Gelfand-Mazur theorem

The Gelfand-Mazur theorem (after Israel Gelfand and Stanisław Mazur ) is one of the starting points of the theory of Banach algebras . It says that the only - is Banach algebra that is an oblique body . ${\displaystyle \mathbb {C} }$ ${\displaystyle \mathbb {C} }$

Lemma about the spectrum

Be one - Banach algebra with one element . Then for each one , so do not be inverted is. ${\displaystyle A}$ ${\displaystyle \mathbb {C} }$ ${\displaystyle 1}$ ${\displaystyle a\in A}$ ${\displaystyle \lambda \in {\mathbb {C} }}$ ${\displaystyle a-\lambda 1}$

One calls the set of all for which is not invertible, also the spectrum of . This statement can be formulated more concisely in such a way that the spectrum of an element of a -Banach algebra with one element is not empty. ${\displaystyle \lambda \in {\mathbb {C} }}$ ${\displaystyle a-\lambda 1}$ ${\displaystyle a}$ ${\displaystyle \mathbb {C} }$

proof

The proof consists of an interplay of functional analysis ( Hahn-Banach theorem ) and function theory ( Liouville theorem ):

We assume that it is invertible for each . Then applies to different from one another ${\displaystyle a-\lambda 1}$ ${\displaystyle \lambda \in {\mathbb {C} }}$ ${\displaystyle \lambda ,\mu \in {\mathbb {C} }}$

${\displaystyle (a-\lambda 1)^{-1}(\lambda -\mu )(a-\mu 1)^{-1}=(a-\lambda 1)^{-1}((a-\mu 1)-(a-\lambda 1))(a-\mu 1)^{-1}=(a-\lambda 1)^{-1}-(a-\mu 1)^{-1}}$

Now apply any one and divide the above equation through . It follows ${\displaystyle f\in A'}$ ${\displaystyle \lambda -\mu }$

${\displaystyle {\frac {f((a-\lambda 1)^{-1})-f((a-\mu 1)^{-1})}{\lambda -\mu }}=f((a-\lambda 1)^{-1}(a-\mu 1)^{-1})}$.

The right side exists for reasons of continuity for , because the algebraic operations including inversion in are continuous and is continuous. Hence the function is holomorphic on whole . It disappears in infinity because and is continuous. Therefore this function is bounded and constant according to Liouville's theorem, so it must be equal to. Since was arbitrary, it follows from Hahn-Banach's theorem that , but that cannot be the case for an invertible element. This contradiction ends the proof. ${\displaystyle \mu \rightarrow \lambda }$ ${\displaystyle A}$ ${\displaystyle f}$ ${\displaystyle \lambda \mapsto f((a-\lambda 1)^{-1})}$ ${\displaystyle \mathbb {C} }$ ${\displaystyle \lim _{|\lambda |\rightarrow \infty }\|(a-\lambda 1)^{-1}\|=0}$ ${\displaystyle f}$ ${\displaystyle \mathbb {C} }$ ${\displaystyle 0}$ ${\displaystyle f\in A'}$ ${\displaystyle (a-\lambda 1)^{-1}=0}$

Gelfand-Mazur theorem

If the - Banach algebra is an inclined body , then it is . ${\displaystyle \mathbb {C} }$ ${\displaystyle A}$ ${\displaystyle A\cong {\mathbb {C} }}$

If , according to the above lemma, there is a such that is not invertible. Since the only non-invertible element is in an inclined body, must be. So every element of is a multiple of one, and the assertion follows. ${\displaystyle a\in A}$ ${\displaystyle \lambda \in {\mathbb {C} }}$ ${\displaystyle a-\lambda 1}$ ${\displaystyle 0}$ ${\displaystyle a=\lambda 1}$ ${\displaystyle A}$ ${\displaystyle {\mathbb {C} }}$

See also

• Frobenius theorem (real division algebras)

swell

• RV Kadison , JR Ringrose : Fundamentals of the Theory of Operator Algebras , 1983
• R. Meise, D. Vogt: Introduction to Functional Analysis , Vieweg (1992)