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!



Uniform inequality for an analytic perturbation
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 $C\cdot (\rho+\delta)^n$.


