# Are Sobolev spaces on non-compact manifolds separable?

If $M$ is a Riemannian manifold that is not compact, is it true that the Sobolev spaces on $M$, $W^{k,p}(M)$, still be separable (for $p < \infty$)?

• Just curious: Are the Sobolev spaces separable when $M$ is compact? Note that the Holder spaces on $[0,1]$ are not separable. Jan 8 '16 at 14:55
• In case conventions differ, could you please remind us of the definition of $W^{k,p}(M)$ and the norm on it? I assume that it means something like: derivatives up to and including order $k$ belong to $L^p$, with the norm being more or less the $L^p$ norm of the $k$'th derivative Jan 8 '16 at 15:19
• Two comments: On a compact manifold you can take e.g. the eigenfunctions of the Laplacian to show $W^{k,2}(M)$ is separable. Presumably this also works for other $p$. On $\mathbb{R}^n$, the Hermite functions (en.wikipedia.org/wiki/Hermite_polynomials#Hermite_functions) are a basis for $L^2(\mathbb{R})$. I think that the span of the Hermite functions is dense in Schwartz functions, which shows that general $W^{k,p}(\mathbb{R}^n)$ is separable. Jan 8 '16 at 15:20
• @YemonChoi That's exactly it.
– Kies
Jan 8 '16 at 15:40
• @IgorBelegradek Yes, as Otis wrote, the eigenfunctions of apprioriate elliptic operator will do.
– Kies
Jan 8 '16 at 15:41

Step 1 There exists measurable sections $e_1, e_2, \dotsc, e_m$, where $m = \dim M$, of $TM$ (measurable functions mapping a point $x$ to a vector of its tangent plane $T_xM$) such that for each $x \in M$, $e_1 (x), e_2 (x), \dotsc, e_m (x)$ forms an orthonormal basis of the tangent space $T_xM$.
Step 2 For $\alpha \in \mathbb{N}^m$, define to be $\partial^\alpha u$ to be the partial derivative in the basis $e_1, e_2, \dotsc, e_m$ The map $$A: u \in W^{k, p} (M) \longmapsto A (u) = (\partial^\alpha u)_{\vert \alpha \vert \le k} \in L^p (M)^\nu,$$ (where $\nu = \# \{\alpha \in \mathbb{N}^n : \vert \alpha \vert \le k\}$) is continuous and its inverse defined on $A (W^{k, p} (M))$ is also continuous.
Step 3 The subspace $A (W^{k, p} (M))$ is separable as a subset of the separable space $L^p (M)^\nu$ (since $M$ is a separable measure space because it is a second countable topological space). Since $A (W^{k, p} (M))$ is isomorphic to the space $W^{k, p} (M)$, the space $W^{k, p} (M)$ is separable.
• I guess by $\nu$ you mean the number of derivative components of order less than or equal to $k$. Feb 24 '16 at 17:04