# Is $H^1(M) \subset L^2(M) \subset H^{-1}(M)$ a Hilbert triple for $M$ a manifold with boundary?

Is $H^1(M) \subset L^2(M) \subset H^{-1}(M)$ a Hilbert triple for $M$ a manifold with boundary? What smoothness is required of the boundary? I would be grateful for some references to this.

I think a sufficient condition is that (it is highly probable this is not necessary): Whenever the smoothness of $\partial M$ supports a Poincaré type inequality: $$\|u\|_{L^2(M)}\leq c\|\nabla u\|_{L^2(M)},$$ then the Hilbert triple $$H^1(M)/\mathbb{R}\subset L^2(M) \subset H^{-1}(M),\quad \text{ or }\quad H^1_0(M)\subset L^2_0(M) \subset H^{-1}(M)$$ holds.

The first inclusion is dense due to the fact that $C^{\infty}$ is dense in $L^2$.

Second inclusion holds if Poincaré inequality on this manifold holds for $H^1(M)$: for $\phi\in L^2(M)$ $$L_{\phi}(u) := \int_M \phi u \leq \|\phi\|_{L^2(M)}\|u\|_{L^2(M)} \leq c\|\phi\|_{L^2(M)}\|\nabla u\|_{L^2(M)},$$ thus $\phi\in H^{-1}(M)$.

I checked Evans's PDE book, the proof of Poincaré inequality relies on Rellich–Kondrachov compact embedding theorem, which normally assumes the boundary is Lipschitz. In Sobolev met Poincaré, the assumption is weakened to twisted cone condition for a domain (see 9.2), I am guessing this condition should be able to be extended to a compact manifold.

• Thanks for the answer. Is the first inclusion really trivial? Is it obvious that the inclusion is dense? – maximumtag May 30 '13 at 11:33
• @maximumtag $C^{\infty}$ is dense in $L^p$. :) – Shuhao Cao May 30 '13 at 19:29
• I know that's true for the standard open bounded set case but was unsure in the hypersurface case. I guess I should not worry too much :P – maximumtag May 30 '13 at 20:33
• @maximumtag The density result definitely holds for smooth and compact manifold with a boundary, for it looks like $\mathbb{R}^n$ locally. But it is not trivial though, roughly the proof should be: $u = \sum_{i} \phi_i u$, where $\{\phi_i\}$ are a partition of unity, also $\phi_i$ has support within an open neighborhood so that we can fix a chart on this, then we can initiate the standard density in $\mathbb{R}^n$. The existence of this partition of unity relies on compactness I believe. – Shuhao Cao May 30 '13 at 20:47
• Thank you. Presumably smoothness of the manifold can be weakened. Any idea where to find a proof of this result? I guess one is expected to just patch it together like you said. – maximumtag May 30 '13 at 21:58