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.

# 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

#### Related Solutions

Here is a (modified) argument which I used repeatedly in a paper of mine. Possibly, this is some form of type and cotype argument (I am not too familiar with these notions), but maybe it is elementary enough for you:

I will make extensive use of the Khintchine inequality, which states that if $\varepsilon_1, \dots, \varepsilon_N$ are independent and uniformly distributed in $\{1, -1\}$, then we have
$$
\mathbb{E} \left|\sum_{n=1}^N \varepsilon_n x_n\right|^p
\asymp \left(\sum_{n=1}^N |x_n|^2\right)^p ,
$$
for arbitrary $x_1,\dots, x_N \in \Bbb{C}$. Here, $\Bbb{E}$ denotes the expected value with respect to $\varepsilon = (\varepsilon_1,\dots, \varepsilon_N)$. It is crucial that the implied constants above are **independent** of $x_1,\dots, x_N$ and of $N$; they only depend on the choice of $p \in (0,\infty)$.

Now, assume that $\mathcal{F} : L^p \to L^q$ is an isomorphism. Let us first rule out the case $q = \infty$: If we had $q=\infty$, then there would be some $f \in L^p$ with $\widehat{f} \equiv 1$, where I write $\widehat{f} = \mathcal{F}f$. But we also have $\widehat{\delta_0} \equiv 1$, where we are using the formalism of tempered distributions, and where $\delta_0$ is the Dirac measure at the origin. But this implies $f = \delta_0$ as tempered distributions, which is absurd.

Hence $q \in (0,\infty)$. Now, let $\varepsilon_1,\dots, \varepsilon_N$ as above. For $n \in \Bbb{N}$, let $y_n := (n,0,\dots,0) \in \Bbb{R}^d$, and set $f_n := T_{y_n} 1_{(0,1)^d}$, where $(T_y f)(x) = f(x-y)$. Note that $\widehat{f_n} = M_{-y_n} \widehat{f}$ for $f := 1_{(0,1)}$, where $M_\xi f(x) = e^{2\pi i \langle x, \xi \rangle} f(x)$ denotes the modulation of $f$. Finally, observe that the supports of $f_n, f_m$ are disjoint for $n \neq m$. All in all, these properties imply \begin{align*} N^{q/p} = \Bbb{E} \, \left\| \sum_{n=1}^N \varepsilon_n \cdot 1_{(n,n+1)} \right\|_{L^p}^q & \asymp \Bbb{E} \, \left\| \sum_{n=1}^N \varepsilon_n \cdot \mathcal{F}[ 1_{(n,n+1)} ] \right\|_{L^q}^q \\ & = \Bbb{E} \int \left| \sum_{n=1}^N \varepsilon_n e^{2\pi i \langle y_n, \xi\rangle} \right|^q \cdot |\widehat{f}(\xi)|^q \, d\xi \\ (\text{Fubini's theorem}) & = \int \Bbb{E} \left| \sum_{n=1}^N \varepsilon_n e^{2\pi i \langle y_n, \xi\rangle} \right|^q \cdot |\widehat{f}(\xi)|^q \, d\xi \\ (\text{Khintchine inequality}) & \asymp N^{q/2} \cdot \| \widehat{f} \|_{L^q}^q \\ & \asymp N^{q/2} \cdot \|f\|_{L^p}^q = N^{q/2}, \end{align*} where the implied constants are independent of $N \in \Bbb{N}$. This can only hold for $p = 2$.

Now, let us show that also $q = 2$. To see this, note for arbitrary $f \in L^p = L^2$ by Plancherel's theorem that $$ \|\widehat{f} \|_{L^2} = \| f \|_{L^2} = \| f \|_{L^p} \asymp \|\widehat{f}\|_{L^q} . $$ Thus, we get for arbitrary $L^2$ functions $g \in L^2$ that $\|g\|_{L^2} \asymp \| g \|_{L^q}$, which easily implies $q = 2$, by considering e.g. $g = \sum_{n=1}^N f_n$ with $f_n$ as above.

## Best Answer

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 $f^{(r)}=\mathcal{F}^{-1}((-2i\pi y)^r \hat{f})$.