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FUNCTIONAL ANALYSIS1
Douglas N. Arnold2


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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
Subspaces and quotient spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
Basic properties of Hilbert spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
II. Linear Operators and Functionals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
The Hahn“Banach Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .10
Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
III. Fundamental Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
The Open Mapping Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .14
The Uniform Boundedness Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
The Closed Range Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
IV. Weak Topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
The weak topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
The weak* topology. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .19
V. Compact Operators and their Spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
Hilbert“Schmidt operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
Compact operators. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .23
Spectral Theorem for compact self-adjoint operators . . . . . . . . . . . . . . . . . . . . . . 26
The spectrum of a general compact operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
VI. Introduction to General Spectral Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
The spectrum and resolvent in a Banach algebra . . . . . . . . . . . . . . . . . . . . . . . . . 31
Spectral Theorem for bounded self-adjoint operators . . . . . . . . . . . . . . . . . . . . . . 35



1 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.
2 Department of Mathematics, Penn State University, University Park, PA 16802.

Email: dna@math.psu.edu. Web: http://www.math.psu.edu/dna/.
1
2

I. Vector spaces and their topology

Basic de¬nitions: (1) Norm and seminorm on vector spaces (real or complex). A norm
de¬nes 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
de¬nite, symmetric bilinear form on X —X ’ R. In the complex case it is positive de¬nite,
Hermitian symmetric, sesquilinear form X — X ’ C. 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

1 2
’ x ’ y 2 ).
(x, y) = ( x+y
4
For complex spaces it is

1 2 2 2
’ i x ’ iy 2 ).
’ x’y
(x, y) = ( x+y + i x + iy
4
In inner product spaces we also have the parallelogram law:
2 2 2
+ y 2 ).
+ x’y
x+y = 2( x

This gives a criterion for a normed space to be an inner product space. Any norm coming
from an inner product satis¬es the parallelogram law and, conversely, if a norm satis¬es the
parallelogram law, we can show (but not so easily) that the polarization identity de¬nes
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 xn in a TVS is Cauchy if
for every neighborhood U of 0 there exists N such that xm ’ xn ∈ U for all m, n ≥ N .
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 X be a vector space and suppose
that a family { · ± }±∈A of seminorms on X is given which are su¬cient in the sense that
± { x ± = 0} = 0. Then the topology generated by the sets { x ± < r}, ± ∈ A, r > 0,
makes X a TVS. A sequence (or net) xn converges to x i¬ xn ’ x ± ’ 0 for all ±. Note
that, a fortiori, | xn ± ’ x ± | ’ 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 de¬ne a metric

x’y n
2’n
d(x, y) = ,
1+ x’y n
n
3

so the topology is metrizable.
Examples: (0) On Rn or Cn we may put the lp norm, 1 ¤ p ¤ ∞, or the weighted
lp 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 p = 2 is it a Hilbert space.
(2) If „¦ is a subset of Rn (or, more generally, any Hausdor¬ space) we may de¬ne the
space Cb („¦) of bounded continuous functions with the supremum norm. It is a Banach
space. If X is compact this is simply the space C(„¦) of continuous functions on „¦.
(3) For simplicity, consider the unit interval, and de¬ne C n ([0, 1]) and C n,± ([0, 1]),
n ∈ N, ± ∈ (0, 1]. Both are Banach spaces with the natural norms. C 0,1 is the space of
Lipschitz functions. C([0, 1]) ‚ C 0,± ‚ C 0,β ‚ C 1 ([0, 1]) if 0 < ± ¤ β ¤ 1.
(4) For 1 ¤ p < ∞ and „¦ an open or closed subspace of Rn (or, more generally, a σ-¬nite
measure space), we have the space Lp („¦) of equivalence classes of measurable p-th power
integrable functions (with equivalence being equality o¬ a set of measure zero), and for
p = ∞ equivalence classes of essentially bounded functions (bounded after modi¬cation
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 Lp -seminorm zero, this
is a normed linear space, and the Riesz-Fischer theorem asserts that it is a Banach space.
L2 is a Hilbert space. If meas(„¦) < ∞, then Lp („¦) ‚ Lq („¦) if 1 ¤ q ¤ p ¤ ∞.
(5) The sequence space lp , 1 ¤ p ¤ ∞ is an example of (4) in the case where the
measure space is N with the counting measure. Each is a Banach space. l2 is a Hilbert
space. lp ‚ lq if 1 ¤ p ¤ q ¤ ∞ (note the inequality is reversed from the previous example).
The subspace c0 of sequences tending to 0 is a closed subspace of l∞ .
(6) If „¦ is an open set in Rn (or any Hausdor¬ space), we can equip C(„¦) with the
norms f ’ |f (x)| indexed by x ∈ „¦. 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 Rn we can equip C(„¦) with the norms f ’ f L∞ (K) indexed
by compact subsets of „¦, thus de¬ning the topology of uniform convergence on compact
subsets. We get the same toplogy by using only the countably many compact sets

Kn = {x ∈ „¦ : |x| ¤ n, dist(x, ‚„¦) ≥ 1/n}.

The topology is complete.
(8) In the previous example, in the case „¦ is a region in C, and we take complex-
valued functions, we may consider the subspace H(„¦) of holomorbarphic functions. By
Weierstrass™s theorem it is a closed subspace, hence itself a complete TVS.
(9) If f, g ∈ L1 (I), I = (0, 1) and

1 1
f (x)φ(x) dx = ’ g(x)φ (x) dx,
0 0
4

for all in¬nitely di¬erentiable φ with support contained in I (so φ is identically zero near
0 and 1), then we say that f is weakly di¬erentiable and that f = g. We can then de¬ne
the Sobolev space Wp (I) = { f ∈ Lp (I) : f ∈ Lp (I) }, with the norm
1


1/p
1 1
|f (x)|p dx + |f (x)|p dx
f = .
1
Wp (I)
0 0

¯
This is a larger space than C 1 (I), but still incorporates ¬rst order di¬erentiability of f .
The case p = 2 is particularly useful, because it allows us to deal with di¬erentiability
in a Hilbert space context. Sobolev spaces can be extended to measure any degree of
di¬erentiability (even fractional), and can be de¬ned on arbitrary domains in Rn .

Subspaces and quotient spaces.
If X is a vector space and S a subspace, we may de¬ne the vector space X/S of cosets.
If X is normed, we may de¬ne

= inf x ’ s
u = inf x X, or equivalently x
¯ X.
X/S X/S
x∈u s∈S

This is a seminorm, and is a norm i¬ S is closed.

Theorem. If X is a Banach space and S is a closed subspace then S is a Banach space
and X/S is a Banach space.

Sketch. Suppose xn is a sequence of elements of X for which the cosets xn are Cauchy.
¯
’n’1
We can take a subsequence with xn ’ xn+1 X/S ¤ 2
¯ ¯ , n = 1, 2, . . . . Set s1 = 0, de¬ne
s2 ∈ S such that x1 ’(x2 +s2 ) X ¤ 1/2, de¬ne s3 ∈ S such that (x2 +s2 )’(x3 +s3 ) X ¤
1/4, . . . . Then {xn + sn } is Cauchy in X . . .

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

Theorem. If X is a normed linear space and S is a closed subspace such that S is a
Banach space and X/S is a Banach space, then X is a Banach space.

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

Theorem. If S is a closed subspace of a Banach space and V is a ¬nite dimensional
subspace, then S + V is closed.

Sketch. We easily pass to the case V is one-dimensional and V © S = 0. We then have that
S +V is algebraically a direct sum and it is enough to show that the projections S +V ’ S
and S + V ’ V are continuous (since then a Cauchy sequence in S + V will lead to a
Cauchy sequence in each of the closed subspaces, and so to a convergent subsequence).
Now the projection π : X ’ X/S restricts to a 1-1 map on V so an isomorphism of V onto
¯ ¯ ¯
its image V . Let µ : V ’ V be the continuous inverse. Since π(S + V ) ‚ V , we may form
the composition µ —¦ π|S+V : S + V ’ V and it is continuous. But it is just the projection
onto V . The projection onto S is id ’ µ —¦ π, so it is also continuous.
5

Note. The sum of closed subspaces of a Banach space need not be closed. For a coun-
terexample (in a separable Hilbert space), let S1 be the vector space of all real sequences
(xn )∞ for which xn = 0 if n is odd, and S2 be the sequences for which x2n = nx2n’1 ,
n=1
n = 1, 2, . . . . Clearly X1 = l2 © S1 and X2 = l2 © S2 are closed subspaces of l2 , the space
of square integrable sequences (they are de¬ned 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 S1 and S2 :

(x1 , x2 , . . . ) = (0, x2 ’ x1 , 0, x4 ’ 2x3 , 0, x6 ’ 3x5 , . . . ) + (x1 , x1 , x3 , 2x3 , x5 , 3x5 , . . . ).

If a sequence has all but ¬nitely many terms zero, so do the two summands. Thus all
such sequences belong to X1 + X2 , showing that X1 + X2 is dense in l2 . Now consider the
sequence (1, 0, 1/2, 0, 1/3, . . . ) ∈ l2 . Its only decomposition as elements of S1 and S2 is

(1, 0, 1/2, 0, 1/3, 0, . . . ) = (0, ’1, 0, ’1, 0, ’1, . . . ) + (1, 1, 1/2, 1, 1/3, 1, . . . ),

and so it does not belong to X1 + X2 . Thus X1 + X2 is not closed in l2 .

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 X be a Hilbert space, K a closed convex subset, and x ∈ X.
Then there exists a unique x ∈ K such that
¯

x ’ x = inf x ’ y .
¯
y∈K



Proof. Translating, we may assume that x = 0, and so we must show that there is a unique
element of K of minimal norm. Let d = inf y∈K y and chose xn ∈ K with xn ’ d.
Then the parallelogram law gives
2 2
xn ’ xm 1 1 xn + xm 1 1
2 2 2 2
’ d2 ,
’ ¤
= xn + xm xn + xm
2 2 2 2 2 2
where we have used convexity to infer that (xn + xm )/2 ∈ K. Thus xn is a Cauchy
sequence and so has a limit x, which must belong to K, since K is closed. Since the norm
¯
is continuous, x = limn xn = d.
¯
For uniqueness, note that if x = x = d, then (¯ + x)/2 = d and the parallelogram
¯ ˜ x˜
law gives
2 2 2 2
= 2d2 + 2d2 ’ 4d2 = 0.
x’x ’ x+x
¯˜ =2 x
¯ +2 x
˜ ¯˜

The unique nearest element to x in K is often denoted PK x, and referred to as the
projection of x onto K. It satis¬es PK —¦ PK = PK , the de¬nition of a projection. This
terminology is especially used when K is a closed linear subspace of X, in which case PK
is a linear projection operator.
6

Projection and orthogonality. If S is any subset of a Hilbert space X, let

S ⊥ = { x ∈ X : x, s = 0 for all s ∈ S }.

Then S ⊥ is a closed subspace of X. We obviously have S © S ⊥ = 0 and S ‚ S ⊥⊥ .
Claim: If S is a closed subspace of X, x ∈ X, and PS x the projection of x onto S, then
x ’ PS x ∈ S ⊥ . Indeed, if s ∈ S is arbitrary and t ∈ R, then
2 2 2
’ 2t(x ’ PS x, s) + t2 s 2 ,
x ’ PS x ¤ x ’ PS x ’ ts = x ’ PS x

so the quadratic polynomial on the right hand side has a minimum at t = 0. Setting the
derivative there to 0 gives (x ’ PS x, s) = 0.
Thus we can write any x ∈ X as s + s⊥ with s ∈ S and s⊥ ∈ S ⊥ (namely s = PS x,
s⊥ = x ’ PS x). Such a decomposition is certainly unique (if s + s⊥ were another one we
¯¯
would have s ’ s = s ’ s ∈ S © S = 0.) We clearly have x 2 = s 2 + s⊥ 2 .
⊥ ⊥ ⊥
¯¯
An immediate corollary is that S ⊥⊥ = S for S a closed subspace, since if x ∈ S ⊥⊥ we
can write it as s + s⊥ , whence s⊥ ∈ S ⊥ © S ⊥⊥ = 0, i.e., x ∈ S. We thus see that the
decomposition
x = (I ’ PS )x + PS x
is the (unique) decomposition of x into elements of S ⊥ and S ⊥⊥ . Thus PS ⊥ = I ’ PS . For
any subset S of X, S ⊥⊥ is the smallest closed subspace containing S.

Orthonormal sets and bases in Hilbert space.
Let e1 , e2 , . . . , eN be orthonormal elements of a Hilbert space X, and let S be their
span. Then n x, en en ∈ S and x ’ n x, en en ⊥ S, so n x, en en = PS x. But
N
2
= n=1 x, en 2 , so
n x, en en

N
2 2
¤x
x, en
n=1

(Bessel™s inequality). Now let E be an orthonormal set of arbitrary cardinality. It follows
from Bessel™s inequality that for > 0 and x ∈ X, { e ∈ E : x, e ≥ } is ¬nite, and
hence that { e ∈ E : x, e > 0 } is countable. We can thus extend Bessel™s inequality to
an arbitrary orthonormal set:
x, e 2 ¤ x 2 ,
e∈E

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 E is an
arbitary set and f : E ’ X a function mapping into a Hilbert space (or any normed linear
space or even TVS), we say

() f (e) = x
e∈E
7

if the net e∈F f (e), indexed by the ¬nite subsets F of E, converges to x. In other words,
( ) holds if, for any neighborhood U of the origin, there is a ¬nite set F0 ‚ E such that
x ’ e∈F f (e) ∈ U whenever F is a ¬nite subset of E containing F0 . In the case E = N,
this is equivalent to absolute convergence of a series. Note that if e∈E f (e) converges,
then for all there is a ¬nite F0 such that if F1 and F2 are ¬nite supersets of F0 , then
e∈F1 f (e) ’ e∈F2 f (e) ¤ . It follows easily that each of the sets { e ∈ E | f (e) ≥
1/n } is ¬nite, and hence, f (e) = 0 for all but countably many e ∈ E.

Lemma. If E is an orthonormal subset of a Hilbert space X and x ∈ X, then

x, e e
e∈E

converges.

Proof. We may order the elements e1 , e2 , . . . of E for which x, e = 0. Note that

N N
2
| x, en |2 ¤ x 2 .
x, en en =
n=1 n=1

N
This shows that the partial sums sN = n=1 x, en en form a Cauchy sequence, and so

converge to an element n=1 x, en en of X. As an exercise in applying the de¬nition,

we show that x, e e = n=1 x, en en . Given > 0 pick N large enough that
e∈E
∞ 2
n=N +1 | x, en | < . If M > N and F is a ¬nite subset of E containing e1 , . . . , eN ,
then
M
2
x, en en ’ ¤.
x, e e
n=1 e∈F

Letting M tend to in¬nity,

2
x, en en ’ ¤,
x, e e
n=1 e∈F


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 in¬nite dimensional spaces they are
di¬cult or impossible to construct and almost never used. Another notion of basis is much
8

more useful, namely one that uses the topology to allow in¬nite 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 de¬nition 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 E is such an orthonormal basis, and x ∈ X,
then
x= x, e e.
e∈E

Indeed, we know that the sum on the right exists in X and it is easy to check that its inner
product with any e0 ∈ E is x, e0 . Thus y := x ’ e∈E x, e e is orthogonal to E, and if it
weren™t zero, then we could adjoin y/ y to E to get a larger orthonormal set.
ce e for some ce ∈ R,
Thus we™ve shown that any element x of X can be expressed as
all but countably many of which are 0. It is easily seen that this determines the ce uniquely,
namely ce = x, e , and that x 2 = c2 . e

The notion of orthonormal basis allows us to de¬ne a Hilbert space dimension, namely
the cardinality of any orthonormal basis. To know that this is well de¬ned, we need to check
that any two bases have the same cardinality. If one is ¬nite, this is trivial. Otherwise,
let E and F be two in¬nite orthonormal bases. For each 0 = x ∈ X, the inner product
x, e = 0 for at least one e ∈ E. Thus

F‚ { f ∈ F : f, e = 0 },
e∈E


i.e., F is contained in the union of card E countable sets. Therefore card F ¤ „µ0 card E =
card E.
If S is any set, we de¬ne a particular Hilbert space l2 (S) as the set of functions c : S ’ R
which are zero o¬ a countable set and such that s∈S c2 < ∞. We thus see that via a basis,
s
any Hilbert space can be put into a norm-preserving (and so inner-product-preserving)
linear bijection (or Hilbert space isomorphism) with an l2 (S). Thus, up to isomorphism,
there is just one Hilbert space for each cardinality. In particular there is only one in¬nite
dimensional separable Hilbert space (up to isometry).
Example: The best known example of an orthonormal basis in an in¬nite Hilbert space
is the set of functions en = exp(2πinθ) which form a basis for complex-valued L2 ([0, 1]).
(They are obviously orthonormal, and they are a maximal orthonormal set by the Weier-
strass approximation Theorem. Thus an arbitrary L2 function has an L2 convergent
Fourier series

ˆ
f (n)e2πinθ ,
f (θ) =
n=’∞

1
ˆ
with f (n) = f, en = 0 f (θ)e’2πinθ dθ. Thus from the Hilbert space point of view, the
theory of Fourier series is rather simple. More di¬cult analysis comes in when we consider
convergence in other topologies (pointwise, uniform, almost everywhere, Lp , C 1 , . . . ).
9

Schauder bases. An orthonormal basis in a Hilbert space is a special example of a
Schauder basis. A subset E of a Banach space X is called a Schauder basis if for every
x ∈ X there is a unique function c : E ’ R such that x = e∈E ce e. Schauder constructed
a useful Schauder basis for C([0, 1]), and there is useful Schauder bases in many other
separable Banach spaces. In 1973 Per En¬‚o settled a long-standing open question by
proving that there exist separable Banach spaces with no Schauder bases.


II. Linear Operators and Functionals

B(X, Y ) = bounded linear operators between normed linear spaces X and Y . 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. If X is a normed linear space and Y is a Banach space, then B(X, Y ) is a
Banach space with the norm

Tx Y
T = sup .
B(X,Y )
xX
0=x∈X




Proof. It is easy to check that B(X, Y ) is a normed linear space, and the only issue is to
show that it is complete.
Suppose that Tn is a Cauchy sequence in B(X, Y ). Then for each x ∈ X Tn x is Cauchy
in the complete space Y , so there exists T x ∈ Y with Tn x ’ T x. Clearly T : X ’ Y is
linear. Is it bounded? The real sequence Tn is Cauchy, hence bounded, say Tn ¤ K.
It follows that T ¤ K, and so T ∈ B(X, Y ). To conclude the proof, we need to show
that Tn ’ T ’ 0. We have

Tn ’ T = sup Tn x ’ T x = sup Tn x ’ Tm x
lim
x ¤1 m’∞
x ¤1

= sup lim sup Tn x ’ Tm x ¤ lim sup Tn ’ Tm .
x ¤1 m’∞ m’∞


Thus lim supn’∞ Tn ’ T = 0.

If T ∈ B(X, Y ) and U ∈ B(Y, Z), then U T = U —¦ T ∈ B(X, Z) and U T B(X,Z) ¤
U B(Y,Z) T B(X,Y ) . In particular, B(X) := B(X, X) 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 X — := B(X, R) (or B(X, C) for complex vector spaces). It is a Banach
space (whether X is or not).
10

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., Lp for 0 < p < 1, with no non-zero continuous linear
functionals.)

Hahn-Banach. If f is a bounded linear functional on a subspace of a normed linear space,
then f 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 X a vector space, we say that
p : X ’ R is sublinear if p(x + y) ¤ p(x) + p(y) and p(±x) = ±p(x) for x, y ∈ X, ± ≥ 0.

Generalized Hahn-Banach. Let X be a vector space, p : X ’ R a sublinear functional,
S a subspace of X, and f : S ’ R a linear function satisfying f (x) ¤ p(x) for all x ∈ S,
then f can be extended to X so that the same inequality holds for all x ∈ X.

Sketch. It su¬ces to extend f to the space spanned by S and one element x0 ∈ X \ S,
preserving the inequality, since if we can do that we can complete the proof with Zorn™s
lemma.
We need to de¬ne f (x0 ) such that f (tx0 + s) ¤ p(tx0 + s) for all t ∈ R, s ∈ S. The case
t = 0 is known and it is easy to use homogeneity to restrict to t = ±1. Thus we need to
¬nd a value f (x0 ) ∈ R such that

f (s) ’ p(’x0 + s) ¤ f (x0 ) ¤ p(x0 + s) ’ f (s) for all s ∈ S.

Now it is easy to check that for any s1 , s2 ∈ S, f (s1 ) ’ p(’x0 + s1 ) ¤ p(x0 + s2 ) ’ f (s2 ),
and so such an f (x0 ) exists.

Corollary. If X is a normed linear space and x ∈ X, then there exists f ∈ X — of norm
1 such that f (x) = x .

Corollary. If X is a normed linear space, S a closed subspace, and x ∈ X, then there
exists f ∈ X — of norm 1 such that f (x) = x X/S .
¯

Duality. If X and Y are normed linear spaces and T : X ’ Y , then we get a natural
map T — : Y — ’ X — by T — f (x) = f (T x) for all f ∈ Y — , x ∈ X. In particular, if
T ∈ B(X, Y ), then T — ∈ B(Y — , X — ). In fact, T — B(Y — ,X — ) = T B(X,Y ) . To prove
this, note that |T — f (x)| = |f (T x)| ¤ f T x . Therefore T — f ¤ f T , so T —
is indeed bounded, with T — ¤ T . Also, given any y ∈ Y , we can ¬nd g ∈ Y —
such that |g(y)| = y , g = 1. Applying this with y = T x (x ∈ X arbitrary), gives
T x = |g(T x)| = |T — gx| ¤ T — g x = T — x . This shows that T ¤ T — . Note
that if T ∈ B(X, Y ), U ∈ B(Y, Z), then (U T )— = T — U — .
If X is a Banach space and S a subset, let

S a = { f ∈ X — | f (s) = 0 ∀s ∈ S }
11

denote the annihilator of S. If V is a subset of X — , we similarly set
a
V = { x ∈ X | f (x) = 0 ∀f ∈ V }.

Note the distinction between V a , which is a subset of X —— and a V , which is a subset of
X. All annihilators are closed subspaces.
It is easy to see that S ‚ T ‚ X implies that T a ‚ S a , and V ‚ W ‚ X — implies that
W ‚ a V . Obviously S ‚ a (S a ) if S ‚ X and V ‚ (a V )a if V ‚ X — . The Hahn-Banach
a

theorem implies that S = a (S a ) in case S is a closed subspace of X (but it can happen
(a V )a for V a closed subspace of X — . For S ‚ X arbitrary, a (S a ) is the smallest
that V
closed subspace of X containing the subset S, namely the closure of the span of S.
Now suppose that T : X ’ Y is a bounded linear operator between Banach spaces. Let
g ∈ Y — . Then g(T x) = 0 ∀x ∈ X ⇐’ T — g(x) = 0 ∀x ∈ X ⇐’ T — g = 0. I.e.,

R(T )a = N (T — ).

∀f ∈ Y — ⇐’ T — f (x) = 0 ∀f ∈ Y — , or
Similarly, for x ∈ X, T x = 0 ⇐’ f (T x) = 0

R(T — ) = N (T ).
a


Taking annihilators gives two more results:

R(T ) = a N (T — ), R(T — ) ‚ N (T )a .

In particular we see that T — is injective i¬ T has dense range; and T is injective if T — has
dense range.
Note: we will have further results in this direction once we introduce the weak*-topology
on X — . In particular, (a S)a is the weak* closure of a subspace S of X — and T is injective
i¬ T — has weak* dense range.

Dual of a subspace. An important case is when T is the inclusion map i : S ’ X,
where S is a closed subspace of X. Then r = i— : X — ’ S — is just the restriction map:
rf (s) = f (s). Hahn-Banach tells us that r is surjective. Obviously N (r) = S a . Thus we
have a canonical isomorphism r : X — /S a ’ S — . In fact, the Hahn-Banach theorem shows
¯
that it is an isometry. Via this isometry one often identi¬es X — /S a with S — .

Dual of a quotient space. Next, consider the projection map π : X ’ X/S where S is
a closed subspace. We then have π — : (X/S)— ’ X — . Since π is surjective, this map is
injective. It is easy to see that the range is contained in S a . In fact we now show that π —
maps (X/S)— onto S a , hence provides a canonical isomorphism of S a with (X/S)— . Indeed,
if f ∈ S a , then we have a splitting f = g —¦ π with g ∈ (X/S)— (just de¬ne g(c) = f (x)
where x is any element of the coset c). Thus f = π — g is indeed in the range of π — . This
correspondence is again an isometry.

Dual of a Hilbert space. The identi¬cation of dual spaces can be quite tricky. The case
of Hilbert spaces is easy.
12

Riesz Representation Theorem. If X is a real Hilbert space, de¬ne j : X ’ X — by
jy (x) = x, y . This map is a linear isometry of X onto X — . For a complex Hilbert space
it is a conjugate linear isometry (it satis¬es j±y = ±jy ).
¯

Proof. It is easy to see that j is an isometry of X into X — and the main issue is to show
that any f ∈ X — can be written as jy for some y. We may assume that f = 0, so N (f ) is
a proper closed subspace of X. Let y0 ∈ [N (f )]⊥ be of norm 1 and set y = (f y0 )y0 . For
all x ∈ X, we clearly have that (f y0 )x ’ (f x)y0 ∈ N (f ), so

jy (x) = x, (f y0 )y0 = (f y0 )x, y0 = (f x)y0 , y0 = f x.


Via the map j we can de¬ne an inner product on X — , so it is again a Hilbert space.
Note that if S is a closed subspace of X, then x ∈ S ⊥ ⇐’ js ∈ S a . The Riesz map j
is sometimes used to identify X and X — . Under this identi¬cation there is no distinction
between S ⊥ and S a .

Dual of C(„¦). Note: there are two quite distinct theorems referred to as the Riesz
Representation Theorem. The proceeding is the easy one. The hard one identi¬es the
dual of C(„¦) where „¦ is a compact subset of Rn (this can be generalized considerably). It
states that there is an isometry between C(„¦)— and the space of ¬nite signed measures on
„¦. (A ¬nite signed measure is a set function of the form µ = µ1 ’ µ2 where µi is a ¬nite
measure, and we view such as a functional on C(X) by f ’ „¦ f dµ1 ’ „¦ f dµ2 .) This
is the real-valued case; in the complex-valued case the isometry is with complex measures
µ + i» where µ and » are ¬nite signed measures.

Dual of C 1 . It is easy to deduce a representation for an arbitrary element of the dual
of, e.g., C 1 ([0, 1]). The map f ’ (f, f ) is an isometry of C 1 onto a closed subspace of
C — C. By the Hahn-Banach Theorem, every element of (C 1 )— extends to a functional on
C — C, which is easily seen to be of the form

(f, g) ’ f dµ + g dν

where µ and and ν are signed measures ((X — Y )— = X — — Y — with the obvious identi¬-
cations). Thus any continuous linear functional on C 1 can be written

f’ f dµ + f dν.

In this representation the measures µ and » are not unique.

Dual of Lp . H¨lder™s inequality states that if 1 ¤ p ¤ ∞, q = p/(p ’ 1), then
o

fg ¤ f g
Lp Lq
13

for all f ∈ Lp , g ∈ Lq . This shows that the map g ’ »g :

»g (f ) = f g,

maps Lq linearly into (Lp )— with »g (Lp )— ¤ g Lq . The choice f = sign(g)|g|q’1 shows
that there is equality. In fact, if p < ∞, » is a linear isometry of Lq onto (Lp )— . For p = ∞
it is an isometric injection, but not in general surjective. Thus the dual of Lp is Lq for p
¬nite. The dual of L∞ is a very big space, much bigger than L1 and rarely used.
Dual of c0 . The above considerations apply to the dual of the sequence spaces lp . Let
us now show that the dual of c0 is l1 . For any c = (cn ) ∈ c0 and d = (dn ) ∈ l1 , we de¬ne
»d (c) = cn dn . Clearly
|»d (c)| ¤ sup |cn | |dn | = c d l1 ,
c0

¤d
so »d l1 . Taking
c—
0
sign(dn ), n ¤ N
cn =
0, n > N,
we see that equality holds. Thus » : l1 ’ c— is an isometric injection. We now show that it
0
(n)

is onto. Given f ∈ c0 , de¬ne dn = f (e ) where e(n) is the usual unit sequence em = δmn .
(n)

Let sn = sign(dn ). Then |dn | = f (sn e(n) ), so
N N N
f (sn e(n) ) = f ( sn e(n) ) ¤ f .
|dn | =
n=0 n=0 n=0
Letting N ’ ∞ we conclude that d ∈ l1 . Now by construction »d agrees with f on all
sequences with only ¬nitely many nonzeros. But these are dense in c0 , so f = »d .
The bidual. If X is any normed linear space, we have a natural map i : X ’ X —— given
by
f ∈ X —.
x ∈ X,
ix (f ) = f (x),
Clearly ix ¤ f and, by the Hahn-Banach theorem, equality holds. Thus X may be
˜
identi¬ed as a subspace of the Banach space X —— . If we de¬ne X as the closure of i(X) in
˜
X —— , then X is isometrically embedded as a dense subspace of the Banach space X. This
˜
determines X up to isometry, and is what we de¬ne as the completion of X. Thus any
normed linear space has a completion.
If i is onto, i.e., if X is isomorphic with X —— via this identi¬cation, we say that X is
re¬‚exive (which can only happen is X is complete). In particular, one can check that if X
is a Hilbert space and j : X ’ X — is the Riesz isomorphism, and j — : X — ’ X —— the Riesz
isomorphism for X — , then i = j — —¦ j, so X is re¬‚exive.
Similarly, the canonical isometries of Lq onto (Lp )— and then Lp onto (Lq )— compose to
give the natural map of Lp into its bidual, and we conclude that Lp (and lp ) is re¬‚exive
for 1 < p < ∞. None of L1 , l1 , L∞ , l∞ , c0 , or C(X) are re¬‚exive.
If X is re¬‚exive, then i(a S) = S a for S ‚ X — . In other words, if we identify X and
X —— , the distinction between the two kinds of annihilators disappears. In particular, for
re¬‚exive Banach spaces, R(T — ) = N (T )a and T is injective i¬ T — has dense range.
14

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 M = n∈N Fn with Fn a closed
subset which does not contain any open set. In particular, F0 is a proper closed set, so
there exists x0 ∈ M , 0 ∈ (0, 1) such that E(x0 , 0 ) ‚ M \ F0 . Since no ball is contained in
F1 , there exists x1 ∈ E(x0 , 0 /2) and 1 ∈ (0, 0 /2) such that E(x1 , 1 ) ‚ M \ F1 . In this
way we get a nested sequence of balls such that the nth ball has radius at most 2’n and is
disjoint from Fn . 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 Fn .

The Open Mapping Theorem. The Open Mapping Theorem follows from the Baire
Category Theorem and the following lemma.

Lemma. Let T : X ’ Y be a bounded linear operator between Banach spaces. If
E(0Y , r) ‚ T (E(0X , 1)) for some r > 0, then E(0Y , r) ‚ T (B(0X , 2)).

Proof. Let U = T (E(0X , 1)). Let y ∈ Y , y < r. There exists y0 ∈ U with y’y0 ¤ r/2.
By homogeneity, there exists y1 ∈ 1 U such that y ’ y0 ’ y1 ¤ r/4, y2 ∈ 1 U such that
2 4
1
y ’ y0 ’ y1 ’ y2 ¤ r/8, etc. Take xn ∈ 2n U such that T xn = yn , and let x = n xn ∈ X.
Then x ¤ 2 and T x = yn = y.

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 su¬cient 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 T of a ball about 0 contains some ball
about 0. The sets T (E(0, n)) cover Y , 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 T .

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.
15

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 xn converges to x in X then T xn converges
to T x. By using the closed graph theorem, we get to assume as well that T xn is converging
to some y in Y and we need only show that y = T x.

Proof. Let G = { (x, T x) | x ∈ X } denote the graph. Then the composition G ‚ X — Y ’
X is a bounded linear operator between Banach spaces given by (x, T x) ’ x. It is
clearly one-to-one and onto, so the inverse is continuous by Banach™s theorem. But the
composition X ’ G ‚ X — Y ’ Y is simply the T , so T is continuous.

Banach™s theorem leads immediately to this useful characterization of closed imbeddings
of Banach spaces.

Theorem. Let T : X ’ Y be a bounded linear map between Banach spaces. Then T is
one-to-one and has closed range if and only if there exists a positive number c such that

x ¤ c Tx ∀x ∈ X.


Proof. If the inequality holds, then T is clearly one-to-one, and if T xn is a Cauchy sequence
in R(T ), then xn is Cauchy, and hence xn converges to some x, so T xn converges to T x.
Thus the inequality implies that R(T ) is closed.
For the other direction, suppose that T is one-to-one with closed range and consider the
map T ’1 : R(T ) ’ X. It is the inverse of a bounded isomorphism, so is itself bounded.
The inequality follows immediately (with c the norm of T ’1 ).

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 X and Y are Banach spaces and S ‚
B(X, Y ). If supT ∈S T (x) Y < ∞ for all x ∈ X, then supT ∈S T < ∞.

Proof. One of the closed sets { x | |fn (x)| ¤ N ∀n } must contain E(x0 , r) for some x0 ∈
X, r > 0. Then, if x < r, |fn (x)| ¤ |fn (x + x0 ) ’ fn (x0 )| ¤ N + sup |fn (x0 )| = M , with
M independent of n. This shows that the fn are uniformly bounded (by M/r).

In words: a set of linear operators between Banach spaces which is bounded pointwise
is norm bounded.
16

The uniform boundedness theorem is often a way to generate counterexamples. A
typical example comes from the theory of Fourier series. For f : R ’ C continuous and
1-periodic the nth partial sum of the Fourier series for f is
n 1 1
’2πikt 2πiks
f (t)Dn (s ’ t) dt
fn (s) = f (t)e dt e =
’1 ’1
k=’n

where
n
e2πikx .
Dn (x) =
k=’n

Writing z = e2πis , we have
n
z n+1/2 ’ z ’n’1/2
2n
’1
’n z sin(2n + 1)πx
k
Dn (s) = z =z = = .
z 1/2 ’ z ’1/2
z’1 sin πx
k=’n

This is the Dirichlet kernel, a C ∞ periodic function. In particular, the value of the nth
partial sum of the Fourier series of f at 0 is
1
Tn f := fn (0) = f (t)Dn (t) dt.
’1

We think of Tn as a linear functional on the Banach space of 1-periodic continuous function
endowed with the sup norm. Clearly
1
Tn ¤ Cn := |Dn (t)| dt.
’1

In fact this is an equality. If g(t) = sign Dn (t), then sup |g| = 1 and Tn g = Cn . Actually,
g is not continuous, so to make this argument correct, we approximate g by a continuous
functions, and thereby prove the norm equality. Now one can calculate that |Dn | ’ ∞
as n ’ ∞. By the uniform boundedness theorem we may conclude that there exists a
continuous periodic function for whose Fourier series diverges at t = 0.

The Closed Range Theorem. We now apply the Open Mapping Theorem to better
understand the relationship between T and T — . The property of having a closed range
is signi¬cant to the structure of an operator between Banach spaces. If T : X ’ Y has
a closed range Z (which is then itself a Banach space), then T factors as the projection
X ’ X/ N (T ), the isomorphism X/ N (T ) ’ Z, and the inclusion Z ‚ Y . The Closed
Range Theorem says that T has a closed range if and only if T — does.

Theorem. Let T : X ’ Y be a bounded linear operator between Banach spaces. Then T
is invertible i¬ T — is.

Proof. If S = T ’1 : Y ’ X exists, then ST = IX and T S = IY , so T — S — = IX — and
S — T — = IY — , which shows that T — is invertible.
17

Conversely, if T — is invertible, then it is open, so there is a number c > 0 such that
T — BY — (0, 1) contains BX — (0, c). Thus, for x ∈ X

|(T — f )x|
|f (T x)| =
Tx = sup sup
f ∈BY — (0,1) f ∈BY — (0,1)

≥ |g(x)| = c x .
sup
g∈BX — (0,c)


The existence of c > 0 such that T x ≥ c x ∀x ∈ X is equivalent to the statement that
T is injective with closed range. But since T — is injective, T has dense range.

Lemma. Let T : X ’ Y be a linear map between Banach spaces such that T — is an
injection with closed range. Then T is a surjection.

Proof. Let E be the closed unit ball of X and F = T E. It su¬ces to show that F contains
a ball around the origin, since then, by the lemma used to prove the Open Mapping
Theorem, T is onto.
There exists c > 0 such that T — f ≥ c f for all f ∈ Y — . We shall show that F
contains the ball of radius c around the origin in Y . Otherwise there exists y ∈ Y , y ¤ c,
y ∈ F . Since F is a closed convex set we can ¬nd a functional f ∈ Y — such that |f (T x)| ¤ ±
/
for all x ∈ E and f (y) > ±. Thus f > ±/c, but

T — f = sup |T — f (x)| = sup |f (T x)| ¤ ±.
x∈E x∈E

This is a contradiction.

Closed Range Theorem. Let T : X ’ Y be a bounded linear operator between Banach
spaces. Then T has closed range if and only if T — does.

Proof. 1) R(T ) closed =’ R(T — ) closed.
¯
Let Z = R(T ). Then T : X/ N (T ) ’ Z is an isomorphism (Inverse Mapping Theorem).
The diagram
T
’’’ Y
’’
X
¦
¦ ¦∪
π ¦

=
X/ N (T ) ’ ’ ’ Z.
’’
¯
T

commutes. Taking adjoints,
T—
X— Y—
←’’
’’
¦
¦ ¦π
∪¦

=
N (T )a ← ’ ’ Y — /Z a .
’’
¯
T—

This shows that R(T — ) = N (T )a .
18

2) R(T — ) closed =’ R(T ) closed.
Let Z = R(T ) (so Z a = N (T — )) and let S be the range restriction of T , S : X ’ Z.
The adjoint is S — : Y — /Z a ’ X — , the lifting of T — to Y — /Z a . Now R(S — ) = R(T — ), is
closed, and S — is an injection. We wish to show that S is onto Z. Thus the theorem follows
from the preceding lemma.

IV. Weak Topologies

The weak topology. Let X be a Banach space. For each f ∈ X — the map x ’ |f (x)| is
a seminorm on X, and the set of all such seminorms, as f varies over X — , is su¬cient by
the Hahn-Banach Theorem. Therefore we can endow X with a new TVS structure from
this family of seminorms. This is called the weak topology on X. In particular, xn ’ x
w
weakly (written xn ’ x) i¬ f (xn ) ’ f (x) for all f ∈ X — . Thus the weak topology is

weaker than the norm topology, but all the elements of X — remain continuous when X is
endowed with the weak topology (it is by de¬nition the weakest topology for which all the
elements of X — are continuous).
Note that the open sets of the weak topology are rather big. If U is an weak neighbor-
hood of 0 in an in¬nite dimensional Banach space then, by de¬nition, there exists > 0
and ¬nitely many functionals fn ∈ X — such that { x | |fn (x)| < ∀N } is contained in U .
Thus U contains the in¬nite dimensional closed subspace N (f1 ) © . . . © N (fn ).
w
If xn ’ x weakly, then, viewing the xn as linear functionals on X — (via the canonical

embedding of X into X —— ), we see that the sequence of real numbers obtained by applying
the xn to any f ∈ X — is convergent and hence bounded uniformly in n. By the Uniform
Boundedness Principle, it follows that the xn 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 xn are small in norm, then their weak limit is too.
w
Theorem. If xn ’ x in some Banach space, then x ¤ lim inf xn .

n’∞

Proof. Take f ∈ X — of norm 1 such that f (x) = x . Then f (xn ) ¤ xn , and taking the
lim inf 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 x does not belong to
the norm closure of a convex set E, then there is a weak neighborhood of x which doesn™t
intersect E. This follows immediately from the following convex separation theorem.
19

Theorem. Let E be a nonempty closed convex subset of a Banach space X and x a point
in the complement of E. Then there exists f ∈ X — such that f (x) < inf y∈E f (y).

In fact we shall prove a stronger result:

Theorem. Let E and F be disjoint, nonempty, convex subsets of a Banach space X with
F open. Then there exists f ∈ X — such that f (x) < inf y∈E f (y) for all x ∈ F .

(The previous result follows by taking F to be any ball about x disjoint from E.)

Proof. This is a consequence of the generalized Hahn-Banach Theorem. Pick x0 ∈ E and
y0 ∈ F and set z0 = x0 ’ y0 and G = F ’ E + z0 . Then G is a convex open set containing
0 but not containing z0 . (The convexity of G follows directly from that of E and F ; the
fact that G is open follows from the representation of G = y∈E F ’ y + z0 as a union of
open sets; obviously 0 = y0 ’ x0 + z0 ∈ G, and z0 ∈ G since E and F and disjoint.)
/
Since G is open and convex and contains 0, for each x ∈ X, { t > 0 | t’1 x ∈ G } is a
nonempty open semi-in¬nite interval. De¬ne p(x) ∈ [0, ∞) to be the left endpoint of this
interval. By de¬nition p is positively homogeneous. Since G is convex, t’1 x ∈ G and
s’1 y ∈ G imply that

t ’1 s ’1
(t + s)’1 (x + y) = s y ∈ G,
t x+
s+t s+t

whence p is subadditive. Thus p is a sublinear functional. Moreover, G = { x ∈ X | p(x) <
1 }.
De¬ne a linear functional f on X0 := Rz0 by f (z0 ) = 1. Then f (tz0 ) = t ¤ tp(z0 ) =
p(tz0 ) for t ≥ 0 and f (tz0 ) < 0 ¤ p(tz0 ) for t < 0. Thus f is a linear functional on X0
satisfying f (x) ¤ p(x) there. By Hahn-Banach we can extend f to a linear functional on
X satisfying the same inequality. This implies that f is bounded (by 1) on the open set
G, so f belongs to X — .
If x ∈ F , y ∈ E, then x ’ y + z0 ∈ G, so f (x) ’ f (y) + 1 = f (x ’ y + z0 ) < 1,
or f (x) < f (y). Therefore supx∈F f (x) ¤ inf y∈E f (y). Since f (F ) is an open interval,
f (˜) < supx∈F f (x) for all x ∈ F , and so we have the theorem.
x ˜

The weak* topology. On the dual space X — we have two new topologies. We may endow
it with the weak topology, the weakest one such that all functionals in X —— are continuous,
or we may endow it with the topology generated by all the seminorms f ’ f (x), x ∈ X.
(This is obviously a su¬cient family of functionals.) The last is called the weak* topology
and is a weaker topology than the weak topology. If X is re¬‚exive, the weak and weak*
topologies coincide.
Examples of weak and weak* convergence: 1) Consider weak convergence in Lp („¦)
where „¦ is a bounded subset of Rn . From the characterization of the dual of Lp we see
that
w— w w
fn ’’ f in L∞ =’ fn ’ f weak in Lp =’ fn ’ f weak in Lq
’ ’ ’
20

w—
whenever 1 ¤ q ¤ p < ∞. In particular we claim that the complex exponentials e2πinx ’’


0 in L ([0, 1]) as n ’ ∞. This is simply the statement that
1
g(x)e2πinx dx = 0,
lim
n’∞ 0

for all g ∈ L1 ([0, 1]), i.e., that the Fourier coe¬cients of an L1 tend to 0, which is known as
the Riemann“Lebesgue Lemma. (Proof: certainly true if g is a trigonometric polynomial.
The trig polynomials are dense in C([0, 1]) by the Weierstrass Approximation Theorem,
and C([0, 1]) is dense in L1 ([0, 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 in¬nity. As a very simple example,
it is easy to see that the unit vectors in lp converge weakly to zero for 1 < p < ∞ (and
weak* in l∞ , but not weakly in l1 ). As a more interesting example, let fn ∈ Lp (R) be
a sequence of function which are uniformly bounded in Lp , and for which fn |[’n,n] ≡ 0.
Then we claim that fn ’ 0 weakly in Lp if 1 < p < ∞. Thus we have to show that

lim fn g dx = 0
n’∞ R

for all g ∈ Lq . Let Sn = { x ∈ R | |x| ≥ n }. Then limn |g|q dx = 0 (by the dominated
Sn
convergence theorem). But

| fn g dx| = | fn g dx| ¤ fn ¤C g ’ 0.
g
Lp Lq (Sn ) Lq (Sn )
Sn
R

The same proof shows that if the fn are uniformly bounded they tend to 0 in L∞ weak*.
Note that the characteristic functions χ[n,n+1] do not tend to zero weakly in L1 however.
3) Consider the measure φn = 2nχ[’1/n,1/n] dx. Formally φn tends to the delta function
δ0 as n ’ ∞. Using the weak* topology on C([’1, 1]) this convergence becomes precise:
w—
φn ’’ δ0 .


Theorem (Alaoglu). The unit ball in X — is weak* compact.

Proof. For x ∈ X, let Ix = { t ∈ R : |t| ¤ x }, and set „¦ = Πx∈X Ix . Recall that this
Cartesian product is nothing but the set of all functions f on X with f (x) ∈ Ix for all
x. This set is endowed with the Cartesian product topology, namely the weakest topology
such that for all x ∈ X, the functions f ’ f (x) (from „¦ to Ix ) are continuous. Tychono¬™s
Theorem states that „¦ is compact with this topology.
Now let E be the unit ball in X — . Then E ‚ „¦ and the topology thereby induced on
E is precisely the weak* topology. Now for each pair x, y ∈ X and each c ∈ R, de¬ne
Fx,y (f ) = f (x) + f (y) ’ f (x + y), Gx,c = f (cx) ’ cf (x). These are continuous functions
on „¦ and
’1
G’1 (0).
Fx,y (0) ©
E= c,x
x,y∈X x∈X
c∈R

Thus E is a closed subset of a compact set, and therefore compact itself.
21

w—
Corollary. If fn ’’ f in X — , then f ¤ lim inf fn
’ X— .
n’∞


Proof. Let C = lim inf fn and let > 0 be arbitrary. Then there exists a subsequence
(also denoted fn ) with fn ¤ C + . The ball of radius C + being weak* compact, and
so weak* closed, f < C + . Since was arbitrary, this gives the result.

On X —— the weak* topology is that induced by the functionals in X — .

Theorem. The unit ball of X is weak* dense in the unit ball of X —— .

Proof. Let z belong to the unit ball of X —— . We need to show that for any f1 , . . . , fn ∈ X —
of norm 1, and any > 0, the set

{ w ∈ X —— | |(w ’ z)(fi )| < , i = 1, . . . , n }

contains a point of the unit ball of X. (Since any neighborhood of z contains a set of this
form.)
It is enough to show that there exists y ∈ X with y < 1 + such that (y ’ z)(fi ) = 0
for each i. Because then y/(1 + ) belongs to the closed unit ball of X, and

|((1 + )’1 y ’ z)(fi )| = |((1 + )’1 y ’ y)(fi )| ¤ ((1 + )’1 y ’ y = y <.
1+

Let S be the span of the fi in X — . Since S is ¬nite dimensional the canonical map
X ’ S — is surjective. (This is equivalent to saying that if the null space of a linear
functional g contains the intersection of the null spaces of a ¬nite set of linear functionals
gi , then g is a linear combination of the gi , which is a simple, purely algebraic result.
[Proof: The nullspace of the map (g1 , . . . , gn ) : X ’ Rn is contained in the nullspace of g,
so g = T —¦ (g1 , . . . , gn ) for some linear T : Rn ’ R.]) Consequently X/a S is isometrically
isomorphic to S.
In particular z|S concides with y + a S for some y ∈ X. Since z ¤ 1, and we can
S—
choose the coset representative y with y ¤ 1 + as claimed.

Corollary. The closed unit ball of a Banach space X is weakly compact if and only if X
is re¬‚exive.

Proof. If the closed unit ball of X is weakly compact, then it is weak* compact when
viewed as a subset of X —— . Thus the ball is weak* closed, and so, by the previous theorem,
the embedding of the the unit ball of X contains the ball of X —— . It follows that the
embedding of X is all of X —— .
The reverse direction is immediate from the Alaoglu theorem.
22

V. Compact Operators and their Spectra

Hilbert“Schmidt operators.

Lemma. Suppose that { ei } and { ei } are two orthonormal bases for a separable Hilbert
˜
space X, and T ∈ B(X). Then

| T ei , ej |2 = | T ei , ej |2 .
˜˜
i,j i,j



| w, ej |2 = w 2 , so
Proof. For all w ∈ X, j


T — ej
| T ei , ej |2 = 2 2
T ei = .
i,j i j


But
T — ei | T — ei , ej |2 =
2 2
= ˜ T ej
˜ .
i i,j j



De¬nition. If T ∈ B(X) de¬ne T by
2

2
| T ei , ej |2 = 2
T = T ei
2
i,j i


where { ei } is any orthonormal basis for X. T is called a Hilbert“Schmidt operator if
T 2 < ∞, and T 2 is called the Hilbert“Schmidt norm of T .

We have just seen that if T is Hilbert“Schmidt, then so is T — and their Hilbert“Schmidt
norms coincide.

T¤T
Proposition. 2.


Proof. Let x = ci ei be an arbitrary element of X. Then
2
2
Tx = cj T ej , ei .
i j


By Cauchy“Schwarz

cj T ej , ei |2 ¤ c2 · | T ej , ei |2 = x 2
| T ej , ei |2 .
| j
j j j j


Summing on i gives the result.
23

Proposition. Let „¦ be an open subset of Rn and K ∈ L2 („¦ — „¦). De¬ne

for all x ∈ „¦.
TK u(x) = K(x, y)u(y) dy,
„¦

Then TK de¬nes a Hilbert“Schmidt operator on L2 („¦) and TK =K L2 .
2

Proof. For x ∈ „¦, set Kx (y) = K(x, y). By Fubini™s theorem, Kx ∈ L2 („¦) for almost all
x ∈ „¦, and
2 2
K = Kx dx.
L2

Now, TK u(x) = Kx , u , so, if { ei } is an orthonormal basis, then

2 2
|(TK ei )(x)|2 dx = | Kx , ei |2 dx
TK = TK ei =
2
i i i

| Kx , ei |2 dx = 2 2
= Kx dx = K L2 .
i




Compact operators.

De¬nition. 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 T has ¬nite rank (dim R(T ) < ∞), then T is compact.
Recall the following characterization of precompact sets in a metric space, which is often
useful.

Proposition. Let M be a metric space. Then the following are equivalent:
(1) M is precompact.
(2) For all > 0 there exist ¬nitely many sets of diameter at most which cover M .
(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 X and Y be Banach spaces and Bc (X, Y ) the space of compact linear
operators from X to Y . Then Bc (X, Y ) is a closed subspace of B(X, Y ).

Proof. Suppose Tn ∈ Bc (X, Y ), T ∈ B(X, Y ), Tn ’ T ’ 0. We must show that T is
compact. Thus we must show that T (E) is precompact in Y , where E is the unit ball in
X. For this, it is enough to show that for any > 0 there are ¬nitely many balls Ui of
radius in Y such that
T (E) ‚ Ui .
i
24

Choose n large enough that T ’ Tn ¤ /2, and let V1 , V2 , . . . , Vn be ¬nitely many balls
of radius /2 which cover Tn E. For each i let Ui be the ball of radius with the same
center as Vi .

It follows that closure of the ¬nite rank operators in B(X, Y ) is contained in Bc (X, Y ).
In general, this may be a strict inclusion, but if Y is a Hilbert space, it is equality. To
prove this, choose an orthonormal basis for Y , and consider the ¬nite rank operators of the
form P T where P is the orthogonal projection of Y onto the span of ¬nitely many basis
elements. Using the fact that T E is compact (E the unit ball of X) and that P = 1, we
can ¬nd for any > 0, an operator P of this form with supx∈E (P T ’ T )x ¤ .
The next result is obvious but useful.

Theorem. Let X and Y be Banach spaces and T ∈ Bc (X, Y ). If Z is another Banach
space and S ∈ B(Y, Z) then ST is compact. If S ∈ B(Z, X), then T S is compact. If
X = Y , then Bc (X) := Bc (X, X) is a two-sided ideal in B(X).

Theorem. Let X and Y be Banach spaces and T ∈ B(X, Y ). Then T is compact if and
only if T — is compact.

Proof. Let E be the unit ball in X and F the unit ball in Y — . Suppose that T is compact.
Given > 0 we must exhibit ¬nitely many sets of diameter at most which cover T — F .
First choose m sets of diameter at most /3 which cover T E, and let T xi belong to the ith
set. Also, let I1 , . . . , In be n intervals of length /3 which cover the interval [’ T , T ].
For any m-tuple (j1 , . . . , jm ) of integers with 1 ¤ ji ¤ n we de¬ne the set

{ f ∈ F | f (T xi ) ∈ Iji , i = 1, . . . , m }.

These sets clearly cover F , so there images under T — cover T — F , so it su¬ces to show that
the images have diameter at most . Indeed, if f and g belong to the set above, and x is any
element of E, pick i such that T x ’ T xi ¤ /3. We know that f (T xi ) ’ g(T xi ) ¤ /3.
Thus

|(T — f ’ T — g)(x)| = |(f ’ g)(T x)|
¤ |f (T x) ’ f (T xi )| + |g(T x) ’ g(T xi )| + |(f ’ g)(T xi )| ¤ .

This shows that T compact =’ T — compact. Conversely, suppose that T — : Y — ’ X —
is compact. Then T —— maps the unit ball of X —— into a precompact subset of Y —— . But the
unit ball of X may be viewed as a subset of the unit ball of its bidual, and the restriction
of T —— to the unit ball of X coincides with T there. Thus T maps the unit ball of X to a
precompact set.

Theorem. If T is a compact operator from a Banach space to itself, then N (1 ’ T ) is
¬nite dimensional and R(1 ’ T ) is closed.

Proof. T is a compact operator that restricts to the identity on N (1 ’ T ). Hence the
closed unit ball in N (1 ’ T ) is compact, whence the dimension of N (1 ’ T ) is ¬nite.
25

Now any ¬nite dimensional subspace is complemented (see below), so there exists a
closed subspace M of X such that N (1 ’ T ) + M = X and N (1 ’ T ) © M = 0. Let
S = (1 ’ T )|M , so S is injective and R(S) = R(1 ’ T ). We will show that for some
c > 0, Sx ≥ c x for all x ∈ M , which will imply that R(S) is closed. If the desired
inequality doesn™t hold for any c > 0, we can choose xn ∈ M of norm 1 with Sxn ’ 0.
After passing to a subsequence we may arrange that also T xn converges to some x0 ∈ X.
It follows that xn ’ x0 , so x0 ∈ M and Sx0 = 0. Therefore x0 = 0 which is impossible
(since xn = 1).

In the proof we used the ¬rst part of the following lemma. We say that a closed
subspace N is complemented in a Banach space X if there is another closed subspace such
that M • N = X.

Lemma. A ¬nite dimensional or ¬nite codimensional closed subspace of a Banach space
is complemented.

Proof. If M is a ¬nite dimensional subspace, choose a basis x1 , . . . , xn and de¬ne a linear
functionals φi : M ’ R by φi (xj ) = δij . Extend the φi to be bounded linear functionals
on X. Then we can take N = N (φ1 ) © . . . © N (φn ).
If M is ¬nite codimensional, we can take N 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. If T is a compact operator from a Banach space to itself, » a non-zero complex
number, and n a positive integer, then N [(»1’T )n ] is ¬nite dimensional and R[(»1’T )n ]
is closed.

Proof. Expanding we see that (»1 ’ T )n = »n (1 ’ S) for some compact operator S, 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 l2 for which the matrix entries are square-
summable.

Theorem. A Hilbert“Schmidt operator on a separable Hilbert space is compact.

Proof. Let { ei } be an orthonormal basis. Let T be a given Hilbert“Schmidt operator
2
< ∞). De¬ne Tn by Tn ei = T ei if i ¤ n, Tn ei = 0 otherwise. Then
(so i T ei

T ’ Tn ¤ T ’ Tn 2 = i=n+1 T ei 2 ’ 0.
26

Spectral Theorem for compact self-adjoint operators. In this section we assume
that X is a complex Hilbert space. If T : X ’ X is a bounded linear operator, we view T —
as a map from X ’ X via the Riesz isometry between X and X — . That is, T — is de¬ned
by
T — x, y = x, T y .
In the case of a ¬nite dimensional complex Hilbert space, T can be represented by a
complex square matrix, and T — 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 X = L2 ([0, 1]), and de¬ne T u(x) = x u(x) for
u ∈ L2 . Then T is clearly bounded and self-adjoint. But it is easy to see that T does not
have any eigenvalues.

Spectral Theorem for Compact Self-Adjoint Operators in Hilbert Space. Let
T be a compact self-adjoint operator in a Hilbert space X. Then there is an orthonormal
basis consisting of eigenvectors of T .

Before proceeding to the proof we prove one lemma.

Lemma. If T is a self-adjoint operator on a Hilbert space, then

T = sup | T x, x |.

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