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Let $T$ be a bounded linear operator acting on a complex Banach space. Suppose that $T$ has spectral radius strictly less than $1$. If we introduce an analytic perturbation to $T$, $s\mapsto T_s$ for $|s|<\epsilon$ (with $T_0 = T$), then by upper-semicontinuity of the spectrum, assuming that $\epsilon$ is sufficiently small, the spectral radius of each $T_s$ is less than $\rho$ for some $\rho <1$. My question is the following:

Is it possible to find $K>0$ and $\rho <\rho '<1$ such that $$\|T_s^n\|\le K (\rho ')^n,$$ for all $|s|<\epsilon$ and $n\in \mathbb{Z}_{\ge 0}$?

This certainly seems like it should be true but I can't find a proof - I think I'm missing something obvious. Any help would be greatly appreciated - cheers!

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Continuity of the perturbation (in the norm operator topology, is it what you mean?) is enough. Denote $T_s=T+A$ where $\|A\|<\varepsilon$ (this is so for small enough $s$ by norm continuity.) Choose any $\rho$ strictly between spectral radius of $T$ and 1 and fix $n_0$ such that $\|T^n\|\leqslant \rho^n$ whenever $n\geqslant n_0$. We have $(T+A)^n=\sum_{R_i\in \{T,A\}} R_1 R_2\dots R_n.$ Consider the product $R_1\dots R_n$ containing exactly $k$ multiples $A$. Its norm may be estimated from above as $\rho^{n-k-n_0(k+1)} \varepsilon^k M^{n_0(k+1)}$, where $M=\|T\|$ (to see this, partition all letters $T$ in the word $R_1\dots R_n$ onto at most $k+1$ connected pieces and estimate each of them: $\|T^m\|\leqslant \rho^m$ for $m\geqslant n_0$ and $\|T^m\|\leqslant M^m$ for $m<n_0$.) Totally we get the estimate like $C\cdot\rho^{n-k}\delta^k$ for small $\delta=\varepsilon (M/\rho)^{n_0}$.Summing up over all words give the norm estimate $C\cdot (\rho+\delta)^n$.

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  • $\begingroup$ @Zestylemonzi you are welcome! $\endgroup$ – Fedor Petrov Dec 28 '18 at 11:48

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