Algebra of functions

function algebra

A semi-simple commutative Banach algebra $ A $, realized as an algebra of continuous functions on the space of maximal ideals $ \mathfrak M $. If $ a \in A $ and if $ f $ is some function defined on the spectrum of the element $ a $( i.e. on the set of values of the function $ \widehat{a} = a $), then $ f(a) $ is some function on $ \mathfrak M $. Clearly, it is not necessarily true that $ f(a) \in A $. If, however, $ f $ is an entire function, then $ f(a) \in A $ for any $ a \in A $. The use of the Cauchy integral formula permits a considerable strengthening of this result: If the function $ f $ is analytic in some neighbourhood of the spectrum of the element $ a $, then $ f(a) \in A $ and the mapping $ f \rightarrow f(a) $ is a homomorphism of the algebra of functions which are analytic in some neighbourhood of the spectrum of $ a \in A $ into $ A $. This proposition is valid for non-semi-simple commutative Banach algebras as well. Moreover, this class of functions that are analytic in a neighbourhood of the spectrum of a given element cannot be enlarged, in general. For example, if $ A = L _ {1} ( \mathbf Z ) $ and $ f(a) \in A $ for all $ a \in A $ with spectrum in the interval $ [0, 1] $, then $ f $ is analytic in some neighbourhood of this interval.

In a few cases $ f(a) $ can also be defined for multi-valued analytic functions $ f $, but such a definition has inherent difficulties. Thus, let $ A $ be the algebra of continuous functions in the disc $ | z | \leq 1 $ that are analytic in the disc $ | z | < 1 $ and that satisfy the condition $ f ^ { \prime } (0) = 0 $. The unit disc is naturally identified with the space of maximal ideals of $ A $. The function $ f _ {1} (z) = z $, which is continuous on the space of maximal ideals, does not belong to $ A $, but is a solution of the quadratic equation

$$ f _ {1} ^ { 2 } - z ^ {2} = 0 , $$

where $ z ^ {2} \in A $.

If $ A $ is a semi-simple algebra with space of maximal ideals $ X $, if $ f \in C(X) $ and if

$$ p (f) \equiv f ^ {n} + a _ {1} f ^ {n-1 } + \dots + a _ {n} = 0 , \ a _ {i} \in A , $$

with $ p ^ \prime (f) \in \epsilon (A) $, the group of units of $ A $( a simple root), then $ f \in A $. Similarly, if $ f \in C(X) $ and if $ \mathop{\rm exp} (f) \in A $, then $ f \in A $.

A function algebra is said to be a uniformly-convergent algebra (or uniform algebra) if the norm in this algebra defines a notion of convergence equivalent to the uniform convergence of the functions $ \widehat{a} $ on the space of maximal ideals. If $ \| a ^ {2} \| = \| a \| ^ {2} $ for all $ a \in A $, then $ A $ is a uniform algebra. The general example of a uniform algebra is a closed subalgebra of the algebra of bounded continuous functions on some topological space, provided with the natural sup-norm.

If $ A $ is a uniform algebra and if its space of maximal ideals is metrizable, then among the boundaries (not only the closed ones) there is a minimal boundary $ \Gamma _ {0} $, the closure of which is the Shilov boundary. The set $ \Gamma _ {0} $ consists of "peak points" : $ x _ {0} $ is a peak point if there exists a function $ f \in A $ such that $ | f(x) | < | f ( x _ {0} ) | $ for all $ x \neq x _ {0} $. In the present case any point in the space of maximal ideals has a representing measure concentrated on $ \Gamma _ {0} $.

A function algebra is said to be analytic if all functions of this algebra that vanish on a non-empty open subset of the space of maximal ideals vanish identically. Algebras that are analytic with respect to the boundary are defined in a similar manner. Any analytic algebra is analytic with respect to the Shilov boundary; the converse is usually not true.

A function algebra $ A $ is said to be regular if, for any closed set $ F $ in the space $ X $ of maximal ideals of $ A $ and for any point $ x _ {0} $ not contained in $ F $, it is possible to find a function $ f \in A $ such that $ f(x) = 1 $ for all $ x \in F $ and $ f( x _ {0} ) = 0 $. All regular algebras are normal, i.e. for any pair of non-intersecting closed sets $ F, F _ {0} \in X $ there exists an element $ f \in A $ such that $ f(x) = 1 $ for all $ x \in F $ and $ f( x ) = 0 $ for all $ x \in F _ {0} $. In a regular algebra, for any finite open covering $ \\{ U _ {i} \\} $, $ 1 \leq i \leq m $, of the space $ X $ there exists a partition of unity belonging to $ A $, i.e. a system of functions $ f _ {1} \dots f _ {n} \in A $ for which

$$ f _ {1} (x) + \dots + f _ {n} (x) \equiv 1 $$

and

$$ f _ {i} (x) = 0 \ \textrm{ if } x \notin U _ {i} . $$

A function $ g $ is said to belong locally to the function algebra $ A $ if for any point $ x _ {0} \in X $ there exists a neighbourhood in which this function coincides with some function of the algebra. Any function which locally belongs to a regular algebra is itself an element of this algebra.

An element $ f $ of a function algebra is called real if $ \widehat{f} (x) $ is real for all $ x \in X $. If $ A $ is an algebra with real generators $ f _ \alpha $ and if

$$ \int\limits _ {- \infty } ^ { {+ } \infty } \mathop{\rm ln} \| \mathop{\rm exp} ( it f _ \alpha ) \| \ \frac{dt}{1 + t ^ {2} } < \infty $$

for all $ f _ \alpha $, then $ A $ is regular.

An ideal in a Banach algebra is said to be primary if it is contained in only one maximal ideal. If $ A $ is a regular function algebra, then each maximal ideal $ x _ {0} $ contains a smallest closed primary ideal $ J ( x _ {0} ) $ which is contained in any closed primary ideal contained in $ x _ {0} $. The ideal $ J (x _ {0} ) $ is the closure of the ideal formed by the functions $ f \in A $ that vanish in some neighbourhood (depending on $ f $) of $ x _ {0} \in X $.

In the algebra of absolutely convergent Fourier series with an adjoined identity any maximal ideal coincides with the corresponding primary ideal.

Let $ A $ be a closed subalgebra of the algebra $ C(X) $, where $ X $ is a compactum (which does not necessarily coincide with the space of maximal ideals of $ A $). Let $ A $ separate the points of $ X $, i.e. for any two different points $ x _ {1} , x _ {2} \in X $ there exists a function $ f $ in $ A $ for which $ f (x _ {1} ) \neq f (x _ {2} ) $. The algebra $ A $ is called symmetric if both a function $ {f(x) } $ and the function $ \overline{ {f (x) }}\; $ belong to it. According to the Stone–Weierstrass theorem, if $ A $ is symmetric, then $ A = C(X) $. The algebra $ A $ called anti-symmetric if it follows from the conditions $ f , \overline{f}\; \in A $ that $ f $ is a constant function. In particular, algebras of analytic functions are anti-symmetric. A subset $ S \subset X $ is called a set of anti-symmetry (with respect to the algebra $ A $) if any function $ f \in A $ that is real on $ S $ is constant on this set. It follows from this definition that the algebra $ A $ is anti-symmetric if the whole set $ X $ is a set of anti-symmetry. In the general case the space $ X $ can be represented as the union of non-intersecting, closed, maximal sets of anti-symmetry. Each maximal set of anti-symmetry is an intersection of peak sets (a set $ P $ is called a peak set if there exists a function $ f \in A $ such that $ f \mid _ {P} =1 $ and $ | f(x) | < 1 $ if $ x \notin P $). It follows that the restriction $ A \mid _ {Y} $ of the algebra $ A $ to a maximal set of anti-symmetry is a closed (anti-symmetric) subalgebra of the algebra $ C(Y) $. If $ X $ is the space of maximal ideals of the algebra $ A $, the maximal sets of anti-symmetry are connected. If a continuous function is such that on each maximal set of anti-symmetry it coincides with some function in the algebra $ A $, then the function itself belongs to $ A $. This generalization of the Stone–Weierstrass theorem makes it possible, in principle, to reduce the study of arbitrary uniform algebras to the study of anti-symmetric algebras $ A $. However, the study of arbitrary algebras $ A $ cannot be reduced to the study of analytic algebras: There exists an example of an algebra of type $ R(X) $( a closed subalgebra of the algebra $ C(X) $) which does not coincide with $ C(X) $, and is anti-symmetric and regular.

Let $ \mathop{\rm Re} (A) $ be the real space of functions of the form $ \mathop{\rm Re} (f) $, where $ f \in A $; if $ \mathop{\rm Re} (A) $ is an algebra or if $ \mathop{\rm Re} (A) $ is closed in $ C(X) $, then $ A = C(X) $. The space $ X $ can be regarded as a part of the space of maximal ideals of the algebra $ A $; accordingly, not only the ordinary topology of the space of maximal ideals, but also the metric induced by the imbedding of $ X $ into the dual space to $ A $ can be considered on $ X $. The distance in the sense of this metric will be denoted by $ \rho _ {A} $. For any points $ x _ {1} , x _ {2} \in X $ the inequality $ \rho _ {A} (x _ {1} , x _ {2} ) \leq 2 $ is valid; the relation $ \rho _ {A} (x _ {1} , x _ {2} ) < 2 $ is an equivalence relation, and the equivalence classes are known as Gleason parts. If $ X $ is the disc $ | z | \leq 1 $ and $ A $ is the closed subalgebra in $ C(X) $ consisting of the functions analytic in $ | z | < 1 $, then the metric $ \rho _ {A} $ is non-Euclidean, and the one-point sets on the circle and in the interior of the disc serve as the Gleason parts. Gleason parts do not necessarily have an analytic structure: Any $ \sigma $- compact completely-regular space is homeomorphic to the Gleason part of the space of maximal ideals of some algebra, such that the restriction of the algebra to this part contains all bounded continuous functions. The fact that two points belong to the same Gleason part can be described in terms of the representing measures on the Shilov boundary: Two such points have two mutually absolutely continuous representing measures with bounded derivatives. An algebra for which $ \mathop{\rm Re} ( A \mid _ \Gamma ) $ is dense in $ C ( \Gamma ) $ is called a Dirichlet algebra; if $ P $ is a Gleason part in the space of maximal ideals of a Dirichlet algebra which contains more than one point, then there exists a continuous one-to-one mapping $ \psi $ of the disc $ | z | < 1 $ into $ P $ such that for any function $ f \in A $ the function $ f ( \psi (z) ) $ is analytic in $ | z | < 1 $. Thus, $ P $ has a structure with respect to which the functions $ f \in A $ are analytic; the mapping $ \psi $ is not a homeomorphism in general if $ P $ is endowed with the ordinary topology of the space of maximal ideals, but $ \psi $ is a homeomorphism if $ P $ is endowed with the metric $ \rho _ {A} $.

For references see Banach algebra.

#### Comments[edit]

#### References[edit]

[a1] |  T.W. Gamelin, "Uniform algebras" , Prentice-Hall (1969)

|

[a2] |  E.L. Stout, "The theory of uniform algebras" , Bogden & Quigley (1971)