Let us consider the space $L_2(\mathbb{R} \times [0,1]; \mathbb{R}^n)$, i.e functions taking values in $\mathbb{R}^n$ and in $L_2$ . Suppose $T$ is a bounded linear operator defined as follows:
$Tf(x,y)=M(x,y)f(x+y,y)$, where $M(x,y)$ is an $n \times n$ matrix.
Suppose we also know that $M(x,y) \to M_\infty$ as $|x| \to \infty$ for every $y \in [0,1]$ where $M_\infty$ is a constant matrix. Suppose the spectral radius of $M_{\infty} > 1$.
$M(x,y)$ also has the properties $M(x,0)=M_\infty$ and $M(x,y)=M(ax,ay)$ for $a>0$ and such that $ay \in [0,1]$
Can we prove something like the spectral radius of $T > 1$
Thank you.
Yes. It can be proved.
Note: I have replaced $M_{\infty} \mbox{ with } M_0$ for ease of typing.
Let $$Tf(x,y) = M(x,y)f(x+y,y)$$
Define $M_n(x,y) = M(x,y)M(x+y,y)\ldots M(x+(n-1)y,y)$
$$T^n f(x,y) = M_n(x,y)f(x+ny,y)$$
Norm of $T^n,||T^n||_{op} = \sup\{{\frac{||T^nf||}{|f|}},|f|\neq0\}$
Consider the case $y=0$.
$$M_n(x,0) = M_0^n$$ $$\frac{||T^nf||}{|f|} = \frac{||M_0^n||\times|f(x,0)|}{|f(x,0)|} = ||M_0^n||$$
So, $$||T^n||_{op} \geq ||M_0^n||$$ $$\rho(T) = \lim_{n\to \infty}||T^n||_{op}^{\frac{1}{n}} \geq \lim_{n \to \infty}||M_0^n||^{\frac{1}{n}} = \rho(M_0)$$
So, $$\rho(T) \geq \rho(M_0)$$
Hope it is what you are looking for.