norm of a product of matrices Suppose for every $x \in \mathbb{R}$ and $y \in [0,1]$, $M(x,y)$ is an $n$ by $n$ matrix  and suppose that for every $y \in [0,1]$, $M(x,y) \to M_\infty$ as $|x| \to \infty$, where $M_\infty$ is a constant matrix (where the norm is the operator norm considering the matrix as a linear operator). 
Also, suppose that spectral radius of $M_\infty > 1$. (i.e we also have that $||M_\infty|| >1$)
Define the product $M_n(x,y) $ as 
$M_n(x,y)=M(x,y)M(x+y,y) M(x+2y,y) \ldots M(x+(n-1)y,y)$. Would it be possible to prove something like 
$$ \lim_{n \to \infty} \bigg(\mbox{sup} ||M_n(x,y)||\bigg)^{1/n} \geq ||M_\infty||$$
where the supremum is taken over all $x \in \mathbb{R}$ and all $y \in [0,1]$. 
Thank you. 
EDIT: What I was looking for is an inequality of the form 
$$ \lim_{n \to \infty} \bigg(\mbox{sup} ||M_n(x,y)||\bigg)^{1/n} \geq spr(M_\infty) $$ which is completely answered by Martin Argerami below. 
 A: No. Take $M(x,y)=M_\infty$ for all $x,y$, where $M_\infty$ is a non-zero nilpotent matrix. For large $n$, we have $M_n(x,y)=0$.
A: (note that in your definition your are using $n$ for two completely different things)
Suppose  that $M(x,y)\to M_\infty$ in norm. Then $M_n(x,y)\to M_\infty^n$ and so
$\|M_n(x,y)\|\to\| M_\infty^n\|$. This implies that $\sup \|M_n(x,y)\|\geq\|M_\infty^n\|$. Then
$$\tag{1}
\lim_n\left(\sup\|M_n(x,y)\|\right)^{1/n}\geq\lim_n\|M_\infty^n\|^{1/n}=\text{spr}(M_\infty).
$$
In particular your equality holds if $M_\infty$ is normal. 
To see that you cannot expect to improve a lot on that, consider first $M(x,y)=S$ for all $x,y$, where $S$ is the forward shift. Then $M_n(x,y)=0$ for all sufficiently big $n$, so your inequality cannot hold (still $(1)$ holds, as $\text{spr}(S)=0$).
A: In order to prove something resembling what you are aiming to, the an inequality relation between $M(x,y)$ and $M_{\infty}$ or their norms must be known.
Suppose $M(x,y) = M_{\infty}(1-e^{-x})$. Then $||M(x,y)||<||M_{\infty}|| \quad \forall(x,y)\in \mathbb D$, where $\mathbb {D=R}\times[0,1]$ is the domain you are dealing with.
So, $||M_n(x,y)||^{\frac{1}{n}} <||M_{\infty}||$. If you replace $1-e^{-x}$ by $1+e^{-x}$, you will get the bound you are asking for. 
You need to know how $M(x,y) \to M_{\infty}$. Without it, such a bound seems impossible to exist, let aside prove.
