# proof of Neumann series in Banach algebras

Let x be an element of a Banach algebra with identity, $\|x\|<1$. By applying the properties of the Norm in a Banach algebra, we see that the partial sums form a Cauchy sequence: $\|\sum_{n=l}^{m}x^{n}\|\leq\sum_{n=l}^{m}\|x\|^{n}\to 0$ for $l,m\to\infty$ (as is well known from real analysis), so by completeness of the Banach Algebra, the series converges to some element $y=\sum_{n=0}^{\infty}x^{n}$.

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, $\|x^{m+1}\|\leq\|x\|^{m+1}$, so ${\lim}_{m}x^{m+1}=0$.

Thus, by taking the limit $m\to\infty$ 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 $y\leq\frac{1}{1-\|y\|}$ again follows by applying the geometric series for real numbers.

Title proof of Neumann series in Banach algebras ProofOfNeumannSeriesInBanachAlgebras 2013-03-22 17:32:40 2013-03-22 17:32:40 FunctorSalad (18100) FunctorSalad (18100) 5 FunctorSalad (18100) Proof msc 46H05