Download Chapter 7 Spectral Theory Of Linear Operators In Normed Spaces

Chapter 7 Spectral Theory Of Linear Operators In Normed Spaces
7.6 Banach Algebra
7.6-1 Definition. An algebra A over a field k is a vector space A over k such
that for all x,yA, a unique product xyA is defined with the properties:
(1) (xy)z = x(yz)
(2) (x+y)z = xz + yz
(3) x(y+z) = xy + xz
(4) (xy) = (x)y = x(y)
for all x,y,zA and scalar k.
A is called an algebra with unity if there exists eA such that ex = xe = x for all
xA. A is called a commutative algebra if ab = ba for all elements a and b in A.
A subalgebra of A is a vector subspace of A that is closed under the product.
Note. If A has an identity, then the identity is unique.
7.6-2 Definition. A normed algebra A is a normed space which is an algebra
such that ||xy||  ||x|| ||y|| for all x,yA. If A has an identity e, then || e || = 1.
The Banach algebra A is a complete normed algebra which is complete
considered as a normed space.
Remark. The product in a normed algebra A is a continuous mapping of AA
into A.
Proof. Left to the reader.
Examples.
7.6-3 Space R and C. The real line and the complex plane C are commutative
Banach algebra with identity e = 1.
7.6-4 Space C[a, b]. The space C[a, b] is a commutative Banach algebra with
identity e = 1, the product xy being defined as usual (xy)(t) = x(t)y(t) for all
tC[a, b] and the norm is the maximum norm. The subspace of C[a, b]
consisting of all polynomials is a commutative normed algebra with identity e
= 1.
Proof. Left to the reader.
7.6-5 Space B(X). The Banach space B(X) of all bounded linear operators on a
complex Banach space X ≠ {0} is a Banach algebra with identity I, the
multiplication being composition of operators. B(X) is not commutative,
unless dim X = 1.
Proof. Left to the reader.
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7.6-6 Definition. An element x in an algebra A is said to be invertible if it
has an inverse in A, that is if there is x -1A such that xx -1 = x -1x = e.
Note. If xA s invertible, then its inverse is unique.
7.6-7 Definition. Let A be a complex Banach algebra with identity. Then the
resolvent set of aA is (a) = { C : ( e –a )-1 exists in A }. The spectrum
set of a is, (a) = C - (a). Any (a) is called a spectral value of a.
Note that (a) = { C : ( e –a ) -1 does not exist in A }.
7.7 Further Properties of Banach Algebras
7.7-1 Theorem. Let A be a complex Banach algebra with unity e. If xA with
|| x || < 1, then ( e – x ) is invertible, and ( e – x ) -1 = e +

x
j
.
j 1
Proof. Since A is a Banach algebra and xA, then for any jN, we have || x j || 

 || x ||
||x|| || x j-1 ||  ………….. || x || j. Then
j
converges, because || x || < 1.
j 0

Therefore,
 || x j || converges. That means e +
j 0

x
j
converges absolutely.
j 1

However, A is complete, then e +
x

j
j 1
converges, say e +
x
j
= s. Since
j 1
( e – x ) ( e + x + x 2 + …. + x n) = ( e + x + x 2 + …. + x n) ( e – x ) = e – x n+1,
then as n → ∞, we have ( e – x ) s = s ( e – x ) = e , because || x || < 1 and the
multiplication is continuous. Therefore, s = ( e – x ) -1, and so ( e – x ) -1 = e +

x
j
.
j 1
7.7-2 Proposition. If A is a complex Banach algebra with identity e, then the set
G = { xA: x -1 exists } is a group.
Proof. Left to the reader.
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Proof. Suppose that X/ is separable. Then X/ contains a countable dense subset
M. Let M1 = { fM : || f || = 1} and U/ = { f X/ : || f || = 1}. We show that
___
M 1 = U/. Let fU/. Then fX/, but M is dense in X/ then there is a sequence
(gn) of elements in M such that gn  f as n . Hence by removing all gn = 0
from this sequence we have,
___
lim || gg nn || =
n 
f
||f ||
= 1. However,
gn
|| g n ||
M1 for
___
all n, then f M 1 . Therefor, M 1 = U/, and so U/ contains a countable dense
subset of M1, say it is (fn).
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