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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.

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Also, here is a link to the original problem I am trying to solve math.stackexchange.com/questions/247861/… –  Shibi Vasudevan Dec 13 '12 at 1:25
Your original question has a much simpler answer. See that link for details. –  dexter04 Dec 13 '12 at 6:06

3 Answers 3

up vote 1 down vote accepted

(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 $\text{spr}(S)=0$).

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Thanks for your answer. I have edited the question a bit to include the condition on the spectral radius of $ M_\infty $ –  Shibi Vasudevan Dec 13 '12 at 1:03
Thank you so much for your answer. –  Shibi Vasudevan Dec 13 '12 at 4:59
@Martin could you please explain $M_n(x,y)\to M_\infty^n$. Also how does $\|M_n(x,y)\|\to\| M_\infty^n\|$ imply $\sup \|M_n(x,y)\|\geq\|M_\infty^n\|$, because it could be approching the end limit from below also, isn't it? –  dineshdileep Dec 13 '12 at 5:37
For the limit, note that $n$ is fixed, and the limit of the product is the product of the limit. For your second question, the fact that you have the limit means that equality is at least guaranteed; from there, you can only get bigger; just do a couple examples with functions. –  Martin Argerami Dec 13 '12 at 13:18

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 $M_n(x,y)=0$.

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Thanks for your answer. My actual problem is here(if you would take a look) math.stackexchange.com/questions/247861/… –  Shibi Vasudevan Dec 13 '12 at 0:49
Also, I have edited this question. Thanks. –  Shibi Vasudevan Dec 13 '12 at 1:04

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.

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Thanks. I have added a condition on the spectrum of $M_\infty$ –  Shibi Vasudevan Dec 13 '12 at 1:04
My previous answer was blatantly wrong, see the edit. –  Martin Argerami Dec 13 '12 at 4:44

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