Download FUNCTIONAL ANALYSIS 1. Topological Vector Spaces Definition 1

FUNCTIONAL ANALYSIS
HUIQIANG JIANG
1. Topological Vector Spaces
Definition 1. A vector space X with topology τ is a called a topological vector
space if
(a). single point sets are closed,
(b). the vector space operations are continuous.
Given a ∈ X, the translation operator Ta : x 7−→ x + a is a homeomorphism of
X onto itself. Hence the vector topology τ is translation invariant.
Similarly, a metric d on a vector space X is called invariant if d (x + z, y + z) =
d (x, y) for all x, y, z ∈ X.
Theorem 1. Every topological vector space is a Hausdorff space.
Proof. We only need to separate 0 and a ∈ X with a 6= 0. Let W = X\ {a}
which is a neighborhood of 0. Since the addition is continuous, there exists V1 , V2 ,
neighborhoods of 0, such that V1 + V2 ⊂ W . Then a − V2 is a neighborhood of a
and V1 ∩ (a − V2 ) = ∅.
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Using a similar proof, one can show
Theorem 2. Suppose K and C are disjoint subsets of a topological vector space,
K is compact and C is closed. Then 0 has a neighborhood V such that (K + V ) ∩
(C + V ) = ∅.
Theorem 3. Let (X, τ ) be a topological vector space. Every neighborhood of 0
contains a balanced neighborhood of 0.
Proof. Let 0 ∈ U ∈ τ . Since scalar multiplication is continuous, ∃δ > 0, and
0 ∈ V ∈ τ such that αV ⊂ U whenever |α| < δ. Let W = ∪|α|<δ αV , then W is
balanced and W ⊂ U .
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Remark 1. Every convex neighborhood of 0 contains a balanced convex neighborhood of 0.
A subset E of a topological vector space is said to be bounded if for any neighborhood V of 0 in X, there exists s > 0 such that E ⊂ tV for every t > s.
Definition 2. A local base of (X, τ ) is a collection B of neighborhoods of 0 s.t.
every neighborhood of 0 contains a member of B.
Theorem 4. Let (X, τ ) be a topological vector space and 0 ∈ V ∈ τ .
(a). If 0 < r1 < r2 < · · · and rn → ∞ as n → ∞, then
∞
[
X=
rn V.
n=1
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HUIQIANG JIANG
(b). Every compact subset K of X is bounded.
(c). If δ1 > δ2 > · · · and δn → 0 as n → ∞ and if V is bounded, then the collection
{δn V : n = 1, 2, 3, · · · }
is a local base for X.
1.1. Linear Mappings. Let X, Y be vector spaces over K. A mapping Λ : X → Y
is said to be linear if
Λ (αx1 + βx2 ) = αΛ (x1 ) + βΛ (x2 )
holds for any x1 , x2 ∈ X and α, β ∈ K. Linear mappings of X to K are called linear
functionals.
The set Λ−1 ({0}) = N (Λ) is a subspace of X, called the null space of Λ. The
following theorem is a direct consequence of the definition of linear mappings.
Theorem 5. Let X, Y be topological vector spaces. If Λ : X → Y is linear and
continuous at 0, then Λ is uniformly continuous in the following sense: For every
0 ∈ W ∈ τY , there exists 0 ∈ V ∈ τX such that
y − x ∈ V implies Λy − Λx ∈ W.
Theorem 6. Let Λ be a linear functional on a topological vector space X. Assume
Λx 6= 0 for some x ∈ X. Then the following four properties are equivalent:
(a) Λ is continuous.
(b) N (Λ) is closed.
(c) N (Λ) is not dense in X.
(d) Λ is bounded in some neighborhood V of 0.
Proof. (a)⇒(b)⇒(c) is trivial. Assuming (c) holds, there exists open set V ⊂
c
(N (Λ)) . Hence, there exists x ∈ X and a balanced neighborhood V of 0 such that
(x + V ) ∩ N (Λ) = ∅.
If Λ is unbounded in V , then ΛV = K and so Λy = −Λx for some y ∈ V , and hence
x + y ∈ N (Λ) which is a contradiction. Assuming (d) holds, then |Λx| < M for all
x ∈ V , hence for any ε > 0, we have
|Λx| < ε
for any x ∈
ε
MV
. So Λ is continuous at the origin and hence continuous.
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1.2. Finite Dimensional Spaces. If X is an n-dimensional topological vector
space over K, then every basis of X induces an isomorphism of X onto Kn .
Theorem 7. Any isomorphism of Kn onto n-dimensional topological vector space
over K is a homeomorphism.
Theorem 8. Every locally compact topological vector space X has finite dimension.
1.3. Metrizable Spaces. A topological vector space (X, τ ) is said to be metrizable
if there is a metric on d which is compatible with τ . A metrizable topological vector
space has a countable local base and the inverse is also true.
FUNCTIONAL ANALYSIS
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Theorem 9. If X is a topological vector space with a countable local base, then
there is a metric d on X such that
(a). d is compatible with the topology of X,
(b). the open balls centered at 0 are balanced, and
(c). d is invariant.
If in addition, X is locally convex, then d can be chosen so as to satisfy (abc) and
also
(d). all open balls are convex.
Remark 2. A sequence {xn } in a metrizable topological vector space X is a dCauchy sequence if and only if it is a τ -Cauchy sequence.
A linear mapping Λ : X → Y is said to be bounded if Λ maps bounded sets into
bounded sets.
Theorem 10. Suppose X and Y are topological vector spaces, X is metrizable and
Λ : X → Y is linear. Then the following four properties are equivalent:
(a) Λ is continuous.
(b) Λ is bounded.
(c) If xn → 0, then {Λxn } is bounded.
(d) If xn → 0, then Λxn → 0.
Proof. (a) ⇒ (b): Let E be a bounded set in X and 0 ∈ W ∈ τY . Since Λ is
continuous, there exists 0 ∈ V ∈ τX such that ΛV ⊂ W . Now since E is bounded,
we have E ⊂ tV for large t, and hence Λ (E) ⊂ tΛV ⊂ tW for large t and hence
Λ (E) is bounded in Y .
(b) ⇒ (c): If xn → 0, we only need to check that {xn } is bounded. For any
balanced 0 ∈ W ∈ τX , there exists N such that xn ∈ W for n > N . Now since
X = ∪ (kW ), there exists K ≥ 1 such that xk ∈ KW for any 1 ≤ k ≤ N . And
hence {xn } ⊂ KW and so {xn } is bounded.
(c) ⇒ (d): If xn → 0, then there exists positive scalar γn → ∞ such that
γn xn → 0. Now (c) implies {Λ (γn xn )} is bounded, and hence for any 0 ∈ W ∈ τY
balanced, there exists t > 0, Λ (γn xn ) ∈ tW , i.e.,
t
Λxn ∈
W ⊂W
γn
if γn > t, hence Λxn → 0.
(d) ⇒ (a): If Λ is not continuous at 0, then there is 0 ∈ W ∈ τY , such that
Λ−1 (W ) contains no nbhd of 0 in X. Hence there exists xn → 0 such that Λxn 6∈ W
which contradicts (d).
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1.4. Seminorms and Local Convexity.
Definition 3. A seminorm on a vector space X is a real-valued function p on X
such that
(a) p (x + y) ≤ p (x) + p (y) and
(b) p (αx) = |α| p (x) for all x, y ∈ X and α ∈ K.
A family P of seminorms on X is said to be separating if to each x 6= 0, corresponds at least one p ∈ P with p (x) 6= 0.
Definition 4. A subset A ⊂ X is said to be absorbing if ∪t>0 (tA) = X.
Hence, any neighborhood of 0 in a topological vector space is absorbing.
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HUIQIANG JIANG
Definition 5. Let A ⊂ X be absorbing. The Minkowski functional µA of A is
defined by
©
ª
µA (x) = inf t > 0 : t−1 x ∈ A .
The connection between seminorm and Minkowski functional can be seen in the
next two theorems.
Theorem 11. Suppose p is a seminorm on a vector space X. Then
(a) p (0) = 0.
(b) |p (x) − p (y)| ≤ p (x − y).
(c) p (x) ≥ 0.
(d) {x : p (x) = 0} is a subspace of X.
(e) The set B = {x : p (x) < 1} is convex, balanced, absorbing and p = µB .
Proof. (a) p (0) = 0 · p (0) = 0.
(b) p (x) ≤ p (y) + p (x − y), hence p (x) − p (y) ≤ p (x − y), similarly, p (y) −
p (x) ≤ p (y − x) = p (x − y).
(c) Follows from (b) with y = 0.
(d) If p (x) = p (y) = 0 and α, β ∈ K, then
0 ≤ p (αx + βy) ≤ p (αx) + p (βy) = 0.
(e) It is easy to verify that
¡ B is¢ convex, balanced and absorbing. Now for x ∈ X
and s > p (x), we have p s−1 x < 1 which implies s−1 x ∈ B, hence µB (x) ≤ s
and
¡ we¢have µB (x) ≤ p (x). On the other hand, for any 0 < t ≤ p (x), we have
p t−1 x ≥ 1 so t−1 x 6∈ B, hence µB (x) ≥ p (x).
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Theorem 12. Suppose A ⊂ X is convex, balanced and absorbing, then µA is a
seminorm.
Proof. We need only to check subadditivity. Let x, y ∈ X. Foe any ε > 0, define
t = µA (x) + ε and s = µA (y) + ε, then xt and ys ∈ A, hence
x+y
t x
s y
=
+
∈ A.
s+t
s+t t
s+ts
So we have µA (x + y) ≤ s + t. Since ε can be arbitrary, we have µA (x + y) ≤
µA (x) + µA (y).
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If P is a separating family of seminorms on a vector space X, then P defines a
locally convex topology on X with the property that every p ∈ P is continuous.
Theorem 13. Suppose P is a separating family of seminorms on a vector space
X. Associate to each p ∈ P and each n ∈ N, the set
¾
½
1
.
V (p, n) = x : p (x) <
n
Let B be the collection of all finite intersections of the sets V (p, n). Then B is a
convex balanced local base for a topology τ on X which turns X into a locally convex
space such that
(a) every p ∈ P is continuous, and
(b) a set E ⊂ X is bounded iff every p ∈ P is bounded on E.
FUNCTIONAL ANALYSIS
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∞
Remark 3. If P = {pi }i=1 is a countable separating family of seminorms on a
vector space X. Then P induces a topology with a countable local base, hence τ is
metrizable. Indeed, an invariant metric can be defined: for any x, y ∈ X,
d (x, y) = max
i
ci pi (x − y)
1 + pi (x − y)
where {ci } is some fixed sequence of positive numbers which converges to 0 as i →
∞. One can verify that d is compatible with τ .
Theorem 14. A topological vector space is normable iff the origin has a convex
bounded neighborhood.
Proof. ”⇒” Trivial.
”⇐” If the origin has a convex bounded neighborhood, it also has a convex
balanced bounded neighborhood which we denote by U . Define for each x ∈ X,
kxk = µU (x) .
If x 6= 0, then x 6∈ rU for some r > 0, hence kxk ≥ r. So k·k is indeed a norm.
Since {x : kxk < r} = rU , the norm topology coincides with the given one.
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1.5. Spaces C (Ω) and C ∞ (Ω). Let Ω ⊂ Rn be a nonempty open set and let Ki
be compact sets such that Ki lies in the interior of Ki+1 and Ω = ∪Ki . For each
k ∈ N, we can define a seminorm on C (Ω) by
pk (f ) = sup {|f (x)| : x ∈ Kn } .
Then the countable family {pk } is separating, hence {pk } induces a topology for
C (Ω).
Similarly, For each k ∈ N, a seminorm pk on C ∞ (Ω) can be defined by
pk (f ) = max {|Dα f (x)| : x ∈ Kk , |α| ≤ k} .
The countable family {pk } is separating, hence {pk } induces a topology for C ∞ (Ω).
Definition 6. X is said to be an F -space if its topology τ is induced by a complete
invariant metric d. X is said to be a Fréchet space if X is a locally convex F -space.
Theorem 15. C (Ω)and C ∞ (Ω) are Fréchet spaces. Moreover, C ∞ (Ω) has the
Heine-Borel property, i.e., every bounded closed subset is compact.
Proof. Let X be C (Ω) or C ∞ (Ω). For any x, y ∈ X, let
d (x, y) = max
i
ci pi (x − y)
1 + pi (x − y)
where {ci } is some fixed sequence of positive numbers which converges to 0 as
i → ∞. Then d is an invariant metric which is complete. Since X is locally convex
with this topology, X is a Fréchet space. It is an excercise to show that C ∞ (Ω)
has the Heine-Borel property.
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HUIQIANG JIANG
1.6. Normed Spaces and Linear Operators. Let X and Y be two normed
spaces. We use L (X, Y ) to denote linear operators between X and Y . A linear
operator L ∈ L (X, Y ) is bounded if there is a constant C > 0 such that
kLxkY ≤ C kxkX
holds for all x ∈ X. We use B (X, Y ) to denote all bounded linear operators between
X and Y . From Theorem 10, a linear operator L ∈ L (X, Y ) is bounded if and only
if it is continuous.
We define for each L ∈ B (X, Y ), the operator norm
kLk = sup kLxkY .
kxk≤1
Then for each x ∈ X,
kLxkY ≤ kLk kxkX .
Theorem 16. B (X, Y ) equipped with the operator norm is a normed space. And
if Y is a Banach space, then B (X, Y ) is also a Banach space.
Proof. It is standard to verify that B (X, Y ) is a normed space. Now if Y is a
Banach space and {Ln } ⊂ B (X, Y ) is a Cauchy sequence, then for each x ∈ X,
since
kLn x − Lm xk ≤ kLn − Lm k kxk ,
{Ln x} ⊂ Y is a Cauchy sequence. From the completness of Y , there exists Lx :=
limn→∞ Ln x. One can verify that L is a linear operator. Now since {Ln } ⊂ B (X, Y )
is a Cauchy sequence, there exists K > 0 such that
kLn − Lm k ≤ 1
for m, n ≥ K. Fix m ≥ K, we have for each x,
kLx − Lm xk = lim kLn x − Lm xk ≤ kxk .
n→∞
Hence for each x,
kLxk ≤ (kLm k + 1) kxk ,
i.e., L ∈ B (X, Y ) .
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When Y = K, the scalar field, we write B (X, K) = X ∗ . The Banach space X ∗
is called the dual space of X.
1.7. Quotient Space.
Definition 7. Let N be a subspace of a vector space X. For every x ∈ X, let
π (x) = x + N ≡ [x] be the coset of N that contains x. The colloection of all cosets
of N form a vector space X/N , called the quotient space of X modulo N .
Let X be a normed space and N be a subspace of X. We can define for [x] ∈
X/N ,
p ([x]) = inf kx + yk .
y∈N
Then p is a seminorm and we have
Theorem 17. p is a norm on X/N if and only if N is a closed subspace of X.
FUNCTIONAL ANALYSIS
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Proof. ”⇒” If N is not a closed subspace of X. Then there exists xn ∈ N , and
x ∈ N C such that kxn − xk → 0 as n → ∞. Then p ([x]) = 0 but [x] 6= 0.
”⇐” If N is a closed subspace of X and if x 6∈ N , we need to show p ([x]) 6= 0.
If p ([x]) = 0, then there exists xn ∈ N such that kxn − xk → 0 as n → ∞. Since
N is closed, we have x ∈ N which is a contradiction.
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Lemma 1. A normed space X is a Banach space if and only if
∞
X
kxn k < ∞ implies the convergence of
n=1
∞
X
xn .
n=1
Proof. ”⇒” Trivial.
”⇐” Let {an } be a Cauchy sequence in X. Then there exists a subsequence
{ank } such that
°
°
1
°an
− ank ° < k ,
k+1
2
°
P∞ °
then n=1 °ank+1 − ank ° < ∞ which implies the convergence of
∞
X
¡
¢
ank+1 − ank = b.
n=1
Hence, {ank } converges to b + an1 which implies the whole sequence {an } converges
to the same limit.
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Theorem 18. Let N be a closed subspace of a banach space. Then X/N is a
Banach space with the norm p.
Proof. Let
∞
X
k[xn ]k < ∞.
n=1
Then there exists yn such that k[xn ]k ≥ kxn + yn k −
∞
X
So
so
P∞
n=1
P∞
1
2n ,
hence
kxn + yn k < ∞.
n=1
xn + yn converges to x for some x ∈ X. Now
° n
° °"
#°
n
°X
° ° X
°
°
° °
°
[xk ] − [x]° = °
xk − x °
°
°
° °
°
k=1
° n
° k=1
°X
°
°
°
≤°
(xk + yk ) − x° → 0,
°
°
k=1
n=1 [xn ] converges to [x]. Lemma 1 implies that X/N is a Banach space.
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Theorem 19. Let p be a semi-norm of a vector space and N = {x : p (x) = 0}.
Then X/N is a normed space with p̃ such that p̃ ([x]) = p (x).
2. Baire Category Theory and its Application
In this chapter, we discuss Baire Category theory and its application to functional
analysis.
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HUIQIANG JIANG
2.1. Baire’s Theorem.
Definition 8. Let S be a topological space. A set E ⊂ S is said to be nowhere
dense if the closure E has empty interior. The sets of the first category in S are
those that are countable unions of nowhere dense sets. Any subset S that is not of
the first category is said to be of the second category in S.
Proposition 1. (a) If A ⊂ B and B is of the first category in S, so is A.
(b) Any countable union of sets of the first category is of the first category.
(c) Any closed set E ⊂ S whose interior is empty is of the first category in S.
(d) If h is a homeomorphism of S onto S and if E ⊂ S, then E and h (E) have the
same category in S.
Theorem 20 (Baire’s Theorem). Let S be a nonempty complete metric space.
Then the intersection of a countable family of dense open sets in S is dense.
Proof. Let {Vi } be a countable family of dense open sets in S. Let V ⊂ S be
an arbitrary nonempty open set. Since V1 is dense, there exists a ball Br1 (x1 ) ⊂
Br1 (x1 ) ⊂ V1 ∩ V with r1 < 1. Similarly, for each n ≥ 2, there exists a ball
Brn (x1 ) ⊂ Brn (x1 ) ⊂ Vn ∩ Brn−1 (x1 ) with rn < n1 . Then the centers {xi } form
a Cauchy sequence. Since S is complete, xi converges to some x∗ ∈ V ∩ (∩∞
i=1 Vn ).
Hence ∩∞
V
is
dense.
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i=1 n
A direct consequence of Baire’s theorem:
Corollary 1. Every nonempty complete metric space S is a set of second category.
Proof. If not, there exists Ei , each of which is a closed nowhere dense set, such that
C
S = ∪∞
i=1 Ei . Let Vi = Ei . Then {Vi } is a countable family of dense open sets in
∞
S with ∩i=1 Vn = ∅ which contradicts to the Baire’s theorem.
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Theorem 21 (Banach). The set of functions in C [0, 1] that are not differentiable
at any point of [0, 1] is a dense Gδ subset.
The proof of Theorem 21 is a typical application of Baire’s theorem: Given a
complete metric space X, we want to prove the existence of an element x ∈ X
with certain property P. The basic idea of Baire category method is to approximate
property P with Pn . And if we can show that
(a) The sets Gn = {x ∈ X : x satisfies Pn } are open and dense;
(b) x has the property P if x satifies all the properties Pn .
∞
Then each x ∈ ∩∞
n=1 Gn has the property P and ∩n=1 Gn is a dense Gδ set.
2.2. Equicontinuity and Banach Steinhaus Theorem.
Definition 9. Suppose X and Y are topological vector spaces and Γ is a family
of linear mappings from X into Y . We say that Γ is equicontinuous if for every
neighborhood W of 0 in Y , there corresponds a neighborhood V of 0 in X such that
Λ (V ) ⊂ W for all Λ ∈ Γ. If X and Y are normed spaces, then Γ is equicontinuous
iff
sup kΛk < ∞.
Λ∈Γ
Theorem 22 (Banach-Steinhaus). Suppose X and Y are topological vector spaces
and Γ is a family of continuous linear mappings from X into Y . Let B be the set
of all x ∈ X such that Γ (x) = {Λx : x ∈ Γ} is bounded in Y . Then either B is of
the first category or B = X and Γ is equicontinuous.
FUNCTIONAL ANALYSIS
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The following version of Banach-Steinhaus Theorem is called uniform boundedness principle.
Theorem 23 (Banach-Steinhaus). Let Γ be a family of continuous linear mappings
from a Banach space X into a normed space Y . If for every x ∈ X,
sup kΛxk < ∞,
Λ∈Γ
then supΛ∈Γ kΛk < ∞.
Proof. Let
½
E=
¾
x : sup kΛxk ≤ 1 .
Λ∈Γ
Since
X=
∞
[
nE
n=1
and X is complete, Baire’s theorem implies that E has an interior point x0 . Then
E − x0 is a neighborhood of 0, hence there exists ε > 0, such that Bε (0) ⊂ E − x0 .
Hence, for any 0 6= x ∈ Bε (0) and Λ ∈ Γ, we have
kΛxk ≤ kΛ (x + x0 )k + kΛx0 k
≤ 1 + sup kΛx0 k .
Λ∈Γ
Hence, we have
sup kΛk ≤
Λ∈Γ
1 + supΛ∈Γ kΛx0 k
.
ε
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Another collary of the Banach Steinhaus theorem is the continuity of limits of
sequences of continuous linear mapping.
Theorem 24. If {Λn } is a sequence of continuous linear mappings from a Banach
space X into a normed space Y , and if
Λx = lim Λn x
n→∞
exists for every x ∈ X, then Λ is continuous.
Proof. Banach Steinhaus theorem implies the equicontinuity of {Λn }, hence Λ is
bounded and continuous.
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2.3. The Open Mapping Theorem. Let X, Y be two topological spaces. We
say f : X → Y is open at point p ∈ X if f (V ) contains a nbhd of f (p) whenever
V is a nbhd of p. We say that f is open if f (U ) is open in Y whenever U is open
in X. A linear mapping between two topological vector spaces is open iff it is open
at the origin.
Theorem 25. Let X and Y be Banach spaces and Λ ∈ B (X, Y ) is onto, then Λ
is an open mapping.
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HUIQIANG JIANG
Proof. It suffices to show that Λ (B1 (0)) contains Bδ (0) for some δ > 0. We break
its proof into two parts:
³
´
Step 1, Λ B 12 (0) contains Bδ (0) for some δ > 0. Since Λ is onto,
Y =
∞
[
Λ (Bk (0)) .
k=1
And since Y is complete, there exists k such that Λ (Bk (0)) contains an open ball
Br (y0 ) for some y0 ∈ Y and r > 0. Hence,
Br (0) = Br (y0 ) − y0 ⊂ Λ (Bk (0)) − Λ (Bk (0)) ⊂ Λ (B2k (0)).
³
´
r
Hence, Bδ (0) ⊂ Λ B 12 (0) if we take δ = 4k
.
³
´
³
´
Step 2, Λ B 12 (0) ⊂ Λ (B1 (0)). Fix y1 ∈ Λ B 12 (0) . Assume n ≥ 1 and
³
´
yn ∈ Λ B 21n (0) has been chosen. Step 1 implies that for any r > 0, Λ (Br (0))
contains a nbhd of 0. Hence
µ
³
´¶
³
´
1
yn − Λ B n+1
(0)
∩ Λ B 21n (0)
2
³
is nonempty, and we can choose xn ∈ B 21n (0) and yn+1 ∈ Λ B
1
2n+1
´
(0) s.t.
yn − yn+1 = Λxn .
Sum over n, we have
Ã
y1 − yn+1 = Λ
n
X
!
xk
→ Λx
k=1
P∞
where x = k=1 xk ∈ B1 (0) is well defined since kxk k < 21k . On the other hand,
the continuity of Λ implies yn → 0 as n → ∞, so we have y1 = Λx ∈ Λ (B1 (0)). ¤
Remark 4. The open mapping theorem holds when X, Y are F -space.
Corollary 2. Let X and Y be Banach spaces and Λ ∈ B (X, Y ) is onto and oneto-one, then there exists positive constants a, b such that
a kxk ≤ kΛxk ≤ b kxk .
Proof. Open mapping theorem implies the inverse is also continuous and hence
bounded.
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2.4. The Closed Graph Theorem. Let f : X → Y , the graph of f is a subset
of X × Y.
Theorem 26. If X is a topological space, Y is a Hausdorff space and f : X → Y
is continuous, then the graph of f is closed.
Proof. Let
G = {(x, f (x) : x ∈ X)} ⊂ X × Y
be the graph of f . For any (x0 , y0 ) ∈ GC , we have y0 6= f (x0 ). Since Y is Hausdorff,
y0 and f (x0 ) have disjoint nbhds V and W in Y . Since f is continuous x0 has a
neighborhood U s.t. f (U ) ⊂ W . Then U × V ⊂ GC is a nbhd of (x0 , y0 ). Hence
GC is open and G is closed.
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FUNCTIONAL ANALYSIS
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Theorem 27 (The closed graph theorem). Let X, Y be Banach spaces and Λ :
X → Y be linear. If the graph of Λ is closed in X × Y , then Λ is continuous.
Proof. It is easy to verify that X × Y is a normed space if we define for any (x, y),
k(x, y)k = kxk + kyk .
And the product topology of X × Y is compatible with such norm. Since Λ is linear
and its graph G is closed, G is a closed subspace of X × Y , and hence a Banach
space too. Now we define π1 : G → X and π2 : X × Y → Y by
π1 (x, Λx) = x, π2 (x, y) = y.
Since π1 is continuous, linear bijection, the open mapping theorem implies that π1−1
is continuous. Since π2 is also continuous, we have Λ = π2 ◦ π1−1 is continuous. ¤
Remark 5. The closed graph theorem holds when X, Y are F -spaces.
Proposition 2. Let X, Y be two metric spaces and Λ : X → Y . Then The graph
G of Λ is closed if and only if for any sequence {xn }, x = lim xn and y = lim Λxn
implies y = Λx.
Proof. ”⇒” x = lim xn and y = lim Λxn implies (x, y) = lim (xn , Λxn ). Since
(xn , Λxn ) ∈ G, we have (x, y) ∈ G if G is closed. Hence y = Λx.
”⇐” Let (x, y) = lim (xn , Λxn ), then x = lim xn and y = lim Λxn , hence y = Λx.
So we have (x, y) ∈ G and G is closed.
¤
2.5. Bilinear Mappings.
Definition 10. Let X, Y, Z be vector spaces and B : X × Y → Z. B is said to be
bilinear if it is linear with respect to each of its variables.
Theorem 28. Let X be a Banach space and Y, Z be normed spaces. If B : X ×Y →
Z is bilinear and B is continuous with respect to each of its variables, then B is
continuous.
Proof. Let (x0 , y0 ) = lim (xn , yn ). We define for each n, a map bn : X → Z by
bn (x) = B (x, yn ). Then bn is continuous and for each fixed x ∈ X,
lim bn (x) = B (x, y0 ) .
n→∞
Hence {bn } is equi-continuous from Banach-Steinhauss theorem. Now
kB (xn , yn ) − B (x0 , y0 )k = kbn (xn − x0 ) + B (x0 , yn − y0 )k
≤ kbn (xn − x0 )k + kB (x0 , yn − y0 )k
≤ sup kbn k kxn − x0 k + kB (x0 , yn − y0 )k → 0,
hence B is continuous.
¤
3. Hahn-Banach Theorems
Most of the theorems in this chapter holds for locally convex topological vector
spaces, although we will only present proofs in the setting of normed spaces only.
We first recall Zorn’s lemma which is equivalent to axiom of choice.
Lemma 2 (Zorn’s Lemma). Every partially ordered set, in which every chain (i.e.
totally ordered subset) has an upper bound, contains at least one maximal element.
12
HUIQIANG JIANG
Theorem 29 (Hahn Banach). Suppose M is a subspace of a real vector space X
and p : X → R satisfies, for any x, y ∈ X, t ≥ 0,
p (x + y) ≤ p (x) + p (y) , p (tx) = tp (x) .
If f : M → R is linear and f (x) ≤ p (x) on M . Then there exists a linear
Λ : X → R such that Λx = f (x) on M and
−p (−x) ≤ Λx ≤ p (x) for x ∈ X.
Proof. If M 6= X, we choose x1 ∈ X\M and let M1 = {x + tx1 : x ∈ M and t ∈ R}.
For any x, y ∈ M , we have
f (x) + f (y) = f (x + y) ≤ p (x + y) ≤ p (x − x1 ) + p (y + x1 ) ,
i.e.
f (x) − p (x − x1 ) ≤ p (y + x1 ) − f (y) .
Let
α = sup {f (x) − p (x − x1 )} .
x∈M
Then
f (x) − α ≤ p (x − x1 ) , f (x) + α ≤ p (x + x1 ) ,
Define f1 on M1 by
f1 (x + tx1 ) = f (x) + tα for any x ∈ M and t ∈ R.
Then we have for t > 0,
³ x
´
f1 (x + tx1 ) = f (x) + t sup {f (y) − p (y − x1 )} ≥ −tp − − x1 = −p (−x − tx1 )
t
y∈M
³ ³x´
´
³x
´
f1 (x + tx1 ) = f (x) + tα = t f
+ α ≤ tp
+ x1 = p (x + tx1 )
t
t
hence −p (−x) ≤ Λx ≤ p (x) for x ∈ M1 .The inequality when t < 0 can be proved
similarly.
Now we consider the family of pairs (M 0 , f 0 ) such that M 0 is a subspace containing M and f 0 = f on M . We can assign a partial order by declaring (M 0 , f 0 ) ≤
(M 00 , f 00 ) if M 0 ⊂ M 00 and f 0 = f 00 on M 0 . Then every totally ordered subset has an
upper bound. Zorn’s lemma implies the existence of a maximal element (M0 , f0 ),
then M0 = X, otherwise we can extend (M0 , f0 ) which leads to a contradiction. ¤
Theorem 30. Suppose M is a subspace of a vector space X and p : X → R is a
seminorm. If f is a linear functional on M such that |f (x)| ≤ p (x) on M . Then
f extends to a linear functional Λ such that |Λx| ≤ p (x) on X.
Proof. The case K = R follows from Theorem 29. If K = C, we can still view X
as a real vector space, and we define u = Re f which is a real linear function on M
and |u (x)| ≤ p (x). One can verify that
f (x) = u (x) − iu (ix) .
Theorem 29 implies the existence of a real linear function U on X such that U = u
on M and U (x) ≤ p (x) on X. Now we define
Λx = U (x) − iU (ix) ,
FUNCTIONAL ANALYSIS
13
then Λ is complex linear and Λ = f on M . Finally for any x ∈ X, there exists
|α| = 1, such that
|Λx| = αΛ (x) = U (αx) ≤ p (αx) = p (x) .
¤
Corollary 3. If X is a normed space and x0 ∈ X. Then there exists Λ ∈ X ∗ such
that Λx0 = kx0 k and kΛk = 1.
Corollary 4. If X is a normed space and x ∈ X. Then
kxk =
sup
Λx.
Λ∈X ∗ ,kΛk≤1
Corollary 5. If f is a continuous linear functional on a subspace M of a normed
(locally convex) space X, then there exists Λ ∈ X ∗ such that Λ = f on M . We can
require kΛk = kf k if X is normed.
Theorem 31. Suppose A, B are disjoint, nonempty, convex subsets of a normed
(locally convex) space X.
(a) If A is open, then there exists Λ ∈ X ∗ and γ ∈ R such that
Re Λx < γ ≤ Re Λy for any x ∈ A, y ∈ B.
(b) If A is compact and B is closed, then there exists Λ ∈ X ∗ and γ1 , γ2 ∈ R such
that
Re Λx < γ1 < γ2 < Re Λy for any x ∈ A, y ∈ B.
Proof. We only prove when X is a real vector space, the complex case follows easily.
(a) Let a0 ∈ A, b0 ∈ B be fixed and x0 = b0 − a0 . Let C = A − B + x0 . Then C
is a convex nbhd of 0. Let p be the Minkowski functional of C, i.e.
n
o
x
p (x) = µC (x) = inf t > 0 : ∈ C .
t
Then p satisfies, for any x, y ∈ X, t ≥ 0,
p (x + y) ≤ p (x) + p (y) , p (tx) = tp (x) .
Moreover, since x0 6∈ C, p (x0 ) ≥ 1. Define f (tx0 ) = t on the subspace M =
{tx0 : t ∈ R}. Then f ≤ p on M . Hahn-Banach Theorem implies the existence of
an extension Λ such that Λ ≤ p. In particular, Λ ≤ 1 on C and Λ ≥ −1 on −C, so
we have |Λx| ≤ 1 on C ∩ (−C). Hence Λ ∈ X ∗ .Now if a ∈ A and b ∈ B, we have
Λa − Λb + 1 = Λ (a − b + x0 ) ≤ p (a − b + x0 ) < 1
since a − b + x0 ∈ C and C is open. Hence Λa < Λb, so ΛA and ΛB are disjoint
convex subsets of R. ΛA is open, so (a) follows by setting γ = inf ΛB.
(b) There is a convex nbhd V of 0 in X such that (A + V ) ∩ B = ∅. Apply (a)
to A + V and B, the result follows since ΛA is a compact subset of Λ (A + V ). ¤
Corollary 6. Suppose M is a subspace of a normed (locally convex) space X and
x0 ∈ X\M . Then there exists Λ ∈ X ∗ such that Λx0 = 1 and Λx = 0 for x ∈ M .
Proof. Theorem 31 implies the existence of Λ ∈ X ∗ such that
Re Λx < γ1 < γ2 < Re Λx0
for any x ∈ M . Since M is a subspace, ΛM is a subspace and Re ΛM < γ1 implies
1
ΛM = {0}. Then Λx
Λ is the linear functional satisfying all requirements.
¤
0
14
HUIQIANG JIANG
Corollary 7. Suppose B is a convex, balanced, closed set in a normed (locally
convex) space X, x0 ∈ X\B. Then there exists Λ ∈ X ∗ such that |Λx| ≤ 1 for all
x ∈ B, but Λx0 > 1.
A space X is said to be separable if there is a countable dense set in X. For
example, Lp [0, 1] is separable if 1 ≤ p < ∞ while L∞ [0, 1] is not separable.
Theorem 32. Let X be a normed space. If X ∗ is separable, then X is separable.
Proof. Let {Λn } be a countable dense subset in X ∗ . Then for each n, there exists
xn ∈ X such that
1
kxn k ≤ 1 and Λn xn ≥ kΛn k .
2
Let X0 be the linear space spanned by {xn }, then X0 is separable. We claim X0 is
dense in X. If not, there exists Λ ∈ X ∗ such that Λx = 0 for x ∈ X0 while Λ 6= 0.
Let Λnk → Λ in X ∗ , we have
|Λnk xnk | = |Λxnk − Λnk xnk | ≤ kΛ − Λnk k → 0.
On the other hand, Λnk xnk ≥ 12 kΛn k → 12 kΛk, which is a contradiction.
Hence X0 is dense in X, since X0 is separable, X must be separable too.
¤
4. Duality in Banach Spaces
Let X be a normed space. For any x∗ ∈ X ∗ and x ∈ X, we write
x∗ (x) = hx, x∗ i .
X ∗ is a Banach space with
kx∗ k = sup hx, x∗ i .
kxk≤1
Let τ1 , τ2 be two topologies on a set X. We say that τ1 is weaker than τ2 , or τ2 is
stronger than τ1 , if τ1 ⊂ τ2 . In this situation, the identity map on X is continuous
from (X, τ2 ) to (X, τ1 ) and is an open mapping from (X, τ1 ) to (X, τ2 ).
The weak topology on X is the weakest topology so that each x∗ ∈ X ∗ is
continuous. Especially, the weak topology is weaker than the original topology.
And the original topology is also called the strong topology of X.
We say that a sequence {xn } ⊂ X converges weakly to x0 if for any x∗ ∈ X ∗ ,
hxn , x∗ i → hx0 , x∗ i as n → ∞.
And we write
xn * x0 as n → ∞.
∗
Any x ∈ X induces fx ∈ (X ∗ ) = X ∗∗ such that
fx (x∗ ) = hx, x∗ i .
And we have kfx k = kxk.
Weak star topology in X ∗ is the weakest topology in X ∗ so that fx is continuous
for each x ∈ X. Weak star topology in X ∗ is weaker than its original topology. We
say that a sequence {x∗n } ⊂ X ∗ converges weakly star to x∗0 if for any x ∈ X,
hx, x∗n i → hx, x∗0 i as n → ∞.
And we write
∗
x∗n * x∗0 as n → ∞.
FUNCTIONAL ANALYSIS
15
Proposition 3. Strong convergence in X implies weak convergence. Strong convergence in X ∗ implies weak-* convergence.
Example 1. sin (nx) * 0 in Lp [0, 1], 1 ≤ p < ∞.
¡
¢∗
∗
Example 2. sin (nx) * 0 in L∞ [0, 1] = L1 [0, 1] .
Theorem 33. Let xn * x0 in a normed space X. Then
sup kxn k < ∞ and kx0 k ≤ lim inf kxn k .
Proof. X ∗ is a Banach space. For any x∗ ∈ X ∗ , we have
fxn (x∗ ) → fx0 (x∗ ) .
Uniform boundedness principle implies that kxn k = kfxn k < M for some M > 0.
Next, if kx0 k > lim inf kxn k, then there exists a subsequence {xnk } such that
kxnk k ≤ L < kx0 k .
From Corollary 7, there exists Λ ∈ X ∗ such that |Λx| ≤ 1 for all kxk ≤ L, but
Λx0 > 1, a contradiction.
¤
Let’s check the weak convergence in C (K) where K is a compact metric space.
Theorem 34. Let K be a compact metric space. Then fn * f in C (K) iff
(a) fn (x) → f (x) for each x ∈ K and
(b) {fn } is bounded.
Proof. ”⇒”: Easy.
”⇐”: Follows from Lebesgue dominated convergence theorem.
¤
Theorem 35. Let X be a normed space. If E ⊂ X and if for any Λ ∈ X ∗ ,
sup Λx < ∞,
x∈E
then
sup kxk < ∞.
x∈E
Proof. We view E as a subset of X ∗∗ , then the theorem follows from BanachSteinhauss Theorem.
¤
Theorem 36. Suppose E is a convex subset of a normed space X, then the weak
closure of E,
E w = E.
Proof. E w is strongly closed, hence E ⊂ E w . Now if E 6= E w , then there exists
x0 ∈ X, x0 6∈ E. Theorem 31 implies the existence of Λ ∈ X ∗ such that
Re Λx0 < γ < Re Λx
holds for any x ∈ E. Hence, x0 6∈ E w and we conclude E w ⊂ E.
¤
Corollary 8 (Mazur’s Lemma). Let xn * x0 in a normed space X. Then there
exists a sequence {yi }, each yi is a convex combination of finitely many xn such
that yi → x0 .
Proof. Let H be the convex hull of the set of all xn . Then H w = H. Hence x0 ∈ H
and there exists a sequence {yi } ⊂ H such that yi → x0 .
¤
16
HUIQIANG JIANG
Theorem 37 (Banach-Alaoglu Theorem). Let X be a separable normed space.
Then every bounded sequence in X ∗ has a weakly-* convergent subsequence.
Proof. Let {xn } be a countable dense subset of X. Let {x∗n } ⊂ X ∗ be bounded.
For each l ∈ N,
{hxl , x∗n i}
is bounded in the scalar field, so there exists a subsequence
© which
ª converges. Using
diagonal sequence argument, there exists a subsequence x∗nk such that for each
l ∈ N,
®
­
lim xl , x∗nk
k→∞
exists. Since {xn } is dense in X, for each x ∈ X,
®
­
lim x, x∗nk
k→∞
exists and
­
®
Λx = lim x, x∗nk
k→∞
defines a continuous linear functonal such that x∗nk * Λ.
¤
If we drop the separability assumption, we have
Theorem 38 (Banach-Alaoglu Theorem). Let X be a normed space. Then the
closed unit ball in X ∗ is weakly-* compact.
Please check Theorem 3.15 in the textbook for its proof.
Theorem 39. Let X be a locally compact metric space and let {µn } be a sequence
of finite positive Borel measures on X such that sup µn (X) < ∞. Then there exists
a subsequence {µnk } and a finite positive Borel measure µ0 such that
µ (X) ≤ lim inf µnk (X)
k→∞
and for every f ∈ C0 (X)
Z
Z
lim
k→∞
X
f dµnk =
f dµ0 .
X
Let ϕ : X → X ∗∗ be the imbedding such that for any x ∈ X, ϕ (x) = fx ∈ X ∗∗ ,
i.e., for any x∗ ∈ X ∗ ,
hx∗ , ϕ (x)i = hx, x∗ i .
Definition 11. We say a Banach space X is reflexive if X ∗∗ = ϕ (X).
Let (X, µ) be a σ-finite measure space. Then Lp (µ) ia reflexive if 1 < p < ∞.
Theorem 40. A closed subspace Y of a reflective Banach space X is reflective.
Proof. Let y0∗∗ ∈ Y ∗∗ . We define a linear functional y1∗∗ on X ∗ such that for any
x∗ ∈ X ∗ ,
hx∗ , y1∗∗ i = hx∗ |Y , y0∗∗ i .
Then y1∗∗ ∈ X ∗∗ and since X is reflexive, there exists y0 ∈ X such that
hy0 , x∗ i = hx∗ , y1∗∗ i = hx∗ |Y , y0∗∗ i .
If y0 6∈ Y , there exists x∗ ∈ X ∗ such that hy0 , x∗ i = ky0 k and x∗ |Y = 0, a contradiction. Hence y0 ∈ Y . Now for any y ∗ ∈ Y ∗ , let x∗ ∈ X be any extension of y ∗ ,
we have
hy0 , y ∗ i = hy0 , x∗ i = hx∗ |Y , y0∗∗ i = hy ∗ , y0∗∗ i .
FUNCTIONAL ANALYSIS
Hence the map ϕ : Y → Y ∗∗ is onto and Y is reflexive.
17
¤
Theorem 41. Let X be a Banach space. X is reflexive iff X ∗ is reflexive.
Proof. If X is reflexive. For any x∗∗∗
∈ X ∗∗∗ , we define x∗0 ∈ X ∗ such that
0
hx, x∗0 i = hϕ (x) , x∗∗∗
0 i
Then for any x∗∗ ∈ X ∗∗ ,
­
®
hx∗0 , x∗∗ i = ϕ−1 (x∗∗ ) , x∗0 = hx∗∗ , x∗∗∗
0 i.
Hence X ∗ is reflexive.
Now if X ∗ is reflexive, then X ∗∗ is reflexive, and since X is isomorphic to ϕ (X),
a closed subspace of X ∗∗ . We conclude X is reflexive.
¤
Theorem 42. A bounded sequence in a reflexive space X has a weakly convergent
subsequence.
Proof. Let {xn } be a bounded sequence in X and X0 be the closure of the linear
space spanned by {xn }. Then X0 is a closed subspace of X and hence reflexive.
We notice that X0 is separable and so is X0∗∗ , hence X0∗ is also separable. Hence,
there exists a subsequence xnk and x0 ∈ X0 , such that for any x∗0 ∈ X0∗ ,
hxnk , x∗0 i → hx0 , x∗0 i .
Now for any x∗ ∈ X ∗ , x∗ |X0 ∈ X0∗ , hence we have
hxnk , x∗ i = hxnk , x∗ |X0 i → hx0 , x∗ |X0 i = hx0 , x∗ i .
¤
Remark 6. Actually, a Banach space X is reflexive iff it is weakly locally sequentially compact. (Eberlein-Shmulyan)
Theorem 43. Let (X, µ) be a σ-finite measure space and 1 < p < ∞. Let {fn } ⊂
Lp (µ) be bounded. Then there exists a subsequence {fnk } and a function f0 ∈ Lp (µ)
such that
kf0 kp ≤ lim inf kfnk kp
k→∞
and for any g ∈ Lq (µ),
Z
lim
k→∞
X
Z
fnk gdµ =
f gdµ.
X
Next, we discuss annihilators of subspaces in Banach spaces.
Definition 12. Let X be a Banach space and M ⊂ X, N ⊂ X ∗ be subspaces.
Their annihilators M ⊥ and ⊥ N are defined as follows
M ⊥ = {x∗ ∈ X ∗ : hx, x∗ i = 0 for any x ∈ M } ,
⊥
N = {x ∈ X : hx, x∗ i = 0 for any x∗ ∈ N } .
Theorem 44. Let X be a Banach space and M ⊂ X, N ⊂ X ∗ be subspaces.
(i). M ⊥ is a weak-* closed subspace of X ∗ ;
(ii). ⊥ N¡ is a¢ closed subspace of X;
(iii). ⊥ M ⊥ is the norm closure of M in X;
¡
¢⊥
is the weak-* closure of N in X ∗ .
(iv). ⊥ N
18
HUIQIANG JIANG
Proof. (i), (ii) trivial.
¡
¢
(iii). If x ∈ M , then hx, x∗ i = 0 for any x∗ ∈ M ⊥ , hence x ∈⊥ M ⊥ . So
¡
¢
¡
¢
¡
¢
M ⊂⊥ M ⊥ . Now since ⊥ M ⊥ is norm closed, M ⊂⊥ M ⊥ . On the other
¢
¡
hand, if x 6∈ M , then there exists x∗ ∈ M ⊥ , x∗ (x) = kxk 6= 0, hence x 6∈⊥ M ⊥ ,
¡
¢
so we have ⊥ M ⊥ ⊂ M .
(iv). It can be proved similarly. Here we need to apply Hahn Banach Theorem
to X ∗ with weak-* topology which is a locally convex space.
¤
With the help of annihilators, we can describe the dual space of subspaces and
quotient spaces.
Theorem 45. Let M be a closed subspace of a Banach space X.
(i). The Hahn Banach Theorem extends each m∗ ∈ M ∗ to a functional x∗ ∈ X ∗ .
Define
σx∗ = x∗ + M ⊥ .
Then σ defines an isometric isomorphism from M ∗ onto X ∗ /M ⊥ .
(ii). Let π : X → X/M = Y be the quotient map. For each y ∗ ∈ Y ∗ , define
τ y∗ = y∗ π ∈ X ∗ .
Then τ defines an isometric isomorphism from Y ∗ onto M ⊥ .
5. Adjoint operators
Theorem 46. Suppose X, Y are normed spaces. To each T ∈ B (X, Y ) corresponds
a unique T ∗ ∈ B (Y ∗ , X ∗ ) such that for any x ∈ X, y ∗ ∈ Y ∗ ,
hT x, y ∗ i = hx, T ∗ y ∗ i .
Moreover,
kT k = kT ∗ k .
Proof. T ∗ y ∗ = y ∗ ◦ T ∈ X ∗ , hence T ∗ : Y ∗ → X ∗ and it is straight forward to
verify that T ∗ is linear. Now,
kT k = sup {kT xk : x ∈ X, kxk ≤ 1}
= sup {hT x, y ∗ i : x ∈ X, y ∗ ∈ Y ∗ , kxk ≤ 1, ky ∗ k ≤ 1}
= sup {hx, T ∗ y ∗ i : x ∈ X, y ∗ ∈ Y ∗ , kxk ≤ 1, ky ∗ k ≤ 1}
= sup {kT ∗ y ∗ k : y ∗ ∈ Y ∗ , ky ∗ k ≤ 1}
= kT ∗ k .
Hence T ∗ ∈ B (Y ∗ , X ∗ ). The uniqueness is trivial.
Definition 13. T ∗ is called the adjoint operator of T .
We recall that
N (T ) = {x ∈ X : T x = 0} ,
R (T ) = {T x : x ∈ X} .
Theorem 47. Suppose X and Y are Banach spaces. Then
(5.1)
⊥
N (T ∗ ) = R (T ) ,
N (T ) =⊥ R (T ∗ ) .
¤
FUNCTIONAL ANALYSIS
19
Proof. y ∗ ∈ N (T ∗ ) iff T ∗ y ∗ = 0 iff hT x, y ∗ i = hx, T ∗ y ∗ i = 0 for any x ∈ X iff
⊥
y ∗ ∈ R (T ) .
x ∈ N (T ) iff T x = y iff hT x, y ∗ i = hx, T ∗ y ∗ i = 0 for any y ∗ ∈ Y ∗ iff x ∈⊥
R (T ∗ ).
¤
Corollary 9. Suppose X and Y are Banach spaces.
(i) N (T ∗ ) is weak-* closed in Y ∗ .
(ii) R (T ) is dense in Y iff T ∗ is one-to-one.
(iii). T is one-to-one iff R (T ∗ ) is weak-* dense in X ∗ .
Theorem 48. Suppose U and V are open unit balls of Banach spaces X and Y
respectively. Suppose T ∈ B (X, Y ) and δ > 0. then the implications (a) → (b) →
(c) → (d) hold among the following statements:
(a) kT ∗ y ∗ k ≥ δ ky ∗ k for every y ∗ ∈ Y ∗ .
(b) δV ⊂ T (U ).
(c), δV ⊂ T (U ).
(d). T (X) = Y .
Moreover, if (d) holds, then (a) holds for some δ > 0.
Proof. (a) ⇒ (b): Let y0 6∈ T (U ). Since T (U ) is convex, balanced and closed,
there exists y ∗ ∈ Y ∗ ,
hy0 , y ∗ i > 1 and |hy, y ∗ i| ≤ 1 for any y ∈ T (U ).
Hence, for any x ∈ U ,
|hx, T ∗ y ∗ i| = |hT x, y ∗ i| ≤ 1
which implies kT ∗ y ∗ k ≤ 1 and hence ky ∗ k ≤ 1δ kT ∗ y ∗ k ≤ 1δ . Now
1 < hy0 , y ∗ i ≤ ky0 k ky ∗ k ≤
1
ky0 k ,
δ
so we have ky0 k > δ.
(b) ⇒ (c): We assume δ = 1, then we have V ⊂ T (U ) and hence for any r > 0,
rV ⊂ T (rU ). Let y1 ∈ V and 0 < ε < 1 − ky1 k. When yk is chosen, k ≥ 1. There
exists kxk k ≤ kyk k, such that
ε
kyk − T xk k ≤ k .
2
And we define yk+1 = yk − T xk . Since
∞
∞
X
X
kxk k ≤ ky1 k +
kyk k ≤ ky1 k + ε < 1,
we have x =
P∞
k=1
k=1
k=2
xk ∈ U and
T x = lim
n
X
n→∞
k=1
T xk = lim
n→∞
n
X
(yk − yk+1 ) = y1 .
k=1
(c) ⇒ (d): Trivial.
Finally, if (d) holds, then the open mapping theorem implies that for some δ > 0,
δV ⊂ T (U ). Hence, for any y ∗ ∈ Y ∗ ,
kT ∗ y ∗ k = sup {|hx, T ∗ y ∗ i| : x ∈ U }
= sup {|hT x, y ∗ i| : x ∈ U }
≥ sup {|hy, y ∗ i| : y ∈ δV } = δ ky ∗ k .
20
HUIQIANG JIANG
¤
Theorem 49 (Closed Range Theorem). Let X, Y be Banach Spaces and T ∈
B (X, Y ). Then the following three statements are equivalent:
(a) R (T ) is closed in Y ;
(b) R (T ∗ ) is weak-* closed in X ∗ ;
(c) R (T ∗ ) is closed in X ∗ .
⊥
Proof. (a) ⇒ (b): Since N (T ) =⊥ R (T ∗ ), N (T ) is the weak-* closure of R (T ∗ ),
⊥
⊥
so we only need to show N (T ) ⊂ R (T ∗ ). Let x∗ ∈ N (T ) . We define a linear
functional Λ on R (T ) by
ΛT x = hx, x∗ i
which is well-defined since
T x = T y ⇒ x − y ∈ N (T ) ⇒ ΛT x − ΛT y = hx − y, x∗ i = 0.
Since R (T ) is closed, open mapping theorem implies T : X → R (T ) is open. Hence
∃K > 0, for any y ∈ R (T ), there exists x ∈ X, such that y = T x and kxk ≤ K kyk.
Hence,
|Λy| = |hx, x∗ i| ≤ kx∗ k kxk ≤ K kx∗ k kyk
∗
and Λ ∈ (R (T )) . By the Hahn Banach Theorem, there exists an extension y ∗ ∈
Y ∗ . Since for any x ∈ X,
hx, T ∗ y ∗ i = hT x, y ∗ i = ΛT x = hx, x∗ i ,
we have x∗ = T ∗ y ∗ ∈ R (T ∗ ).
(b) ⇒ (c): Trivial.
(c) ⇒ (a): Let Z = R (T ). Define S ∈ B (X, Z) by Sx = T x. Since R (S) is
dense in Z, S ∗ ∈ B (Z ∗ , X ∗ ) is one-to-one. If z ∗ ∈ Z ∗ , there exists an extension y ∗
of z ∗ , and for any x ∈ X,
hx, T ∗ y ∗ i = hT x, y ∗ i = hSx, z ∗ i = hx, S ∗ z ∗ i .
Since R (T ∗ ) is closed, R (S ∗ ) is closed. Applying open mapping theorem to the
one-to-one mapping S ∗ : Z ∗ → R (S ∗ ), ∃c > 0, for any z ∗ ∈ Z ∗ ,
c kz ∗ k ≤ kS ∗ z ∗ k .
Hence S : X → Z is an open mapping. So we have R (S) = Z and hence R (T ) =
R (S) is closed.
¤
Corollary 10. Let X, Y be Banach Spaces and T ∈ B (X, Y ). Then R (T ) = Y iff
T ∗ is one-to-one and R (T ∗ ) is closed.
⊥
Proof. If R (T ) = Y , then N (T ∗ ) = R (T ) = {0} and hence T ∗ is one-to-one.
And since R (T ) is closed, R (T ∗ ) is closed.
If T ∗ is one-to-one and R (T ∗ ) is closed, then R (T ) is dense in Y and closed,
hence R (T ) = Y .
¤
FUNCTIONAL ANALYSIS
21
6. Compact operators
Let X, Y be Banach Spaces and U ⊂ X be the unit ball in X. T ∈ L (X, Y ) is
said to be compact if T (U ) is compact. Since compact set is bounded, compact
operators are bounded.
Let X be a Banach Space. B (X) = B (X, X) is an algebra such that for any
S, T ∈ B (X), we can define ST ∈ B (X) such that (ST ) (x) = S (T x).
It is easy to check kST k ≤ kSk kT k.
Given T ∈ B (X), we can define T 0 = I, and for any integer k ≥ 1, T k = T T k−1 .
We say T ∈ B (X) is invertible if there exists S ∈ B (X) such that ST = T S = I,
and we write S = T −1 .
Proposition 4. T ∈ B (X) is invertible iff N (T ) = {0} and R (T ) = X.
Proof. This follows from open mapping theorem.
¤
Definition 14. The spectrum σ (T ) is the set of all scalars λ such that T − λI is
not invertible.
λ ∈ σ (T ) iff T − λI is not one-to-one or R (T − λI) 6= X.
Definition 15. λ is said to be an eigenvalue of T if T −λI is not one-to-one. Each
0 6= x ∈ N (T − λI) is said to be an eigenvector of T .
If x is an eigenvector corresponds to eigenvalue λ, then T x = λx.
We recall that every locally compact topological vector space X has finite dimension. When X is a normed space, this is a consequence of Riesz’s lemma.
Theorem 50 (Riesz’s Lemma). Let X be a normed space and Y be a closed subspace of X. If Y 6= X, then for any 0 < r < 1, there exists x0 6∈ Y , such that
kx0 k = 1 and dist (x0 , Y ) > r.
Proof. Let x 6∈ Y , then dist (x, Y ) > 0. Hence for any r ∈ (0, 1), there exists
yr ∈ Y ,
1
dist (x, Y ) ≤ kx − yr k < dist (x, Y ) .
r
Let
x − yr
x0 =
,
kx − yr k
then for any y ∈ Y ,
1
kx − yr − kx − yr k yk
kx − yr k
1
≥
dist (x, Y ) > r.
kx − yr k
kx0 − yk =
Hence dist (x0 , Y ) > r.
¤
Corollary 11. Every locally compact normed space X has finite dimension.
Proof. If dim X = ∞. For any r ∈ (0, 1), we can use Riesz’s lemma to construct a
sequence {xn }, such that kxn k = 1 for each n and kxn − xm k > r if n 6= m. Hence
the unit sphere is not totally bounded.
¤
22
HUIQIANG JIANG
Theorem 51. Let X, Y be Banach Spaces.
(a). If T ∈ B (X, Y ) and dim (R (T )) < ∞, then T is compact.
(b). If T ∈ B (X, Y ) is compact and R (T ) is closed, then dim (R (T )) < ∞.
(c). The compact operators form a closed subspace of B (X, Y ).
(d). If T ∈ B (X) is compact and λ 6= 0, then dim (N (T − λI)) < ∞.
(e). If dim X = ∞, and T ∈ B (X) is compact, then 0 ∈ σ (T ).
(f ). If S, T ∈ B (X) and T is compact, so are ST and T S.
Proof. (a). Trivial.
(b). T is an open mapping from X onto R (T ), compactness of T implies that
R (T ) is locally compact, hence dim (R (T )) < ∞.
(c). Compact operators form a subspace of B (X, Y ). Let {Tn } be a sequence of
compact operators in B (X, Y ) and Tn → T0 . Let {xn } be a bounded sequence in
X, since Tn is compact for each n, using a diagonalisation argument, there exists a
subsequence {xnk } such that Tl xnk converges for each l ∈ N. Hence T0 xnk is also
a Cauchy sequence. So T0 is compact.
(d). Y = N (T − λI) is closed and the restriction of T on Y is a compact
operator with range Y , hence dim (N (T − λI)) < ∞.
(e). If 0 6∈ σ (T ), then R (T ) = X, compactness of T implies that dim X < ∞.
A contradiction.
(f). Trivial.
¤
We recall Ascoli’s theorem
Theorem 52 (Ascoli). Suppose X is a compact space and A ⊂ C (X) is pointwisely
bounded and equicontinuous, then A is totally bounded in C (X).
Theorem 53. Suppose X, Y are Banach spaces and T ∈ B (X, Y ), then T is
compact if and only if T ∗ is compact.
Proof. Suppose T is compact. Let U be the unit ball of X. Let {yn∗ } be a sequence in the unit ball of Y ∗ . Then yn∗ restricted nto T (U )ois pointwisely bounded
and equicontinuous, hence, by Ascoli’s theorem, yn∗ |T (U ) is totally bounded in
³
´
´
n
o
³
C T (U ) . So there exists a y0∗ ∈ C T (U ) and a subsequence yn∗ k |T (U ) such
that
yn∗ k → y0∗ uniformly on T (U ).
Now
°
°
°D
E°
° ∗ ∗
°
°
°
°T yni − T ∗ yn∗ j ° = sup ° T x, yn∗ i − yn∗ j ° → 0.
x∈U
n
o
Hence, T ∗ yn∗ j converges in X ∗ . Hence T ∗ is compact.
Suppose T ∗ is compact, then T ∗∗ is compact. Let ϕ : X → X ∗∗ and ψ : Y → Y ∗∗
be the isometry embedding. Then
hy ∗ , ψT xi = hT x, y ∗ i = hx, T ∗ y ∗ i = hT ∗ y ∗ , ϕxi = hy ∗ , T ∗∗ ϕxi
holds for any x ∈ X, y ∗ ∈ Y ∗ . Hence, ψT = T ∗∗ ϕ. Since ϕ (U ) ⊂ U ∗∗ , we have
ψT U = T ∗∗ ϕU ⊂ T ∗∗ (U ∗∗ )
which is totally bounded, hence ψT U is totally bounded and T U is totally bounded.
So T is compact.
¤
FUNCTIONAL ANALYSIS
23
Definition 16. Suppose M is a closed subspace of a topological vector space X. If
there exists a closed subspace N of X such that
X = M + N and M ∩ N = {0} ,
then M is said to be complemented in X and X is said to be the direct sum of M
and N , and we write
X = M ⊕ N.
n
Lemma 3. Let {ek }k=1 be a basis for an n-dimensional normed space M . Then
there exists c1 , c2 > 0, such that
v
v
° n
°
u n
u n
°X
°
uX
uX
°
°
2
2
c1 t
|αk | ≤ °
αk ek ° ≤ c2 t
|αk |
°
°
k=1
k=1
k=1
holds for any
x=
n
X
αk ek .
k=1
Moreover, the map A : Kn → M defined by
A (α1 , α2 , · · · , αn ) =
n
X
αk ek
k=1
is a homeomorphism.
Proof. First, we have
°
°
n
n
°X
° X
°
°
αk ek ° ≤
|αk | kek k
°
°
°
k=1
k=1
v
v
v
u n
u n
u n
X
u
uX
uX
2
2
2
kek k t
|αk | ≡ c2 t
|αk | .
≤t
k=1
k=1
k=1
Next, we consider the unit sphere in Kn
(
S=
n
(α1 , α2 , · · · , αn ) ∈ K :
n
X
)
2
|αk | = 1 .
k=1
We define function f on S such that
°
°
n
°X
°
°
°
f (α1 , α2 , · · · , αn ) = °
αk ek ° .
°
°
k=1
Since
|f (α1 , α2 , · · · , αn ) − f (β1 , β2 , · · · , βn )|
¯°
° °
° ¯ ¯°
°¯
n
n
n
¯° X
° °X
° ¯ ¯° X
°¯
¯°
° °
° ¯ ¯°
°¯
≤ ¯°
αk ek ° − °
βk ek °¯ ≤ ¯°
(αk − βk ) ek °¯
¯°
° °
° ¯ ¯°
°¯
k=1
k=1
k=1
v
u n
uX
2
|αk − βk | ,
≤ c2 t
k=1
24
HUIQIANG JIANG
f is Lipschitz continuous on compact set S. Let
c1 = min f (y) > 0,
y∈S
we have
v
°
°
u n
n
°
°X
uX
°
°
2
t
α
e
≥
c
|αk | .
°
k k°
1
°
°
k=1
k=1
Since
c1 |y1 − y2 | ≤ kA (y1 ) − A (y2 )k ≤ c2 |y1 − y2 | ,
−1
both A and A
are continuous and hence A is a homeomorphism.
¤
Lemma 4. Let M be a closed subspace of a normed space X.
(i) If dim M < ∞, then M is complemented in X.
(ii) If dim (X/M ) < ∞, then M is complemented in X.
n
Proof. (i) Let {ek }k=1 be a basis for M . Then every x ∈ M has a unique representation
n
X
x=
αk (x) ek .
k=1
Here each αk is a continuous linear functional on M which, by Hahn-Banach Theorem, can be extended to a continuous linear functional on X. Let
N=
n
\
N (αk ) .
k=1
Then one can verify X = M ⊕ N .
n
(ii) Let π : X → X/M be the quotient map and let {ek }k=1 be a basis for X/M .
n
Pick xk ∈ X such that πxk = ek and let N be the vector space spanned by {xk }k=1 .
Then X = M ⊕ N .
¤
Theorem 54. If X is a Banach space, T ∈ B (X) is compact and λ 6= 0, then
T − λI has closed range.
Proof. Let N = N (T − λI), then dim N < ∞. Hence there exists a closed subspace
M of X, such that X = M ⊕ N . Define S ∈ B (M, X) by Sx = T x − λx. Then S
is one-to-one and R (S) = R (T − λI).
Next, we claim the existence of r > 0 such that
r kxk ≤ kSxk for all x ∈ M.
If not, there exists {xn } ⊂ M , such that kxn k = 1 and Sxn → 0. Since T is
compact, passing to a subsequence if necessary, T xn → x0 as n → ∞, hence
λxn → x0 ∈ M . Now
Sx0 = lim λSxn = 0.
implies x0 = 0 which contradicts to kx0 k = lim kλxn k = |λ|.
Now for any Cauchy sequence {yn } ⊂ R (S), there exists {xn } ⊂ S such that
Sxn = yn . Then {xn } is a Cauchy sequence in M , and hence converges to some
x0 ∈ M . So we have yn → Sx0 ∈ R (S). Hence R (S) is complete and closed. ¤
n
Lemma 5. Suppose X is a Banach space, T ∈ B (X), {λi }i=1 are distinct eigenn
n
values of T with eigenvectors {ei }i=1 . Then {ei }i=1 are linearly independent.
FUNCTIONAL ANALYSIS
n
Proof. If {ei }i=1 are linearly dependent, we assume en =
25
Pn−1
i=1
αi ei . Since
(A − λi I) (A − λj I) = (A − λj I) (A − λi I) ,
we have
(λn − λ1 ) · · · (λn − λn−1 ) en = (A − λ1 ) · · · (A − λn−1 ) en
=
n−1
X
αi (A − λ1 ) · · · (A − λn−1 ) ei = 0,
i=1
a contradiction.
¤
Theorem 55. Suppose X is a Banach space, T ∈ B (X) is compact, and λ 6= 0 is
an eigenvalue of T . Then R (T − λI) 6= X.
Proof. If for some λ0 ∈ E, R (T − λ0 I) = X. Let S = T − λ0 I and define Mn to
be the null space of S n . Then we have Mn ⊂ Mn+1 and SMn+1 = Mn for each
n ≥ 1. Let x1 ∈ M1 and x1 6= 0. Since R (S) = X, there exists xn ∈ X, such that
Sxn+1 = xn for each n ≥ 1. Hence, xn ∈ Mn but xn 6∈ Mn−1 for any n ≥ 2.
Riesz’s lemma implies the existence of yn ∈ Mn such that kyn k = 1 and
dist (yn , Mn−1 ) > 21 for any n ≥ 2. For any 2 ≤ m < n,
kT yn − T ym k = kλ0 yn + Syn − Sym − λ0 ym k ≥ |λ0 | dist (yn , Mn−1 ) >
|λ0 |
.
2
Hence, {T yn } has no convergent subsequence, a contradiction.
¤
Theorem 56. Suppose X is a Banach space, T ∈ B (X) is compact, r > 0 and E
is a set of eigenvalues λ of T such that |λ| > r. Then E is a finite set.
∞
Proof. If E is an infinite set, we assume {λn }n=1 ⊂ E be distinct. Then there exists
n
en 6= 0 such that T en = λn en . Let Mn be the space spanned by {ek }k=1 . Then
we have Mn ⊂ Mn+1 for each n ≥ 1 and (T − λn+1 I) Mn+1 ⊂ Mn . Moreover,
en+1 ∈ Mn+1 \Mn . Riesz’s lemma implies the existence of yn ∈ Mn such that
kyn k = 1 and dist (yn , Mn−1 ) > 12 for any n ≥ 2. For any 2 ≤ m < n,
kT yn − T ym k = kλn yn + (T − λn I) yn − (T − λm ) Sym − λm ym k ≥ |λn | dist (yn , Mn−1 ) >
Hence, {T yn } has no convergent subsequence, a contradiction.
¤
Lemma 6. Let M0 be a closed subspace of a locally convex space Y and Σ be the
space of all continuous linear functionals on Y that annihilate M0 , then
dim Y /M0 ≤ dim Σ.
n
Proof. For any n ≤ dim Y /M0 , there exists linearly independent vectors {yi }i=1 ⊂
k
Y \M0 . Let Mk be the smallest subspace containing M0 and {yi }i=1 . There exists,
for each k, a continuous linear functional Λk such that Λk yk = 1 and Λk y = 0 for
n
any y ∈ Mk−1 . Since {Λk }k=1 are linearly independent, we have dim Σ ≥ n. Hence
dim Y /M0 ≤ dim Σ.
¤
Theorem 57 (Riesz-Schauder). Suppose X is a Banach space and T ∈ B (X) is
compact.
(a) If λ 6= 0, then the four numbers
α = dim N (T − λI) , β = dim X/R (T − λI) ,
α∗ = dim N (T ∗ − λI) , β ∗ = dim X/R (T ∗ − λI)
r
.
2
26
HUIQIANG JIANG
are equal and finite.
(b) If λ 6= 0 and λ ∈ σ (T ), then λ is an eigenvalue of T and of T ∗ .
(c) σ (T ) is compact, at most countable, and has at most one limit point, namely 0.
Proof. (a) Applying Lemma 6 with Y = X, M0 = R (T − λI), we have
⊥
β = dim X/R (T − λI) ≤ dim R (T − λI) = dim N (T ∗ − λI) = α∗ .
Next, let Y = X ∗ with its weak-* topology, and M0 = R (T ∗ − λI), then M0 is
weak-* closed. Hence Lemma 6 implies
β ∗ = dim X/R (T ∗ − λI) ≤ dim⊥ R (T ∗ − λI) = dim N (T − λI) = α.
Now we show α ≤ β, if not, we have β < α < ∞. Then there exists closed
subspaces E and F ,
X = N (T − λI) ⊕ E and X = R (T − λI) ⊕ F.
For each x ∈ X, there exists a unique representation x = x1 + x2 with x1 ∈
N (T − λI) and x2 ∈ E. Define π : X → N (T − λI) so that πx = x1 . Then π
is continuous since its graph is closed. Since dim F < dim N (T − λI), there exists
a map φ of N (T − λI) onto F such that φ (x0 ) = 0 for some x0 6= 0. Define
Φ ∈ B (X) by
Φx = T x + φπx.
Then Φ is compact. For x ∈ E,
(Φ − λI) x = (T − λI) x,
hence (Φ − λI) E = R (T − λI). For x ∈ N (T − λI),
(Φ − λI) x = φx,
hence (Φ − λI) N (T − λI) = F , hence R (Φ − λI) = X. On the other hand,
(Φ − λI) x0 = 0,
so λ is an eigenvalue of Φ and R (Φ − λI) 6= X, a contradiction.
Finally, since T ∗ ∈ B (X ∗ ) is compact, α∗ ≤ β ∗ .
(b) follows from (a).
(c) σ (T ) is at most countable, and has at most one possible limit point 0. If
dim X = ∞, 0 ∈ σ (T ), hence σ (T ) is compact; If dim X < ∞, then σ (T ) is finite
and hence compact too.
¤
Theorem 58. Suppose X is a Banach space and T ∈ B (X) is compact.
(1) If λ 6= µ, T x = λx, and T ∗ x∗ = µx∗ , then hx, x∗ i = 0.
(2) Let λ 6= 0 be an eigenvalue of T . The equation
(T − λI) x = y
has a solution iff hy, y ∗ i = 0 for any y ∗ such that T ∗ y ∗ = λy ∗ . The equation
(T ∗ − λI) x∗ = y ∗
has a solution iff hy, y ∗ i = 0 for any y such that T y = λy.
Proof. (1)
µ hx, x∗ i = hx, µx∗ i = hx, T ∗ x∗ i = hT x, x∗ i = hλx, x∗ i = λ hx, x∗ i ,
λ 6= µ implies hx, x∗ i = 0.
(2) This follows from Theorem 5.1.
¤
FUNCTIONAL ANALYSIS
27
7. Bounded operators on Hilbert spaces
7.1. Basic Concepts.
Definition 17. A complex vector space H is called an inner product space if there
exists a map (·, ·) : H × H → C such that for any x, y, z ∈ H and α ∈ C
(a) (x, y) = (y, x).
(b) (x + y, z) = (x, z) + (y, z).
(c) (αx, y) = α (x, y) .
(d) (x, x) ≥ 0.
(e) (x, x) = 0 iff x = 0.
The map (·, ·) is called the inner product.
Similarly, we can define real inner product space.
1
We define kxk = (x, x) 2 for any x ∈ H.
Theorem 59 (Cauchy-Schwarz). If x, y ∈ H, then
|(x, y)| ≤ kxk kyk .
Theorem 60. k·k is a norm for H.
Proof. One can verify
kx + yk ≤ kxk + kyk .
¤
Hence, any inner product space H is a normed space. One can verify that the
inner product is continuous under the topology induced by the norm.
We say H is a Hilbert space if it is complete under such norm.
Example 3. l2 and L2 (X, µ) are Hilbert space with their natural inner product.
Lemma 7. If H is a real inner product space, then for any x, y ∈ H,
´
1³
2
2
(7.1)
(x, y) =
kx + yk − kx − yk .
4
If H is a complex inner product space, then for any x, y ∈ H,
´
1³
2
2
2
2
(7.2)
(x, y) =
kx + yk − kx − yk + i kx + iyk − i kx − iyk .
4
Theorem 61. Let H be an inner product space. Then for any x, y ∈ H, we have
the parallelogram law
³
´
2
2
2
2
kx + yk + kx − yk = 2 kxk + kyk .
On the other hand, any normed space satisfying parallelogram law has an inner
product structure.
Theorem 62. Let X be a normed space satisfying parallelogram law. Then (·, ·)
defined in (7.1), (7.2) is an inner product.
Proof. We only prove the real case.
(a) (x, y) = (y, x). (d) (x, x) ≥ 0. (e) (x, x) = 0 iff x = 0.
28
HUIQIANG JIANG
(b) (x + y, z) = (x, z) + (y, z). To see this,
(x, z) + (y, z)
´
1³
2
2
2
2
=
kx + zk − kx − zk + ky + zk − ky − zk
4Ã
°
°2 °
°2 !
°
°x + y
°
1 °
x+y
°
°
°
=
+ z° − °
− z°
°
°
2
2
2
µ
¶
x+y
=2
,z .
2
It is easy to verify (0, z) = 0, hence the above identity implies (x, z) = 2
Hence
µ
¶
x+y
(x, z) + (y, z) = 2
, z = (x + y, z) .
2
(c) (tx, y) = t (x, y). To see this, given x, y ∈ X, we define for any t ∈ R,
¡x
2,z
¢
.
f (t) = (tx, y) .
Then f is continuous and (b) implies for any t1 , t2 ∈ R
f (t1 + t2 ) = f (t1 ) + f (t2 ) .
However, from the Lemma below, such f must satisfy f (t) = tf (1). hence (tx, y) =
t (x, y).
¤
Lemma 8. Let f be a real continuous function on R such that for any t1 , t2 ∈ R
f (t1 + t2 ) = f (t1 ) + f (t2 ) .
Then f (t) = tf (1) holds for any t ∈ R.
Proof. First, for any n ∈ N,
f (nt) = nf (t) for any t ∈ R.
Taking n = 2 and t = 0, we have f (0) = 0. Now for any positive rational number
m
n , we have
µ ¶
³n´
1
n
f
= nf
= f (1) .
m
m
m
And since
³n´
³ n´
f
+f −
= f (0) = 0,
m
m
¡ n¢
n
we have f − m = − m
f (1). Hence for any r ∈ Q, f (r) = rf (1). Since f is
continuous, f (t) = tf (1) holds for any t ∈ R.
¤
7.2. Orthogonal Projection. Let H be a Hilbert space.
We say x, y ∈ H are orthogonal to each other if (x, y) = 0, and we write x⊥y.
Let E, F ⊂ H, we write E⊥F if (x, y) = 0 for any x ∈ E and y ∈ F . And we write
E ⊥ = {y ∈ H : (x, y) = 0 for any x ∈ E}.
Here are some basic properties:
(i) x⊥y iff y⊥x.
(ii) x⊥H iff x = 0.
(iii) E ⊂ F implies F ⊥ ⊂ E ⊥ .
FUNCTIONAL ANALYSIS
29
(iv) For any E ⊂ H, x ∈ E ∩ E ⊥ implies x = 0.
(v) If x⊥y, then
kx + yk = kxk + kyk .
(iv) For any E ⊂ H, E ⊥ is a closed subspace of H.
Lemma 9. Let E be a nonempty closed convex set in H. Then for any x ∈ H,
there exists a unique x0 ∈ E such that
d (x, E) = kx − x0 k .
Proof. Let xn ∈ E be such that
lim kx − xn k = d (x, E) .
n→∞
From parallelogram identity, for any n, m,
°
°
°
°2
°
° xm − xn °2
°
° = kxn − xk2 + kxm − xk2 − 2 ° xm + xn − x°
2°
°
°
°
°
2
2
2
2
≤ kxn − xk + kxm − xk − 2d (x, H)
n
since xm +x
∈ E. Hence {xn } is a Cauchy sequence. Since E is complete, xn → x0
2
as n → ∞ for some x0 ∈ E. And we have d (x, H) = kx − x0 k. If y0 ∈ E satisfies
d (x, H) = kx − y0 k, then parallelogram identity implies
°
°
° x0 − y0 °2
°
° ≤ 0,
2°
2 °
hence x0 = y0 .
¤
Lemma 10. Let E be a subspace of H, x ∈ H, and x0 ∈ E. Then d (x, E) =
kx − x0 k implies x − x0 ⊥E.
Proof. For any y ∈ E, y 6= 0 and λ ∈ K, x0 + λy ∈ E, hence
¡
¢
2
2
2
2
d2 ≤ kx − x0 − λyk = kx − x0 k − 2 Re λ̄ (x − x0 ) , y + |λ| kyk .
Let λ =
(x−x0 ,y)
,
kyk2
we have
2
2
d2 ≤ d2 − |λ| kyk .
Hence λ = 0, i.e. x − x0 ⊥y. So we have x − x0 ⊥E.
¤
Theorem 63. Let E be a closed subspace of H. Then H = E ⊕ E ⊥ . E ⊥ is called
the orthogonal complement of E.
Proof. E ⊥ is closed and E ∩ E ⊥ = {0}. Now for any x ∈ H, let x0 ∈ E be such
that d (x, E) = kx − x0 k. Then x1 = x − x0 ∈ E ⊥ . Hence E + E ⊥ = H.
¤
¡ ¢⊥
Corollary 12. Let E be a closed subspace of H. Then E ⊥ = E.
Proof.
¡ ¢⊥
H = E ⊕ E⊥ = E⊥ ⊕ E⊥.
¡ ¢⊥
¡ ¢⊥
Since E ⊂ E ⊥ , we conclude E ⊥ = E.
¤
Definition 18. Let E be a closed subspace of H and x ∈ H. x0 is said to be the
orthogonal projection of x on E if x − x0 ∈ E ⊥ . And we say PE : x 7→ x0 is a
projector on H.
30
HUIQIANG JIANG
Theorem 64. Let E be a closed subspace of H and P = PE be the projector. Then
P is a bounded linear operator satisfying
P = P 2 (idempotent),
(P x, y) = (x, P y) (symmetric).
Conversely, a bounded linear operator P on a Hilbert space is a projector upon
M = R (P ) if it is idempotent and symmetric.
Proof. The first part is easy. Now let P be an idempotent, symmetric and bounded
linear operator on H and M = R (P ). x ∈ M iff for some y ∈ H, x = P y = P 2 y =
P x. Hence M is closed. For any x ∈ M , we have P x = x = PM x. And for any
x ∈ M ⊥ , PM x = 0. On the other hand,
¡
¢
(P x, P x) = x, P 2 x = 0,
hence P x = 0. Since H = E ⊕ E ⊥ , P = PM .
¤
7.3. Orthonormal Base. Let H be a Hilbert space. A set S of vectors in H is
called an orthogonal set if x⊥y for any x, y ∈ H. If in addition, kxk = 1 for any
x ∈ S, then S is called an orthonormal set. An orthonormal set S of a Hilbert
space X is called a complete orthonormal system or an orthonormal base of X, if
no orthonormal set of X consists S as a proper subset.
Theorem 65. A nontrivial Hilbert space has at least one complete ortonormal base.
Moreover, if S is an orthonormal set, there is an orthonormal base containing S.
Proof. By Zorn’s lemma.
¤
Theorem 66 (Parseval’s Identity). Let S = {eα }α∈A be an orthonormal base of a
Hilbert space H. For any x ∈ H, we define its Fourier coefficients
xα = (x, eα ) ,
then we have
x=
X
2
xα eα and kxk =
α∈A
X
2
|xα | .
α∈A
Proof. We first show Bessel’s inequality
X
2
2
(7.3)
|xα | ≤ kxk .
α∈A
Pn
Let
be any finite subset of A. Then k=1 xαk eαk is the projection of x
n
upon span {eαk }k=1 . Hence
n
X
2
2
|xαk | ≤ kxk
n
{αk }k=1
k=1
This implies (7.3), and in particularly at most countably many xα are nonzero, say
∞
{αk }k=1 . Let
∞
X
xαk eαk
x0 =
k=1
0
which is well defined. Then x − x ⊥eα for any α ∈ A. Since S is complete,
x − x0 = 0.
¤
FUNCTIONAL ANALYSIS
31
Theorem 67 (Schmidt’s Orthogonalization). Given a finite or countably infinite
sequence of linearly independent vectors {xn }. Then there exists an orthonormal
set which has the same cardinal number and spans the same linear space as {xn }.
Proof.
x1
,
kx1 k
x2 − (x2 , e1 ) e1
e2 =
,
kx2 − (x2 , e1 ) e1 k
Pk
xk+1 − j=1 (xk+1 , ej ) ej
°.
ek+1 = °
Pk
°
°
°xk+1 − j=1 (xk+1 , ej ) ej °
e1 =
¤
Theorem 68. Let H be a separable Hilbert space. Then H has an orthonormal
base consisting of at most countable elements.
Proof. Let {xn } be a dense subset, we can reduce it to a linearly independent
subset. Applying Schmidt’s Orthogonalization, we obtain an orthonormal set which
is countable and span a linear space containing all {xn }, hence X.
¤
Corollary 13. Every separable Hilbert space H is isomorphic and isometric to Kn
or l2 .
7.4. Riesz Representation. Next, we look at the dual space H ∗ of H.
Theorem 69 (Riesz representation theorem). There is a conjugate-linear isometry
y → Λ of H onto H ∗ , given by
Λx = (x, y) , x ∈ H.
Proof. For any y ∈ H,
Λx = (x, y) , x ∈ H
defines a continuous linear functional Λ with kΛk = kyk. We only need to check
every Λ ∈ H ∗ can be represented this way. We can take y = 0 if Λ = 0. Now
⊥
assume Λ 6= 0, then there exists z ∈ N (Λ) , z 6= 0. Since for any x ∈ H,
(Λx) z − (Λz) x ∈ N (Λ) ,
we have
(Λx) (z, z) − (Λz) (x, z) = 0.
So we have for any x ∈ H,
Λx =
h
if we choose y =
1
(z,z) Λz
(Λz) (x, z)
= (x, y)
(z, z)
i
z.
Theorem 70. Any Hilbert space H is reflexive.
¤
32
HUIQIANG JIANG
Proof. Let ϕ : H → H ∗∗ be the natural imbedding. Then
hx∗ , ϕ (x)i = hx, x∗ i = (x, T x∗ ) .
We need to show ϕ is onto. Let T : H ∗ → H be the conjugate-linear isometry in
Riesz representation theorem, then H ∗ is a Hilbert space with inner product
(x∗ , y ∗ ) = (T y ∗ , T x∗ ) for any x∗ , y ∗ ∈ H ∗ .
Applying Riesz representation theorem to H ∗ , for any x∗∗ ∈ H ∗∗ , there exists
y ∗ ∈ H ∗ such that
hx∗ , x∗∗ i = (x∗ , y ∗ ) = (T y ∗ , T x∗ ) = hx∗ , ϕ (T y ∗ )i .
Hence x∗∗ = ϕ (T y ∗ ).
¤
7.5. Adjoint operators. In this section, we assume all Hilbert spaces are complex
linear spaces. Let H, G be two complex Hilbert spaces. Let T ∈ B (H, G). We
could define adjoint operator in B (H ∗ , G∗ ). Since H ∗ , G∗ can be represented by
H, G respectively, we could instead define adjoint operator T ∗ ∈ B (G, H).
Theorem 71. Let f be a complex valued functional defined on the product space
H × G such that
(1) Sesqui-linearity:
f (α1 x1 + α2 x2 , y) = α1 f (x1 , y) + α2 f (x2 , y) ,
f (x, β1 y1 + β2 y2 ) = β1 f (x, y1 ) + β2 f (x, y2 ) .
(2) Boundedness:
M=
sup
|f (x, y)| < ∞.
x∈H,y∈G,kxk=kyk=1
Then there exists a unique S ∈ B (G, H) satisfying for any x ∈ H, y ∈ G,
f (x, y) = (x, Sy) .
Moreover, kSk = M .
Proof. Since for any x ∈ H, y ∈ G,
|f (x, y)| ≤ M kxk kyk .
For each y ∈ G, the mapping x 7→ f (x, y) defines a bounded linear functional on
H. Riesz representation theorem implies the existence of a unique element, denoted
by Sy ∈ H such that
f (x, y) = (x, Sy) .
One can verify that S is linear and since
2
kSyk = f (Sy, y) ≤ M kSyk kyk ,
we have kSyk ≤ M kyk for any y ∈ G. So S ∈ B (G, H) with kSk ≤ M . Finally,
|f (x, y)| = |(x, Sy)| ≤ kxk kSyk ≤ kSk kxk kyk ,
hence kSk ≥ M . So we have kSk = M .
¤
Let T ∈ B (H, G). Then (T x, y) is sesquilinear and bounded on H × G, so
there exists an adjoint operator T ∗ ∈ B (G, H), such that kT k = kT ∗ k and for any
x ∈ H, y ∈ G,
(T x, y) = (x, T ∗ y) .
FUNCTIONAL ANALYSIS
33
Theorem 72. Let H, G, K be Hilbert spaces, T, S ∈ B (H, G), R ∈ B (K, H) and
α, β ∈ C. Then
∗
(i) (T ∗ ) = T ;
2
∗ 2
(ii) kT k = kT k = kT ∗ T k;
∗
(iii) (αS + βT ) = ᾱS ∗ + β̄T ∗ ;
∗
(iv) (T R) = R∗ T ∗ ;
−1
T¢is invertible if and only if T ∗ is invertible. If T is invertible, then (T ∗ ) =
¡(v)−1
∗
.
T
If T ∈ B (H). Then
(vi) λ ∈ σ (T ) iff λ̄ ∈ σ (T ∗ ).
(vii) Let T x = λx and T ∗ y = µy. If λ 6= µ̄, then (x, y) = 0.
∗
Proof. (i) (T x, y) = (x, T ∗ y), hence (y, T x) = (T ∗ y, x) so (T ∗ ) = T .
(ii) First, kT ∗ T k ≤ kT k kT ∗ k. On the other hand, for any x ∈ H, kxk = 1, we have
2
kT xk = (T x, T x) = (x, T ∗ T x) ≤ kT ∗ T k .
2
Hence kT k ≤ kT ∗ T k.
(iii) Trivial.
(iv) Trivial.
¡
¢∗
¡
¢∗
(v) T T −1 = IG and T −1 T = IH implies T −1 T ∗ = IG and T ∗ T −1 = IH .
(vi) Trivial.
(vii) λ (x, y) = µ̄ (x, y).
¤
Theorem 73. Let H, G be Hilbert spaces and T ∈ B (H, G). Then
⊥
⊥
⊥
⊥
N (T ) = R (T ∗ ) , N (T ∗ ) = R (T ) ,
R (T ) = N (T ∗ ) , R (T ∗ ) = N (T ) .
Proof. For any x ∈ N (T ), (T x, y) = (x, T ∗ y) = 0 holds for any y ∈ G, hence
⊥
⊥
x ∈ R (T ∗ ) . On the other hand, for any x ∈ R (T ∗ ) , (x, T ∗ y) = (T x, y) = 0
holds for any y ∈ G. Let y = T x, we have T x = 0, i.e., x ∈ N (T ). The other three
identities follows from the first one.
¤
Definition 19. Let T ∈ B (H). We say T is an self-adjoint (hermitian) operator
if T = T ∗ .
Remark 7. Projectors are self-adjoint.
Theorem 74. Let T ∈ B (H). Then T is self-adjoint iff (T x, x) ∈ R for any
x ∈ H.
Proof. ”⇒” trivial.
”⇐” For any x, y ∈ H,
1
(T x, y) = [(T (x + y) , x + y) − (T (x − y) , x − y)]
4
1
+ i [(T (x + iy) , x + iy) − (T (x − iy) , x − iy)] .
4
If (T x, x) ∈ R for any x ∈ H, then (T x, y) = (T y, x) = (x, T y). Hence T = T ∗ .
¤
Theorem 75. The set of self-adjoint operators in B (H) is closed in B (H). And
any real linear combination of self-adjoint operators is self-adjoint.
34
HUIQIANG JIANG
Definition 20. An operator T ∈ B (H) is said to be normal if T ∗ T = T T ∗ .
Self-adjoint operators are normal.
Lemma 11. If T ∈ B (H) and if (T x, x) = 0 for every x ∈ H, then T = 0.
Proof. For any x, y ∈ H,
1
(T x, y) = [(T (x + y) , x + y) − (T (x − y) , x − y)]
4
1
+ i [(T (x + iy) , x + iy) − (T (x − iy) , x − iy)] .
4
¤
Theorem 76. An operator T ∈ B (H) is normal iff kT xk = kT ∗ xk for every
x ∈ H. A normal operator has the following properties:
(i) N (T ) = N (T ∗ );
(ii) R (T ) is dense in H iff T is one to one;
(iii) T is invertible iff there exists δ > 0 such that kT xk ≥ δ kxk for every x ∈ H.
(iv) If T x = αx for some x ∈ H, α ∈ C, then T ∗ x = ᾱx.
(v) If α, β are distinctive eigenvalues of T , then the corresponding eigenspaces are
orthogonal to each other.
Proof. From Lemma 11, T is normal iff for any x ∈ H
(T ∗ T x, x) = (T T ∗ x, x)
which is equivalent to
(T x, T x) = (T ∗ x, T ∗ x) ,
∗
i.e. kT xk = kT xk.
(i) Follows from kT xk = kT ∗ xk.
⊥
(ii) N (T ) = N (T ∗ ) = R (T ) .
(iii) ”⇒” Open mapping Theorem. ”⇐” one to one implies R (T ) is dense and
kT xk ≥ δ kxk for every x ∈ H implies R (T ) is closed. Hence R (T ) = H and T is
invertible.
∗
(iv) (T − αI) = T ∗ − ᾱI, then apply part (i).
(v) Let T x = αx, T y = βy. Then
¡
¢
α (x, y) = (T x, y) = (x, T ∗ y) = x, β̄y = β (x, y) .
¤
Definition 21. An operator T ∈ B (H) is said to be unitary if T ∗ T = T T ∗ = I.
Unitary operators are normal.
Theorem 77. Let U ∈ B (H). Then the following three statements are equivalent:
(a) U is unitary.
(b) R (U ) = H and (U x, U y) = (x, y) for all x, y ∈ H.
(c) R (U ) = H and kU xk = kxk for every x ∈ H.
Proof. ”(a)⇒(b)” Trivial.
”(b)⇒(c)” Trivial.
”(c)⇒(a)” (c) implies that U is invertible. Now kU xk = kxk implies U ∗ U = I,
hence U ∗ = U −1 and U U ∗ = I.
¤
FUNCTIONAL ANALYSIS
35
8. Banach Algebra
Definition 22. A complex algebra is a vector space A over the complex field C in
which a multiplication is defined that satisfies for any x, y, z ∈ A and α ∈ C,
x (yz) = (xy) z,
(x + y) z = xz + yz, x (y + z) = xy + xz
and
α (xy) = (αx) y = x (αy) .
If in addition, A is a Banach space with respect to a norm satisfies the multiplicative
inequality
kxyk ≤ kxk kyk for any x, y ∈ A
and if A contains a unit element e with kek = 1 such that
xe = ex = x for any x ∈ A,
then A is called a Banach algebra.
Remark 8. In Banach algebra, unit element is unique.
Remark 9. Multiplication is continuous in Banach algebra.
Example 4. Let C (K) be the Banach space of all continuous functions on a
nonempty compact Hausdorff space K with the supremum norm. Then C (K) is a
commutative algebra with multiplication (f g) (p) = f (p) g (p).
Example 5. Let X be a Banach space. Then B (X) is a Banach algebra with
T S = T ◦ S and unit e = I. Moreover, every subalgebra of B (X) containing I is a
Banach algebra. If dim X = n < ∞, then B (X) is isomorphic to the algebra of all
complex n by n matrices. If n ≥ 2, then B (X) is not commutative.
8.1. Complex Homomorphisms.
Definition 23. Suppose A is a complex algebra and φ is a linear functional on A
which is not identically 0. If
φ (xy) = φ (x) φ (y) for any x, y ∈ A,
then φ is called a complex homomorphism on A.
An element x ∈ A is said to be invertible if it has an inverse in A, and we write
x−1 x = xx−1 = e.
The inverse of x is unique if x is invertible.
Proposition 5. If φ is a complex homomorphism on a complex algebra A with unit
e, then φ (e) = 1 and φ (x) 6= 0 for every invertible x ∈ A.
Proof. For some y ∈ A, φ (y) 6= 0. Since φ (y) = φ (ye) = φ (y) φ (e), φ (e) = 1.
Now if x is invertible, we have
¡
¢
¡
¢
1 = φ (e) = φ xx−1 = φ (x) φ x−1 ,
hence φ (x) 6= 0.
The inverse of the proposition is also true.
¤
36
HUIQIANG JIANG
Theorem 78 (Gleanson, kahane, Zelazko). If φ is a linear functional on a complex
algebra A with unit e such that φ (e) = 1 and φ (x) 6= 0 for every invertible x ∈ A,
then φ is a complex homomorphism on A.
Next, we discuss some basic properties of Banach algebra.
Theorem 79. Suppose A is a Banach algebra, x ∈ A, kxk < 1. Then
(a) e − x is invertible and
∞
X
−1
(e − x) =
xn ,
n=0
here we define x0 = e,
(b)
°
°
°
°
−1
°(e − x) − e − x° ≤
2
kxk
,
1 − kxk
(c) |φ (x)| < 1 if φ is a complex homomorphism on A.
Proof. (a) Let
sn = e + x + x2 + · · · + xn .
n
Since kxn k ≤ kxk and kxk < 1, sn is a Cauchy sequence which converges to some
s0 ∈ A. Now
sn (e − x) = (e − x) sn = e − xn+1 → e as n → ∞.
Since multiplication is continuous, s0 (e − x) = (e − x) s0 = e, hence e − x is invertible.
(b)
°
°
°
°
−1
°(e − x) − e − x° = ks0 − e − xk
°
°
∞
∞
2
°X
° X
kxk
°
n
n°
.
=°
x °≤
kxk =
°
°
1 − kxk
n=2
n=2
(c) For any λ ∈ C such that |λ| ≥ 1, e − λ−1 x is invertible, hence
¡
¢
1 − λ−1 φ (x) = φ e − λ−1 x 6= 0.
So φ (x) 6= λ.
¤
8.2. Basic Properties of Spectra. We assume that A is a Banach algebra from
now on and we define G = G (A) to be the set of all invertible elements of A.
One can verify that G forms a group.
Definition 24. Let x ∈ A. The spectrum σ (x) of x is the set of all complex
numbers λ such that λe − x is not invertible. The complement of σ (x) is called the
resolvent set of x.
Definition 25. The spectral radius of x ∈ A is the number
ρ (x) = sup {|λ| : λ ∈ σ (x)} .
Theorem 80. Let x ∈ G (A). If h ∈ A satisfies
°
°−1
khk < °x−1 ° ,
FUNCTIONAL ANALYSIS
37
then x + h ∈ G (A),
(x + h)
−1
=
∞
X
¡ −1 ¢n −1
−x h x
n=0
and
° −1 °2
°
°
°x ° khk
°
−1
−1 °
.
°(x + h) − x ° ≤
1 − kx−1 k khk
Proof. Since
° −1 ° ° −1 °
°x h° ≤ °x ° khk < 1,
¡
¢
we have e + x−1 h ∈ G (A) and hence x + h = x e + x−1 h ∈ G (A). Moreover, we
have
∞
¡
¢−1 −1 X
¡ −1 ¢n −1
−1
−1
(x + h) = e + x h
x =
−x h x
n=0
So we have
°
°
∞
°
° °X
¡ −1 ¢n −1 °
°
°
°
−1
−1 °
−x h x °
°(x + h) − x ° = °
°
°
n=1
∞
∞
°
°X
° −1 °n X
° −1 °n+1
n
° x h° ≤
°x °
≤ °x−1 °
khk
n=1
° −1 °2
°x ° khk
.
=
1 − kx−1 k khk
n=1
¤
Theorem 81. G (A) is an open subset of A and the map x 7→ x−1 is a homeomorphism of G (A) onto G (A).
Proof. From Theorem 80, G (A) is open and the mapping ϕ : x 7→ x−1 is continuous. The mapping ϕ is also invertable with continuous ϕ−1 = ϕ, so ϕ is a
homeomorphism.
¤
Theorem 82. If A is a Banach algebra and x ∈ A, then the spectrum σ (x) of x
is compact and nonempty.
¡
¢
Proof. If λ > kxk then λe − x = λ e − λ−1 x ∈ G (A) and so λ 6∈ σ (x). Hence
σ (x) is bounded. Now we define g : λ 7→ λe − x, then g is continuous, hence
¡
¢C
σ (x) = g −1 (G (A))
is closed because G (A) is open. Now let f ∈ A∗ to be
chosen later. If σ (x) = ∅, then
³
´
−1
f (λe − x)
is defined on the whole complex plane C. For any λ0 ∈ C, whenever,
1
°,
|λ − λ0 | < °
°
−1 °
°(λ0 e − x) °
we have
°
°
°
−1 °
°(λ − λ0 ) (λ0 e − x) ° < 1,
38
HUIQIANG JIANG
and hence
(λe − x)
−1
−1
= ((λ − λ0 ) e + λ0 e − x)
³
´−1
−1
−1
= (λ0 e − x)
e + (λ − λ0 ) (λ0 e − x)
= (λ0 e − x)
−1
∞
X
n
−n
(−1) (λ0 e − x)
n
(λ − λ0 )
n=0
=
∞
X
n
(−1) (λ0 e − x)
−(n+1)
n
(λ − λ0 ) .
n=0
Since f is continuous,
∞
³
´ X
³
´
−1
n
−(n+1)
n
f (λe − x)
=
(−1) f (λ0 e − x)
(λ − λ0 ) ,
³
n=0
´
³
´
−1
−1
i.e., f (λe − x)
is analytic at any λ0 ∈ C. Hence, f (λe − x)
is an entire
function of λ. On the other hand, whenever |λ| > kxk, we have
(λe − x)
−1
∞
³
x ´−1 X xn
= λ−1 e −
=
,
λ
λn+1
n=0
hence
∞
n
¯ ³
°
°
´¯
X
kxk
¯
°
−1 ¯
−1 °
¯f (λe − x)
¯ ≤ kf k °(λe − x) ° ≤ kf k
n+1
n=0 |λ|
=
kf k
.
|λ| − kxk
³
´
−1
So f (λe − x)
is bounded on C and tends to 0 as |λ| → ∞. Liouville theorem
³
´
−1
implies that f (λe − x)
≡ 0. This is a contradiction if we choose f so that
°
³
´ °
°
−1
−1 °
f (λ0 e − x)
= °(λ0 e − x) ° 6= 0 for some λ0 ∈ C.
¤
Lemma 12. If A is a Banach algebra and x ∈ A, then
1
lim kxn k n
n→∞
exists and
1
1
lim kxn k n = inf kxn k n .
n→∞
n≥1
Proof. Let
1
r = inf kxn k n ,
n≥1
then
1
lim inf kxn k n ≥ r.
n→∞
For any ε > 0, there exists m, such that
1
kxm k m ≤ r + ε.
FUNCTIONAL ANALYSIS
39
And for any n ≥ 1, we can write n = mk + m̃ for some integers k and m̃ such that
0 ≤ m̃ < m, so
³
´ mk
1
1
n ° m̃ ° 1
°x ° n ≤ r + 2ε
kxn k n ≤ kxm k m
if n is sufficiently large. So we have
1
lim sup kxn k n ≤ r + 2ε.
n→∞
Letting ε → 0, we have
1
lim sup kxn k n ≤ r.
n→∞
¤
Theorem 83. If A is a Banach algebra and x ∈ A, then the spectral radius ρ (x)
of x satisfies
1
ρ (x) = lim kxn k n .
n→∞
1
n
Proof. For any |λ| > limn→∞ kxn k , we have
³ x ´n
lim
=0
n→∞ λ
and also the root test implies that
∞ ³ ´n
X
x
λ
n=0
converges. Since
(λe − x)
n ³ ´k
X
x
λ
k=0
we have
(λe − x)
µ
³ x ´n+1 ¶
=λ e−
,
λ
∞ ³ ´k
X
x
k=0
hence
−1
(λe − x)
=
λ
∞
X
= λe
λ−(n+1) xn .
n=0
λe − x ∈ G (A) and λ 6∈ σ (x). So
1
ρ (x) ≤ lim kxn k n .
n→∞
On the other hand, for any f ∈ A∗ ,
³
´
−1
f (λe − x)
C
is analytic on σ (x) , hence on {λ ∈ C : |λ| > ρ (x)}. And we have the Laurent
series
∞
³
´ X
−1
f (λe − x)
=
λ−(n+1) f (xn )
n=0
on {λ ∈ C : |λ| > ρ (x)}. Hence, for any r > ρ (x),
∞
X
|f (xn )|
< ∞.
rn+1
n=0
40
HUIQIANG JIANG
Now we define a series
© x n ª∞
rn
, then for any f ∈ A∗ ,
¯ µ n ¶¯
¯
x ¯¯
sup ¯¯f
< ∞,
rn ¯
n≥1
n=1
hence the uniformly boundedness principle implies
° n°
°x °
°
sup °
° n ° < ∞,
n≥1 r
hence
1
r > lim kxn k n .
n→∞
So we have
1
ρ (x) ≥ lim kxn k n .
n→∞
¤
Theorem 84 (Gelfand-Mazur). If A is a Banach algebra in which every nonzero
element is invertible, then A is isometrically isomorphic to C.
Proof. Let x ∈ A. Since σ (x) is nonempty, there exists λ ∈ σ (x), and hence
λe − x = 0 since every nonzero element is invertible. So σ (x) consists of exactly
one point which we denote by λ (x). The mapping x → λ (x) is an isomorphism of
A onto C which is also an isometry since |λ (x)| = kxk.
¤
Theorem 85. Let H be a Hilbert space.
(a) If U ∈ B (H) is unitary and λ ∈ σ (U ), then |λ| = 1.
(b) If S ∈ B (H) is self-adjoint and λ ∈ σ (S), then λ ∈ R.
Proof. (a) Since kU k = 1, ρ (U ) ≤ 1, hence |λ| ≤ 1 for any λ ∈ σ (U ). On the other
hand, if |λ| < 1, then kλU ∗ k < 1, hence
λI − U = −U (I − λU ∗ )
is invertible, so λ 6∈ σ (U ).
(b) Suppose λ = α + iβ ∈ σ (S). Let Sλ = S − λI, then for any x ∈ H,
2
kSλ xk = (Sx − λx, Sx − λx)
= (Sx − αx − iβx, Sx − αx − iβx)
2
2
2
2
= kSx − αxk + β 2 kxk − iβ (x, Sx − αx) + iβ (Sx − αx, x)
= kSx − αxk + β 2 kxk .
Hence kSλ xk ≥ |β| kxk. If β 6= 0, then Sλ is invertible. Hence we must have β = 0
and λ ∈ R.
¤
9. Vector valued functions
Definition 26. Let X be a vector space and E ⊂ X. The convex hull of E, denoted
by co (E) is the intersection of all convex subsets of X containing E.
Hence, co (E) is the smallest convex subsets of X containing E. It is not difficult
to see that co (E) is the set of all finite convex combinations of members of E.
If X is a topological vector space, then the closed convex hull of E is the closure of
co (E), denoted by co (E). One can verify that co (E) is closed using the continuity
of vector operations,.
FUNCTIONAL ANALYSIS
41
Theorem 86 (Mazur). Let X be a Banach (Frechet) space and K ⊂ X be compact.
Then co (K) is compact.
Proof. We only need to show that co (K) is totally bounded. For any ε > 0, there
N
exists {xn }n=1 such that
N
³
[
ε´
K⊂
B xn ,
.
2
n=1
Hence,
³
´
³ ε´
N
co (K) ⊂ co {xn }n=1 + B 0,
,
2
³
´
N
M
Since co {xn }n=1 is totally bounded, there exists {ym }m=1 ∈ X, such that
³
co
N
{xn }n=1
´
⊂
³
ε´
B ym ,
.
2
m=1
M
[
Hence
M
[
co (K) ⊂
B (ym , ε) .
m=1
So co (K) is totally bounded which implies co (K) is totally bounded. Since co (K)
¤
is complete, we conclude co (K) is compact.
Now we consider vector valued integration.
Theorem 87. Let X be a Banach space and µ be a Borel probability measure on a
compact metric space E. Let f : E → X be continuous. Then there exists a unique
y ∈ X, such that for any Λ ∈ X ∗ ,
Z
Λy =
Λf dµ.
E
Moreover, y ∈ co (f (E)) and we write
Z
y=
f dµ.
E
Proof. We regard X as a real vector space. Uniqueness of y follows from HahnBanach Theorem. Since f (E) is compact, Mazur’s Theorem implies that K =
n
co (f (E)) is compact. Let L = {Λk }k=1 be a finite subset of X ∗ and
½
¾
Z
EL = y ∈ K : Λk y =
Λk f dµ, 1 ≤ k ≤ n .
E
Each EL is closed and hence compact. If no EL is empty, then the collection of
all EL has the finite intersection property and hence the intersection of all EL is
nonempty and the existence of y is established. So we only need to prove that
EL 6= ∅ for any finite L ⊂ X ∗ . Regard L = (Λ1 , · · · , Λn ) as a mapping from X into
Rn , then L (f (E)) is compact. Define
Z
mi =
Λi f dµ and m = (m1 , · · · , mn ) ,
E
42
HUIQIANG JIANG
n
we claim m ∈ co (L (f (E))). If t ∈ Rn is not in the hull, then there exists {ci }i=1 ⊂
R such that for any u ∈ L (f (E)),
n
X
ci ui <
i=1
n
X
ci ti .
i=1
Hence, for any a ∈ E,
n
X
ci Λi f (a) <
i=1
n
X
ci ti .
i=1
Integrating over E, we have
n
X
ci mi <
i=1
n
X
ci ti ,
i=1
so t 6= m. Hence, m ∈ co (L (f (E))). Since L is linear, there exists y ∈ co (f (E)),
such that Ly = m. Hence y ∈ EL and EL 6= ∅.
¤
Remark 10. This integral can also be established using Riemann sums.
Remark 11. The requirement of µ being a probability measure can be dropped.
Theorem 88. Let X be a Banach space and µ be a positive Borel measure on a
compact metric space E. Let f : E → X be continuous. Then
°Z
° Z
°
°
° f dµ° ≤
kf k dµ.
°
°
E
E
R
Proof. Let y = E f dµ. We can choose Λ ∈ X ∗ such that Λy = kyk with kΛk = 1.
Hence for any a ∈ E,
Λf (a) ≤ kf (a)k .
Integrating over E, we have
°Z
°
¯Z
¯ Z
°
°
¯
¯
° f dµ° = kyk = |Λy| = ¯ Λf dµ¯ ≤
kf k dµ.
°
°
¯
¯
E
E
E
¤
Next, we discuss vector valued holomorphic functions.
Definition 27. Let Ω be an open set in C and let X be a Banach space over C.
(a) A function f : Ω → X is said to be weakly holomorphic in Ω if Λf is holomorphic
for every Λ ∈ X ∗ .
(b) A function f : Ω → X is said to be strongly holomorphic in Ω if
lim
w→z
f (w) − f (z)
w−z
exists for every z ∈ Ω.
The index of point z ∈ C w.r.t. a closed path Γ that doesn’t pass through z will
be denoted IndΓ (z). Hence
Z
1
dζ
.
IndΓ (z) =
2πi Γ ζ − z
Here we assume closed paths are piecewise continuously differentiable.
FUNCTIONAL ANALYSIS
43
Theorem 89. Let Ω be an open set in C and let X be a Banach space over C.
Assume that f : Ω → X is weakly holomorphic in Ω, then we have
(i) f is continous in Ω.
(ii) The Cauchy theorem and Cauchy formula hold: If Γ is a closed path in Ω such
that IndΓ (w) = 0 for any w 6∈ Ω, then
Z
f (ζ) dζ = 0,
Γ
and
Z
1
f (ζ) dζ
2πi Γ ζ − z
if z ∈ Ω and IndΓ (z) = 1. If Γ1 , Γ2 are closed paths in Ω such that IndΓ1 (w) =
IndΓ2 (w) = 0 for any w 6∈ Ω, then
Z
Z
f (ζ) dζ =
f (ζ) dζ.
f (z) =
Γ1
Γ2
(iii) f is strongly holomorphic in Ω.
Proof. (a) Assume 0 ∈ Ω, we shall prove that f is continuous at 0. Let r > 0 be
such that B̄2r (0) ⊂ Ω and Γ = ∂B2r (0) with positive orientation. For any Λ ∈ X ∗ ,
since Λf is holomorphic, for any z ∈ B̄r (0) \ {0},
Z
Z
Λf (z) − Λf (0)
1
(Λf (ζ) − Λf (0)) dζ
1
Λf (ζ) dζ
=
=
.
z
2πi Γ
(ζ − z) ζ
2πi Γ (ζ − z) ζ
Let M (Λ) = maxz∈Γ |Λf (z)|, then we have
¯ µ
¯
¶¯ ¯
¯
¯ ¯
¯
¯Λ f (z) − f (0) ¯ = ¯ Λf (z) − Λf (0) ¯ ≤ M (Λ) .
¯
¯
¯
¯
z
z
r
n
o
(0)
Hence f (z)−f
: z ∈ B̄r (0) \ {0} is weakly bounded, and hence strongly bounded.
z
So there exists K, such that for any 0 < z ≤ r,
kf (z) − f (0)k ≤ K |z| ,
so f is continuous at 0.
(b) Follows from Hahn Banach theorem.
(c) With the same setting as the proof of part (a), define
Z
1
f (ζ) dζ
y=
.
2πi Γ ζ 2
We have
1
f (z) − f (0)
=
z
2πi
Z
f (ζ) dζ
(ζ − z) ζ
Z
1
f (ζ) dζ
=y+z
.
2πi Γ (ζ − z) ζ 2
Γ
Since f is continuous and hence bounded on Γ,
lim
z→0
f (z) − f (0)
= y.
z
¤
Finally, we prove the Liouville theorem for vector valued holomorphic function.
44
HUIQIANG JIANG
Theorem 90. Let X be a Banach space over C. Assume that f : C → X is weakly
holomorphic and f (C) is weakly bounded in X, then f is constant.
Proof. For any Λ ∈ X ∗ , Λf is holomorphic and bounded in C, hence it is constant
from Liouville theorem. Hahn-Banach theorem implies that f is constant.
¤
10. Commutative Banach Algebras
10.1. Proper Ideals. A complex algebra A is said to be commutative if xy = yx
for any x, y ∈ A.
Definition 28. A subset J of a commutative complex algebra A is said to be an
ideal if
(a) J is a subspace of A and
(b) xy ∈ J whenever x ∈ A and y ∈ J.
If J 6= A, J is a proper ideal. Maximal ideals are proper ideals which are not
contained in any larger proper ideal.
Proposition 6. (a) No proper ideal of A contains any invertible elements of A.
(b) If J is an ideal in a commutative Banach algebra A, then its closure J¯ is also
an ideal.
Theorem 91. (a) If A is a commutative complex algebra with unit, then every
proper ideal of A is contained in a maximal ideal of A.
(b) If A is a commutative Banach algebra, then every maximal ideal of A is closed.
Proof. (a) Zorn’s Lemma of Haudorff’s maximality theorem.
(b) Let M be a maximal ideal. Then M is an ideal containing no invertible element
since G (A) is open. Henc M = M .
¤
Let A, B be two commutative Banach algebras. Let φ : A → B be a homomorphism, i.e.,
φ (xy) = φ (x) φ (y)
−1
holds for any x, y ∈ A. Then φ (0) is an ideal in A.
Conversely, suppose J is a proper closed ideal in A and π : A → A/J is the
quotient map.
Theorem 92. A/J is a Banach algebra and π is a homomorphism.
Proof. If x0 − x ∈ J and y 0 − y ∈ J, then
x0 y 0 − xy = x0 (y 0 − y) + (x0 − x) y ∈ J.
Hence π (x) π (y) = π (xy) is well defined. So A/J is a complex algebra and π is a
homomorphism. A/J is a Banach space. Since kπ (x)k ≤ kxk for any x ∈ A, π is
continuous. Suppose x1 , x2 ∈ A and δ > 0. Then there exists y1 , y2 ∈ J, such that
kxi + yi k ≤ kπ (xi )k + δ, i = 1, 2.
Since
(x1 + y1 ) (x2 + y2 ) ∈ x1 x2 + J,
kπ (x1 ) π (x2 )k ≤ kx1 + y1 k kx2 + y2 k ≤ (kπ (x1 )k + δ) (kπ (x2 )k + δ) .
Hence,
kπ (x1 ) π (x2 )k ≤ kπ (x1 )k kπ (x2 )k
FUNCTIONAL ANALYSIS
45
and A/J is a Banach algebra. Finally, we verify that π (e) is the unit of A/J. First,
π (e) π (x) = π (x) π (e) = π (x) , for any x ∈ A.
Since π (e) 6= 0,
kπ (e)k = kπ (e) π (e)k ≤ kπ (e)k kπ (e)k = kπ (e)k
2
implies kπ (e)k ≥ 1. Since kπ (e)k ≤ kek = 1, we conclude kπ (e)k = 1.
¤
Theorem 93. Let A be a commutative Banach algebra and let ∆ be the set of all
complex homomorphisms of A.
(a) Every maximal ideal of A is the kernel of some h ∈ ∆.
(b) If h ∈ ∆, then h−1 (0) is a maximal ideal of A.
(c) An element x ∈ A is invertible in A iff h (x) 6= 0 for every h ∈ ∆.
(d) An element x ∈ A is invertible in A iff x lies in no proper ideal of A.
(e) λ ∈ σ (x) iff h (x) = λ for some h ∈ ∆.
Proof. (a) Let M be a maximal ideal of A. Then A/M is a Banach algebra and π
is a homomorphism. For any a ∈ A\M , define
J = {ax + y : x ∈ A, y ∈ M } ,
then J is an ideal larger than M , hence J = A. So there exist x ∈ A, y ∈ M
s.t. ax + y = e, hence π (a) is invertible in A\M . Gelfand-Mazur theorem implies
the existence of an isomorphism j : A\M → C. Let h = j ◦ π, then h ∈ ∆ and
M = h−1 (0).
(b) If h ∈ ∆, then h−1 (0) is an ideal. It is maximal because it has codimension
1.
¡
¢
(c) If x is invertible, the h (x) h x−1 = h (e) = 1, hence h (x) 6= 0. If x is not
invertible, then {ax : a ∈ A} is a proper ideal which lies in a maximal one, and
hence in the kernel of some h ∈ ∆.
(d) It is equivalent to (c).
(e) λ ∈ σ (x) iff λe − x is not invertible iff h (λe − x) = 0 for some h ∈ ∆ iff
h (x) = λ for some h ∈ ∆.
¤
Theorem 94 (Wiener’s Lemma). Suppose f is a function on Rn , and
X
X
f (x) =
am eim·x ,
|am | < ∞.
m∈Zn
m∈Zn
n
If f (x) 6= 0 for every x ∈ R , then
X
X
1
=
bm eim·x with
|bm | < ∞.
f (x)
n
n
m∈Z
m∈Z
Proof. Let A be the set of all functions of the form in the theorem. Define for any
f ∈ A,
X
|am | .
kf k =
m∈Zn
Then A is a commutative Banach algebra with pointwise multiplication with unit
the constant function 1. For any x ∈ Rn , the map f → f (x) is a complex homomorphism. Now we show that any complex homomorphism on A is of this form.
ixk
Let h be a complex homomorphism.
° −1 ° For 1 ≤ k ≤ n, define gk (x) = e , then
−1
−ixk
°
°
, so we have kgk k = gk
= 1. Hence,
gk = e
¯ ¡
¢¯
|h (gk )| ≤ 1 and ¯h g −1 ¯ ≤ 1
k
46
HUIQIANG JIANG
and so |h (gk )| = 1. Hence there exist yk , 1 ≤ k ≤ n such that
h (gk ) = eiyk .
Let y = (y1 , y2 , · · · yn ). Then for any trigonometric polynomial P , we have
h (P ) = P (y) .
Since h is continuous on A, we conclude for any f ∈ A,
h (y) = f (y) .
The theorem is then a consequence of part (c) of Theorem 93.
¤
10.2. Gelfand Transforms.
Definition 29. Let ∆ be the set of all complex homomorphisms of a commutative
Banach algebra A. For any x ∈ A, the formula
x̂ (h) = h (x)
defines a function x̂ : ∆ → C, and x̂ is called the Gelfand transform of x. And the
map x → x̂ is also called Gelfand transform.
Let  = {x̂ : x ∈ A}. We define the Gelfand topology of ∆ as the weak toplogy
induced by Â, i.e., the weakest topology that makes every x̂ continuous. Then
 ⊂ C (∆), the algebra of all complex continuous functions on ∆. Theorem 93
implies that there is a one to one correspondence of maximal ideal J of A and
h ∈ ∆. So ∆ is also called maximal ideal space of A.
The radical of A, dented by rad A, is the intersection of all maximal ideals of A.
A is called semisimple if rad A = 0.
Theorem 95. Let ∆ be the maximal ideal space of a commutative Banach algebra
A.
(a) ∆ is a compact Hausdorff space.
(b) The Gelfand transform is a homomorphism of A onto a subalgebra  ⊂ C (∆),
whose kernel is rad A. The Gelfand transform is an isomorphism iff A is semisimple.
(c) For each x ∈ A, the range of x̂ is the spectrum σ (x). Hence,
kx̂k∞ = ρ (x) ≤ kxk
and x ∈ rad A iff ρ (x) = 0.
Proof. (b) It is easy to see that the Gelfand transform is linear. For every x, y ∈ A,
and h ∈ ∆,
x
cy (h) = h (xy) = h (x) h (y) = x̂ (h) ŷ (h) ,
hence the Gelfand transform is a homomorphism of A onto a subalgebra  ⊂ C (∆).
x in the kernel of the Gelfand transform iff x̂ = 0 iff h (x) = 0 for every h ∈ ∆ iff x
belongs every maximal ideal iff x ∈ rad A.
(c) λ is in the range of x̂ iff h (x) = λ for some h ∈ ∆ iff x ∈ σ (x). Finally,
x ∈ rad A iff x̂ = 0 iff ρ (x) = 0.
(a) Let K be the closed unit ball of A∗ . Then ∆ ⊂ K and the Gelfand topology is
equivalent to the restriction of the weak-* topology of A∗ . Banach-Alaoglu theorem
implies that K is weak-* compact, hence we only need to show that ∆ is closed.
Let Λ0 be in the weak* closure of ∆. We need to show Λ0 ∈ ∆.
FUNCTIONAL ANALYSIS
47
Fix x, y ∈ A and ε > 0, we define
W = {Λ ∈ A∗ : |Λzi − Λ0 zi | < ε 1 ≤ i ≤ 4}
where z1 = e, z2 = x, z3 = y and z4 = xy. Then W is a weak* nbhd of Λ0 and
therefore there exists h ∈ ∆ ∩ W . Hence,
|1 − Λ0 e| = |h (e) − Λ0 e| < ε
and
|Λ0 (xy) − Λ0 (x) Λ0 (y)|
≤ |Λ0 (xy) − h (xy)| + |h (x) h (y) − Λ0 (x) Λ0 (y)|
= |Λ0 (xy) − h (xy)| + |(h (x) − Λ0 (x)) h (y)| + |Λ0 (x) (h (y) − Λ0 (y))|
≤ ε (1 + |h (y)| + |Λ0 (x)|) ≤ ε (1 + kyk + |Λ0 (x)|) .
Letting ε → 0, we have
Λ0 (xy) = Λ0 (x) Λ0 (y) for any x, y ∈ A
and Λ0 (e) = 1. Hence Λ0 ∈ ∆.
¤
10.3. Involutions.
Definition 30. A mapping x → x∗ of a complex algebra A into A is called an
involution on A if for any x, y ∈ A and λ ∈ C,
∗
(x + y) = x∗ + y ∗ ,
∗
(λx) = λ̄x∗ ,
∗
(xy) = y ∗ x∗ ,
x∗∗ = x.
x ∈ A is said to be self-adjoint if x = x∗ .
Example 6. Let A = B (H) where H is a Hilbert space. Then x → x∗ is an
involution.
Theorem 96. Let A be a Banach algebra with an involution.
(a) x + x∗ , i (x − x∗ ) and xx∗ are self-adjoint for any x ∈ A.
(b) Any x ∈ A has a unique representation x = u + iv where u, v ∈ A are selfadjoint.
(c) The unit e is self-adjoint.
¡
¢∗
−1
(d) x ∈ G (A) iff x∗ ∈ G (A). And (x∗ ) = x−1 .
(e) λ ∈ σ (x) iff λ̄ ∈ σ (x∗ ).
Proof. (b) 2u = x + x∗ , 2v = i (x∗ − x). If x = u0 + iv 0 where u0 , v 0 ∈ A are
self-adjoint. Let w = v − v 0 . Then w and iw are self-adjoint. Hence,
∗
iw = (iw) = −iw∗ = −iw,
hence w = 0 and v = v 0 , u = u0 .
(c) e = ee∗ .
¤
∗
Definition 31. A Banach algebra A with an involution x → x that satisfies
2
kxx∗ k = kxk
for every x ∈ A is called C ∗ -algebra.
48
HUIQIANG JIANG
Proposition 7. Let A be a C ∗ -algebra. Then kxk = kx∗ k and kxx∗ k = kxk kx∗ k.
Theorem 97 (Gelfand-Naimark). Let A be a commutative C ∗ -algebra with maximal ideal space ∆. The Gelfand transform is an isometric isomorphism of A onto
C (∆). Moreover, for any x ∈ A, h ∈ ∆,
h (x∗ ) = h (x),
ˆ
i.e., (x∗ ) = x̂.
Proof. Let u ∈ A be self-adjoint and h ∈ ∆. Let h (u) = α + iβ where α, β ∈ R.
For any t ∈ R, we define z = u + ite, then
h (z) = α + i (β + t) , zz ∗ = u2 + t2 e.
Hence,
2
2
2
2
α2 + (β + t) = |h (z)| ≤ kzk = kzz ∗ k ≤ kuk + t2 .
Hence,
2
α2 + β 2 + 2βt ≤ kuk
for any t ∈ R which implies β = 0. So h (u) ∈ R.
For any x ∈ A, we write x = u + iv s.t. u = u∗ and v = v ∗ . Hence x∗ = u − iv,
and for any h ∈ ∆,
h (x∗ ) = h (x),
ˆ
i.e., (x∗ ) = x̂ and so  is closed under complex conjugation. Stone-Weierstrass
Theorem
that  is dense in C (∆). If x ∈ A and y = xx∗ , then y = y ∗ and
° 2 ° implies
2
m
2
°
°
so y = kyk , by induction, ky m k = kyk for m = 2n . Hence kx̂k∞ = kŷk∞ =
2
∗
kyk = kxx k = kxk so Gelfand transform is an isometry. Hence  is closed in
C (∆) which implies  = C (∆).
¤
Theorem 98. Let A be a commutative C ∗ -algebra which contains an element x
such that the polynomials in x and x∗ are dense in A. Then the formula
c = f ◦ x̂
Ψf
defines an isometric isomorphism Ψ of C (σ (x)) onto A which satisfies
Ψf¯ = (Ψf )
∗
for every f ∈ C (σ (x)). Moreover, if f (λ) = λ on σ (x), then Ψf = x.
Proof. x̂ is a continuous function on ∆ with range σ (x). If x̂ (h1 ) = x̂ (h2 ) for some
h1 , h2 ∈ ∆, then h1 (x) = h2 (x) and h1 (x∗ ) = h2 (x∗ ). If y is a polynomial in x
and x∗ , we then have h1 (y) = h2 (y). Since h is continuous and the polynomials in
x and x∗ are dense in A, we have h1 = h2 on A. So x̂ defines a one-to-one map from
∆ onto σ (x). Since ∆ is compact, x̂ is a homeomorphism. Here we used the fact
that if a space is both compact and Hausdorff, then no finer topology on the space
is compact, and no coarser topology is Hausdorff. Hence Ψ defines an isometric
isomorphism of C (σ (x)) onto A which satisfies
Ψf¯ = (Ψf )
∗
c = f ◦ x̂ = x̂ and hence
for every f ∈ C (σ (x)). If f (λ) = λ on σ (x), then Ψf
Ψf = x.
¤
FUNCTIONAL ANALYSIS
49
10.4. Application to Noncommutative Algebras.
Definition 32. Let S be a subset of a Banach algebra A, the centralizer of S is
the set
Γ (S) = {x ∈ A : xs = sx for every x ∈ S} .
Proposition 8. Let S be a subset of a Banach algebra A.
(a) Γ (S) is a closed subalgebra of A.
(b) S ⊂ Γ (Γ (S)).
(c) If S commutes, then Γ (Γ (S)) commutes.
Proof. (c) If S commutes, then S ⊂ Γ (S) and Γ (S) ⊃ Γ (Γ (S)). Hence for any
x, y ∈ Γ (Γ (S)) ⊂ Γ (S), xy = yx.
¤
Theorem 99. Suppose A is a Banach algebra, S ⊂ A, S commutes and B =
Γ (Γ (S)). Then B is a commutative Banach algebra, S ⊂ B and σA (x) = σB (x)
for every x ∈ B.
Proof. B is a commutative Banach algebra since e ∈ B. Suppose x ∈ B is invertible
in A, then for every y ∈ Γ (S), xy = yx, so y = x−1 yx and hence x−1 y = yx−1 , so
x−1 ∈ Γ (Γ (S)) = B. Hence σA (x) = σB (x) for every x ∈ B.
¤
Theorem 100. Suppose A is a Banach algebra, x, y ∈ A and xy = yx. Then
σ (x + y) ⊂ σ (x) + σ (y) and σ (xy) ⊂ σ (x) σ (y) .
Proof. Let B = Γ (Γ ({x, y})). Then σA (z) = σB (z) for every z ∈ B. For any
z ∈ B, σB (z) is the range of the Gelfand transform ẑ. Since
x[
+ y = x̂ + ŷ and x
cy = x̂ŷ,
we have
σ (x + y) ⊂ σ (x) + σ (y) and σ (xy) ⊂ σ (x) σ (y) .
¤
Definition 33. Let A be an algebra with an involution. x ∈ A is said to be normal
if x∗ x = xx∗ . A subset S ⊂ A is said to be normal if S commutes and if x∗ ∈ S
whenever x ∈ S.
Theorem 101. Suppose A is a Banach algebra with an involution and B is a
normal subset of A that is maximal w.r.t. being normal. Then B is a closed
commutative subalgebra of A and σA (x) = σB (x) for every x ∈ B.
Proof. First, we show that x ∈ B iff xx∗ = x∗ x and xy = yx for every y ∈ B.
Suppose xx∗ = x∗ x and xy = yx for every y ∈ B, since B is normal, for any y ∈ B,
xy ∗ = y ∗ x and hence x∗ y = yx∗ . So B ∪ {x, x∗ } is normal. The maximality of B
implies x ∈ B.
Second, we show that B is a commutative subalgebra of A. If x, y ∈ B, then
∗
∗
(xy) (xy) = xyy ∗ x∗ = (xy) (xy) and for every z ∈ B, (xy) z = z (xy), hence
xy ∈ B. Similarly, x + y ∈ B.
Third, we show B is closed. Let xn ∈ B and xn → x. For every y ∈ B, since
∗
xn y = yxn for each n, we have xy = yx. Hence for every y ∈ B, x∗ y = (y ∗ x) =
∗
(xy ∗ ) = yx∗ . In particular, x∗ xn = xn x∗ for each n, hence x∗ x = xx∗ . So we have
x ∈ B.
Finally, if x ∈ B is invertible in A, then since x is normal, x−1 is normal. For
every y ∈ B, xy = yx implies x−1 y = yx−1 hence x−1 ∈ B.
¤
50
HUIQIANG JIANG
11. Spectral Theorem
Definition 34. Let M be a σ-algebra in a set Ω and let H be a Hilbert space. A
resolution of the identity on M is a mapping
E : M → B (H)
satisfying
(a) E (∅) = 0, E (Ω) = I.
(b) Each E (ω) is a projector.
(c) E (ω 0 ∩ ω 00 ) = E (ω 0 ) E (ω 00 ).
(d) If ω 0 ∩ ω 00 = ∅, then E (ω 0 ∪ ω 00 ) = E (ω 0 ) + E (ω 00 ).
(e) For every x, y ∈ H, the set function Ex,y defined by
Ex,y (ω) = (E (ω) x, y)
is a complex measure on M.
Proposition 9. If E is a resolution of the identity, and if x ∈ H, then
ω → E (ω) x
is a countably additive H-valued measure on M.
P∞
Proof. Let x ∈ H and ω = n=1 ωn be a disjoint union. Then for any y ∈ H,
∞
X
(E (ωn ) x, y) = (E (ω) x, y) ,
n=1
hence
E (ω) x =
∞
X
E (ωn ) x.
n=1
¤
Definition 35. Let E be a resolution of the identity. Let f be a complex measurable
function on Ω, the essential range of f is the smallest closed subset of C that
contains f (p) for almost all p ∈ Ω, i.e., for all p ∈ Ω except those lie in some set
ω ∈ M with E (ω) = 0. We say that f is essentially bounded if its essential range
is bounded. And kf k∞ is the largest value of |λ| as λ runs through the essential
range of f . We use L∞ (E) to denote the Banach algebra of essentially bounded
functions. Here we identify f and g if kf − gk∞ = 0.
Theorem 102. Let E be a resolution of the identity. There exists an isometric
*-isomorphism Ψ of the Banach algebra L∞ (E) onto a closed normal subalgebra A
of B (H) such that for any x, y ∈ H and f ∈ L∞ (E),
Z
(Ψ (f ) x, y) =
f dEx,y .
Ω
And we write
Z
Ψ (f ) =
f dE.
Ω
Moreover,
Z
2
2
kΨ (f ) xk =
|f | dEx,x
Ω
and an operator Q ∈ B (H) commutes with every E (ω) iff Q commutes with every
Ψ (f ).
FUNCTIONAL ANALYSIS
51
Remark 12. Here Ψ is a *-isomorphism means that Ψ is one-to-one, linear, multiplicative and that
¡ ¢
∗
Ψ f¯ = (Ψ (f )) .
Proof. Please check the textbook for the proof. For example,
¡
¢
2
∗
kΨ (f ) xk = (Ψ (f ) x, Ψ (f ) x) = (Ψ (f )) Ψ (f ) x, x
´ Z
¡ ¡ ¢
¢ ³ ³ 2´
2
|f | dEx,x .
= Ψ f¯ Ψ (f ) x, x = Ψ |f | x, x =
Ω
¤
Theorem 103 (The Spectral Theorem). If A is a closed normal subalgebra of
B (H) which contains the identity operator I and if ∆ is the maximal ideal space of
A, then
(a) There exists a unique resolution E of the identity on the Borel subsets of ∆
which satisfies
Z
T =
T̂ dE
∆
for every T ∈ A, where T̂ is the Gelfand transform of T .
(b) The inverse of the Gelfand transform extends to an isometric *-isomorphism Φ
of the algebra L∞ (E) onto a closed subalgebra B of B (H), given by
Z
Φf =
f dE.
∆
(c) B is the closure of the set of all finite linear combinations of the projectors
E (ω).
(d) If ω ⊂ ∆ is open and nonempty, then E (ω) 6= 0.
(e) An operator S ∈ B (H) commutes with every T ∈ A iff S commutes with every
projector E (ω).
Proof. (ab) We first prove the uniqueness. Since A is a commutative C ∗ -algebra
with maximal ideal space ∆, Gelfand-Naimark Thm implies that the Gelfand transform is an isometric isomorphism of A onto C (∆). For any x, y ∈ H, T ∈ A,
Z
(T x, y) =
T̂ dEx,y
∆
where Ex,y is a Borel measure. Since T̂ runs through C (∆), the uniqueness of Ex,y
follows from the uniqueness of Riesz representation. For any Borel set ω, since
(E (ω) x, y) = Ex,y (ω) ,
E (ω) is uniquely determined.
Now we prove the existence of E. For any x, y ∈ H, T̂ → (T x, y) is a bounded
linear functional on C (∆), The Riesz representation theorem implies the existence
of a unique regular complex Borel measure µx,y on ∆ such that
Z
(T x, y) =
T̂ dµx,y .
∆
For each bounded Borel function f , we can define Φf ∈ B (H) such that
Z
((Φf ) x, y) =
f dµx,y
∆
52
HUIQIANG JIANG
In particular, ΦT̂ = ΦT , hence Φ is an extension of the inverse of the Gelfand
transform.
Φ is linear. Since T is self-adjoint iff T̂ is real valued,
Z
Z
T̂ dµy,x ,
T̂ dµx,y = (T x, y) = (x, T y) = (T y, x) =
∆
∆
so µx,y = µy,x . Hence
¡¡
¢
¢
Φf¯ x, y =
Z
Z
f¯dµx,y =
∆
f dµy,x = ((Φf ) y, x) = (x, (Φf ) y) ,
∆
∗
i.e., Φf¯ = (Φf ) .
c = Ŝ T̂ , hence
If S, T ∈ A, then ST
Z
Z
Ŝ T̂ dµx,y = (ST x, y) =
ŜdµT x,y .
∆
∆
For any bounded Borel functions f on ∆, we have
Z
Z
f dµT x,y
f T̂ dµx,y = (ST x, y) =
∆
∆
Z
= ((Φf ) T x, y) = (T x, z) =
T̂ dµx,z
∆
∗
where z = (Φf ) y. Hence, for any bounded Borel functions f, g on ∆, we have
Z
Z
f gdµx,y =
gdµx,z = ((Φg) x, z)
∆
∆
¡
∗ ¢
= (Φg) x, (Φf ) y = ((Φf ) (Φg) x, y) .
Hence
Φ (f g) = Φ (f ) Φ (g) .
For any Borel set ω of ∆, we define
E (ω) = Φ (χω ) .
For any ω, ω 0 ,
E (ω ∩ ω 0 ) = E (ω) E (ω 0 ) .
Hence E (ω) = E (ω) E (ω). Since χω is real, E (ω) is also self-adjoint. Hence, E (ω)
is a projector. E (∅) = 0, E (∆) = I. E is finite additive. For any x, y ∈ H,
(E (ω) x, y) = µx,y (ω)
is a Borel measure. Hence, E is a resolution of identity. Moreover,
Z
Φ (f ) =
f dE
∆
and
kΦ (f )k = kf k∞ .
(c) Every f ∈ L∞ (E) is a uniform limit of simple functions.
(d) ∆ is a compact Hausdorff spaces are hence normal, if ω ⊂ ∆ is open and
nonempty, Uryson’s lemma implies the existence of a continuous function f on ∆
s.t. f 6≡ 0 and supp f ⊂ ω. Since  = C (∆), there exists T ∈ H, T̂ = f . If
E (ω) = 0, then µx,y (ω) = 0 for any x, y ∈ H. Since
Z
(T x, y) =
f dµx,y ,
∆
FUNCTIONAL ANALYSIS
53
T = 0. Hence T̂ = f = 0, a contradiction.
(e) For any given S ∈ B (H) and for any x, y ∈ H, T ∈ A, denote z = S ∗ y, we
have
Z
(ST x, y) = (T x, z) =
T̂ dEx,z ,
∆
Z
(T Sx, y) =
T̂ dESx,y .
∆
So ST = T S for any T ∈ A iff Ex,z (ω) = ESx,y (ω) for any x, y ∈ H and for every
Borel set ω, iff for any x, y ∈ H and for every Borel set ω,
(E (ω) x, S ∗ y) = (E (ω) Sx, y) ,
iff SE (ω) = E (ω) S for every Borel set ω.
¤
Lemma 13. Let A be a closed Banach subalgebra of a Banach algebra B. If x ∈ A,
then the boundary of σA (x) lies in σB (x).
Proof. Let x ∈ A. Since A ⊂ B, σB (x) ⊂ σA (x). Assume that there exists
C
C
λ ∈ ∂σA (x), but λ 6∈ σB (x). There exists λn ∈ (σA (x)) ⊂ (σB (x)) such that
−1
−1
λn → λ. Since inverse is continuous, (λn e − x) → (λe − x) in B. In particular,
there exists M > 0 s.t. for any n ≥ 1,
°
°
°
−1 °
°(λn e − x) ° ≤ M.
1
Choose N large so that |λN − λ| < M
, then
°
° °
°
°
° °
−1
−1 °
°e − (λn e − x) (λe − x)° = °(λn − λ) (λn e − x) ° < 1,
hence
−1
(λn e − x)
h
i
−1
(λe − x) = e − e − (λn e − x) (λe − x)
is invertible in A, hence λe − x is invertible in A which is a contradiction.
¤
Corollary 14. (i) Let T ∈ B (H) be self-adjoint and A be a *-subalgebra of B (H)
containing T . Then σA (T ) = σB (T ).
(ii) Let T ∈ B (H) be normal and A be a *-subalgebra of B (H) containing T . Then
σA (T ) = σB (T ).
Proof. (i) Since T is self-adjoint, σA (x) is real, hence ∂σA (T ) = σA (T ) ⊂ σB (T ),
which implies σA (T ) = σB (T ).
(ii) If T is normal and λ 6∈ σB (T ), then (λI − T ) (λI − T ∗ ) is invertible in B
and self-adjoint, hence (λI − T ) (λI − T ∗ ) is invertible in A and hence
−1
(λI − T )
−1
= (λI − T ∗ ) [(λI − T ) (λI − T ∗ )]
∈ A.
So λ 6∈ σA (T ). Hence σA (T ) ⊂ σB (T ), which implies σA (T ) = σB (T ).
¤
Theorem 104 (Spectral Decomposition of Normal Operator). If T ∈ B (H) and
T is normal, then there exists a unique resolution of the identity E on the Borel
subsets of σ (T ) which satisfies
Z
T =
λdE (λ) .
σ(T )
Further, every projection E (ω) commutes with every S ∈ B (H) which commutes
with T .
54
HUIQIANG JIANG
Proof. Let A be the smallest closed subalgebra of B (H) that contains I, T and T ∗
and ∆ be the maximal ideal space of A. Since the polynomials in T and T ∗ are
dense in A, theorem 98 implies that ∆ is homeomorphism to σA (T ). Since T is
normal, and A is a *-subalgebra of B (H) containing T , we have σA (T ) = σ (T ).
The existence of E follows from theorem 103.
¤
Corollary 15. Let T ∈ B (H) be a nonzero normal operator. Then T has a nontrivial invariant subspace.
Remark 13. It is an open problem whether every nonnormal T ∈ B (H) has a
nontrivial invariant subspace if H is an infinite-dimensional separable Hilbert space.
Theorem 105. A normal T ∈ B (H) is
(a) self-adjoint iff σ (T ) is real.
(b) unitary iff σ (T ) lies on the unit circle.
∧
Proof. (a) Let A be as in the spectral decomposition. T̂ (λ) = λ and (T ∗ ) (λ) = λ̄
on σ (T ). Hence, T = T ∗ iff λ = λ̄ on σ (T ), i.e., σ (T ) is real. And T T ∗ = I iff
λλ̄ = 1 on σ (T ), i.e., σ (T ) lies on the unit circle.
¤
Theorem 106. Suppose E is the spectral decomposition of a normal T ∈ B (H),
λ0 ∈ σ (T ) and E0 = E ({λ0 }). Then
(a) N (T − λ0 I) = R (E0 ).
(b) λ0 is an eigenvalue of T iff E0 6= 0, and
(c) every isolated point of σ (T ) is an eigenvalue of T .
(d) If σ (T ) = {λ1 , λ2 , · · · } is a countable set, then every x ∈ H has a unique
expansion of the form
∞
X
x=
xi
i=1
where T xi = λi xi . Also xi ⊥xj if i 6= j.
Theorem 107. A normal operator T ∈ B (H) is compact iff it satisfies the following two conditions:
(a) σ (T ) has no limit point except possibly 0,
(b) If λ 6= 0, then dim N (T − λ0 I) < ∞.
Definition 36. T ∈ B (H) is called a positive operator if (T x, x) ≥ 0 for any
x ∈ H.
Theorem 108. T ∈ B (H) is a positive operator iff T = T ∗ and σ (T ) ∈ [0, ∞).
Proof. If (T x, x) ≥ 0 for any x ∈ H, then T = T ∗ and hence σ (T ) ∈ R. Now for
any λ > 0,
2
λ kxk = (λx, x) ≤ ((T + λI) x, x) ≤ k(T + λI) xk kxk ,
hence k(T + λI) xk ≥ λ kxk. So T + λI is invertible and hence −λ 6∈ σ (T ). Hence
σ (T ) ∈ [0, ∞).
On the other hand, if T = T ∗ and σ (T ) ∈ [0, ∞), then for any x ∈ H,
Z
(T x, x) =
λdEx,x (λ) ≥ 0,
σ(T )
since Ex,x is a positive measure.
¤
FUNCTIONAL ANALYSIS
55
Theorem 109. Every positive T ∈ B (H) has a unique positive square root S ∈
B (H). If T is invertible, then so is S.
Proof. Let A be any closed normal subalgebra of B (H) containing I and T and ∆
be the maximal ideal space of A. Then  = C (∆). Since T is positive, σ (T ) =
T̂ (∆) ∈ [0, ∞), so there exists unique S ∈ A, such that
p
Ŝ = T̂ .
Hence S 2 = T . Since S is normal and σ (S) = Ŝ (∆) ∈ [0, ∞), S = S ∗ and S is
positive. To see the uniqueness, if S12 = T and S1 is positive, let A0 be the smallest
closed normal subalgebra of B (H) containing I and T and A1 be the smallest closed
normal subalgebra of B (H) containing I and S1 , then T ∈ A1 , hence A0 ⊂ A1 ,
hence S1 ∈ A0 which is uniquely defined.
¤
12. Test functions and distributions
Let Ω be a nonempty open set in Rn . The test function space D (Ω) = C0∞ (Ω)
consists of all infinitely differentiable functions with compact support. A sequence
{fk } ⊂ D (Ω) is said to converge to f ∈ D (Ω) if the following two conditions hold:
(a) There is a compact set K ⊂ Ω containing the supports of all fk , k ≥ 1;
(b) For each multi-index α, Dα fk → Dα f uniformly.
With this definition, we can show that D (Ω) is a complete locally convex topological vector space satisfying the Heine–Borel property.
Remark 14. D (Ω) can be viewed as a subspace of the Fréchet space C ∞ (Ω) defined
in Theorem 15. However, the topology of D (Ω) inherited from C ∞ (Ω) is different
from what we are defining here. Actually, the inherited topology of D (Ω) is not
complete. For example, let Ω = Rn and ϕ (x) be the standard mollifier supported
on the unit ball, then
m
X
1
fm (x) =
ϕ (|x| − k)
k
k=1
is a Cauchy sequence w.r.t. the topology of C ∞ (Ω) whose limit is not in D (Ω).
The topology of D (Ω) can be defined precisely. For every cpt K ⊂ Ω, we use
τK to denote the topology of DK (Ω) ⊂ C ∞ (Ω) equipped with the Fréchet space
topology defined in Theorem 15. The topology τK on DK (Ω) agrees with the
topology defined by a family of norms
kf kN = max kDα f kL∞ , N ≥ 0.
|α|≤N
Let β be the collection of all convex balanced sets W ⊂ D (Ω) such that DK (Ω) ∩
W ∈ τK for every compact K ⊂ Ω. Let τ be the collection of all unions of sets of
the form φ + W with φ ∈ D (Ω) and W ∈ β.
Theorem 110. τ is a topology in D (Ω) and β is a local base for τ . The topology
τ makes D (Ω) into a locally convex topological vector space.
Proof. To show that τ is a topology, it suffices to show that V1 , V2 ∈ τ implies
V1 ∩ V2 ∈ τ . For every φ ∈ V1 ∩ V2 , there exists φi and Wi , i = 1, 2 such that
φ ∈ φi + Wi ⊂ Vi , i = 1, 2.
56
HUIQIANG JIANG
Choose K so that K contains supports of φ1 , φ2 and φ. Since DK (Ω) ∩ W is open
in DK , we have
φ − φi ∈ (1 − δi ) Wi
for some δi > 0. The convexity of Wi implies that
φ − φi + δi Wi ⊂ (1 − δi ) Wi + δi Wi ⊂ Wi
and so
φ + δi Wi ⊂ φi + Wi ⊂ Vi .
Let W = (δ1 W1 ) ∩ (δ2 W2 ), then φ + W ⊂ V1 ∩ V2 . Here W ∈ β since Wi is convex
and balanced. Hence τ is a topology and β is a local base for τ .
Now we verify that D (Ω) is a topological vector space. First, for any φi ∈ D (Ω),
i = 1, 2, such that φ1 6= φ2 , one can verify that
W = {φ ∈ D (Ω) : kφk0 < kφ1 − φ2 k0 } ∈ β
and φ1 6∈ φ2 + W . Hence, single point sets are closed.
Addition is continuous.
Scalar multiplication is continuous.
¤
Theorem 111. (a) A convex balanced subset V of D (Ω) is open iff V ∈ β.
(b) The topology τK of any DK (Ω) coincides with the subspace topology that DK
inherits from D (Ω).
(c) If E is bounded subset of D (Ω), then E ⊂ DK (Ω) for some compact K ⊂ Ω
and for each N ≥ 0, there exists MN < ∞ s.t.,
kφkN ≤ MN for any φ ∈ E.
(d) D (Ω) has the Heine-Borel property.
(e) If {φi } is a Cauchy sequence in D (Ω), then {φi } ⊂ DK (Ω) for some compact
K ⊂ Ω and
lim kφi − φj kN = 0 for any N ≥ 0.
i,j→∞
(f ) If φi → 0, then {φi } ⊂ DK (Ω) for some compact K ⊂ Ω and Dα φi → 0
uniformly for every multi-index α.
(g) In D (Ω), every Cauchy sequence converges.
Proof. (a) Let V ∈ τ and K ⊂ Ω be cpt. For any φ ∈ DK ∩ V , there exists W ∈ β,
φ + W ⊂ V . Hence
φ + (DK ∩ W ) ⊂ DK ∩ V.
So we have DK ∩ V ∈ τK for each V ∈ τ and each compact K ⊂ Ω. (a) follows.
(b) From the proof of (a), we only need to check that each E ∈ τK is relative
open in DK ⊂ D (Ω). For each φ ∈ E, there exists N ≥ 0 and δ > 0 s.t.
{ψ ∈ DK : kψ − φkN < δ} ⊂ E.
Put W = {ψ ∈ D (Ω) : kψkN < δ}, then W ∈ β and W ∩ DK ⊂ E.
(c) Consider a set E ⊂ D (Ω) which lies in no DK . Then there exists {φm } ⊂ E
and {xm } ⊂ Ω s.t. φm (xm ) 6= 0 and {xm } has no limit points. Let
¾
½
1
|φm (xm )| , m ≥ 1 .
W = φ ∈ D (Ω) : |φ (xm )| <
m
FUNCTIONAL ANALYSIS
57
Then W ∈ β. Since φm 6∈ mW , E is not bounded. Hence, If E is bounded subset of
D (Ω), then E ⊂ DK (Ω) for some compact K ⊂ Ω and so E is bounded in DK (Ω)
and so for each N ≥ 0, there exists MN < ∞ s.t.,
kφkN ≤ MN for any φ ∈ E.
(d) (e) (f) (g) follows.
¤
Theorem 112. Suppose Λ is a linear mapping of D (Ω) into a locally convex space
Y . Then the following statements and equivalent:
(a) Λ is continuous.
(b) Λ is bounded.
(c) If φi → 0 in D (Ω) then Λφi → 0 in Y .
(d) The restriction of Λ to every DK are continuous.
Proof. ”(a)→(b)”.
”(b)→(c)”: If φi → 0 in D (Ω), then {φi } ⊂ DK for some K compact. Since DK
is metrizable, Λφi → 0 in Y . See Theorem 10.
”(c)→(d)”: See Theorem 10.
”(d)→(a)”: Let U be a convex balanced nbhd of 0 in Y and put V = Λ−1 (U ).
Then V is convex and balanced. Since V is open in D (Ω) iff V ∩ DK is open for
every K compact, we have (a) ⇔ (d).
¤
Theorem 113. Every differential operator Dα : D (Ω) → D (Ω) is continuous.
Proof. This is because Dα : DK (Ω) → DK (Ω) is continuous.
¤
Definition 37. A continuous linear functional on D (Ω) is called a distribution.
The space of all distributions in Ω is denoted by D0 (Ω).
Theorem 114. If Λ is a linear functional on D (Ω), the following two conditions
are equivalent:
(a) Λ ∈ D0 (Ω).
(b) For every compact K ⊂ Ω, there exists N ≥ 0 and C > 0 s.t.
|Λφ| ≤ C kφkN
holds for every φ ∈ DK (Ω).
Remark 15. If Λ is such that one N will do for all K, then the smallest such N
is called the order of Λ.
Example 7. δx is a distribution of order 0.
Example 8. Every locally integrable complex function is a distribution of order 0.
Example 9. Any locally finite Borel measure is a distribution of order 0.
The space of distribution D (Ω) is a topological vector space with weak-* topology, and it has many other nice structures.
Definition 38. If α is a multi-index and Λ ∈ D0 (Ω), Dα Λ ∈ D0 (Ω) is a linear
functional defined by the formula
|α|
(Dα Λ) (φ) = (−1)
Λ (Dα φ) , φ ∈ D (Ω) .
Proposition 10. For any multi-indices α, β and for any Λ ∈ D0 (Ω),
Dβ Dα Λ = Dα Dβ Λ = Dα+β Λ.
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HUIQIANG JIANG
Remark 16. If f has continuous partial derivatives of all orders up to N , then
Dα Λf = ΛDα f for any |α| ≤ N.
Remark 17. Sobolev spaces.