Let $A:\mathcal{D}(A)\to\mathcal{H}$ be a self-adjoint operator in a Hilbert space $\mathcal{H}$. Furthermore, assume that its spectrum satisfies $\sigma(A)\in [a,\infty)$ for some $a>0$. Then, using the spectral theorem, we can define the operator $A^{s}:\mathcal{D}(A^{s})\to\mathcal{H}$ for all $s\in\mathbb{R}$ via
$$A^{s}=\int_{\sigma(A)}\lambda^{s}\,\mathrm{d}P(\lambda),$$
where $P$ denotes the spectral measure of $A$.
Is it true that $0\notin\sigma(A^{s})$ for all $s\in\mathbb{R}$?
Yes. $$ \|A^{s}x\|^2=\int_a^\infty |\lambda^{s}|^2 d\|P(\lambda)x\|^2 \ge |a^s|^2\int_a^{\infty} d\|P(\lambda)x\|^2=|a^s|^2\|x\|^2. $$