1
$\begingroup$

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.

1 Answers 1

0

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.

  • 0
    Thank you so much for the answer. But ||T^n f(x,0)||=||M_0^n f(x,0)||. How did you get that this is equal to ||M_0^n|| |f(x,0)|2012-12-13