Normed and Banach Algebras

Submultiplicative Norm

A norm over a vector space is submultiplicative if for all .

Normed Algebra & Banach Algebra

Let be an algebra (over ) equipped with a submultiplicative norm, then is called a normed algebra. If it is also a Banach space (i.e. complete), then it is called a Banach algebra.

e.g.

• Suppose is a Banach space, and is the algebra of bounded linear operators on with operator norm . Then is a Banach algebra.

Ideals

Theorem

Let be a Banach algebra, be a closed ideal in , then is a Banach algebra, with the norm defined as for all .

Spectrum

Recall that the set of invertible elements in a algebra forms a group:

An element of a ring is called a unit if it is invertible with respect to multiplication. The set of invertible elements is a group called the the group of units in and denotes .

Link to original

Here, for any unital algebra , we denote the set of invertible elements in as .

Spectrum

Suppose is a normed algebra. Then the spectrum of is the set

e.g.

• Suppose , the algebra of complex matrices, then the spectrum of is the set of all eigenvalues of .
• is a compact Hilbert space, then the spectrum of , for which is the algebra of all continuous maps , is the range of .
• Consider the algebra . The spectrum of is the whole .

Proposition

The spectrum is a closed subset of .

Neumann Theorem

Let be a unital Banach algebra, and such that . Then is invertible and

Theorem

Let be a unital Banach algebra, then is open in , and the mapping , is Fréchet differentiable.

Lemma

If is an element of a unital Banach algebra , then the spectrum is a closed subset of and for any .

Proof We first prove that by contradiction. Assume for some . Then, by the Neumann theorem, we have is invertible, hence is also invertible. This contradicts the definition of . Now we show that is closed. Define the map , , which is continuous. Then . Since is open in , it follows that is closed in .

Lemma

If is an element of a unital Banach algebra , then the map , is differentiable.

Proof Observe that , which is a composition of two differentiable maps, hence is differentiable.

Gelfand Theorem

If is an element of a unital Banach algebra , then the spectrum is nonempty.

Proof Assume that (i.e. is invertible for all ) and we shall obtain a contradiction. For all with , we have , so Neumann theorem implies that is invertible, and Therefore, where the first inequality follows from the triangle inequality. Consequently, we have Moreover, from the above lemma, we know that , is differentiable, so it is continuous on . Since is compact,