Polynomial decay of any order implies exponential decay for operators norm

Question

Context and Motivation

Consider a Banach space $X$ and a linear bounded operator $T \in \mathcal{L}(X)$. We have established that for each non-negative integer $k$, there exists a constant $C_k$ dependent solely on $k$, such that:

$$ \|T^n\| \leq \frac{C_k}{n^{k}}, \quad \forall n \in \mathbb{N}_0. $$

Question

Is it possible to conclude that the norms of the operator's powers exhibit exponential decay? In other words, do there exist constants $M > 0$ and $\omega > 0$ such that:

$$ \|T^n\| \leq M e^{-\omega n}, \quad \forall n \in \mathbb{N}_0? $$

My Approach

I have attempted to follow a proof similar to the one utilized when demonstrating the exponential bound of a $C_0$-semigroup. My approach is as follows:

For any given $k \geq 1$, let's find an $N$ such that $\frac{C_k}{N^k} < 1$. Now, we can express any power greater than $N$ as $n = sN + m$, where $s\in \mathbb{N}$ and $m \in \{0,1,2, \dots , N-1\}$. Consequently, we obtain:

$$ \|T^n\| \leq \|T^N\|^s \|T^m\| \leq M e^{s \log \left( \frac{C_k}{N^k} \right)} = M e^{-\omega s}, $$

where $\omega = -\log \left( \frac{C_k}{N^k} \right) > 0$. However, I encounter difficulty in recovering the original value of $n$ from this point onwards.

Side Note

I have reviewed the discussion in What Lies Between Polynomial Decay of Any Order and Exponential Decay. It's worth noting that their example of $k^{\log k}$ involves a function that is not linear, so I believe it may not directly apply to my specific question.

Answer 1

The proof in OP is correct. Another argument could be as follows. For $m>C_1$ we have $\|T^m\|<1.$ Therefore the spectral radius of $T^m$ is less than $1.$ Thus the spectral radius $r_0$ of $T$ is less than $1.$ Let $r=(1+r_0)/2.$ Then $r<1$ and $$r_0=\lim_n\| T^n\|^{1/n}<r$$ We get $$\|T^n\|\le r^n, \quad n\ge n_0$$ for some $n_0.$ This implies the conclusion with $\omega =-\log r$ and $M=\max_{0\le j<n_0}\|T^j\|$

Answer 2

I will post the answer in order to finish correctly my approach but the other solution posted by Ryszard Szwarc is really inspiring.

For any given $k \geq 1$, let's find an $N$ such that $\frac{C_k}{N^k} < 1$. Now, we can express any power greater than $N$ as $n = sN + m$, where $s\in \mathbb{N}$ and $m \in \{0,1,2, \dots , N-1\}$. Consequently, we obtain:

$$ \|T^n\| \leq \|T^N\|^s \|T^m\| \leq \max_{m\in \{0,\dots,N-1\}} \|T^m\| e^{s \log \left( \frac{C_k}{N^k} \right)} = M e^{-\omega(N-1)} e^{-\omega Ns} \leq M e^{-\omega (Ns+m)} = M e^{-\omega n} , $$

where $\omega = -\frac{\log \left( \frac{C_k}{N^k} \right)}{N} > 0$ and $M = \max_{m\in \{0,\dots,N-1\}} \|T^m\| e^{\omega(N-1)}$.