# Alternative definitions of Sobolev spaces on non-compact Riemannian manifolds

SHORT VERSION: Does the Meyers-Serrin theorem hold on complete, non-compact Riemannian manifolds, i.e. $W^{k,p}(M) = H^{k,p}(M)$? My guess is that this holds for the special case $k=1$ (and all $p\geq 1$), and can fail to hold in general, but I didn't find any explicit reference in the literature.

LONG VERSION: As it is well known (?), many facts about Sobolev spaces that are well known on $\mathbb{R}^n$ are no longer true when the base space is a complete, non-compact Riemannian manifold (e.g. $C^\infty_c(M)$ is not necessarily dense in $H^{k}(M)$, for $k \geq 2$).

On $\mathbb{R}^n$, Sobolev spaces are usually defined in two ways. I'm mainly interested in the case $p=2$ and $k=1$, so I will stick to this case.

The first definition is as follows:

$$W^{1,2}(\mathbb{R}^n):=\{u \in L^2(\mathbb{R}^n) \mid D_i u \in L^2(\mathbb{R}^n),\; \forall i =1,\ldots,n\},$$

where $D_i u$ is a derivative in the sense of distributions. $W^{1,2}(\mathbb{R}^n)$ is equipped with the norm

$$\|u\|_{W^{1,2}(\mathbb{R}^n)}^2 = \|u\|^2_{L^2(\mathbb{R}^n)} + \sum_{i=1}^n \| D_i u \|^2_{L^2(\mathbb{R}^n)}.$$

The second definition is as the completion

$$H^{1,2}(\mathbb{R}^n):=\overline{C^\infty(\mathbb{R}^n) \cap W^{1,2}(\mathbb{R}^n)},$$

with the norm defined above.

It is a textbook result (Meyers-Serrin theorem) that $W^{1,2}(\mathbb{R}^n) = H^{1,2}(\mathbb{R}^n)$, i.e. the two definitions are equivalent. Actually $W^{k,p}(\mathbb{R}^n) = H^{k,p}(\mathbb{R}^n)$ for all $k \in \mathbb{N},p \geq 1$.

What happens on manifolds? Let $M$ be a Riemannian manifold (not compact, in general, but always without boundary and complete). Hebey, in his book (Chapter 3), defines Sobolev space $H^{1,2}(M)$, in the non-compact case, as the closure of the space $C^\infty(M)$ of smooth functions with the norm

$$\| u\|_{W^{1,2}(M)}^2:= \|u\|^2_{L^2(M)} + \int_M g(\nabla u,\nabla u) d\mu.$$

The fact that he does not present a definition similar to $W^{1,2}(\mathbb{R}^n)$ for manifolds made me wonder. For a Riemannian manifold $M$ and an $L^2_{\mathrm{loc}}(M)$ function, the weak gradient $\nabla u$ is the following linear functional on smooth, compactly supported sections:

$$\nabla u[ X] := (u, \nabla^* X)_{L^2(M)}, \qquad \forall X \in \Gamma_c^{\infty}(TM),$$

where $\nabla^* X = - \mathrm{div}(X)$ is the formal adjoint. If there exists an $L^2$ section $V$ of $TM$ such that

$$(u,\nabla^* X) = \int_M g(V,X) d\omega,$$

then we say that $V = \nabla u \in L^2(TM)$. Then, we can define

$$W^{1,2}(M) := \{ u \in L^2(M) \mid \nabla u \in L^2(TM) \}$$,

equipped with the norm

$$\| u\|^2_{W^{1,2}(M)} : = \|u\|^2_{L^2(M)} + \int_M g(\nabla u,\nabla u)d\omega.$$

Does this definition make sense? Why did Hebey not mention it in his book (while he mentions both the possible definitions and their equivalence in the case of $M=\mathbb{R}^n$)? Are the two definitions equivalent? Is there a standard reference where the definition of $W^{1,2}(M)$ as above is presented ?

• Does mathoverflow.net/questions/126419/… answer your question? – Willie Wong Jul 14 '16 at 20:27
• Not completely. I believe that, in the special case $k=1$, the answer should be yes. – Raziel Jul 14 '16 at 20:41
• The way I read Michor's answer is that as long as the curvature tensor is bounded, then the answer is yes for $k = 1$. Do you specifically care about the case without any control whatsoever on the geometry? – Willie Wong Jul 14 '16 at 20:44
• Yes, precisely. – Raziel Jul 14 '16 at 21:38
• Let me also mention that I believe the result is true for $k=1$ because there always exist "good" cut-off functions with controlled gradient, while this does not hold for higher order derivatives (without some bounded geometry assumption). – Raziel Jul 15 '16 at 10:48

If $M$ is a complete Riemannian manifold, then \begin{equation}W^{1,p}(M) = H^p(M) := \overline{C^\infty(M) \cap W^{1,p}(M)} = \overline{C^\infty_c(M)}.\end{equation}
If $$(M,g)$$ is complete, then $$\overline{C^\infty_c(M)}^{||\cdot||_{p,k}}=L^p_k(M)$$ for any $$1\leq p< \infty$$ and $$k\in \{0,1,2,\cdots\}$$, where $$L^p_k(M)$$ is the space of functions $$u\in L^p(M)$$ such that $$Pu=f\in L^p(M)$$ (in the weak sense) whenever $$P$$ is a smooth differential operator of order $$\leq k$$.
• Is it clear that always $W^{k,p}(M) \subset L_k^p(M)$? It seems that the result you stated characterize the closure of smooth compactly supported functions with respect to the Sobolev norms. Does the condition $||\nabla u ||_p < \infty$ imply that $Pu \in L^p(M)$ for any smooth differential operator $P$ of order $1$? For example, you may choose the coefficients of a smooth differential operator $P$ to be unbounded since the manifold itself is noncompact. – Jesse Railo Aug 23 '19 at 22:52