This is an exercise of the book Analysis III of Amann and Escher:

For $s\ge 0$ define $H^s:=\{u\in L_2:\Lambda^s\hat u\in L_2\}$ and $(u|v)_{H^s}:=(\Lambda^s\hat u|\hat v)_{L_2}$ where $(\cdot|\cdot)_{L_2}$ is the inner product in $L_2$. Show that $(H^s,(\cdot|\cdot)_{H^s})$ is a Hilbert space.

*Note: here $L_2$ stay for $L_2(\Bbb R^n,\Bbb C)$, $\hat u$ is the Fourier transform of $u$ and $\Lambda^s(x):=(1+|x|^2)^{s/2}$. Also $\mathcal S$ stay for the Schwartz space. Can you check if the proof below is correct?

(1) Showing that $(\cdot|\cdot)_{H^s}$ is indeed an inner product in $H^s$:

We have that $$ (u|v)_{H^s}=\int\Lambda^s(x)\hat u(x)\bar{\hat v}(x)\, dx=(\hat u|\overline{\Lambda ^s}\hat v) \\=\overline{(\overline{\Lambda ^s}\hat v|\hat u)}=(\Lambda^s\bar{\hat v}|\bar{\hat u}) =(\bar v|\bar u)_{H^s}=\overline{(v|u)_{H^s}}\tag1 $$ because $$ (\bar v|\bar u)_{H^s}=\int\Lambda^s(x)\hat{\bar v}(x)\overline{\hat{\bar u}(x)}dx\\ \hat{\bar v}(x)=\int\bar v(t)e^{-i\langle x,t\rangle}dt=\overline{\int v(t)e^{i\langle x,t\rangle}dt} =\overline{\hat v(-x)}\\ \overline{\hat{\bar u}(x)}=\hat u(-x)\\ \therefore\quad (\bar v|\bar u)_{H^s}=\int\Lambda^s(x)\hat u(-x)\overline{\hat v(-x)}dx =\int\Lambda^s(x)\hat u(x)\bar{\hat v}(x)\, dx=(u|v)_{H^s}\tag2 $$ and $\Lambda^s(x)=\Lambda^s(-x)$. And also is clear that $(u|u)_{H^s}=0$ if and only if $u=0$, so $(\cdot|\cdot)_{H^s}$ is a sesquilinear map. By last from the integral definition its easy to see that $$ |(u|v)_{H^s}|\le\|\Lambda^s\hat u\|_2\|\hat v\|_2=\|\Lambda^s\hat u\|_2\|v\|_2<\infty,\quad u,v\in H^s\tag3 $$ so $(\cdot|\cdot)_{H^s}$ is a well-defined inner product on $H^s$.

(2) Showing that $H^s$ is complete:

Now we want to show that if $\{u_j\}$ is a Cauchy sequence in $H^s$ then there is some $u\in H^s$ such that $u_j\to u$ in $H^s$.

Because $\{u_j\}$ is a convergent sequence in $L_2$ also, then there is some $u\in L_2$ such that $u_j\to u$ in $L_2$. Hence $\{u_j\}$ must converge to $u$ in $H^s$, so we want to show that $$ \|u_j-u\|_2\to 0\implies\|u_j-u\|_{H^s}\to 0\tag4 $$ But from the Cauchy-Schwarz inequality we knows that $$ \|u_j-u\|_{H^s}^2\le\|\Lambda^s\widehat{(u_j-u)}\|_2\|u_j-u\|_2\tag5 $$ so it will be enough to show that $\|\Lambda^s\hat u_j-\Lambda^s\hat u\|_2$ is a bounded sequence.

We knows that the map $T_s h:=\Lambda^s \hat h$ is linear and continuous in $\mathcal S$, and because $\mathcal S$ is dense in $L_2$ then we knows that there is a unique continuous extension of $T_s$ in $L_2$. Then $T_s u_j$ converges to $T_s u\in L_2$, so we only need to show that $T_s u=\Lambda^s\hat u$.

We knows that if a sequence $\{u_j\}$ converges to $u$ in $L_2$ then there is a subsequence of representatives $\{u^*_{j_k}\}$ that converges point-wise (almost everywhere) to any representative $u^*$ of $u$, thus $T_s u^*_{j_k}=\Lambda^s \hat u^*_{j_k}$ converges point-wise almost everywhere to $\Lambda^s \hat u^*$, from where we can conclude that $T_s u=\Lambda^s\hat u$, so we find that $\lim_j \Lambda^s u_j=\Lambda^s u$, finishing the proof for the completeness of $H^s$.

EDIT: the above proof is not right because I didn't shown that if $\{u_j\}$ is Cauchy in $H^s$ then it is also Cauchy in $L_2$, I just assumed that this would be true without any proof. I will try to fix this soon.

THE FIX: note that $$ \begin{align} \|u_j-u_k\|^2_{H^s}&=|(\Lambda^s(\widehat{ u_j-u_k})| \widehat{ u_j-u_k})_{L_2}| \\&=|(\Lambda^{s/2}(\widehat{ u_j-u_k})|\Lambda^{s/2}(\widehat{ u_j-u_k}))_{L_2}|\\ &=\|\Lambda^{s/2}(\widehat{ u_j-u_k})\|_2^2 \end{align} $$ Also note that $\Lambda^{s}(x)=(1+|x|^2)^{s/2}\ge 1$ for all $s\ge 0$ and all $x\in\Bbb R^n$, so its easy to see that $\|u_j-u_k\|_{H^s}\ge\|\widehat{u_j-u_k}\|_2=\|u_j-u_k\|_2$, hence if $\{u_j\}$ converges in $H^s$ then $\{u_j\}$ converges in $L_2$.

  • 1
    $\begingroup$ $H^s$ is by definition isomorphic to $L^2$ so it is an Hilbert space, the isomorphism is to send $v \in L^2$ to $u=\mathcal{F}^{-1}[(1+|x|^2)^{-s/2} v] \in H^s$, this as well as the backward isomorphism is well-defined because $(1+|x|^2)^{-s/2} v \in L^2$ thus the Fourier inversion theorem applies and $u \in L^2$ . Also it is $(u|v)_{H^s}:=(\Lambda^s\hat u|\Lambda^s \hat v)_{L_2}=(\Lambda^{2s}\hat u| \hat v)_{L_2}$ where $\Lambda^s v = (1+|x|^2)^{s/2} v$ $\endgroup$ – reuns Jun 11 at 21:43
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    $\begingroup$ For $s < 0$ it is still by definition isomorphic to $L^2$ just that you need the Fourier transform of distributions. $\endgroup$ – reuns Jun 11 at 21:53
  • $\begingroup$ Could you explain better what you did to get (5), please? $\endgroup$ – Danilo Gregorin Aug 8 at 16:48
  • $\begingroup$ @Danilo note that $$\|u-v\|_{H^s}^2=(u-v|u-v)_{H^s}=(\Lambda^s(\hat u-\hat v)|\hat u-\hat v)_{L_2}$$ Now apply the linearity of the Fourier transform together with it isometry properties in $L_2$ and finally the Cauchy-Schwarz inequality. $\endgroup$ – Masacroso Aug 8 at 19:38

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