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?

  • $\begingroup$ One can take $s=1$ to demonstrate. That's enough. Thanks. $\endgroup$
    – DreamAR
    Dec 8, 2021 at 4:14
  • $\begingroup$ Don't confuse the $L^2$ inner product with the $H^s$ inner product. $\endgroup$
    – reuns
    Dec 8, 2021 at 5:00
  • 1
    $\begingroup$ 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. $\endgroup$
    – reuns
    Dec 8, 2021 at 5:34
  • 1
    $\begingroup$ 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). $\endgroup$
    – reuns
    Dec 8, 2021 at 5:38
  • 1
    $\begingroup$ See these questions: math.stackexchange.com/questions/740355/… or math.stackexchange.com/questions/896249/… $\endgroup$
    – daw
    Dec 8, 2021 at 10:40


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy