Does a operator norm exist to bound each element in sequence of bounded linear operators.

Assume we have a sequence, $(O_n)$, of bounded linear operators which map from $X$ to $X$ a finite dimensional Banach space. Also assume each operators has the same spectral radius, and $\rho(O_1)<1$.

Does an operator norm $\|\cdot\|$ exist such that $\|O_n\|<\alpha<1$ for all $n$. Just to clarify I want to know if a norm exists that does not depend on $n$.

I know that for a single operator, $O$, that for any $\epsilon>0$ there exists an operator norm $\|\cdot\|_{\epsilon}$, such that $\|O\|_{\epsilon}\le \rho(O)+\epsilon$. Now if $\rho(O)<1$. and since $\epsilon>0$ can be arbitrary we can pick an operator norm $\|\cdot\|_{\epsilon}$ such that a $\|O\|_{\epsilon}\le\alpha<1$.

I'm just not sure if you can construct the single operator norm such that $\|O_n\|<\alpha<1$ for all $n$.

• Did you mean operator norm ($\|AB\|\le\|A\|\|B\|$)? – Martín-Blas Pérez Pinilla Mar 21 '17 at 9:50
No. Regardless of the operator norm chosen, $\lim\limits_{n\to\infty}\left\|\begin{bmatrix}0&n\\0&0\end{bmatrix}\right\|=\lim\limits_{n\to\infty}n\left\|\begin{bmatrix}0&1\\0&0\end{bmatrix}\right\|=\infty.$