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I am reading the following paper :

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 .

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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 $\delta>0$.

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