1
$\begingroup$

Wikipedia defines the Sobolev Space: $H^{s,p}(\mathbb{R}^n)= \left\{f \in L^p(\mathbb{R}^n): \mathcal{F}^{-1}[(1+|k|^2)^{\frac{s}{2}} \mathcal{F}f] \in L^p(\mathbb{R}^n) \right\}$

Where $s \in \mathbb{R}$, $\mathcal{F}$ denotes the Fourier Transform. The article goes on to say that in the case $p=2$, $H^{s,2}=H^s$ is a Hilbert Space. We have the norm:

$||f||_{H^s}= ||\mathcal{F}^{-1}[(1+|k|^2)^{\frac{s}{2}} \mathcal{F}f]||_{L^2}$

My question is this: If $H^s$ is a Hilbert Space, it has an inner product. Typically (at least in my experience), the inner product is introduced first and it subsequently is found to induce a norm, but here we are given the norm and no information about the inner product. Does anyone know what the inner product is and if it induces this particular norm (or perhaps another norm which is equivalent in the sense of continuity and convergence)? Here $H^s$ is of infinite dimension, so not all norms are equivalent.

$\endgroup$
1
  • 2
    $\begingroup$ The norm gives you the inner product (parallelogram law). You can get it through polarization. Moreover, you can actually leave away the $\mathcal F^{-1}$ since the Fourier transform is unitary. The inner product should be given by $\langle f,g\rangle = \langle (1+|k|^2)^{s/2}\mathcal Ff,(1+|k|^2)^{s/2}\mathcal Fg\rangle_{L^2}$. $\endgroup$ Mar 15, 2016 at 23:29

1 Answer 1

3
$\begingroup$

The norm gives you the inner product (parallelogram law). You can get it through polarization. Moreover, you can actually leave away the $\mathcal F^{-1}$ since the Fourier transform is unitary. The inner product should be given by $\langle f,g\rangle = \langle (1+|k|^2)^{s/2}\mathcal Ff,(1+|k|^2)^{s/2}\mathcal Fg\rangle_{L^2}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .