# orthonormal basis or Parseval frame for Sobolev spaces

Consider the uni-variate Sobolev space of order $m$: $$\mathcal{W}_2^m=\{ f:[0,1] \rightarrow \mathbb{R}|f,f^{(1)},\dots ,f^{(m-1)}\, \text{are absolutely continuous and}\, f^{(m)}\in L^2\}.$$ It is known that $\mathcal{W}_2^m$ under some norms, (for example $\|f\|^2_{\mathcal{W}_2^m}=\sum_{q=0}^{m-1}\left( \int f^{(q)}\right)^2 + \int (f^{(m)})^2$), is a reproducing kernel Hilbert space. My questions are:

1) Is there an orthonormal basis or a Parseval frame for $\mathcal{W}_2^m$?

2) What is an extension of $\mathcal{W}_2^m$ to $\mathbb{R}^n$-valued functions? I mean $f:[0,1] \rightarrow \mathbb{R}^n$ with $f(x)=[f_1(x),f_2(x),\dots ,f_n(x)]^{\prime}$.

• Just to be sure: Do you assume $f(0)=f(1)$ or anything of the kind? If you assume periodicity, the Fourier series gives you an orthogonal basis (which you can normalize). If not, it still gives you a large orthogonal set which you might be able to extend to a basis. Oct 20 '14 at 14:46
• I want to write reproducing kernel of $\mathcal{W}^m_2$ by linear combination from orthonormal basis or Parseval frame using Papadakis's theorem. Oct 20 '14 at 15:39

Consider the operator $d^{2m}/dx^{2m}$ with natural boundary conditions $u^{(m)}=u^{(m+1)}=...=u^{(2m-1)}=0$. This operator is self-adjoint in $L^2$, and $W_2^m$ is its form domain. Hence the eigenfunctions of the operator form a basis for both $L^2$ and $W_2^m$.
I do not understand the second question. The obvious extension is to require every component of f to be in $W_2^m$. Do you want something different?