Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am reading the following paper :

http://www2.icmc.usp.br/~sma/cadernos/toc9.1/292.pdf

In the second paragraph the author introduces a new operator $\|x\|_{T,\epsilon}$ , which i don't really understand how it is equivalent to the operator norm and why its greater or equal to $r(T)+\epsilon$ .

I understand that if we have the above result we can see that $r(A+B) \le r(A)+ r(B)$ .

$r(A)$ denotes the spectral radius of the operator .

But basically i don't understand the construction of new norm .

Can some one explain me .

share|improve this question
add comment

1 Answer

up vote 0 down vote accepted

Equivalence of norms follows from the fact that $T$ bounded. More precisely, $$|x|^2\leqslant \sum_{j=0}^m\left(\frac{|T^jx|}{(r(T)+\varepsilon)^j}\right)^2=|x|_{T,\varepsilon}\leqslant |x|^2\sum_{j=0}^m\left(\frac{|T^j|}{(r(T)+\varepsilon)^j}\right)^2.$$

We can write \begin{align} |Tx|_{T,\varepsilon}^2&=\sum_{j=0}^m \left(\frac{|T^{j+1}x|}{(r(T)+\varepsilon)^j}\right)^2\\ &=(r(T)+\varepsilon)\sum_{k=1}^{m+1}\left(\frac{|T^kx|}{(r(T)+\varepsilon)^k}\right)^2\\ &\leqslant (r(T)+\varepsilon)\sum_{k=1}^m\left(\frac{|T^kx|}{(r(T)+\varepsilon)^k}\right)^2 +(r(T)+\varepsilon)|x|\cdot |T^{m+1}|\cdot M^{-(m+1)}\\ &\leqslant (r(T)+\varepsilon)|x|_{T,\varepsilon}-\delta(r(T)+\varepsilon)|x|, \end{align} as $|T^{m+1}|\cdot M^{-(m+1)}=1-\delta$ for some $\delta>0$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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