real analysis - a unique solution ? iteration involving conditional distributions

The transformation $L=TG$ is defined on vectors $x$ with positive coordinates by
$$
Lx(s)=sum_uq(u|s)mathrm{e}^{-r(u)}Mx(u),quadmbox{where}
Mx(s)=prod_ux(u)^{p(u|s)}.
$$
Thus $M$ and $L$ are homogenous and nondecreasing on the positive orthant. This means that one considers vectors $x$ such that $x(s)>0$ for every $s$, that $M(lambda x)=lambda Mx$ and $L(lambda x)=lambda L(x)$ for every positive scalar $lambda$, and that $Mxle Mtilde x$ and $Lxle Ltilde x$ if $xletilde x$ in the sense that $x(s)letilde x(s)$ for every $s$.

For every vector $x$ with positive coordinates, let $u(x)$ and $ell(x)$ denote the supremum and the infimum of its coordinates $x(s)$, hence $ell(x)le x(s)le u(x)$ for every $s$.

Since $p$ is a transition kernel, $displaystylesum_up(u|s)=1$ for every $s$ hence $ell(x)le Mx(s)le u(x)$ for every $s$ and $ell(x)a(s)le Lx(s)le u(x)a(s)$ with
$$
a(s)=sum_uq(u|s)mathrm{e}^{-r(u)}.
$$
More generally, for every positive $t$,
$$
ell(x)ell(a)^{t-1}a(s)le L^tx(s)le u(x)u(a)^{t-1}a(s),
$$
hence
$$
ell(x)ell(a)^{t}le ell(L^tx)le u(L^tx)le u(x)u(a)^{t}.
$$
Furthermore, $u(a)le u(mathrm{e}^{-r})$ and $ell(a)geell(mathrm{e}^{-r})$. Now everything depends on the hypothesis made on $r$.

If $r(s)>0$ for every $s$ (and I believe this is what the OP wanted to write), then $u(a)<1$ hence $L^tx$ converges geometrically to $0$. If $r(s)<0$ for every $s$ (and this is what the OP actually wrote), then $ell(a)>1$ hence $L^tx$ diverges geometrically to $+infty$.

For $(x_t)$ to converge to a nondegenerate limit, one should assume that $r$ has positive and negative coordinates.