Computing the $H^s$ norm of a time-dependent integral operator

Question

Let $T>0$ arbitrary but fixed. Consider $f:\mathbb{R}\to\mathbb{R}$ any function in the Schwartz class. Now define the quantity:$$ J(t):=\int_{\vert \tau T\vert>1}(i\tau)^{-1}e^{it\tau}\widehat{f}(\tau)d\tau. $$ I am reading a book on which the author says that, for any $b,b'\in\mathbb{R}$ such that $b-b'<1$, the latter integral can be "trivially" bound in $H^b(\mathbb{R})$ by: $$ \Vert J\Vert_{H^b(\mathbb{R})}\leq \Vert f\Vert_{H^{b'}}\sup_{\vert \tau T\vert>1}\tau^{-1}(1+\vert \tau\vert)^{b-b'}. $$ I am trying to recover this bound but I haven't succeed, so I am wondering if anyone has any hint. So far I have just tried the obvious, I tried to write the $H^b$ norm in terms of the Fourier Transform of $J$ (by using Plancharel's Theorem) as $$ \Vert J\Vert_{H^b(\mathbb{R})}=\Vert (1+s^2)^{b/2}\widehat{J}(s)\Vert_{L^2(\mathbb{R})}. $$ However, I am now sure how to compute its Fourier Transform since it is a very complicated function. I think that once you write down the Fourier transform of $J$ the idea would be just to somehow multiply and divide by $(1+\tau^2)^{b'/2}$ in the inner integral and then to use Cauchy-Schwarz in order to pull out the $H^{b'}$ norm of $f$.

Answer 1

Again I think I have solved my own question. Please, I would really appreciate any comment if you think that this answer is not sufficiently rigorous.

Answer: First, let us recall that $\mathcal{F}(e^{iat})(s)=\sqrt{2\pi}\delta(s-a)$. From now on I will omit any $2\pi$ factor. Thus, we have $$ \qquad \mathcal{F}_t\big(J(t)\big)(s)=\int_{\vert\tau T\vert>1}(i\tau)^{-1}\widehat{f}(\tau)\int_{\mathbb{R}}e^{i(\tau-s)t}dsd\tau=\int_{\vert\tau T\vert>1}(i\tau)^{-1}\widehat{f}(\tau)\delta(s-\tau)d\tau \qquad (*) $$ Now, define $A:=(-\infty,-1/T)\cup(1/T,+\infty)$ and write the right-hand side of the latter identity as: $$ (*)=\int_{\mathbb{R}}\mathsf{1}_A(\tau)(i\tau)^{-1}\widehat{f}(\tau)\delta(s-\tau)d\tau=(is)^{-1}\mathsf{1}_A(s)\widehat{f}(s), $$ where $\mathsf{1}_A$ denotes the characteristic function of the set $A$. Hence, replacing this identity into the $H^b$ norm of $J(t)$ we obtain $$ \qquad \Vert J\Vert_{H^b}^2=\int_{\mathbb{R}}(1+s^2)^{b}s^{-2}\mathsf{1}_A(s)\widehat{f}(s)^2ds\leq C\Vert(1+s^2)^{b-b'}s^{-2}\mathsf{1}_A(s)\Vert_{L^\infty}\Vert(1+s^2)^{b'/2}\widehat{f}\Vert_{L^2}^2 \qquad (**) $$ Then, by enlarging the constant if necessary, the right-hand side of the latter display can be bound by $$ \Vert f\Vert_{H^{b'}}^2 \sup_{\vert sT\vert>1}s^{-2}(1+\vert s\vert)^{2b-2b'} $$ Finally, by plugging the latter bound into $(**)$ and taking square root in both sides of the inequality we obtain the desired result.

Please let me know if you think the proof is wrong.