I am currently reading a paper which, somewhat indirectly, asserts the following result:
Lemma: Let $\Delta \subset \mathbb{R}^d$ denote the simplex $\{(x_1,\ldots,x_d):\sum_{i=1}^d x_i=1\}$, let $A_1,\ldots,A_N$ be positive $d \times d$ matrices and let $\overline{A}_\ell \colon \Delta \to \Delta$ be the projective transformation induced by $\overline{A}_\ell$, that is if $A=[a_{ij}]_{i,j=1}^d$ then $$\overline{A}\left((x_i)_{i=1}^d\right):=\left(\frac{\sum_{k=1}^d a_{ik}x_k}{\sum_{j,k=1}^d a_{jk}x_k}\right)_{i=1}^d.$$ Then (up to identifying $\Delta$ with a subset of $\mathbb{R}^{d-1}$ by an affine change of co-ordinates) there exists a complex neighbourhood $D\subset \mathbb{C}^{d-1}$ of $\Delta$ such that each $\overline{A}_\ell$ extends to a holomorphic map $\overline{A}_\ell \colon D \to D$ with the property that $\overline{A}_\ell D$ is precompact in $D$.
The case $d=2$ is given in the paper as an example: in a natural way we can identify $\Delta$ with $[0,1]$ and each induced map $\overline{A}_\ell$ with a linear fractional transformation. If we let $D$ be a complex disc centred at $1/2$ with diameter slightly larger than $1$ then each $\overline{A}_i$ maps the interval $[0,1]$ to a proper subinterval and therefore maps $D$ to a disc which is centred somewhere in $(0,1)$, has diameter smaller than one, and is precompact in $D$. However, I do not understand how to generalise this argument to higher dimensions. Is it perhaps more clear to someone better-versed in multivariate complex analysis?
The paper goes on to assert this result for the more general situation where $\mathcal{K}\subseteq \mathbb{R}^d$ is a cone which is mapped strictly inside itself by each $A_\ell$ and where $\Delta$ is the intersection of $\mathcal{K}$ with a suitable hyperplane. This seems even more challenging to me, since $\Delta$ is then not merely a simplex but an arbitrary compact convex set with interior. How might the lemma be proved in this case?
