Denote $$E^{k,m}=\left\{f\in C^m(\mathbb{R}^3)\mid\sup_{x\in\mathbb{R}^3}(1+|x|)^{k+|\alpha|}|\partial^\alpha_xf(x)|<\infty\right\}.$$ Now given $f\in E^{2,m}$ for any fixed $m\in \mathbb{Z}^+$, and $\hat\theta_s(\xi)=|\xi|e^{(s-1)|\xi|^2}$, define $$\hat{W}(\xi)=\int_{\frac 12}^1\hat{\theta}_s(\xi)\sqrt s\hat{f}\left(\sqrt s\xi\right)ds,$$ prove that $W\in E^{2,m}$.
This is some thing I read in a paper where the authors claim one could "easily see" this without giving a proof. Well, I'm not very familiar with singular operators in Harmonic analysis and this seems not so obvious to me.
I'll sketch how far I got and if anyone could tell me what I'm missing it would be very appreciated.
Apply inverse Fourier transform one obtains, $$W(x)=\int_{\frac 12}^1\theta_s\ast\left(\frac 1 sf\left(\frac \cdot{\sqrt s}\right)\right)(x)ds.$$ Direct computation implies $$(1+|x|)^{2+|\alpha|}\partial_x^\alpha W(x)\lesssim (1+|x|)^{2+|\alpha|}\int_{\frac 12}^1\int_{\mathbb{R}^3_y}\frac{\theta_s(x-y)}{(1+|y|)^{2+|\alpha|}}dyds\|f\|_{E^{2,m}}.$$ Here is the problem, first the space integral should be finite and this will depend on $s$,then it should be integrable on $(\frac 12,1)$, finally then take supremum over $x$.
When $s=1$, however, $\hat\theta_s$ loses decay property and $|\xi|$ plays dominant role, then it will not be integrable.
Just take $x=0$ as example, then $$W(0)\lesssim\int_{\frac 12}^1\int_{\mathbb{R}^3}\frac{\theta_s(-y)}{(1+|y|)^{2+|\alpha|}}dads.$$
In the 3-d case if this is integrable, then $\theta_s(-y)$ should have at least $|y|^{-1-\epsilon}$ decay at infinity, for any $\epsilon>0$. But I fail to show this when $s\in (\frac 12,1)$.