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 ?

  • 1
    $\begingroup$ Does mathoverflow.net/questions/126419/… answer your question? $\endgroup$ Jul 14, 2016 at 20:27
  • $\begingroup$ Not completely. I believe that, in the special case $k=1$, the answer should be yes. $\endgroup$
    – Raziel
    Jul 14, 2016 at 20:41
  • $\begingroup$ 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? $\endgroup$ Jul 14, 2016 at 20:44
  • $\begingroup$ Yes, precisely. $\endgroup$
    – Raziel
    Jul 14, 2016 at 21:38
  • $\begingroup$ 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). $\endgroup$
    – Raziel
    Jul 15, 2016 at 10:48

2 Answers 2


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}

This is a special case of Satz 2.3 of the following work (in general it holds for differential forms):

Eichhorn, Jürgen: Sobolev-Räume, Einbettungssätze und Normungleichungen auf offenen Mannigfaltigkeiten [In English: Sobolev spaces, embedding theorems and norm inequalities on open manifolds]. Math. Nachr. 138 (1988), 157–168.


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 $\leq k$.

For a reference for this fact you might have a look at "The Yamabe problem" by Lee and Parker, section 2.

  • $\begingroup$ 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. $\endgroup$ Aug 23, 2019 at 22:52
  • $\begingroup$ Good observation: actually, in the aforementioned reference the result is only stated and not proved. But your remark is perfectly legitimate. $\endgroup$
    – user590533
    Aug 26, 2019 at 6:58

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