I am given a lemma and it states:
Lemma: Let $A$ be a real $n\times n$ matrix. Then given any $e > 0$, there is a $norm$ such that $norm(A) \leq p(A) + e$, where $p(A)$ is the spectral radius of $A$.
They then go on to state that based on this lemma, if $p(A) < 1$, then $norm(A) < 1$ for the correct choice of the norm.
What I'm looking for help on is understanding how they came to that conclusion based only on that lemma. For example, given $e =0.001$, then the lemma states there is a norm such that $norm(A)\leq p(A) + .001$.
Since it is a "or equal" then this means there is the possibility that $norm(A) = p(A) + 0.001$, and in such a case then $norm(A) > p(A)$. So it seems that based on the lemma, then $p(A) < 1$ does not necessarily always imply that $norm(A) < 1$.
I even tried to do a proof of this but still came up with the same result:
$$\begin{align} p(A)<1&\implies p(A)+e<1+e\\ &\implies norm(A)\leq p(A)+e<1+e\\ &\implies norm(A)-e\leq p(A)<1. \end{align} $$
This pretty much says the same thing, if $norm(A) - e \leq p(A)$, then $p(A) < 1$, but if $p(A)$ close to $1$ by a difference less than $e$, then $norm(A) > 1$ in the "equal to" case where $norm(A) - e = p(A)$.
Now of course you could pick a smaller e, but the way the lemma reads it says that for a given $e$, you can find a norm s.t. $norm(A) \leq p(A) + e$, so the "or equal" part will still get you using a counter example I gave similar to the $e = .001$ above
Now I have seen a different proof on wikipedia that shows if $p(A) < 1$ then $norm(A) < 1$, so I am not disputing that fact, but that proof gets into looking at entries of the Jordan Normal Form. So my point is, yes $p(A) < 1$ then $norm(A) < 1$, BUT I don't understand how you can conclude that based only on the Lemma I gave at the beginning. It's these big jumps in rational that confuse me.