I have to evaluate the $H^s$ norm of the Fourier transform of a function, $\hat f(\lambda,t)=\mathscr{F}(f(\xi,t))(\lambda,t)$.
According to the definition (that I know) of the norm of a Sobolev space of fractional order, \begin{equation*} ||\hat f(\lambda,t)||_{H^s }= \int_{\mathbb{R}} \left(1+|\xi|^2\right)^s \big|\widehat{ \hat f(\lambda,t) }\big|^2 d\xi. \end{equation*}
Giving that to me $\hat f(\lambda,t)=a(\lambda,t) \hat f(\lambda,0) $, I was wondering if there is another way to calculate the $H^s$ norm, with $s$ fractional, that does not use $\widehat{ \hat f(\lambda,t) }$.
Getting an exact value for the $H^s$ norm of $\hat{f}$ is probably not possible (unless $a$ has a very nice form). However, we can re-write the norm in such a way that we can get upper bounds using Young's inequality.
First, notice that the Fourier transform and inverse Fourier transform are the same 'up to a sign;' that is, $$\hat{g}(\xi) = \check{g}(-\xi)$$ In particular, $$\widehat{\hat{f}(\lambda,t)}(\xi) = f(-\xi,t).$$ Since the weight $\langle \xi \rangle^{s} = (1 + |\xi|^2)^s$ is invariant under the change of variables $\xi \mapsto -\xi$, we can forget about the minus sign. (Here, $\langle \xi \rangle := \sqrt{1 + |\xi|^2}$ is the so-called 'Japanese bracket.')
Thus, $$\lVert \hat{f}(\lambda,t) \rVert_{H^s(d\lambda)} = \lVert \langle \xi \rangle^s f(\xi, t) \lVert_{L^2(d\xi)} = \lVert f(\xi,t) \rVert_{L^2(\langle \xi \rangle^2s d\xi)}$$ Since $\hat{f}(\lambda, t) = a(\lambda, t) \hat{f}_0(\lambda)$, we have that $$f(\xi, t) = (2\pi)^{d/2}\left(\check{a}(\cdot, t) * f_0\right)(\xi)$$ Thus, by Young's inequality (applied to the weighted $L^2(\langle\xi\rangle^{2s}d\;\xi)$ norm) gives $$\lVert \hat{f}(\lambda,t) \rVert_{H^s(d\lambda)} \le (2\pi)^{d/2} \lVert \check a(\xi, t) \rVert_{L^1(\langle \xi \rangle^{-2s}\;d\xi)} \lVert f_0(\xi) \rVert_{L^2(\langle \xi \rangle^{2s}d\;\xi)}$$ By recognizing the last factor as just $\lVert \hat f_0(\lambda) \rVert_{H^s(d\lambda)}$, we see that the growth/decay of the $H^s$ norm of $f$ is controlled by the growth/decay of $\check{a}$ in a weighted Lebesgue space.