Spectral Radius at the Unique and Globally Attractive Fixed Point of a Specific Type of Mapping

Question

Consider a mapping $f: \mathbb{R}_+^n \rightarrow \mathbb{R}_{++}^n$. If it is monotonic and strictly subhomogeous, then it is contractive under the Thompson’s metric (See Lemma 2.1.7). Here,

1. Monotonic: If $x \leq y$, then $f(x)\leq f(y) \quad (\forall x, y \in \mathbb{R}_+^n)$.
2. Strictly subhomogeous: $f(\alpha x) < \alpha f(x) \quad (\forall \alpha\in(0,1) \text{ and }\forall x \in \mathbb{R}_+^n)$.
3. Contractive: $$\mu(f(x), f(y)) < \mu(x, y),$$ where $\mu$ is the Thompson’s metric on $\mathbb{R}_{++}^n$, defined by $$\mu(x, y) = \max_{i=1}^n\left\{\log (x_i/y_i)\right\}. $$

Using a fixed point theorem on the compact space, one can show the uniqueness and the global attractiveness of the fixed point, $x^\star$, of this type of mapping (See Theorem 2.3). To facilitate the discussion, we may strengthen the assumption by

1. asking $f$ to be a local $L$-contraction at $x^\star$ under the Thompson's metric, i.e., for some $L < 1$ and some neighbourhood $U$ of $x^\star$ $$\mu(f(x), f(y)) \leq L \mu(x, y) \quad (\forall x, y \in U), $$ and;
2. assuming that the Jacobian $J(x)$ is symmetric.

I am interested in the invertability of $I - J(x^*)$. One sufficient condition to ensure the invertability is to show that the spectral radius $\rho(J(x^\star)) < 1$.

Two investgated cases:

1. If $f$ is an affine function, then standard results shows that if $x^\star$ is globally attractive, then $\rho(J) < 1$. Hence the invertability of $I - J(x^\star)$ is guaranteed.

2. If $f$ is an abitrary differentiable function, then this answer shows that if $f$ is a $L$-contraction under the Euclidean norm $\|\cdot\|$ if and only if $\|J(x)\| \leq L \quad (\forall x \in \mathbb{R}_+^n)$. Since we assume that $J(x)$ is symmetric, $\rho(J(x)) = \|J(x)\| \leq L < 1$. However, this conclusion cannot be applied to our case because the local $L$-contraction of $f$ under the Thompson's metric does not imply the local $L$-contraction of $f$ under the Euclidean norm.

Any thoughts on this problem is appreciated!

Answer 1

I think it is true that $\rho(J(x^\ast)) \le L$ under your assumptions: In the following let $\le$ and $\ll$ be intended coordinatewise $\le$ and $<$, respectively, for vectors as well as for matrices. We have $0 \ll x^\ast$, so we can norm $\mathbb{R}^n$ by the Minkowski functional of the order interval $[-x^\ast,x^\ast]$, that is $\|x\|=\min\{\lambda \ge 0: -\lambda x^\ast \le x \le \lambda x^\ast \}$. Let $||| \cdot|||$ denote the corresponding matrix norm. For each $A \in \mathbb{R}^{n \times n}$ with $A \ge 0$ we then have $|||A|||=\|Ax^\ast\|$. For $t$ in a small interval $I=(-\delta, \delta)$ we have $$ f((1+t)x^\ast)=f(x^\ast)+ t J(x^\ast)x^\ast + o(t) \quad (t \to 0). $$ The contraction condition yields $\mu(f(x),f(x^\ast)) \le L\mu(x,x^\ast)$ for $x$ in a neighborhood $U$ of $x^\ast$. W.l.o.g. let $\delta$ be such that $(1+t)x^\ast\in U$ $(t \in (-\delta, \delta))$. Now let $t \in (0,\delta)$. We have $$ \mu(f((1+t)x^\ast),f(x^\ast))\le L\mu((1+t)x^\ast,x^\ast)= L\log(1+t), $$ which yields $$ f((1+t)x^\ast) \le (1+t)^Lf(x^\ast), $$ so $$ t J(x^\ast)x^\ast=f((1+t)x^\ast)-f(x^\ast) + o(t) \le ((1+t)^L-1)f(x^\ast)+o(t)=((1+t)^L-1)x^\ast + o(t). $$ Division by $t$ (recall $t>0$) gives $$ J(x^\ast)x^\ast \le \frac{(1+t)^L-1}{t}x^\ast + \frac{1}{t}o(t) \to Lx^\ast \quad (t \to 0+). $$ Thus $J(x^\ast)x^\ast \le Lx^\ast$. Since $J(x^\ast) \ge 0$ we have $|||J(x^\ast)|||= \|J(x^\ast)x^\ast\| \le \|Lx^\ast\|= L$. By Gelfand's formula $\rho(J(x^\ast))= \inf_{n\ge 1} |||J(x^\ast)^n|||^{1/n} \le |||J(x^\ast)|||\le L$ (so we don't need symmetry). Hope everything is correct, please check.