Let $n\in \mathbb{N}, \,\, s>n/2$ and denote $H^s(\mathbb{R}^n)$ the Sobolev space. Let $u :(0, 1] \to H^s$ be bounded and denote
$$f(x) := \left({\rm e}^{x \Delta + (x^2 \Delta^2 + \Delta)^{\frac 12}}- {\rm e}^{{\rm i}\sqrt{-\Delta}}\right)u(x).$$
I wonder whether we can prove that $\lim_{x\to 0} \|f(x)\|_{H^s} = 0.$
My attempt is to use the dominated convergence theorem on
$$\|f(x)\|^2_{H^s} = \int(1+|\xi|^2)^s\left|{\rm e}^{-x |\xi|^2 + (x^2|\xi|^4 -|\xi|^2)^{\frac 12}}- {\rm e}^{{\rm i}|\xi|}\right|^2 |\widehat{u(x)}(\xi)|^2 d\xi .$$
It is clear that $\left|{\rm e}^{-x |\xi|^2 + (x^2 |\xi|^4 -|\xi|^2)^{\frac 12}}- {\rm e}^{{\rm i}|\xi|}\right| \leq2.$ Howewer, does the boundness of $u(x)$ imply the integrability of
$$\xi \mapsto (1+|\xi|^2)^s \sup_{x} |\widehat{u(x)}(\xi)|?$$
Thank you for any hint
I don't think $\lim_{x \to 0} ||f(x)||_{H^s} = 0$ is true (in general). All we know about $u$ is that $\int (1+|\xi|^2)^s|\widehat{u(x)}(\xi)|^2d\xi$ is bounded. So we can have $||u(x)||_{H^s} = 1$ for each $x$ and $\int_{x^{-100} < \xi < 2x^{-100}} (1+|\xi|^2)^s|\widehat{u(x)}(\xi)|^2d\xi = 1$ if we want (I want this). And then $$\int(1+|\xi|^2)^s\left|{\rm e}^{-x |\xi|^2 + (x^2|\xi|^4 -|\xi|^2)^{\frac 12}}- {\rm e}^{{\rm i}|\xi|}\right|^2 |\widehat{u(x)}(\xi)|^2 d\xi = \int_{x^{-100} < \xi < 2x^{-100}} (1+|\xi|^2)^s\left|{\rm e}^{-x |\xi|^2 + (x^2|\xi|^4 -|\xi|^2)^{\frac 12}}- {\rm e}^{{\rm i}|\xi|}\right|^2 |\widehat{u(x)}(\xi)|^2 d\xi.$$ The whole point is that $\left|{\rm e}^{-x |\xi|^2 + (x^2|\xi|^4 -|\xi|^2)^{\frac 12}}- {\rm e}^{{\rm i}|\xi|}\right|^2$ will basically be $\left|1-e^{i|\xi|}\right|^2$ as $x \downarrow 0$ for $\xi \in [x^{-100},2x^{-100}]$, so we end up with $$\int_{x^{-100} < \xi < 2x^{-100}} (1+|\xi|^2)^s\left|1-e^{i|\xi|}\right|^2 |\widehat{u(x)}(\xi)|^2 d\xi,$$ which has absolutely no reason to tend to $0$. If you wanted me to be more rigorous, instead of just restricting $\xi$ to $[x^{-100},2x^{-100}]$, we could restrict to very small balls around $\xi$ that satisfy $e^{i\xi} = -1$, so that $|1-e^{i|\xi|}| \ge 1.9$, say, and then we end up with $||f(x)||_{H_s}^2 \ge 1.9$ for each $x$.