Green-kernel resolvent identity

If $c$ is a finite positive real and $(a_n{)}_{n\ge 0}$ is nonnegative, then $c{\sum }_{n\ge 0}{a}_n={\sum }_{n\ge 0}c{a}_n$. For partial sums $s_N={\sum }_{n<N}{a}_n$, if ${\sup }_N{s}_N<\mathrm{\infty }$, continuity of multiplication by $c$ gives $c{s}_N\uparrow c{\sup }_N{s}_N$; if ${\sup }_N{s}_N=+\mathrm{\infty }$, the $s_N$ are unbounded and so are $c{s}_N$. By [F5], every product is defined because $c>0$.

Fix $x,y\in E$. For each $z$ with $p(x,z)>0$, [F1, F2] and step 1.1 give $p(x,z)G(z,y)={\sum }_{n\ge 0}p(x,z)p^{(n)}(z,y)$. Terms with $p(x,z)=0$ are omitted in $(PG)(x,y)$; inserting corresponding zero terms in the nonnegative double series is valid because $p^{(n)}(z,y)\le 1$. Apply [F4] to that double series. If $E$ is finite, use a finite listing and pad with zeros; if countably infinite, use a bijection with $\mathbb{N}$ from [F6]. Then [F3] with $m=1$ yields $(PG)(x,y)={\sum }_{z\in E}{\sum }_{n\ge 0}p(x,z)p^{(n)}(z,y)={\sum }_{n\ge 0}{\sum }_{z\in E}p(x,z)p^{(n)}(z,y)={\sum }_{n\ge 0}{p}^{(n+1)}(x,y)={\sum }_{n\ge 1}{p}^{(n)}(x,y)\mathrm{.}$

For each $z$ with $p(z,y)>0$, [F1] and step 1.1 give $G(x,z)p(z,y)={\sum }_{n\ge 0}{p}^{(n)}(x,z)p(z,y)$. Terms with $p(z,y)=0$ are omitted in $(GP)(x,y)$; inserting their zero finite products in the double series introduces no undefined extended-real product. Tonelli [F4], now summing first over $z$, and [F3] with $m=n$ and second time index $1$ give $(GP)(x,y)={\sum }_{z\in E}{\sum }_{n\ge 0}{p}^{(n)}(x,z)p(z,y)={\sum }_{n\ge 0}{\sum }_{z\in E}{p}^{(n)}(x,z)p(z,y)={\sum }_{n\ge 0}{p}^{(n+1)}(x,y)={\sum }_{n\ge 1}{p}^{(n)}(x,y)\mathrm{.}$

By [F1, F8], separating the $n=0$ term in the nonnegative series gives $G(x,y)=p^{(0)}(x,y)+{\sum }_{n\ge 1}{p}^{(n)}(x,y)$. This is a split of nonnegative partial sums, not a subtraction. Using [F7] and steps 2.1 and 2.2 proves both identities. If $E=\mathrm{\varnothing }$, there are no $x,y$ and the claim is vacuous. For a one-state absorbing chain, $G=+\mathrm{\infty }$ and both support-restricted products equal $+\mathrm{\infty }$, so $G=1+\mathrm{\infty }$ is well defined. Zero transition coefficients are always omitted; positive coefficients may multiply $+\mathrm{\infty }$ and produce $+\mathrm{\infty }$. The argument includes deterministic rows and all zero-time endpoints. It uses no AC: the one enumeration of this fixed countable $E$ is part of [F6], and [F4] is proved using finite choice. The lemma states no biconditional.