On powers of operators with spectrum in cantor sets and spectral synthesis

For $\xi \in \big( 0, \frac{1}{2} \big)$, let $E_{\xi}$ be the perfect symmetric set associated with $\xi$, that is $$E_{\xi} = \Big\{ \exp \Big( 2i \pi (1-\xi) \sum_{n = 1}^{+\infty} \epsilon_{n} \xi^{n-1} \Big) : \, \epsilon_{n} = 0 \textrm{ or } 1 \quad (n \geq 1) \Big\}$$ and $$b(\xi) = \frac{\log{\frac{1}{\xi}} - \log{2}}{2\log{\frac{1}{\xi}} - \log{2}}.$$ Let $q\geq 3$ be an integer and $s$ be a nonnegative real number. We show that any invertible operator $T$ on a Banach space with spectrum contained in $E_{1/q}$ that satisfies \begin{eqnarray*} & & \big\| T^{n} \big\| = O \big( n^{s} \big), \,n \rightarrow +\infty \\ & \textrm{and} & \big\| T^{-n} \big\| = O \big( e^{n^{\beta}} \big), \, n \rightarrow +\infty \textrm{ for some } \beta < b(1/q),\end{eqnarray*} also satisfies the stronger property $\big\| T^{-n} \big\| = O \big( n^{s} \big), \, n \rightarrow +\infty.$ We also show that this result is false for $E_\xi$ when $1/\xi$ is not a Pisot number and that the constant $b(1/q)$ is sharp. As a consequence we prove that, if $\omega$ is a submulticative weight such that $\omega(n)=(1+n)^s, \, (n \geq 0)$ and $C^{-1} (1+|n|)^s \leq \omega(-n) \leq C e^{n^{\beta}},\, (n\geq 0)$, for some constants $C>0$ and $\beta < b( 1/q),$ then $E_{1/q}$ satisfies spectral synthesis in the Beurling algebra of all continuous functions $f$ on the unit circle $\mathbb{T}$ such that $\sum_{n = -\infty}^{+\infty} | \widehat{f}(n) | \omega (n) < +\infty$.