# Uniform inequality for an analytic perturbation

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!

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 $$m.) Totally we get the estimate like $$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$$.