# Do we have $H^{-s}\cong H^s$?

In the context of Hilbert spaces, I keep wondering why we distinguish between $$H^{-1}$$ and $$H_0^1.$$ If $$H_0^1$$ is Hilbert space, shouldn't the dual space be $$H_0^1$$ itself? I come to believe that it's because for $$f,g\in H^{-1},$$ $$u\in H_0^1,$$ maybe $$\langle{f,u\rangle}=(v,u)=\langle{g,u\rangle}$$ for certain $$v\in H_0^1$$ by Riesz Representation Theorem, so $$f=g$$ in functional sense (acting on $$H_0^1$$), but $$f\neq g$$ in $$H^{-1}$$ space. That's why we distinguish them.

However, I start to wonder do we have $$H^{-s}\cong H^s$$ then. Recall that $$H^s$$ $$(s\in \mathbb{R})$$is a Hilbert space equipped with inner product: $$(f,g)_s=\int_{\mathbb R^n}(1+|\xi|^2)^s F[f]\overline{F[g]}d\xi.$$ What we know that is $$C_c^\infty$$ is dense in $$H^s,$$ $$H^{-s}$$ is $$H^s$$'s dual space. Then do we have $$H^{-s}\cong H^s?$$

By Riesz Representation Theorem, $$\forall f\in H^{-s},$$ $$v\in H^s,$$ $$\langle{f,v\rangle}=(u,v)=(Af,v)$$ for unique $$u\in H^s.$$ Let $$Af=u.$$ $$A:H^{-s}\rightarrow H^s$$ is a linear operator of course. It's bounded because $$\|Af\|_s^2=\|u\|_s^2=(u,u)=\langle{f,u\rangle}\le \|f\|_{-s}\|u\|_s\,\Rightarrow\, \|Af\|_s\le \|f\|_{-s}.$$ If $$Af=0,$$ then $$\langle{f,v\rangle}=0$$ for all $$v\in H^s,$$ so $$f=0$$ in $$H^{-s},$$ which means that $$A$$ is injective; As $$H^{-s}$$ is $$H^s$$'s dual space, for all $$u\in H^s,$$ $$(u,v)$$ is a linear function on $$v\in H^s,$$ so there should be a $$f\in H^{-s}$$ to have $$\langle{f,v\rangle}=(u,v),$$ which means that $$A$$ is surjective.

That means $$A:H^{-s}\cong H^s.$$ We know that $$H^{s}\subset L^2\subset H^{-s}$$ for $$s>0.$$ I understand that these two things are not in conflict. But it looks weird. Am I right about that $$A$$ is a bijection?

• One can take $s=1$ to demonstrate. That's enough. Thanks. Dec 8, 2021 at 4:14
• Don't confuse the $L^2$ inner product with the $H^s$ inner product. Dec 8, 2021 at 5:00
• Anyway on the Fourier side $H^s$ is naturally isomorphic to $L^2$ through multiplication by $1/(1+|\xi|^s)$ which makes it natural to look at the dual wrt the $L^2$ inner product and - composing with the Fourier transform - gives a natural isomorphism between $H^s$ and $H^{-s}$ (there are many un-natural ones through any orthogonal basis), that you can restate as convolution operators. The Riesz representation is easily understood from the Fourier domain as well. Dec 8, 2021 at 5:34
• The Fourier transform is unitary for the $L^2$ inner product so it is natural to look at the $L^2$ inner product dual both on the Fourier series and the time domain (so that the pairing between $H^s$ and $H^{-s}$ is given through a $L^2$ inner product). Dec 8, 2021 at 5:38
• See these questions: math.stackexchange.com/questions/740355/… or math.stackexchange.com/questions/896249/…
– daw
Dec 8, 2021 at 10:40