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Let $\Omega$ be an open subset of $\mathbb{R}^d$. Under regularity conditions, we know that the $s$-th order Sobolev space $H^s(\Omega)$ with $s \geq d/2$ is a reproducing kernel Hilbert space. In contrast, $H^s(\Omega)$ with $s < d/2$ is only a Hilbert space without the reproducing property.

My question is about the construction of orthonormal basis of $H^s(\Omega)$.

For $s \geq d/2$, the eigenfunction of the reproducing kernel gives us an orthonormal basis, which up to a rescaling of magnitudes is also an orthonormal basis of $L^2(\Omega)$ (which can be equivalently written as $H^0(\Omega)$).

For $s < d/2$, how do we construct an orthonormal basis in a similar fashion, given now there is no reproducing kernel? Moreover, for $0 \leq s_1, s_2 < d/2$, is it possible to align the orthonormal bases of $H^{s_1}(\Omega)$ and $H^{s_2}(\Omega)$ so that they only differ up to a rescaling of magnitudes?

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  • $\begingroup$ I believe that this is not a meaningful question. You see, even though $H^s(\Omega)$ is called a Hilbert space it is technically not, in the sense that the actual Hilbert norm on it is not fixed. It is a sort of infinite dimensional analog of a linear space of finite dimension: you are free to chose any of the (equivalent) scalar products. Naturally, an orthonormal basis would very much depend on the choice. $\endgroup$
    – Alex Gavrilov
    Commented Dec 23, 2018 at 9:32
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    $\begingroup$ @AlexGavrilov I totally agree with you on that the norm and inner product are not unique. We are interested in the "natural" choice of Sobolev norm and inner product $<u, v> = \sum_{i=1}^k \langle D^i u, D^i v \rangle$, especially how to construct the corresponding basis, for example, the eigenvector of the reproducing kernel. Also, for $0 \leq s_1, s_2 < d/2$, is it possible to align the orthonormal bases of $H^{s_1}(\Omega)$ and $H^{s_2}(\Omega)$ so that they only differ up to a rescaling of magnitudes? $\endgroup$
    – Minkov
    Commented Dec 23, 2018 at 9:43
  • $\begingroup$ Even this is not so ``natural'' a choice. What is the big reason to take all of this terms with weight one? (It is not invariant with respect to scaling, for example.) I admit that in this case the question actually has an answer, but I also admit that, personally, I totally would not care about it. $\endgroup$
    – Alex Gavrilov
    Commented Dec 23, 2018 at 12:08

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