Limit in the Sobolev space

Question

Let $A$ be the operator whose symbol in the Fourier domain is $i|\xi|$. Let $u \in H^k(\mathbb{R})$ with $k>0$. Consider now the semi-group $e^{(A-\alpha)s}$. I want to prove that $\lim_{s\rightarrow \infty} e^{(A-\alpha)s}u = 0$. To this end, I write $$\|e^{(A-\alpha)s}u\|_k \leq e^{-\alpha s}\|u\|_k,$$ where $\|e^{(A-\alpha)s}u\|_k^2= \int_{\mathbb{R}}(1+|\xi|^2)^k|e^{(i|\xi|-\alpha)s}\hat{u}(\xi)|^2d\xi$. Thus, It suffices to take the limit in the right side of the inequality when $s$ tends to $\infty$. I want to be sure about this calculus. Thank you in advanced