# proof of Neumann series in Banach algebras

Let x be an element of a Banach algebra^{} with identity, $$. By applying the properties of the Norm in a Banach algebra, we see that the partial sums form a Cauchy sequence^{}: $\parallel {\sum}_{n=l}^{m}{x}^{n}\parallel \le {\sum}_{n=l}^{m}{\parallel x\parallel}^{n}\to 0$ for $l,m\to \mathrm{\infty}$ (as is well known from real analysis), so by completeness of the Banach Algebra, the series converges^{} to some element $y={\sum}_{n=0}^{\mathrm{\infty}}{x}^{n}$.

We observe that for any $m\in \mathbb{N}$,

$$(1-x)\sum _{n=0}^{m}{x}^{n}=\sum _{n=0}^{m}{x}^{n}-\sum _{n=1}^{m+1}{x}^{n}=1-{x}^{m+1}$$ | (1) |

Furthermore, $\parallel {x}^{m+1}\parallel \le {\parallel x\parallel}^{m+1}$, so ${lim}_{m}{x}^{m+1}=0$.

Thus, by taking the limit $m\to \mathrm{\infty}$ on both sides of (1), we get

$$(1-x)y=1$$ |

(We can exchange the limit with the multiplication by $(1-x)$, since the multiplication in Banach algebras is continuous^{})

Since the Banach algebra generated by a single element is commutative and $(1-x)$ and $y$ are both in the Banach algebra generated by $x$, we also get $y(1-x)=1$. Hence, $y={(1-x)}^{-1}$.

As in the first paragraph, the last claim $y\le \frac{1}{1-\parallel y\parallel}$ again follows by applying the geometric series for real numbers.

Title | proof of Neumann series in Banach algebras |
---|---|

Canonical name | ProofOfNeumannSeriesInBanachAlgebras |

Date of creation | 2013-03-22 17:32:40 |

Last modified on | 2013-03-22 17:32:40 |

Owner | FunctorSalad (18100) |

Last modified by | FunctorSalad (18100) |

Numerical id | 5 |

Author | FunctorSalad (18100) |

Entry type | Proof |

Classification | msc 46H05 |