# Show that Fourier transform is isomorphism between $H^r_m (\mathbb R)$ and $H^m_r (\mathbb R)$.

fractional-sobolev-spacesfunctional-analysispartial differential equationsreal-analysissobolev-spaces

Let $$H^m_r (\mathbb R) = \{u \in L^2(\mathbb R): \| \rho^r u \|_{H^m} < \infty \}$$ and $$H^r_m (\mathbb R) = \{u \in L^2(\mathbb R): \| \rho^m u \|_{H^r} < \infty \}$$ where $$\rho(x) = (1+|x|^2)^{\frac 1 2}$$. It has been about a month that I have been showing that Fourier transform is an isomorphism between these two spaces, in every calculation I face the $$\mathcal{F}(\rho^m u)$$ which is complicated. I am assuming there is another way to show it other than normal calculations!Any comment or reference mentioning appreciated.

For $$m,r \in \Bbb{Z}_{\ge 0}$$,

let $$\mathbb{H}^r_m$$ be the Hilbert space with norm $$\|f\|_{\mathbb{H}^m_r}^2 =\sum_{a\le m,b\le r}\|(2i\pi x)^a f^{(b)}\|_{L^2(\Bbb{R})}^2+\|((2i\pi x)^a f)^{(b)}\|_{L^2(\Bbb{R})}^2$$

since $$\mathcal{F}( (2i\pi x)^m f^{(r)}) = ((-2i\pi y)^r\hat{f})^{(m)},\qquad\mathcal{F}( ((2i\pi x)^m f)^{(r)})= ((-2i\pi y)^r \hat{f}^{(m)}$$

we get that $$\mathcal{F}$$ is an isometry from $$\mathbb{H}^m_r$$ to $$\mathbb{H}^r_m$$.

It remains to check that your space $$H^m_r$$ is the same as $$\mathbb{H}^m_r$$.

When $$m,r\in \Bbb{R}_{>0}$$ it is the same idea except that the $$r$$-th derivative is defined as $$f^{(r)}=\mathcal{F}^{-1}((-2i\pi y)^r \hat{f})$$.