# Orthonormal Basis of Multi-Dimensional Sobolev Space of Different Orders without Reproducing Kernel

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?

• 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. – Alex Gavrilov Dec 23 '18 at 9:32
• @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? – Minkov Dec 23 '18 at 9:43
• 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. – Alex Gavrilov Dec 23 '18 at 12:08