# Combinatorial von Neumann series

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Combinatorial von Neumann series are the result of transferring, respectively utilizing, methods from combinatorial number theory onto functional analysis.

## Theorem

Let $X$ be a Banach space (normed vector space), let $T$ and $S$ be each a bounded linear operator on $X$ , that suffice:

$\Vert TS \Vert < 1$ then the following series:

$(\operatorname{id} - TS)^{-1} = (E - TS)^{-1} = \displaystyle\sum\limits_{\nu = 1}^{\infty}~(TS)^{\nu} \geq \displaystyle\sum\limits_{\operatorname{com}_{\nu}}~(TS)^{\nu}$ are convergent in the space $L(X$ , $X)$ in the sense of the operator norm.

#### Proof

The first equation on the left hand side follows by applying the Neumann series and exchanging the product $TS$ with the bounded linear operator $W$ in it; this yields:

$\begin{array}{lll} & (E - W)^{-1} & = \displaystyle\sum\limits_{\nu = 1}^{\infty}~W^{\nu} \\ & & \\ \Longrightarrow & (E - TS)^{-1} & = \displaystyle\sum\limits_{\nu = 1}^{\infty}~(TS)^{\nu} \\ \end{array}$ To see that:

$\displaystyle\sum\limits_{\nu = 1}^{\infty}~(TS)^{\nu} \geq \displaystyle\sum\limits_{\operatorname{com}_{\nu}}~(TS)^{\nu}$ holds, it is only necessary to thin out the series on the left hand side of the equation.

## Theorem

Let $X$ be a Banach space (normed vector space), let $T_{\mu}$ and $S_{\nu}$ be for all $\mu = 1$ , $2$ , $3$ , $\dots$ and $\nu = 1$ , $2$ , $3$ , $\dots$ bounded linear operators on $X$ , which are well ordered and sorted by size, and that suffice:

$\Vert T_{\mu} S_{\nu} \Vert < 1$ for all $\mu = 1$ , $2$ , $3$ , $\dots$ and $\nu = 1$ , $2$ , $3$ , $\dots$ . Then the following series:

$\displaystyle\sum\limits_{\operatorname{com}_{\nu \mu \sigma}}~(T_{\mu} S_{\nu})^{\sigma} \leq \displaystyle\sum\limits_{m = 1 \atop \operatorname{com}_{\nu \mu}}^{\infty}~(T_{\mu} S_{\nu})^{m} \leq \displaystyle\sum\limits_{m = 1}^{\infty}~(T_{\alpha} S_{\beta})^{m}$ are convergent in the space $L(X$ , $X)$ in the sense of the operator norm with the condition:

$\alpha \in \mu$ , $\beta \in \nu$ , $\forall \mu$ , $\forall \nu: \Vert T_{\alpha} S_{\beta} \Vert \geq \Vert T_{\mu} S_{\nu} \Vert$ #### Proof

After the von Neumann series follows for the right hand side:

$(\operatorname{id} - T_{\alpha} S_{\beta})^{-1} = (E - T_{\alpha} S_{\beta})^{-1} = \displaystyle\sum\limits_{m = 1}^{\infty}~(T_{\alpha} S_{\beta})^{m}$ The condition implies immediately the rest of the inequation.