<p> Chapter 7 Spectral Theory Of Linear Operators In Normed Spaces</p><p>7.6 Banach Algebra</p><p>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.</p><p>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.</p><p>Note. If A has an identity, then the identity is unique. </p><p>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.</p><p>Remark. The product in a normed algebra A is a continuous mapping of AA into A. Proof. Left to the reader.</p><p>Examples. 7.6-3 Space R and C. The real line and the complex plane C are commutative Banach algebra with identity e = 1.</p><p>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.</p><p>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.</p><p>1 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.</p><p>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 }.</p><p>7.7 Further Properties of Banach Algebras</p><p>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</p><p>Proof. Since A is a Banach algebra and xA, then for any jN, we have || x j ||  ||x|| || x j-1 ||  ………….. || x || j. Then ||x || j converges, because || x || < 1. j=0  Therefore, ||x j || converges. That means e +  x j converges absolutely. j=0 j1   However, A is complete, then e +  x j converges, say e +  x j = s. Since j1 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.</p><p>2 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 ___ = U / . Let f  U / . Then f  X / , but M is dense in X / then there is a sequence M 1</p><p>(gn) of elements in M such that gn  f as n   . Hence by removing all gn = 0 </p><p> g n f g n from this sequence we have, lim ||g || = = 1. However,  M1 for n n ||f || ||g n || ______/ / all n, then f  M 1 . Therefor, M 1 = U , and so U contains a countable dense subset of M1, say it is (fn).</p><p>3</p>