# 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 ${\displaystyle X}$ be a Banach space (normed vector space), let ${\displaystyle T}$ and ${\displaystyle S}$ be each a bounded linear operator on ${\displaystyle X}$, that suffice:

${\displaystyle \Vert TS \Vert < 1}$

then the following series:

${\displaystyle (\operatorname{id} - TS)^{-1} = (E - TS)^{-1} = \displaystyle\sum\limits_{\nu = 1}^{\infty}~(TS)^{\nu} \geq \displaystyle\sum\limits_{\operatorname{com}_{\nu}}~(TS)^{\nu}}$

are convergent in the space ${\displaystyle L(X}$, ${\displaystyle X)}$ 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 ${\displaystyle TS}$ with the bounded linear operator ${\displaystyle W}$ in it; this yields:

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

To see that:

${\displaystyle \displaystyle\sum\limits_{\nu = 1}^{\infty}~(TS)^{\nu} \geq \displaystyle\sum\limits_{\operatorname{com}_{\nu}}~(TS)^{\nu}}$

holds, it is only necessary to thin out the series on the left hand side of the equation.

## Theorem

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

${\displaystyle \Vert T_{\mu} S_{\nu} \Vert < 1}$

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

${\displaystyle \displaystyle\sum\limits_{\operatorname{com}_{\nu \mu \sigma}}~(T_{\mu} S_{\nu})^{\sigma} \leq \displaystyle\sum\limits_{m = 1 \atop \operatorname{com}_{\nu \mu}}^{\infty}~(T_{\mu} S_{\nu})^{m} \leq \displaystyle\sum\limits_{m = 1}^{\infty}~(T_{\alpha} S_{\beta})^{m}}$

are convergent in the space ${\displaystyle L(X}$, ${\displaystyle X)}$ in the sense of the operator norm with the condition:

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

#### Proof

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

${\displaystyle (\operatorname{id} - T_{\alpha} S_{\beta})^{-1} = (E - T_{\alpha} S_{\beta})^{-1} = \displaystyle\sum\limits_{m = 1}^{\infty}~(T_{\alpha} S_{\beta})^{m}}$

The condition implies immediately the rest of the inequation.