# Show that $H^s$ is a Hilbert space

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 $$L_2$$.

• $H^s$ 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 $\Lambda^s v = (1+|x|^2)^{s/2} v$ – reuns Jun 11 at 21:43
• For $s < 0$ it is still by definition isomorphic to $L^2$ just that you need the Fourier transform of distributions. – reuns Jun 11 at 21:53
• Could you explain better what you did to get (5), please? – Danilo Gregorin Aug 8 at 16:48
• @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. – Masacroso Aug 8 at 19:38