Functional Analysis

36 Pages·1991·0.295 MB·English

by  Arnold D.N.

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
1These 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.
2Department of Mathematics, Penn State University, University Park, PA 16802.
Email: [email protected]. Web: http://www.math.psu.edu/dna/.
1
2
I. Vector spaces and their topology
Basic definitions: (1) Norm and seminorm on vector spaces (real or complex). A norm
defines a Hausdorff 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
definite, symmetric bilinear form on X×X → R. In the complex case it is positive definite,
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
(x,y) = ((cid:107)x+y(cid:107)2 −(cid:107)x−y(cid:107)2).
4
For complex spaces it is
1
(x,y) = ((cid:107)x+y(cid:107)2 +i(cid:107)x+iy(cid:107)2 −(cid:107)x−y(cid:107)2 −i(cid:107)x−iy(cid:107)2).
4
In inner product spaces we also have the parallelogram law:
(cid:107)x+y(cid:107)2 +(cid:107)x−y(cid:107)2 = 2((cid:107)x(cid:107)2 +(cid:107)y(cid:107)2).
This gives a criterion for a normed space to be an inner product space. Any norm coming
from an inner product satisfies the parallelogram law and, conversely, if a norm satisfies the
parallelogram law, we can show (but not so easily) that the polarization identity defines
an inner product, which gives rise to the norm.
(3) A topological vector space is a vector space endowed with a Hausdorff topology such
thatthealgebraicoperationsarecontinuous. NotethatwecanextendthenotionofCauchy
sequence, and therefore of completeness, to a TVS: a sequence x in a TVS is Cauchy if
n
for every neighborhood U of 0 there exists N such that x −x ∈ U for all m,n ≥ N.
m 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 {(cid:107)·(cid:107) } of seminorms on X is given which are sufficient in the sense that
α α∈A
{(cid:107)x(cid:107) = 0} = 0. Then the topology generated by the sets {(cid:107)x(cid:107) < r}, α ∈ A, r > 0,
α α α
m(cid:84)akes X a TVS. A sequence (or net) x converges to x iff (cid:107)x −x(cid:107) → 0 for all α. Note
n n α
that, a fortiori, |(cid:107)x (cid:107) −(cid:107)x(cid:107) | → 0, showing that each seminorm is continuous.
n α α
If the number of seminorms is finite, we may add them to get a norm generating the
same topology. If the number is countable, we may define a metric
(cid:107)x−y(cid:107)
d(x,y) = 2−n n ,
(cid:88) 1+(cid:107)x−y(cid:107)
n
n
3
so the topology is metrizable.
Examples: (0) On Rn or Cn we may put the l norm, 1 ≤ p ≤ ∞, or the weighted
p
l norm with some arbitrary positive weight. All of these norms are equivalent (indeed
p
all norms on a finite dimensional space are equivalent), and generate the same Banach
topology. Only for p = 2 is it a Hilbert space.
(2) If Ω is a subset of Rn (or, more generally, any Hausdorff space) we may define the
space C (Ω) of bounded continuous functions with the supremum norm. It is a Banach
b
space. If X is compact this is simply the space C(Ω) of continuous functions on Ω.
(3) For simplicity, consider the unit interval, and define Cn([0,1]) and Cn,α([0,1]),
n ∈ N, α ∈ (0,1]. Both are Banach spaces with the natural norms. C0,1 is the space of
Lipschitz functions. C([0,1]) ⊂ C0,α ⊂ C0,β ⊂ C1([0,1]) if 0 < α ≤ β ≤ 1.
(4) For 1 ≤ p < ∞ and Ω an open or closed subspace of Rn (or, more generally, a σ-finite
measure space), we have the space Lp(Ω) of equivalence classes of measurable p-th power
integrable functions (with equivalence being equality off a set of measure zero), and for
p = ∞ equivalence classes of essentially bounded functions (bounded after modification
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(Ω) < ∞, then Lp(Ω) ⊂ Lq(Ω) if 1 ≤ q ≤ p ≤ ∞.
(5) The sequence space l , 1 ≤ p ≤ ∞ is an example of (4) in the case where the
p
measure space is N with the counting measure. Each is a Banach space. l is a Hilbert
2
space. l ⊂ l if1 ≤ p ≤ q ≤ ∞(notetheinequalityisreversedfromthepreviousexample).
p q
The subspace c of sequences tending to 0 is a closed subspace of l .
0 ∞
(6) If Ω is an open set in Rn (or any Hausdorff space), we can equip C(Ω) with the
norms f (cid:55)→ |f(x)| indexed by x ∈ Ω. 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 Ω is an open set in Rn we can equip C(Ω) with the norms f (cid:55)→ (cid:107)f(cid:107) indexed
L∞(K)
by compact subsets of Ω, thus defining the topology of uniform convergence on compact
subsets. We get the same toplogy by using only the countably many compact sets
K = {x ∈ Ω : |x| ≤ n, dist(x,∂Ω) ≥ 1/n}.
n
The topology is complete.
(8) In the previous example, in the case Ω is a region in C, and we take complex-
valued functions, we may consider the subspace H(Ω) 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
(cid:90) f(x)φ(x)dx = −(cid:90) g(x)φ(cid:48)(x)dx,
0 0
4
for all infinitely differentiable φ with support contained in I (so φ is identically zero near
(cid:48)
0 and 1), then we say that f is weakly differentiable and that f = g. We can then define
the Sobolev space W1(I) = {f ∈ Lp(I) : f(cid:48) ∈ Lp(I)}, with the norm
p
1 1 1/p
(cid:107)f(cid:107)W1(I) = (cid:18)(cid:90) |f(x)|pdx+(cid:90) |f(cid:48)(x)|pdx(cid:19) .
p
0 0
This is a larger space than C1(I¯), but still incorporates first order differentiability of f.
The case p = 2 is particularly useful, because it allows us to deal with differentiability
in a Hilbert space context. Sobolev spaces can be extended to measure any degree of
differentiability (even fractional), and can be defined on arbitrary domains in Rn.
Subspaces and quotient spaces.
If X is a vector space and S a subspace, we may define the vector space X/S of cosets.
If X is normed, we may define
(cid:107)u(cid:107) = inf (cid:107)x(cid:107) , or equivalently (cid:107)x¯(cid:107) = inf (cid:107)x−s(cid:107) .
X/S X X/S X
x∈u s∈S
This is a seminorm, and is a norm iff 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 x is a sequence of elements of X for which the cosets x¯ are Cauchy.
n n
We can take a subsequence with (cid:107)x¯ −x¯ (cid:107) ≤ 2−n−1, n = 1,2,.... Set s = 0, define
n n+1 X/S 1
s ∈ S such that (cid:107)x −(x +s )(cid:107) ≤ 1/2, define s ∈ S such that (cid:107)(x +s )−(x +s )(cid:107) ≤
2 1 2 2 X 3 2 2 3 3 X
1/4, ... . Then {x +s } is Cauchy in X ... (cid:3)
n n
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 finite 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 → V and it is continuous. But it is just the projection
S+V
onto V. The projection onto S is id−µ◦π, so it is also continuous. (cid:3)
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 S be the vector space of all real sequences
1
∞
(x ) for which x = 0 if n is odd, and S be the sequences for which x = nx ,
n n=1 n 2 2n 2n−1
n = 1,2,.... Clearly X = l ∩S and X = l ∩S are closed subspaces of l , the space
1 2 1 2 2 2 2
of square integrable sequences (they are defined 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 S and S :
1 2
(x ,x ,...) = (0,x −x ,0,x −2x ,0,x −3x ,...)+(x ,x ,x ,2x ,x ,3x ,...).
1 2 2 1 4 3 6 5 1 1 3 3 5 5
If a sequence has all but finitely many terms zero, so do the two summands. Thus all
such sequences belong to X +X , showing that X +X is dense in l . Now consider the
1 2 1 2 2
sequence (1,0,1/2,0,1/3,...) ∈ l . Its only decomposition as elements of S and S is
2 1 2
(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 X +X . Thus X +X is not closed in l .
1 2 1 2 2
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
(cid:107)x−x¯(cid:107) = inf (cid:107)x−y(cid:107).
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 (cid:107)y(cid:107) and chose x ∈ K with (cid:107)x (cid:107) → d.
y∈K n n
Then the parallelogram law gives
x −x 2 1 1 x +x 2 1 1
(cid:13) n m(cid:13) = (cid:107)x (cid:107)2 + (cid:107)x (cid:107)2 −(cid:13) n m(cid:13) ≤ (cid:107)x (cid:107)2 + (cid:107)x (cid:107)2 −d2,
n m n m
(cid:13) 2 (cid:13) 2 2 (cid:13) 2 (cid:13) 2 2
(cid:13) (cid:13) (cid:13) (cid:13)
(cid:13) (cid:13) (cid:13) (cid:13)
where we have used convexity to infer that (x + x )/2 ∈ K. Thus x is a Cauchy
n m n
sequence and so has a limit x¯, which must belong to K, since K is closed. Since the norm
is continuous, (cid:107)x¯(cid:107) = lim (cid:107)x (cid:107) = d.
n n
For uniqueness, note that if (cid:107)x¯(cid:107) = (cid:107)x˜(cid:107) = d, then (cid:107)(x¯+x˜)/2(cid:107) = d and the parallelogram
law gives
(cid:107)x¯−x˜(cid:107)2 = 2(cid:107)x¯(cid:107)2 +2(cid:107)x˜(cid:107)2 −(cid:107)x¯+x˜(cid:107)2 = 2d2 +2d2 −4d2 = 0. (cid:3)
The unique nearest element to x in K is often denoted P x, and referred to as the
K
projection of x onto K. It satisfies P ◦ P = P , the definition of a projection. This
K K K
terminology is especially used when K is a closed linear subspace of X, in which case P
K
is a linear projection operator.
6
Projection and orthogonality. If S is any subset of a Hilbert space X, let
S⊥ = {x ∈ X : (cid:104)x,s(cid:105) = 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 P x the projection of x onto S, then
S
x−P x ∈ S⊥. Indeed, if s ∈ S is arbitrary and t ∈ R, then
S
(cid:107)x−P x(cid:107)2 ≤ (cid:107)x−P x−ts(cid:107)2 = (cid:107)x−P x(cid:107)2 −2t(x−P x,s)+t2(cid:107)s(cid:107)2,
S S S S
so the quadratic polynomial on the right hand side has a minimum at t = 0. Setting the
derivative there to 0 gives (x−P x,s) = 0.
S
Thus we can write any x ∈ X as s + s⊥ with s ∈ S and s⊥ ∈ S⊥ (namely s = P x,
S
s⊥ = x−P x). Such a decomposition is certainly unique (if s¯+s¯⊥ were another one we
S
would have s−s¯= s¯⊥ −s⊥ ∈ S ∩S⊥ = 0.) We clearly have (cid:107)x(cid:107)2 = (cid:107)s(cid:107)2 +(cid:107)s⊥(cid:107)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 −P )x+P x
S S
is the (unique) decomposition of x into elements of S⊥ and S⊥⊥. Thus P = I−P . For
S⊥ S
⊥⊥
any subset S of X, S is the smallest closed subspace containing S.
Orthonormal sets and bases in Hilbert space.
Let e , e , ... , e be orthonormal elements of a Hilbert space X, and let S be their
1 2 N
span. Then (cid:104)x,e (cid:105)e ∈ S and x − (cid:104)x,e (cid:105)e ⊥ S, so (cid:104)x,e (cid:105)e = P x. But
n n n n n n n n n S
(cid:80) (cid:80) (cid:80)
(cid:107) (cid:104)x,e (cid:105)e (cid:107)2 = N (cid:104)x,e (cid:105)2, so
n n n n=1 n
(cid:80) (cid:80)
N
(cid:104)x,e (cid:105)2 ≤ (cid:107)x(cid:107)2
(cid:88) n
n=1
(Bessel’s inequality). Now let E be an orthonormal set of arbitrary cardinality. It follows
from Bessel’s inequality that for (cid:15) > 0 and x ∈ X, {e ∈ E : (cid:104)x,e(cid:105) ≥ (cid:15)} is finite, and
hence that {e ∈ E : (cid:104)x,e(cid:105) > 0} is countable. We can thus extend Bessel’s inequality to
an arbitrary orthonormal set:
(cid:104)x,e(cid:105)2 ≤ (cid:107)x(cid:107)2,
(cid:88)
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
((cid:63)) f(e) = x
(cid:88)
e∈E
7
if the net f(e), indexed by the finite subsets F of E, converges to x. In other words,
e∈F
((cid:63)) holds (cid:80)if, for any neighborhood U of the origin, there is a finite set F ⊂ E such that
0
x− f(e) ∈ U whenever F is a finite subset of E containing F . In the case E = N,
e∈F 0
(cid:80)
this is equivalent to absolute convergence of a series. Note that if f(e) converges,
e∈E
then for all (cid:15) there is a finite F such that if F and F are finite s(cid:80)upersets of F , then
0 1 2 0
(cid:107) f(e)− f(e)(cid:107) ≤ (cid:15). It follows easily that each of the sets {e ∈ E |(cid:107)f(e)(cid:107) ≥
e∈F e∈F
1/(cid:80)n} is1finite, a(cid:80)nd hen2ce, 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
(cid:104)x,e(cid:105)e
(cid:88)
e∈E
converges.
Proof. We may order the elements e , e , ... of E for which (cid:104)x,e(cid:105) =(cid:54) 0. Note that
1 2
N N
(cid:107) (cid:104)x,e (cid:105)e (cid:107)2 = |(cid:104)x,e (cid:105)|2 ≤ (cid:107)x(cid:107)2.
(cid:88) n n (cid:88) n
n=1 n=1
This shows that the partial sums s = N (cid:104)x,e (cid:105)e form a Cauchy sequence, and so
N n=1 n n
converge to an element ∞ (cid:104)x,e (cid:105)e o(cid:80)f X. As an exercise in applying the definition,
n=1 n n
we show that (cid:104)x,e(cid:80)(cid:105)e = ∞ (cid:104)x,e (cid:105)e . Given (cid:15) > 0 pick N large enough that
e∈E n=1 n n
∞ |(cid:104)x,e (cid:80)(cid:105)|2 < (cid:15). If M >(cid:80)N and F is a finite subset of E containing e , ... , e ,
n=N+1 n 1 N
(cid:80)
then
M
(cid:107) (cid:104)x,e (cid:105)e − (cid:104)x,e(cid:105)e(cid:107)2 ≤ (cid:15).
(cid:88) n n (cid:88)
n=1 e∈F
Letting M tend to infinity,
∞
(cid:107) (cid:104)x,e (cid:105)e − (cid:104)x,e(cid:105)e(cid:107)2 ≤ (cid:15),
(cid:88) n n (cid:88)
n=1 e∈F
as required. (cid:3)
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 unionis 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 infinite dimensional spaces they are
difficult 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 infinite 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 definition this is a maximal
orthonormalset. By Zorn’s lemma, any orthonormalset 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 = (cid:104)x,e(cid:105)e.
(cid:88)
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 e ∈ E is (cid:104)x,e (cid:105). Thus y := x− (cid:104)x,e(cid:105)e is orthogonal to E, and if it
0 0 e∈E
weren’t zero, then we could adjoin y/(cid:107)y(cid:107) to E to g(cid:80)et a larger orthonormal set.
Thus we’ve shown that any element x of X can be expressed as c e for some c ∈ R,
e e
(cid:80)
all but countablymany of which are 0. It is easily seen that this determines the c uniquely,
e
namely c = (cid:104)x,e(cid:105), and that (cid:107)x(cid:107)2 = c2.
e e
(cid:80)
The notion of orthonormal basis allows us to define a Hilbert space dimension, namely
thecardinalityofanyorthonormalbasis. Toknowthatthisiswelldefined, weneedtocheck
that any two bases have the same cardinality. If one is finite, this is trivial. Otherwise,
let E and F be two infinite orthonormal bases. For each 0 (cid:54)= x ∈ X, the inner product
(cid:104)x,e(cid:105) =(cid:54) 0 for at least one e ∈ E. Thus
F ⊂ {f ∈ F : (cid:104)f,e(cid:105) =(cid:54) 0},
(cid:91)
e∈E
i.e., F is contained in the union of cardE countable sets. Therefore cardF ≤ ℵ cardE =
0
cardE.
IfS isanyset, wedefineaparticularHilbertspacel2(S)asthesetoffunctionsc : S → R
whicharezerooffacountablesetandsuchthat c2 < ∞. Wethusseethatviaabasis,
s∈S s
(cid:80)
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 infinite
dimensional separable Hilbert space (up to isometry).
Example: The best known example of an orthonormal basis in an infinite Hilbert space
is the set of functions e = exp(2πinθ) which form a basis for complex-valued L2([0,1]).
n
(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
∞
f(θ) = fˆ(n)e2πinθ,
(cid:88)
n=−∞
with fˆ(n) = (cid:104)f,e (cid:105) = 1f(θ)e−2πinθdθ. Thus from the Hilbert space point of view, the
n (cid:82)0
theory of Fourier series is rather simple. More difficult analysis comes in when we consider
convergence in other topologies (pointwise, uniform, almost everywhere, Lp, C1, ... ).
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 = c e. Schauder constructed
e∈E e
(cid:80)
a useful Schauder basis for C([0,1]), and there is useful Schauder bases in many other
separable Banach spaces. In 1973 Per Enflo 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 iff it is bounded on every ball iff it is bounded on some ball iff it is
continuous at every point iff 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
(cid:107)Tx(cid:107)
(cid:107)T(cid:107) = sup Y .
B(X,Y) (cid:107)x(cid:107)
0(cid:54)=x∈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 T is a Cauchy sequence in B(X,Y). Then for each x ∈ X T x is Cauchy
n n
in the complete space Y, so there exists Tx ∈ Y with T x → Tx. Clearly T : X → Y is
n
linear. Is it bounded? The real sequence (cid:107)T (cid:107) is Cauchy, hence bounded, say (cid:107)T (cid:107) ≤ K.
n n
It follows that (cid:107)T(cid:107) ≤ K, and so T ∈ B(X,Y). To conclude the proof, we need to show
that (cid:107)T −T(cid:107) → 0. We have
n
(cid:107)T −T(cid:107) = sup (cid:107)T x−Tx(cid:107) = sup lim (cid:107)T x−T x(cid:107)
n n n m
(cid:107)x(cid:107)≤1 (cid:107)x(cid:107)≤1m→∞
= sup limsup(cid:107)T x−T x(cid:107) ≤ limsup(cid:107)T −T (cid:107).
n m n m
(cid:107)x(cid:107)≤1 m→∞ m→∞
Thus limsup (cid:107)T −T(cid:107) = 0. (cid:3)
n→∞ n
If T ∈ B(X,Y) and U ∈ B(Y,Z), then UT = U ◦ T ∈ B(X,Z) and (cid:107)UT(cid:107) ≤
B(X,Z)
(cid:107)U(cid:107) (cid:107)T(cid:107) . In particular, B(X) := B(X,X) is a Banach algebra, i.e., it has an
B(Y,Z) B(X,Y)
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(αx) = αp(x) for x,y ∈ X, α ≥ 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 suffices to extend f to the space spanned by S and one element x ∈ X \ S,
0
preserving the inequality, since if we can do that we can complete the proof with Zorn’s
lemma.
We need to define f(x ) such that f(tx +s) ≤ p(tx +s) for all t ∈ R, s ∈ S. The case
0 0 0
t = 0 is known and it is easy to use homogeneity to restrict to t = ±1. Thus we need to
find a value f(x ) ∈ R such that
0
f(s)−p(−x +s) ≤ f(x ) ≤ p(x +s)−f(s) for all s ∈ S.
0 0 0
Now it is easy to check that for any s , s ∈ S, f(s )−p(−x +s ) ≤ p(x +s )−f(s ),
1 2 1 0 1 0 2 2
and so such an f(x ) exists. (cid:3)
0
Corollary. If X is a normed linear space and x ∈ X, then there exists f ∈ X∗ of norm
1 such that f(x) = (cid:107)x(cid:107).
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) = (cid:107)x¯(cid:107) .
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(Tx) for all f ∈ Y∗, x ∈ X. In particular, if
T ∈ B(X,Y), then T∗ ∈ B(Y∗,X∗). In fact, (cid:107)T∗(cid:107) = (cid:107)T(cid:107) . To prove
B(Y∗,X∗) B(X,Y)
this, note that |T∗f(x)| = |f(Tx)| ≤ (cid:107)f(cid:107)(cid:107)T(cid:107)(cid:107)x(cid:107). Therefore (cid:107)T∗f(cid:107) ≤ (cid:107)f(cid:107)(cid:107)T(cid:107), so T∗
is indeed bounded, with (cid:107)T∗(cid:107) ≤ (cid:107)T(cid:107). Also, given any y ∈ Y, we can find g ∈ Y∗
such that |g(y)| = (cid:107)y(cid:107), (cid:107)g(cid:107) = 1. Applying this with y = Tx (x ∈ X arbitrary), gives
(cid:107)Tx(cid:107) = |g(Tx)| = |T∗gx| ≤ (cid:107)T∗(cid:107)(cid:107)g(cid:107)(cid:107)x(cid:107) = (cid:107)T∗(cid:107)(cid:107)x(cid:107). This shows that (cid:107)T(cid:107) ≤ (cid:107)T∗(cid:107). Note
that if T ∈ B(X,Y), U ∈ B(Y,Z), then (UT)∗ = T∗U∗.
If X is a Banach space and S a subset, let
Sa = {f ∈ X∗|f(s) = 0 ∀s ∈ S}