Normed and Banach Algebras

Normed Algebra & Banach Algebra

Let be an algebra (over ) equipped with a submultiplicative norm, that is, for all , then is called a normed algebra.
If admits a unit such that , then it is called a unital 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 with unit being the identity operator on .

Proposition

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

Proof We first check submultiplicativity of the norm on (we skip checking it is a norm here). For any and , note that as is an ideal, we have By submultiplicativity of the norm in , we have . Taking infimum over , we get . Hence, the norm is submultiplicative. The unit of has norm because on the one hand we have , on the other hand, submultiplicativity implies , and since is proper, so . To show completeness, we pick a Cauchy sequence in , then one can pick a subsequence such that . Thus we can choose some , so that . Now define , and . Note that , so . Moreover, the sequence is Cauchy in because . Since is complete, there exists such that . Therefore, because the quotient map is a contraction. Since is a subsequence of the original Cauchy sequence , we conclude that (ref. proposition). Hence, is complete.

Remark

The topology induced from the norm on coincides with the quotient topology induced from the quotient map , . In fact, we have

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 unital normed algebra. Then the spectrum of is the set

e.g.

• Suppose , the algebra of complex matrices, then the spectrum of a matrix 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 of polynomials . The spectrum of is the whole .
  

Proposition

The spectrum is a closed subset of .

Neumann Theorem (Murphy, 1990, Theorem 1.2.2)

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,

Gelfand-Mazur Theorem

If is a unital Banach algebra in which every nonzero element is invertible, then .

Beurling's Theorem (Gelfand's Formula)

If is an element of a unital Banach algebra , then the spectral radius of is given by