Let $s\geq 1$, $T>0$, $\epsilon >0$ and $f\in\mathcal{C}^1(0,T,H^{s-1}(\mathbb{T}))\cap \mathcal{C}(0,T,H^{s}(\mathbb{T}))$. Consider the propagator $\exp\left[\displaystyle-\frac{it}{\epsilon}\sqrt{1-\Delta}\right]$ and define $g_\epsilon$ by the formula $$g_\epsilon(t,x) = \int_0^t\exp\left[\displaystyle-\frac{i(t-\tau)}{\epsilon}\sqrt{1-\Delta}\right]f(\tau)d\tau .$$ I want to prove that $\lim_{\epsilon \rightarrow 0}\|g_\epsilon\|_{L^\infty(0,T,H^s)} = 0.$
I tried the following: We have $$\|g_\epsilon(t,.)\|^2_{H^s}:= \sum_{n\in \mathbb{Z}}(1+n^2)^s\left|\int_0^t\exp\left[\displaystyle\frac{i(t-\tau)\sqrt{1+n^2}}{\epsilon}\right]\hat{f}(n,\tau)d\tau\right|^2.$$
Integrating by parts, we find
$$\int_0^t\exp\left[\displaystyle-\frac{i\tau\sqrt{1+n^2}}{\epsilon}\right]\hat{f}(n,\tau)d\tau = \\ \left[\frac{i\epsilon}{\sqrt{1+n^2}}\exp\left[\displaystyle-\frac{i\tau\sqrt{1+n^2}}{\epsilon}\right]\hat{f}(n,\tau)\right]_0^t-\frac{i\epsilon}{\sqrt{1+n^2}}\int_0^t\exp\left[\displaystyle-\frac{i\tau\sqrt{1+n^2}}{\epsilon}\right]\partial_t\hat{f}(n,\tau).$$ The complement is clear for the first term in the previous resulting. For the second term, can we write $$\sum_{n\in \mathbb{Z}}(1+n^2)^s\left|\frac{i\epsilon}{\sqrt{1+n^2}}\int_0^t\exp\left[\displaystyle\frac{-i\tau\sqrt{1+n^2}}{\epsilon}\right]\hat{f}(n,\tau)d\tau\right|^2 \leq \epsilon ^2T^2\|\partial_tf\|_{L^\infty(0,T,H^{s-1})}?$$ It might be trivial, but I need to be sure about this argument.
Thank you in advance!