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 $n$.

Any advice or tips would be greatly helpful.

  • $\begingroup$ Did you mean operator norm ($\|AB\|\le\|A\|\|B\|$)? $\endgroup$ – Martín-Blas Pérez Pinilla Mar 21 '17 at 9:50
  • $\begingroup$ I do mean a operator norm. I have updated the question to be clearer. $\endgroup$ – tinyhippo Mar 21 '17 at 13:36

No. Regardless of the operator norm chosen, $\lim\limits_{n\to\infty}\left\|\begin{bmatrix}0&n\\0&0\end{bmatrix}\right\|=\lim\limits_{n\to\infty}n\left\|\begin{bmatrix}0&1\\0&0\end{bmatrix}\right\|=\infty.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.