Combinatorial von Neumann series

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.