed linear space is a TVS, but there is another, more general operation involving

norms which endows a vector space with a topology. Let

be a vector space and suppose

that a family

of seminorms on

is given which are

su cient

in the sense that

= 0

= 0. Then the topology generated by the sets

< r

r >

makes

a TVS. A sequence (or net)

converges to

;

0 for all

. Note

that, a fortiori,

;

0, showing that each seminorm is continuous.

If the number of seminorms is nite, we may add them to get a norm generating the

same topology. If the number is countable, we may dene a metric

(

) =

;

1 +

;

so the topology is metrizable.

Examples: (0) On

we may put the

norm, 1

, or the weighted

norm with some arbitrary positive weight. All of these norms are equivalent (indeed

all norms on a nite dimensional space are equivalent), and generate the same Banach

topology. Only for

= 2 is it a Hilbert space.

(2) If is a subset of

(or, more generally, any Hausdor space) we may dene the

space

() of bounded continuous functions with the supremum norm. It is a Banach

space. If

is compact this is simply the space

() of continuous functions on .

(3) For simplicity, consider the unit interval, and dene

1]) and

1]),

1]. Both are Banach spaces with the natural norms.

is the space of

Lipschitz functions.

1])

1]) if 0

(4) For 1

p <

and an open or closed subspace of

(or, more generally, a

-nite

measure space), we have the space

() of equivalence classes of measurable

-th power

integrable functions (with equivalence being equality o a set of measure zero), and for

equivalence classes of essentially bounded functions (bounded after modication

on a set of measure zero). For 1

< p <

the triangle inequality is not obvious, it is

Minkowski's inequality. Since we modded out the functions with

-seminorm zero, this

is a normed linear space, and the Riesz-Fischer theorem asserts that it is a Banach space.

is a Hilbert space. If meas()

, then

()

() if 1

(5) The sequence space

, 1

is an example of (4) in the case where the

measure space is

with the counting measure. Each is a Banach space.

is a Hilbert

space.

if 1

(note the inequality is reversed from the previous example).

The subspace

of sequences tending to 0 is a closed subspace of

(6) If is an open set in

(or any Hausdor space), we can equip

() with the

norms

(

)

indexed by

. This makes it a TVS, with the topology being that

of pointwise convergence. It is not complete (pointwise limit of continuous functions may

not be continuous).

(7) If is an open set in

we can equip

() with the norms

(

)

indexed

by compact subsets of , thus dening the topology of uniform convergence on compact

subsets. We get the same toplogy by using only the countably many compact sets

dist(

)

The topology is complete.

(8) In the previous example, in the case is a region in

, and we take complex-

valued functions, we may consider the subspace

() of holomorbarphic functions. By

Weierstrass's theorem it is a closed subspace, hence itself a complete TVS.

(9) If

(

= (0

1) and

(

)

(

)

;

(

)

(

)

for all innitely dierentiable

with support contained in

(so

is identically zero near

0 and 1), then we say that

is weakly dierentiable and that

. We can then dene

the

Sobolev space

(

) =

(

) :

(

)

, with the norm

(

)

(

)

(

)

This is a larger space than

(

), but still incorporates rst order dierentiability of

The case

= 2 is particularly useful, because it allows us to deal with dierentiability

in a Hilbert space context. Sobolev spaces can be extended to measure any degree of

dierentiability (even fractional), and can be dened on arbitrary domains in

Subspaces and quotient spaces.

is a vector space and

a subspace, we may dene the vector space

X=S

of cosets.

is normed, we may dene

X=S

= inf

, or equivalently

X=S

= inf

;

This is a seminorm, and is a norm i

is closed.

Theorem.

is a Banach space and

is a closed subspace then

is a Banach space

and

X=S

is a Banach space.

Sketch.

Suppose

is a sequence of elements of

for which the cosets

are Cauchy.

We can take a subsequence with

;

X=S

;

= 1

:::

. Set

= 0, dene

such that

;

(

)

2, dene

such that

(

)

;

(

)

:::

. Then

is Cauchy in

X :::

A converse is true as well (and easily proved).

Theorem.

is a normed linear space and

is a closed subspace such that

is a

Banach space and

X=S

is a Banach space, then

is a Banach space.

Finite dimensional subspaces are always closed (they're complete). More generally:

Theorem.

is a closed subspace of a Banach space and

is a nite dimensional

subspace, then

is closed.

Sketch.

We easily pass to the case

is one-dimensional and

= 0. We then have that

is algebraically a direct sum and it is enough to show that the projections

and

are continuous (since then a Cauchy sequence in

will lead to a

Cauchy sequence in each of the closed subspaces, and so to a convergent subsequence).

Now the projection

X=S

restricts to a 1-1 map on

so an isomorphism of

onto

its image

. Let

be the continuous inverse. Since

(

)

, we may form

the composition

and it is continuous. But it is just the projection

onto

. The projection onto

;

, so it is also continuous.

Note.

The sum of closed subspaces of a Banach space need not be closed. For a coun-

terexample (in a separable Hilbert space), let

be the vector space of all real sequences

(

)

for which

= 0 if

is odd, and

be the sequences for which

= 1

:::

. Clearly

and

are closed subspaces of

, the space

of square integrable sequences (they are dened as the intersection of the null spaces of

continuous linear functionals). Obviously every sequence can be written in a unique way

as sum of elements of

and

(

:::

) = (0

;

:::

) + (

:::

)

If a sequence has all but nitely many terms zero, so do the two summands. Thus all

such sequences belong to

, showing that

is dense in

. Now consider the

sequence (1

:::

)

. Its only decomposition as elements of

and

:::

) = (0

;

:::

) + (1

:::

)

and so it does not belong to

. Thus

is not closed in

Basic properties of Hilbert spaces.

An essential property of Hilbert space is that the distance of a point to a closed convex

set is alway attained.

Projection Theorem.

Let

be a Hilbert space,

a closed convex subset, and

Then there exists a unique

such that

;

= inf

;

Proof.

Translating, we may assume that

= 0, and so we must show that there is a unique

element of

of minimal norm. Let

= inf

and chose

with

Then the parallelogram law gives

;

= 12

+ 12

;

+ 12

;

where we have used convexity to infer that (

)

. Thus

is a Cauchy

sequence and so has a limit

, which must belong to

, since

is closed. Since the norm

is continuous,

= lim

For uniqueness, note that if

, then

(

+ ~

)

and the parallelogram

law gives

;

= 2

+ 2

;

+ ~

= 2

+ 2

;

= 0

The unique nearest element to

is often denoted

, and referred to as the

projection of

onto

. It satises

, the denition of a projection. This

terminology is especially used when

is a closed linear subspace of

, in which case

is a linear projection operator.

Projection and orthogonality.

is any subset of a Hilbert space

, let

= 0 for all

Then

is a closed subspace of

. We obviously have

= 0 and

Claim: If

is a closed subspace of

, and

the projection of

onto

, then

;

. Indeed, if

is arbitrary and

, then

;

(

;

) +

so the quadratic polynomial on the right hand side has a minimum at

= 0. Setting the

derivative there to 0 gives (

;

) = 0.

Thus we can write any

with

and

(namely

;

). Such a decomposition is certainly unique (if

were another one we

would have

;

= 0.) We clearly have

An immediate corollary is that

for

a closed subspace, since if

can write it as

, whence

= 0, i.e.,

. We thus see that the

decomposition

= (

;

)

is the (unique) decomposition of

into elements of

and

. Thus

;

. For

any subset

is the smallest closed subspace containing

Orthonormal sets and bases in Hilbert space.

Let

:::

be orthonormal elements of a Hilbert space

, and let

be their

span. Then

and

;

, so

. But

, so

(Bessel's inequality). Now let

be an orthonormal set of arbitrary cardinality. It follows

from Bessel's inequality that for

0 and

is nite, and

hence that

is countable. We can thus extend Bessel's inequality to

an arbitrary orthonormal set:

where the sum is just a countable sum of positive terms.

It is useful to extend the notion of sums over sets of arbitrary cardinality. If

is an

arbitary set and

a function mapping into a Hilbert space (or any normed linear

space or even TVS), we say
(

)

(

) =

if the net

(

), indexed by the

nite

subsets

, converges to

. In other words,

(

) holds if, for any neighborhood

of the origin, there is a nite set

such that

;

(

)

whenever

is a nite subset of

containing

. In the case

this is equivalent to absolute convergence of a series. Note that if

(

) converges,

then for all

there is a nite

such that if

and

are nite supersets of

, then

(

)

;

(

)

. It follows easily that each of the sets

(

)

is nite, and hence,

(

) = 0 for all but countably many

Lemma.

is an orthonormal subset of a Hilbert space

and

, then

converges.
Proof.

We may order the elements

:::

for which

= 0. Note that

This shows that the partial sums

form a Cauchy sequence, and so

converge to an element

. As an exercise in applying the denition,

we show that

. Given

0 pick

large enough that

. If

M > N

and

is a nite subset of

containing

:::

then

;

Letting

tend to innity,

;

as required.

Recall the proof that every vector space has a basis. We consider the set of all linearly

independent subsets of the vector space ordered by inclusions, and note that if we have a

totally ordered subset of this set, then the union is a linearly independent subset containing

all its members. Therefore Zorn's lemma implies that there exists a maximal linearly

independent set. It follows directly from the maximality that this set also spans, i.e., is a

basis. In an inner product space we can use the same argument to establish the existence

of an orthonormal basis.

In fact, while bases exist for all vector spaces, for innite dimensional spaces they are

dicult or impossible to construct and almost never used. Another notion of basis is much

more useful, namely one that uses the topology to allow innite linear combinations. To

distinguish ordinary bases from such notions, an ordinary basis is called a Hamel basis.

Here we describe an orthonormal Hilbert space basis. By denition this is a maximal

orthonormal set. By Zorn's lemma, any orthonormal set in a Hilbert space can be extended

to a basis, and so orthonormal bases exist. If

is such an orthonormal basis, and

then

Indeed, we know that the sum on the right exists in

and it is easy to check that its inner

product with any

. Thus

;

is orthogonal to

, and if it

weren't zero, then we could adjoin

to get a larger orthonormal set.

Thus we've shown that any element

can be expressed as

for some

all but countably many of which are 0. It is easily seen that this determines the

uniquely,

namely

, and that

The notion of orthonormal basis allows us to dene a Hilbert space dimension, namely

the cardinality of any orthonormalbasis. To know that this is well dened, we need to check

that any two bases have the same cardinality. If one is nite, this is trivial. Otherwise,

let

and

be two innite orthonormal bases. For each 0

, the inner product

= 0 for at least one

. Thus

= 0

i.e.,

is contained in the union of card

countable sets. Therefore card

card

is any set, we dene a particular Hilbert space

(

) as the set of functions

which are zero o a countable set and such that

. We thus see that via a basis,

any Hilbert space can be put into a norm-preserving (and so inner-product-preserving)

linear bijection (or Hilbert space isomorphism) with an

(

). Thus, up to isomorphism,

there is just one Hilbert space for each cardinality. In particular there is only one innite

dimensional separable Hilbert space (up to isometry).

Example: The best known example of an orthonormal basis in an innite Hilbert space

is the set of functions

= exp(2

) which form a basis for complex-valued

1]).

(They are obviously orthonormal, and they are a maximal orthonormal set by the Weier-

strass approximation Theorem. Thus an arbitrary

function has an

convergent

Fourier series

(

) =

=;1

(

)

with ^

(

) =

(

)

. Thus from the Hilbert space point of view, the

theory of Fourier series is rather simple. More dicult analysis comes in when we consider

convergence in other topologies (pointwise, uniform, almost everywhere,

:::

Schauder bases.

An orthonormal basis in a Hilbert space is a special example of a

Schauder basis. A subset

of a Banach space

is called a Schauder basis if for every

there is a unique function

such that

. Schauder constructed

a useful Schauder basis for

1]), and there is useful Schauder bases in many other

separable Banach spaces. In 1973 Per Eno settled a long-standing open question by

proving that there exist separable Banach spaces with no Schauder bases.

II. Linear Operators and Functionals

(

) = bounded linear operators between normed linear spaces

and

. A linear

operator is bounded i it is bounded on every ball i it is bounded on some ball i it is

continuous at every point i it is continuous at some point.

Theorem.

is a normed linear space and

is a Banach space, then

(

)

is a

Banach space with the norm

(

)

= sup

06=

Proof.

It is easy to check that

(

) is a normed linear space, and the only issue is to

show that it is complete.

Suppose that

is a Cauchy sequence in

(

). Then for each

X T

is Cauchy

in the complete space

, so there exists

with

. Clearly

linear. Is it bounded? The real sequence

is Cauchy, hence bounded, say

It follows that

, and so

(

). To conclude the proof, we need to show

that

;

0. We have

;

= sup

;

= sup

lim

;

= sup

limsup

;

limsup

;

Thus limsup

;

= 0.

(

) and

(

), then

(

) and

(

)

(

)

(

)

. In particular,

(

) :=

(

) is a

Banach algebra

, i.e., it has an

additional \multiplication" operation which makes it a non-commutative algebra, and the

multiplication is continuous.

The dual space is

(

) (or

(

) for complex vector spaces). It is a Banach

space (whether

is or not).

The Hahn{Banach Theorem.

A key theorem for dealing with dual spaces of normed

linear spaces is the Hahn-Banach Theorem. It assures us that the dual space of a nontrivial

normed linear space is itself nontrivial. (Note: the norm is important for this. There exist

topological vector spaces, e.g.,

for 0

< p <

1, with no non-zero continuous linear

functionals.)

Hahn-Banach.

is a bounded linear functional on a subspace of a normed linear space,

then

extends to the whole space with preservation of norm.

Note that there are virtually no hypotheses beyond linearity and existence of a norm.

In fact for some purposes a weaker version is useful. For

a vector space, we say that

is sublinear if

(

)

(

) +

(

) and

(

) =

(

) for

Generalized Hahn-Banach.

Let

be a vector space,

a sublinear functional,

a subspace of

, and

a linear function satisfying

(

)

(

)

for all

then

can be extended to

so that the same inequality holds for all

Sketch.

It suces to extend

to the space spanned by

and one element

preserving the inequality, since if we can do that we can complete the proof with Zorn's

lemma.

We need to dene

(

) such that

(

)

(

) for all

. The case

= 0 is known and it is easy to use homogeneity to restrict to

1. Thus we need to

nd a value

(

)

such that

(

)

;

(

;

)

(

)

(

)

;

(

) for all

Now it is easy to check that for any

(

)

;

(

;

)

(

)

;

(

and so such an

(

) exists.

Corollary.

is a normed linear space and

, then there exists

of norm

1 such that

(

) =

Corollary.

is a normed linear space,

a closed subspace, and

, then there

exists

of norm 1 such that

(

) =

X=S

Duality.

and

are normed linear spaces and

, then we get a natural

map

(

) =

(

) for all

. In particular, if

(

), then

(

). In fact,

(

)

(

)

. To prove

this, note that

(

)

(

)

. Therefore

, so

is indeed bounded, with

. Also, given any

, we can nd

such that

(

)

= 1. Applying this with

(

arbitrary), gives

(

)

. This shows that

. Note

that if

(

), then (

)

is a Banach space and

a subset, let

(

) = 0

denote the

annihilator

. If

is a subset of

, we similarly set

(

) = 0

Note the distinction between

, which is a subset of

and

, which is a subset of

. All annihilators are closed subspaces.

It is easy to see that

implies that

, and

implies that

. Obviously

(

) if

and

(

)

. The Hahn-Banach

theorem implies that

(

) in case

is a closed subspace of

(but it can happen

that

(

)

for

a closed subspace of

. For

arbitrary,

(

) is the smallest

closed subspace of

containing the subset

, namely the closure of the span of

Now suppose that

is a bounded linear operator between Banach spaces. Let

. Then

(

) = 0

(

)

(

) = 0

(

)

= 0. I.e.,

(

)

(

)

Similarly, for

= 0

(

)

(

) = 0

(

)

(

) = 0

, or

(

) =

(

)

Taking annihilators gives two more results:

(

) =

(

)

(

)

(

)

In particular we see that

is injective i

has dense range" and

is injective if

has

dense range.

Note: we will have further results in this direction once we introduce the weak*-topology

. In particular, (

)

is the weak* closure of a subspace

and

is injective

has weak* dense range.

Dual of a subspace.

An important case is when

is the inclusion map

where

is a closed subspace of

. Then

is just the restriction map:

(

) =

(

). Hahn-Banach tells us that

is surjective. Obviously

(

) =

. Thus we

have a canonical isomorphism

. In fact, the Hahn-Banach theorem shows

that it is an isometry. Via this isometry one often identies

with

Dual of a quotient space.

Next, consider the projection map

X=S

where

a closed subspace. We then have

: (

X=S

)

. Since

is surjective, this map is

injective. It is easy to see that the range is contained in

. In fact we now show that

maps (

X=S

)

onto

, hence provides a canonical isomorphism of

with (

X=S

)

. Indeed,

, then we have a splitting

with

(

X=S

)

(just dene

(

) =

(

)

where

is any element of the coset

). Thus

is indeed in the range of

. This

correspondence is again an isometry.

Dual of a Hilbert space.

The identication of dual spaces can be quite tricky. The case

of Hilbert spaces is easy.

Riesz Representation Theorem.

is a real Hilbert space, dene

(

) =

. This map is a linear isometry of

onto

. For a complex Hilbert space

it is a conjugate linear isometry (it satises

Proof.

It is easy to see that

is an isometry of

into

and the main issue is to show

that any

can be written as

for some

. We may assume that

= 0, so

(

) is

a proper closed subspace of

. Let

(

)]

be of norm 1 and set

= (

)

. For

all

, we clearly have that (

)

;

(

)

(

), so

(

) =

(

)

(

)

(

)

fx:

Via the map

we can dene an inner product on

, so it is again a Hilbert space.

Note that if

is a closed subspace of

, then

(

)

. The Riesz map

is sometimes used to identify

and

. Under this identication there is no distinction

between

and

Dual of

()

Note: there are two quite distinct theorems referred to as the Riesz

Representation Theorem. The proceeding is the easy one. The hard one identies the

dual of

() where is a compact subset of

(this can be generalized considerably). It

states that there is an isometry between

()

and the space of nite signed measures on

. (A nite signed measure is a set function of the form

;

where

is a nite

measure, and we view such as a functional on

(

) by

f d

;

f d

.) This

is the real-valued case" in the complex-valued case the isometry is with complex measures

where

and

are nite signed measures.

Dual of

It is easy to deduce a representation for an arbitrary element of the dual

of, e.g.,

1]). The map

(

) is an isometry of

onto a closed subspace of

C C

. By the Hahn-Banach Theorem, every element of (

)

extends to a functional on

C C

, which is easily seen to be of the form

(

)

f d

g d

where

and and

are signed measures ((

X Y

)

with the obvious identi-

cations). Thus any continuous linear functional on

can be written

f d

In this representation the measures

and

are

not

unique.

Dual of

H#older's inequality states that if 1

(

;

1), then

for all

. This shows that the map

(

) =

maps

linearly into (

)

with

(

)

. The choice

= sign(

)

shows

that there is equality. In fact, if

p <

is a linear isometry of

onto (

)

. For

it is an isometric injection, but not in general surjective. Thus the dual of

for

nite. The dual of

is a very big space, much bigger than

and rarely used.

Dual of

The above considerations apply to the dual of the sequence spaces

. Let

us now show that the dual of

. For any

= (

)

and

= (

)

, we dene

(

) =

. Clearly

(

)

sup

. Taking

sign(

)

n > N

we see that equality holds. Thus

is an isometric injection. We now show that it

is onto. Given

, dene

(

)

) where

(

)

is the usual unit sequence

(

)

Let

= sign(

). Then

(

)

), so

(

)

) =

(

)

Letting

we conclude that

. Now by construction

agrees with

on all

sequences with only nitely many nonzeros. But these are dense in

, so

The bidual.

is any normed linear space, we have a natural map

given

(

) =

(

)

X f

Clearly

and, by the Hahn-Banach theorem, equality holds. Thus

may be

identied as a subspace of the Banach space

. If we dene ~

as the closure of

(

) in

, then

is isometrically embedded as a dense subspace of the Banach space ~

. This

determines ~

up to isometry, and is what we dene as the completion of

. Thus any

normed linear space has a completion.

is onto, i.e., if

is isomorphic with

via this identication, we say that

reexive (which can only happen is

is complete). In particular, one can check that if

is a Hilbert space and

is the Riesz isomorphism, and

the Riesz

isomorphism for

, then

, so

is reexive.

Similarly, the canonical isometries of

onto (

)

and then

onto (

)

compose to

give the natural map of

into its bidual, and we conclude that

(and

) is reexive

for 1

< p <

. None of

, or

(

) are reexive.

is reexive, then

(

) =

for

. In other words, if we identify

and

, the distinction between the two kinds of annihilators disappears. In particular, for

reexive Banach spaces,

(

) =

(

)

and

is injective i

has dense range.

III. Fundamental Theorems

The Open Mapping Theorem and the Uniform Boundedness Principle join the Hahn-

Banach Theorem as the \big three". These two are fairly easy consequence of the Baire

Category Theorem.

Baire Category Theorem.

A complete metric space cannot be written as a countable

union of nowhere dense sets.
Sketch of proof.

If the statement were false, we could write

with

a closed

subset which does not contain any open set. In particular,

is a proper closed set, so

there exists

1) such that

(

)

. Since no ball is contained in

, there exists

(

2) and

2) such that

(

)

. In this

way we get a nested sequence of balls such that the

th ball has radius at most 2

;

and is

disjoint from

. It is then easy to check that their centers form a Cauchy sequence and

its limit, which must exist by completeness, can't belong to any

The Open Mapping Theorem.

The Open Mapping Theorem follows from the Baire

Category Theorem and the following lemma.

Lemma.

Let

be a bounded linear operator between Banach spaces. If

)

(

1))

for some

r >

, then

)

(

2))

Proof.

Let

(

1)). Let

< r

. There exists

with

;

By homogeneity, there exists

such that

;

such that

;

8, etc. Take

such that

, and let

Then

2 and

Remark.

The same proof works to prove the statement with 2 replaced by any number

greater than 1. With a small additional argument, we can even replace it with 1 itself.

However the statement above is sucient for our purposes.

Open Mapping Theorem.

A bounded linear surjection between Banach spaces is open.

Proof.

It is enough to show that the image under

of a ball about 0 contains some ball

about 0. The sets

(

)) cover

, so the closure of one of them must contain an open

ball. By the previous result, we can dispense with the closure. The theorem easily follows

using the linearity of

There are two major corollaries of the Open Mapping Theorem, each of which is equiv-

alent to it.

Inverse Mapping Theorem or Banach's Theorem.

The inverse of an invertible

bounded linear operator between Banach spaces is continuous.
Proof.

The map is open, so its inverse is continuous.

Closed Graph Theorem.

A linear operator between Banach spaces is continuous i its

graph is closed.

A map between topological spaces is called closed if its graph is closed. In a general

Hausdor space, this is a weaker property than continuity, but the theorem asserts that

for linear operators between Banach spaces it is equivalent. The usefulness is that a direct

proof of continuity requires us to show that if

converges to

then

converges

. By using the closed graph theorem, we get to assume as well that

is converging

to some

and we need only show that

Proof.

Let

(

xTx

)

denote the graph. Then the composition

X Y

is a bounded linear operator between Banach spaces given by (

xTx

)

. It is

clearly one-to-one and onto, so the inverse is continuous by Banach's theorem. But the

composition

X Y

is simply the

, so

is continuous.

Banach's theorem leads immediately to this useful characterization of closed imbeddings

of Banach spaces.

Theorem.

Let

be a bounded linear map between Banach spaces. Then

one-to-one and has closed range if and only if there exists a positive number

such that

Proof.

If the inequality holds, then

is clearly one-to-one, and if

is a Cauchy sequence

(

), then

is Cauchy, and hence

converges to some

, so

converges to

Thus the inequality implies that

(

) is closed.

For the other direction, suppose that

is one-to-one with closed range and consider the

map

(

)

. It is the inverse of a bounded isomorphism, so is itself bounded.

The inequality follows immediately (with

the norm of

Another useful corollary is that if a Banach space admits a second weaker or stronger

norm under which it is still Banach, then the two norms are equivalent. This follows

directly from Banach's theorem applied to the identity.

The Uniform Boundedness Principle.

The Uniform Boundedness Principle (or the

Banach-Steinhaus Theorem) also comes from the Baire Category Theorem.

Uniform Boundedness Principle.

Suppose that

and

are Banach spaces and

(

)

. If

sup

(

)

for all

, then

sup

Proof.

One of the closed sets

(

)

must contain

(

) for some

r >

0. Then, if

< r

(

)

(

)

;

(

)

+ sup

(

)

, with

independent of

. This shows that the

are uniformly bounded (by

M=r

In words: a set of linear operators between Banach spaces which is bounded pointwise

is norm bounded.

The uniform boundedness theorem is often a way to generate counterexamples. A

typical example comes from the theory of Fourier series. For

continuous and

1-periodic the

th partial sum of the Fourier series for

(

) =

(

)

ikt

dte

iks

(

)

(

;

)

where

(

) =

ikx

Writing

, we have

(

) =

;

1 =

;

= sin(2

+ 1)

sin

x :

This is the Dirichlet kernel, a

periodic function. In particular, the value of the

partial sum of the Fourier series of

at 0 is

(0) =

(

)

(

)

dt:

We think of

as a linear functional on the Banach space of 1-periodic continuous function

endowed with the sup norm. Clearly

(

)

dt:

In fact this is an equality. If

(

) = sign

(

), then sup

= 1 and

. Actually,

is not continuous, so to make this argument correct, we approximate

by a continuous

functions, and thereby prove the norm equality. Now one can calculate that

. By the uniform boundedness theorem we may conclude that there exists a

continuous periodic function for whose Fourier series diverges at

= 0.

The Closed Range Theorem.

We now apply the Open Mapping Theorem to better

understand the relationship between

and

. The property of having a closed range

is signicant to the structure of an operator between Banach spaces. If

has

a closed range

(which is then itself a Banach space), then

factors as the projection

(

), the isomorphism

(

)

, and the inclusion

. The Closed

Range Theorem says that

has a closed range if and only if

does.

Theorem.

Let

be a bounded linear operator between Banach spaces. Then

is invertible i

is.

Proof.

exists, then

and

, so

and

, which shows that

is invertible.

Conversely, if

is invertible, then it is open, so there is a number

c >

0 such that

1) contains

). Thus, for

= sup

(

)

= sup

(

)

sup

)

(

)

The existence of

c >

0 such that

is equivalent to the statement that

is injective with closed range. But since

is injective,

has dense range.

Lemma.

Let

be a linear map between Banach spaces such that

is an

injection with closed range. Then

is a surjection.

Proof.

Let

be the closed unit ball of

and

. It suces to show that

contains

a ball around the origin, since then, by the lemma used to prove the Open Mapping

Theorem,

is onto.

There exists

c >

0 such that

for all

. We shall show that

contains the ball of radius

around the origin in

. Otherwise there exists

y =

. Since

is a closed convex set we can nd a functional

such that

(

)

for all

and

(

)

. Thus

> =c

, but

= sup

(

)

= sup

(

)

This is a contradiction.

Closed Range Theorem.

Let

be a bounded linear operator between Banach

spaces. Then T has closed range if and only if

does.

Proof.

(

) closed =

)

(

) closed.

Let

(

). Then

(

)

is an isomorphism (Inverse Mapping Theorem).

The diagram

;

(

)

;

commutes. Taking adjoints,

;

(

)

;

This shows that

(

) =

(

)

(

) closed =

)

(

) closed.

Let

(

) (so

(

)) and let

be the range restriction of

The adjoint is

, the lifting of

. Now

(

) =

(

), is

closed, and

is an injection. We wish to show that

is onto

. Thus the theorem follows

from the preceding lemma.

IV. Weak Topologies

The weak topology.

Let

be a Banach space. For each

the map

(

)

a seminorm on

, and the set of all such seminorms, as

varies over

, is sucient by

the Hahn-Banach Theorem. Therefore we can endow

with a new TVS structure from

this family of seminorms. This is called the weak topology on

. In particular,

weakly (written

n w

;

) i

(

)

(

) for all

. Thus the weak topology is

weaker than the norm topology, but all the elements of

remain continuous when

endowed with the weak topology (it is by denition the weakest topology for which all the

elements of

are continuous).

Note that the open sets of the weak topology are rather big. If

is an weak neighbor-

hood of 0 in an innite dimensional Banach space then, by denition, there exists

and nitely many functionals

such that

(

)

is contained in

Thus

contains the innite dimensional closed subspace

(

)

:::

(

n w

;

weakly, then, viewing the

as linear functionals on

(via the canonical

embedding of

into

), we see that the sequence of real numbers obtained by applying

the

to any

is convergent and hence bounded uniformly in

. By the Uniform

Boundedness Principle, it follows that the

are bounded.

Theorem.

If a sequence of elements of a Banach space converges weakly, then the sequence

is norm bounded.

On the other hand, if the

are small in norm, then their weak limit is too.

Theorem.

n w

;

in some Banach space, then

liminf

Proof.

Take

of norm 1 such that

(

) =

. Then

(

)

, and taking the

liminf gives the result.

For convex sets (in particular, for subspaces) weak closure coincides with norm closure:

Theorem.

1) The weak closure of a convex set is equal to its norm closure.

2) A convex set is weak closed i it is normed closed.

3) A convex set is weak dense i it is norm dense.

Proof.

The second and third statement obviously follow from the rst, and the weak closure

obviously contains the norm closure. So it remains to show that if

does not belong to

the norm closure of a convex set

, then there is a weak neighborhood of

which doesn't

intersect

. This follows immediately from the following convex separation theorem.

Theorem.

Let

be a nonempty closed convex subset of a Banach space

and

a point

in the complement of

. Then there exists

such that

(

)

inf

(

)

In fact we shall prove a stronger result:

Theorem.

Let

and

be disjoint, nonempty, convex subsets of a Banach space

with

open. Then there exists

such that

(

)

inf

(

)

for all

(The previous result follows by taking

to be any ball about

disjoint from

Proof.

This is a consequence of the generalized Hahn-Banach Theorem. Pick

and

and set

;

and

;

. Then

is a convex open set containing

0 but not containing

. (The convexity of

follows directly from that of

and

" the

fact that

is open follows from the representation of

;

as a union of

open sets" obviously 0 =

;

, and

since

and

and disjoint.)

Since

is open and convex and contains 0, for each

t >

is a

nonempty open semi-innite interval. Dene

(

)

) to be the left endpoint of this

interval. By denition

is positively homogeneous. Since

is convex,

and

imply that

(

)

(

) =

whence

is subadditive. Thus

is a sublinear functional. Moreover,

(

)

Dene a linear functional

(

) = 1. Then

(

) =

(

) =

(

) for

0 and

(

)

(

) for

t <

0. Thus

is a linear functional on

satisfying

(

)

(

) there. By Hahn-Banach we can extend

to a linear functional on

satisfying the same inequality. This implies that

is bounded (by 1) on the open set

, so

belongs to

, then

;

, so

(

)

;

(

) + 1 =

(

;

)

(

)

< f

(

). Therefore sup

(

)

inf

(

). Since

(

) is an open interval,

)

sup

(

) for all ~

, and so we have the theorem.

The weak* topology.

On the dual space

we have two new topologies. We may endow

it with the weak topology, the weakest one such that all functionals in

are continuous,

or we may endow it with the topology generated by all the seminorms

(

(This is obviously a sucient family of functionals.) The last is called the weak* topology

and is a weaker topology than the weak topology. If

is reexive, the weak and weak*

topologies coincide.

Examples of weak and weak* convergence: 1) Consider weak convergence in

()

where is a bounded subset of

. From the characterization of the dual of

we see

that

n w

;

)

n w

;

weak in

)

n w

;

weak in

whenever 1

p <

. In particular we claim that the complex exponentials

inx w

;

0 in

1]) as

. This is simply the statement that

lim

(

)

inx

= 0

for all

1]), i.e., that the Fourier coecients of an

tend to 0, which is known as

the Riemann{Lebesgue Lemma. (Proof: certainly true if

is a trigonometric polynomial.

The trig polynomials are dense in

1]) by the Weierstrass Approximation Theorem,

and

1]) is dense in

1]).) This is one common example of weak convergence

which is not norm convergence, namely weak vanishing by oscillation.

2) Another common situation is weak vanishing to innity. As a very simple example,

it is easy to see that the unit vectors in

converge weakly to zero for 1

< p <

(and

weak* in

, but not weakly in

). As a more interesting example, let

(

) be

a sequence of function which are uniformly bounded in

, and for which

;

]

Then we claim that

0 weakly in

if 1

< p <

. Thus we have to show that

lim

g dx

= 0

for all

. Let

. Then lim

= 0 (by the dominated

convergence theorem). But

g dx

(

)

(

)

The same proof shows that if the

are uniformly bounded they tend to 0 in

weak*.

Note that the characteristic functions

+1]

not

tend to zero weakly in

however.

3) Consider the measure

= 2

]

. Formally

tends to the delta function

. Using the weak* topology on

(

;

1]) this convergence becomes precise:

n w

;

Theorem (Alaoglu).

The unit ball in

is weak* compact.

Proof.

For

, let

, and set = $

. Recall that this

Cartesian product is nothing but the set of all functions

with

(

)

for all

. This set is endowed with the Cartesian product topology, namely the weakest topology

such that for all

, the functions

(

) (from to

) are continuous. Tychono's

Theorem states that is compact with this topology.

Now let

be the unit ball in

. Then

and the topology thereby induced on

is precisely the weak* topology. Now for each pair

and each

, dene

(

) =

(

) +

(

)

;

(

)

;

(

). These are continuous functions

on and

(0)

Thus

is a closed subset of a compact set, and therefore compact itself.

Corollary.

n w

;

, then

liminf

Proof.

Let

= liminf

and let

0 be arbitrary. Then there exists a subsequence

(also denoted

) with

. The ball of radius

being weak* compact, and

so weak* closed,

< C

. Since

was arbitrary, this gives the result.

the weak* topology is that induced by the functionals in

Theorem.

The unit ball of

is weak* dense in the unit ball of

Proof.

Let

belong to the unit ball of

. We need to show that for any

::: f

of norm 1, and any

0, the set

(

;

)(

)

< i

= 1

::: n

contains a point of the unit ball of

. (Since any neighborhood of

contains a set of this

form.)

It is enough to show that there exists

with

1 +

such that (

;

)(

) = 0

for each

. Because then

(1 +

) belongs to the closed unit ball of

, and

((1 +

)

;

)(

)

((1 +

)

;

)(

)

((1 +

)

;

1 +

< :

Let

be the span of the

. Since

is nite dimensional the canonical map

is surjective. (This is equivalent to saying that if the null space of a linear

functional

contains the intersection of the null spaces of a nite set of linear functionals

, then

is a linear combination of the

, which is a simple, purely algebraic result.

Proof: The nullspace of the map (

::: g

) :

is contained in the nullspace of

(

::: g

) for some linear

.]) Consequently

is isometrically

isomorphic to

In particular

concides with

for some

. Since

1, and we can

choose the coset representative

with

1 +

as claimed.

Corollary.

The closed unit ball of a Banach space

is weakly compact if and only if

is reexive.
Proof.

If the closed unit ball of

is weakly compact, then it is weak* compact when

viewed as a subset of

. Thus the ball is weak* closed, and so, by the previous theorem,

the embedding of the the unit ball of

contains the ball of

. It follows that the

embedding of

is all of

The reverse direction is immediate from the Alaoglu theorem.

V. Compact Operators and their Spectra

Hilbert{Schmidt operators.
Lemma.

Suppose that

and

are two orthonormal bases for a separable Hilbert

space

, and

(

)

. Then

Proof.

For all

, so

But

Denition.

(

) dene

where

is any orthonormal basis for

is called a Hilbert{Schmidt operator if

, and

is called the Hilbert{Schmidt norm of

We have just seen that if

is Hilbert{Schmidt, then so is

and their Hilbert{Schmidt

norms coincide.

Proposition.

Proof.

Let

be an arbitrary element of

. Then

By Cauchy{Schwarz

Summing on

gives the result.

Proposition.

Let

be an open subset of

and

( )

. Dene

(

) =

(

)

(

)

for all

Then

denes a Hilbert{Schmidt operator on

()

and

Proof.

For

, set

(

) =

(

). By Fubini's theorem,

() for almost all

, and

dx:

Now,

(

) =

, so, if

is an orthonormal basis, then

(

)(

)

Compact operators.
Denition.

A bounded linear operator between Banach spaces is called compact if it

maps the unit ball (and therefore every bounded set) to a precompact set.

For example, if

has nite rank (dim

(

)

), then

is compact.

Recall the following characterization of precompact sets in a metric space, which is often

useful.

Proposition.

Let

be a metric space. Then the following are equivalent:

(1)

is precompact.

(2)

For all

there exist nitely many sets of diameter at most

which cover

(3)

Every sequence contains a Cauchy subsequence.

Sketch of proof.

(1) =

)

(2) and (3) =

)

(1) are easy. For (2) =

)

(3) use a Cantor

diagonalization argument to extract a Cauchy subsequence.

Theorem.

Let

and

be Banach spaces and

(

)

the space of compact linear

operators from

. Then

(

)

is a closed subspace of

(

)

Proof.

Suppose

(

;

0. We must show that

compact. Thus we must show that

(

) is precompact in

, where

is the unit ball in

. For this, it is enough to show that for any

0 there are nitely many balls

radius

such that

(

)

Choose

large enough that

;

2, and let

::: V

be nitely many balls

of radius

2 which cover

. For each

let

be the ball of radius

with the same

center as

It follows that closure of the nite rank operators in

(

) is contained in

(

In general, this may be a strict inclusion, but if

is a Hilbert space, it is equality. To

prove this, choose an orthonormal basis for

, and consider the nite rank operators of the

form

where

is the orthogonal projection of

onto the span of nitely many basis

elements. Using the fact that

is compact (

the unit ball of

) and that

= 1, we

can nd for any

0, an operator

of this form with sup

(

;

)

The next result is obvious but useful.

Theorem.

Let

and

be Banach spaces and

(

)

. If

is another Banach

space and

(

)

then

is compact. If

(

)

, then

is compact. If

, then

(

) :=

(

)

is a two-sided ideal in

(

)

Theorem.

Let

and

be Banach spaces and

(

)

. Then

is compact if and

only if

is compact.

Proof.

Let

be the unit ball in

and

the unit ball in

. Suppose that

is compact.

Given

0 we must exhibit nitely many sets of diameter at most

which cover

First choose

sets of diameter at most

3 which cover

, and let

belong to the

set. Also, let

::: I

intervals of length

3 which cover the interval

For any

-tuple (

::: j

) of integers with 1

we dene the set

(

)

= 1

::: m

These sets clearly cover

, so there images under

cover

, so it suces to show that

the images have diameter at most

. Indeed, if

and

belong to the set above, and

is any

element of

, pick

such that

;

3. We know that

(

)

;

(

)

Thus

(

;

)(

)

(

;

)(

)

(

)

;

(

)

(

)

;

(

)

(

;

)(

)

This shows that

compact =

)

compact. Conversely, suppose that

is compact. Then

maps the unit ball of

into a precompact subset of

. But the

unit ball of

may be viewed as a subset of the unit ball of its bidual, and the restriction

to the unit ball of

coincides with

there. Thus

maps the unit ball of

to a

precompact set.

Theorem.

is a compact operator from a Banach space to itself, then

(

;

)

nite dimensional and

(

;

)

is closed.

Proof.

is a compact operator that restricts to the identity on

(

;

). Hence the

closed unit ball in

(

;

) is compact, whence the dimension of

(

;

) is nite.

Now any nite dimensional subspace is complemented (see below), so there exists a

closed subspace

such that

(

;

) +

and

(

;

)

= 0. Let

= (

;

)

, so

is injective and

(

) =

(

;

). We will show that for some

c >

for all

, which will imply that

(

) is closed. If the desired

inequality doesn't hold for any

c >

0, we can choose

of norm 1 with

After passing to a subsequence we may arrange that also

converges to some

It follows that

, so

and

= 0. Therefore

= 0 which is impossible

(since

= 1).

In the proof we used the rst part of the following lemma. We say that a closed

subspace

is complemented in a Banach space

if there is another closed subspace such

that

Lemma.

A nite dimensional or nite codimensional closed subspace of a Banach space

is complemented.

Proof.

is a nite dimensional subspace, choose a basis

::: x

and dene a linear

functionals

(

) =

. Extend the

to be bounded linear functionals

. Then we can take

(

)

:::

(

is nite codimensional, we can take

to be the span of a set of nonzero coset

representatives.

A simple generalization of the theorem will be useful when we study the spectrum of

compact operators.

Theorem.

is a compact operator from a Banach space to itself,

a non-zero complex

number, and

a positive integer, then

(

;

)

]

is nite dimensional and

(

;

)

]

is closed.

Proof.

Expanding we see that (

;

)

(

;

) for some compact operator

, so the

result reduces to the previous one.

We close the section with a good source of examples of compact operators, which in-

cludes, for example, any matrix operator on

for which the matrix entries are square-

summable.

Theorem.

A Hilbert{Schmidt operator on a separable Hilbert space is compact.

Proof.

Let

be an orthonormal basis. Let

be a given Hilbert{Schmidt operator

(so

). Dene

= 0 otherwise. Then

;

Spectral Theorem for compact self-adjoint operators.

In this section we assume

that

is a

complex

Hilbert space. If

is a bounded linear operator, we view

as a map from

via the Riesz isometry between

and

. That is,

is dened

xTy

In the case of a nite dimensional complex Hilbert space,

can be represented by a

complex square matrix, and

is represented by its Hermitian transpose.

Recall that a Hermitian symmetric matrix has real eigenvalues and an orthonormal

basis of eigenvectors. For a self-adjoint operator on a Hilbert space, it is easy to see that

any eigenvalues are real, and that eigenvectors corresponding to distinct eigenvalues are

orthogonal. However there may not exist an orthonormal basis of eigenvectors, or even any

nonzero eigenvectors at all. For example, let

1]), and dene

(

) =

(

) for

. Then

is clearly bounded and self-adjoint. But it is easy to see that

does not

have any eigenvalues.

Spectral Theorem for Compact Self-Adjoint Operators in Hilbert Space.

Let

be a compact self-adjoint operator in a Hilbert space

. Then there is an orthonormal

basis consisting of eigenvectors of

Before proceeding to the proof we prove one lemma.

Lemma.

is a self-adjoint operator on a Hilbert space, then

= sup

Txx

Proof.

Let

= sup

Txx

. It is enough to prove that

Txy

for all

and

. We can obviously assume that

and

are nonzero. Moreover, we may

multiply

by a complex number of modulus one, so we can assume that

Txy

0. Then

(

)

;

(

;

)

;

= 4Re

Txy

= 4

Txy

;

) =

)

Now apply this result with

replaced by

and

replaced by

Proof of spectral theorem for compact self-adjoint operators.

We rst show that

has a

nonzero eigenvector. If

= 0, this is obvious, so we assume that

= 0. Choose a sequence

with

= 1 so that

. Since

is self-adjoint,

so we may pass to a subsequence (still denoted

), for which

Since

is compact we may pass to a further subsequence and assume that

Note that

Using the fact that

is self-adjoint and

is real, we get

;

= 0

Since

we infer that

as well, or

= 0. Applying

we have

Ty=

, so

is indeed a nonzero eigenvalue.

To complete the proof, consider the set of all orthonormal subsets of

consisting of

eigenvectors of

. By Zorn's lemma, it has a maximal element

. Let

be the closure

of the span of

. Clearly

, and it follows directly (since

is self-adjoint), that

. Therefore

restricts to a self-adjoint operator on

and thus, unless

= 0,

has an eigenvector in

. But this clearly contradicts the maximality of

(since we can adjoin this element to

to get a larger orthonormal set of eigenvectors).

Thus

= 0, and

is an orthonormal basis.

The following structure result on the set of eigenvalues is generally considered part of

the spectral theorem as well.

Theorem.

is a compact self-adjoint operator on a Hilbert space, then the set of

nonzero eigenvalues of

is either a nite set or a sequence approaching

and the corre-

sponding eigenspaces are all nite dimensional.
Remark.

0 may or may not be an eigenvalue, and its eigenspace may or may not be nite.

Proof.

Let

be an orthonormal basis of eigenvectors, with

. Here

ranges over

some index set

. It suces to show that

is nite for all

0. Then

;

so if

, then

;

. If

were innite, we could then choose a sequence

of unit elements in

whose image under

has no convergent subsequence, which violates

the compactness of

Suppose, for concreteness, that

is an innite dimensional separable Hilbert space and

that

is an orthonormal basis adapted to a compact self-adjoint operator

Then the map

given by

(

) = (

:::

)

is an isometric isomorphism. Moreover, when we use this map to transfer the action of

, i.e., when we consider the operator

UTU

, we see that this operator is simply

multiplication by the bounded sequence (

:::

)

. Thus the spectral theorem says

that every compact self-adjoint

is unitarily equivalent to a multiplication operator on

(An isometric isomorphism of Hilbert spaces is also called a unitary operator. Note that

it is characterized by the property

A useful extension is the spectral theorem for commuting self-adjoint compact operators.

Theorem.

and

are self-adjoint compact operators in a Hilbert space

and

, then there is an orthonormal basis of

whose elements are eigenvectors for both

and

Proof.

For an eigenvalue

, let

denote the corresponding eigenspace of

. If

, then

TSx

STx

, so

. Thus

restricts to a self-adjoint

operator on

, and so there is an orthonormal basis of

{eigenvectors for

. These are

{eigenvectors as well. Taking the union over all the eigenvalues

completes the

construction.

Let

and

be any two self-adjoint operators and set

. Then

(

)

2 and

= (

;

)

). Conversely, if

is any element of

(

), then we

can dene two self-adjoint operators from these formulas and have

. Now

suppose that

is compact and also

normal

, i.e., that

and

commute. Then

and

are compact and commute, and hence we have an orthonormal basis whose elements

are eigenvectors for both

and

, and hence for

. Since the real and imaginary parts

of the eigenvalues are the eigenvalues of

and

, we again see that the eigenvalues form

a sequence tending to zero and all have nite dimensional eigenspaces.

We have thus shown that a compact normal operator admits an orthonormal basis

of eigenvectors. Conversely, if

is an orthonormal basis of eigenvectors of

, then

= 0 if

, which implies that each

is also an eigenvector for

. Thus

for all

, and it follows easily that

is normal. We have thus shown:

Spectral Theorem for compact normal operators.

Let

be a compact operator on a

Hilbert space

. Then there exists an orthonormal basis for

consisting of eigenvectors of

if and only if

is normal. In this case, the set of nonzero eigenvalues form a nite set or

a sequence tending to zero and the eigenspaces corresponding to the nonzero eigenvalues are

nite dimensional. The eigenvalues are all real if and only if the operator is self-adjoint.

The spectrum of a general compact operator.

In this section we derive the structure

of the spectrum of a compact operator (not necessarily self-adjoint or normal) on a complex

Banach space

For any operator

on a complex Banach space, the

resolvent set

(

) consists

of those

such that

;

is invertible, and the spectrum

(

) is the complement.

(

), then

;

may fail to be invertible in several ways. (1) It may be that

(

;

)

= 0, i.e., that

is an eigenvalue of

. In this case we say that

belongs to the

point spectrum

, denoted

(

). (2) If

;

is injective, it may be that its range is

dense but not closed in

. In this case we say that

belongs to the

continuous spectrum

(

). Or (3) it may be that

;

is injective but that its range is not even dense

. This is the

residual spectrum

(

). Clearly we have a decomposition of

into the

disjoint sets

(

), and

(

). As an example of the continuous spectrum,

consider the operator

where the

form an orthonormal basis of a Hilbert

space and the

form a positive sequence tending to 0. Then 0

(

). If

(

Now if

is compact and

is innite dimensional, then 0

(

) (since if

were

invertible, the image of the unit ball would contain an open set, and so couldn't be pre-

compact). From the examples just given, we see that 0 may belong to the point spectrum,

the continuous spectrum, or the residual spectrum. However, we shall show that all other

elements of the spectrum are eigenvalues, i.e., that

(

) =

(

)

, and that, as in the

normal case, the point spectrum consists of a nite set or a sequence approaching zero.

The structure of the spectrum of a compact operator will be deduced from two lemmas.

The rst is purely algebraic. To state it we need some terminology: consider a linear

operator

from a vector space

to itself, and consider the chains of subspaces

0 =

(

)

(

)

(

)

(

)

Either this chain is strictly increasing forever, or there is a least

0 such that

(

) =

(

), in which case only the rst

spaces are distinct and all the others equal the

th one. In the latter case we say that the kernel chain for

stabilizes

. In particular,

the kernel chain stabilizes at 0 i

is injective. Similarly we may consider the chain

(

)

(

)

(

)

(

)

and dene what it means for the range chain to stabilize at

n >

0. (So the range stabilizes

at 0 i

is surjective.) It could happen that neither or only one of these chains stabilizes.

However:

Lemma.

Let

be a linear operator from a vector space

to itself. If the kernel chain

stabilizes at

and the range chain stabilizes at

, then

and

decomposes as the

direct sum of

(

)

and

(

)

Proof.

Suppose

were less than

. Since the range chain stabilizes at

, there exists

with

x =

(

), and then there exists

such that

. Thus

;

(

and, since kernel chain stabilizes at

m < n

(

) =

(

). Thus

, a

contradiction. Thus

. A similar argument, left to the reader, establishes the reverse

inequality.

Now if

(

), then

= 0, whence

= 0. Thus

(

)

(

) = 0.

Given

, let

, so

decomposes as

(

) and

;

(

The second lemma brings in the topology of compact operators.

Lemma.

Let

be a compact operator on a Banach space and

:::

sequence of complex numbers with

inf

. Then the following is impossible: There

exists a strictly increasing chain of closed subspaces

with

(

;

)

for all

Proof.

Suppose such a chain exists. Note that each

for each

. Since

contains an element of norm 1, we may choose

with

2, dist(

) = 1.

m < n

, then

;

(

;

)

and

;

This implies that the sequence (

) has no Cauchy subsequence, which contradicts the

compactness of

We are now ready to prove the result quoted at the beginning of the subsection.

Theorem.

Let

be a compact operator on a Banach space

. Then any nonzero ele-

ment of the spectrum of

is an eigenvalue. Moreover

(

)

is either nite or a sequence

approaching zero.
Proof.

Consider the subspace chains

(

;

)

] and

(

;

)

] (these are closed

subspaces by a previous result). Clearly

;

maps

(

;

)

] into

(

;

)

so the previous lemma implies that the kernel chain stabilizes, say at

. Now

(

;

)

] =

(

;

)

] (since the range is closed), and since these last stabilize, the range chain

stabilizes as well.

Thus we have

(

;

)

]

(

;

)

]. Thus

(

;

)

(

;

)

(

;

)

= 0 =

)

(

;

)

= 0

In other words

(

) =

)

(

Finally we prove the last statement. If it were false we could nd a sequence of

eigenvalues

with inf

0. Let

:::

be corresponding nonzero eigenvectors

and set

= span

::: x

]. These form a strictly increasing chain of subspaces (re-

call that eigenvectors corresponding to distinct eigenvalues are linearly independent) and

(

;

)

, which contradicts the lemma.

The above reasoning also gives us the Fredholm alternative:

Theorem.

Let

be a compact operator on a Banach space

and

a nonzero complex

number. Then either (1)

;

is an isomorphism, or (2) it is neither injective nor

surjective.
Proof.

Since the kernel chain and range chain for

;

stabilize, either they both

stabilize at 0, in which case

is injective and surjective, or neither does, in which case it

is neither.

We close this section with a result which is fundamental to the study of Fredholm

operators.

Theorem.

Let

be a compact operator on a Banach space

and

a nonzero complex

number. Then

dim

(

;

) = dim

(

;

) = codim

(

;

) = codim

(

;

)

Proof.

Let

;

. Since

(

) is closed

(

)]

(

)

(

)

Thus

(

)]

is nite dimensional, so

(

) is nite dimensional, and these two

spaces are of the same dimension. Thus codim

(

) = dim

(

For a general operator

we only have

(

)

(

)

, but, as we now show, when

(

) is closed,

(

) =

(

)

. Indeed,

induces an isomorphism of

(

) onto

(

), and for any

(

)

induces a map

(

) to

. It follows that

for some bounded linear operator

(

), which can be extended to an element of

by Hahn-Banach. But

simply means that

, showing that

(

)

(

)

(and so equality holds) as claimed.

Thus

(

)

(

)

(

)

so codim

(

) = dim

(

)

= dim

(

We complete the theorem by showing that dim

(

)

codim

(

) and dim

(

)

codim

(

). Indeed, since

(

) is closed with nite codimension, it is complemented by a

nite dimensional space

(with dim

= codim

(

). Since

(

) is nite dimensional,

it is complemented by a space

. Let

denote the projection of

onto

(

) which is

a bounded map which to the identity on

(

) and to zero on

. Now if codim

(

)

dim

(

), then there is a linear map of

(

) onto

which is not injective. But then

;

is a compact operator and

;

is easily seen to be surjective. By the Fredholm

alternative, it is injective as well. This implies that

is injective, a contradiction. We

have thus shown that dim

(

)

codim

(

). Since

is compact, the same argument

shows that dim

(

)

codim

(

). This completes the proof.

VI. Introduction to General Spectral Theory

In this section we skim the surface of the spectral theory for a general (not necessarily

compact) operator on a Banach space, before encountering a version of the Spectral The-

orem for a bounded self-adjoint operator in Hilbert space. Our rst results don't require

the full structure of in the algebra of operators on a Banach space, but just an arbitrary

Banach algebra structure, and so we start there.

The spectrum and resolvent in a Banach algebra.

Let

be a Banach algebra with

an identity element denoted

. We assume that the norm in

has been normalized so that

= 1. The two main examples to bear in mind are (1)

(

), where

is some Banach

space" and (2)

(

) endowed with the sup norm, where

is some compact topological

space, the multiplication is just pointwise multiplication of functions, and

is the constant

function 1.

In this set up the resolvent set and spectrum may be dened as before:

(

) =

;

is invertible

(

) =

(

). The

spectral radius

is dened to be

(

) =

sup

(

)

. For

(

), the resolvent is dened as

(

) = (

;

)

Lemma.

with

invertible and

, then

;

is invertible,

(

;

)

(

)

and

(

;

)

;

)

Proof.

(

)

;

)

so the sum converges absolutely and the norm bound holds. Also

(

)

(

;

) =

(

)

;

(

)

and similarly for the product in the reverse order.

As a corollary, we see that if

, then

;

is invertible, i.e.,

(

). In other

words:

Proposition.

(

)

We also see from the lemma that lim

(

)

= 0. Another corollary is that if

(

) and

(

)

, then

;

(

) and

(

;

) =

(

)

Theorem.

The resolvent

(

)

is always open and contains a neighborhood of

and

the spectrum is always non-empty and compact.
Proof.

The above considerations show that the resolvent is open, and so the spectrum is

closed. It is also bounded, so it is compact.

To see that the spectrum is non-empty, let

be arbitrary and dene

(

) =

(

)]. Then

maps

(

) into

, and it is easy to see that it is holomorphic (since we

have the power series expansion

(

;

) =

(

)

]

is suciently small). If

(

) =

, then

is entire. It is also bounded (since it tends to 0

at innity), so Liouville's theorem implies that it is identically zero. Thus for any

(

;

)

] = 0 . This implies that (

;

)

= 0, which is clearly impossible.

Corollary (Gelfand{Mazur).

is a complex Banach division algebra, then

isometrically isomorphic to

Proof.

For each 0

, let

(

). Then

;

is not invertible, and since

is a

division algebra, this means that

. Thus

Now we turn to a bit of \functional calculus." Let

and let

be a complex

function of a complex variable which is holomorphic on the closed disk of radius

about

the origin. Then we make two claims: (1) plugging

into the power series expansion of

denes an element

(

)

" and (2) the complex function

maps the spectrum of

into

the spectrum of

(

). (In fact onto, as we shall show later in the case

is polynomial.)

To prove these claims, note that, by assumption, the radius of convergence of the power

series for

about the origin exceeds

, so we can expand

(

) =

where

. Thus the series

is absolutely convergent in the Banach space

" we call its limit

(

). (This is the

denition

(

). It is a suggestive abuse of

notation to use

to denote the this function, which maps a subset of

into

, as well as

the original complex-valued function of a complex variable.) Now suppose that

(

Then

(

)

;

(

) =

(

;

) = (

;

)

(

;

)

where

;

Note that

, so

converges to some

. Thus

(

)

;

(

) = (

;

)

(

;

)

Now

(

)

;

(

) can't be invertible, because these formulas would then imply that

;

would be invertible as well, but

(

). Thus we have veried that

(

)

;

(

)

for

all

(

Theorem (Spectral Radius Formula).

(

) = lim

= inf

Proof.

(

), then

(

) (which is also evident algebraically), so

This shows that

(

)

inf

Now take

, and consider

(

) =

(

;

)

] =

;

(

)

Then

is clearly holomorphic for

, but we know it extends holomorphically to

> r

(

) and tends to 0 as

tends to innity. Let

(

) =

). Then

extends

analytically to zero with value zero and denes an analytic function on the open ball of

radius 1

(

) about zero, as does, therefore,

(

)

(

)

This shows that for each

(

) and each

(

) is bounded. By the

uniform boundedness principle, the set of elements

are bounded in

, say by

Thus

. This is true for all

(

), so limsup

(

Corollary.

is a Hilbert space and

(

)

a normal operator, then

(

) =

Proof.

= sup

TxTx

= sup

Txx

since

is self-adjoint. Using the normality of

we also get

= sup

TxT

= sup

TxTx

= sup

xTx

= sup

Thus

. Replacing

with

gives,

, and similarly for all powers

of 2. The result thus follows from the spectral radius formula.

As mentioned, we can now show that

maps

(

)

onto

;

(

)

is a polynomial.

Spectral Mapping Theorem.

Let

be a complex Banach algebra with identity,

and let

a polynomial in one variable with complex coe cients. Then

;

(

)

;

(

)

Proof.

We have already shown that

;

(

)

;

(

)

. Now suppose that

;

(

)

By the Fundamental Theorem of Algebra we can factor

;

, so

(

)

;

(

;

)

for some nonzero

and some roots

. Since

(

)

;

is not invertible, it

follows that

;

is not invertible for at least one

. In orther words,

(

), so

(

)

;

(

)

Spectral Theorem for bounded self-adjoint operators in Hilbert space.

We now

restrict to self-adjoint operators on Hilbert space and close with a version of the Spectral

Theorem for this class of operator. We follow Halmos's article \What does the Spec-

tral Theorem Say?" (American Mathematical Monthly 70, 1963) both in the relatively

elementary statement of the theorem and the outline of the proof.

First we note that self-adjoint operators have real spectra (not just real eigenvalues).

Proposition.

is a Hilbert space and

(

)

is self-adjoint, then

(

)

Proof.

(

;

)

(

;

)

so if Im

= 0,

;

is injective with closed range. The same reasoning shows that

(

;

)

;

is injective, so

(

;

) is dense. Thus

(

Spectral Theorem for self-adjoint operators in Hilbert space.

is a complex

Hilbert space and

(

)

is self-adjoint, then there exists a measure space

with

measure

, a bounded measurable function

, and an isometric isomorphism

such that

where

is the operation of multiplication by

. (Here

means

(

)

the space of complex-valued functions on

which are square integrable with respect to the

measure

Sketch of proof.

Let

be a nonzero element of

, and consider the smallest closed sub-

space

containing

for

= 0

:::

, i.e.,

(

)

. Here

is the

space of polynomials in one variable with complex coecients. Both

and its orthogonal

complement are invariant under

(this uses the self-adjointness of

). By a straightfor-

ward application of Zorn's lemma we see that

can be written as a Hilbert space direct

sum of

invariant spaces of the form of

. If we can prove the theorem for each of these

subspaces, we can take direct products to get the result for all of

. Therefore we may

assume from the start that

(

)

for some

. (In other terminology, that

has a

cyclic vector

Now set =

(

), which is a compact subset of the real line, and consider the space

(

), the space of all continuous real-valued functions on . The subspace of

real-valued polynomial functions is dense in

(since any continuous function on can

be extended to the interval

;

(

)

(

)] thanks to Tietze's extension theorem and then

approximated arbitrarily closely by a polynomial thanks to the Weierstrass approximation

theorem). For such a polynomial function,

, dene

(

)

. Clearly

linear and

(

)

;

(

)

by the special form of the spectral radius formula for self-adjoint operators in Hilbert space.

Since

;

(

)

;

(

)

, we have

;

(

)

()

, and thus,

This shows that

is a bounded linear functional on a dense subspace of

and so extends

uniquely to dene a bounded linear functional on

Next we show that

is positive in the sense that

0 for all non-negative functions

. Indeed, if

for some polynomial, then

(

)

(

)

(

)

For an arbitrary non-negative

, we can approximate

uniformly by polynomials

, so

= lim

and

= lim

We now apply the Riesz Representation Theorem for the representation of the linear

functional

. It state that there exists a nite measure on such that

f d

for

(it is a positive measure since

is positive). In particular,

(

)

for all

We now turn to the space

of complex-valued functions on which are square inte-

grable with respect to the measure

. The subspace of complex-valued polynomial func-

tions is dense in

(since the measure is nite, the

norm is dominated by the supremum

norm). For such a polynomial function,

, dene

(

)

. Then

(

)

(

)

(

)

(

)

(

)

Thus

is an isometry of a dense subspace of

into

and so extends to an isometry of

onto a closed subspace of

. In fact,

is onto

itself, since, by the assumption that

is a cyclic vector for

, the range of

is dense.

Finally, dene

(

) =

. If

is a complex polynomial, then (

)(

) =

(

), which is also a polynomial. Thus

TUq

(

)

;

(

)(

)

Thus the bounded operators

and

coincide on a dense subset of

, and hence

they are equal.

For a more precise description of the measure space and the extension to normal oper-

ators, see Zimmer.

MATH 502

Second exam

May 9, 1997

Professor Arnold

Do 3 problems, omit 1. State which problem you omit. Partial credit will be awarded, but

errors and unacknowledged omissions count negatively.

1. Suppose that

are elements in a Hilbert space such that

converges to

weakly and

converges to

kxk

. Prove that

converges to

in norm. Give a

counterexample in Banach space.

2. Let

and

be Banach spaces and set

(

)

(

) =

Prove that

is open in

(

3. Prove that the sum of a closed subspace and a nite dimensional subspace of a Banach

space is closed.

4. Prove that any compact subset of the complex plane is the spectrum of a bounded

linear operator on a Banach algebra.

MATH 502

Second exam

May 9, 1997

Professor Arnold

Do 3 problems, omit 1. State which problem you omit. Partial credit will be awarded, but

errors and unacknowledged omissions count negatively.

1. Suppose that

are elements in a Hilbert space such that

converges to

weakly and

converges to

kxk

. Prove that

converges to

in norm. Give a

counterexample in Banach space.

Under the given hypotheses,

;

kxk

;

kxk

;

For a counterexample in Banach space, consider

, where the

are the

usual sequence space basis elements. Then

= 1 for all

1 and

weakly.

2. Let

and

be Banach spaces and set

(

)

(

) =

Prove that

is open in

(

Since the null-space of

is the annihilator of range of

is one-to-one i

has dense

range. Combining with the closed range theorem, we get that

is one-to-one with closed

range i

has dense, closed range, i.e., i

is surjective. Thus

is the set of operators

which are one-to-one with closed range, which is the set of operators for which there exists

0 such that

ckxk

for all

. If

and

0 is the corresponding constant,

we show that

for all

2. Indeed

(

)

;

ckxk

;

kxk

kxk:

3. Prove that the sum of a closed subspace and a nite dimensional subspace of a Banach

space is closed.

It is enough to consider a closed space

and one element

and show that

is closed. By the Hahn-Banach Theorem there is a bounded linear functional

which

vanishes on

and satises

(

) = 1. Now suppose that

and

. Say

. Then

(

)

(

). Also

;

(

)

and, since

is closed,

;

(

)

. Thus

;

(

)

] +

(

)

Alternate proof:

(

) where

is the projection of the Banach space

onto

X=S

. Since

is nite dimensional,

is nite dimensional, hence closed in

X=S

, and its

inverse image under a continuous map is closed as well.

4. Prove that any compact subset of the complex plane is the spectrum of a bounded

linear operator on a Banach algebra.

Let

be the compact set and let

be the space of continuous complex valued functions

. Dene

(

) =

(

). If

, then dist(

)

0, so 1

(

;

)

is bounded on

, so multliplication by 1

(

;

) is continuous on

, and provides the

inverse of

;

. Thus

(

). Conversely, if

, then we can take

of norm

one with (

;

)

arbitrarily small (e.g., if

is chosen to takes values in 0 1] and vanish

when

;

, then

(

;

)

). This shows that

;

cannot be invertible

i.e.,

(

). Note: there are variants of this construction. E.g., we could choose a dense

sequence of numbers in

and consider the corresponding diagonal matrix on