29
$\begingroup$

Question

Let $M$ be a compact Riemannian manifold with piecewise smooth boundary. Why are smooth functions with vanishing normal derivative dense in $C^\infty(M)$ in the $H^1$ norm?

Here I define $C^\infty(M)$ to be those functions which have all orders of derivative continuous on $M$ and smooth in its interior. For example, $(x\mapsto \sin(\pi x))\in C^\infty([0,1])$ but $(x\mapsto \sqrt{x})\notin C^\infty([0,1])$.

I have crossposted this to MathOverflow.


Background

This is inspired by my belief that the form domain of the Friedrichs extension of the Neumann Laplacian on $M$ is equal to $H^1(M)$. If my belief is wrong, I would certainly accept as answer a counterexample, preferably with some discussion/references.

Here are the approaches I'm exploring.

  • Given $u\in C^\infty(M)$ construct a function that agrees away from an $\epsilon$ neighborhood of $\partial M$, but has been modified to have zero normal derivative. This is described below, but runs into some trouble at corners and with smoothing.
  • Given $u\in C^\infty(M)$, find some $\eta$ supported on an $\epsilon$ neighborhood of $\partial M$ such that $\nabla\eta|_{\partial M}$ is equal to the projection of $\nabla u|_{\partial M}$ onto the normal direction and $\eta$ and $|\nabla\eta|$ uniformly bounded in $\epsilon$. Then $u - \eta$ will be the desired approximation and uniform boundedness will imply $\|u - (u-\eta)\|_{H^1}\to 0$. I'm not sure if this is a search for an integrable harmonic vector field or if it's a constrained optimization problem.
  • Simply show that any $u\in C^\infty(M)$ that is perpendicular to all smooth functions with vanishing normal derivative must be zero. In order to do this, I think it still runs into the same fundamental difficulty as the others, which is constructing functions with vanishing normal derivative supported on an $\epsilon$-neighborhood of the boundary.

(I'm also happy to simply have a reference to follow up. A reference for domains in $\mathbb{R}^n$ should be fine, too, and just a partition of unity away from a Riemannian manifold.)


Current Approach

The approach I'm considering right now is an elaboration on possibility (2) from my list. The sketch is: set the boundary condition that $\nabla\eta$ on the boundary be equal to the normal component of $\nabla u$ and find an integrable harmonic vector field $E$ satisfying that boundary condition. Then cut off $E$ so it is only supported in a neighborhood of $\partial M$, and let $\eta$ be so that $E = \nabla\eta$.

Intuition is that the maximum principle will control $|E|$ and $|\eta|$, so that $\eta$ will be bounded above in the $H^1$ norm by a constant times the volume of the $\epsilon$-neighborhood of $\partial M$.

Another approach is inspired by zhw's comment below: approximate $\nabla u$ among $L^2$ vector fields, multiply it by a cutoff function so it is supported in the interior of $M$, and then integrate it to approximate $u$. This should work with some tweaking in $[0,1]$ but I'm not sure how well it will work in general.


Older work

My approach has been as follows. The intuition is to take an arbitrary smooth function, restrict it to the complement of a collar neighborhood, then extend the restriction to the collar neighborhood so that the value is constant on inward-normal geodesics. However I'm running into issues at corners.

Let $\epsilon > 0$ be such that $\{p\in M\ |\ d(p,\partial M) < \epsilon\}$ is a collar neighborhood of $\partial M$. Let $e_\epsilon(p)$ for $p\in\partial M$ be the smaller of $\epsilon$ or the greatest time parameter such that the inward normal geodesic collides with no other inward normal geodesic. Let $N$ be the set of inward normal vectors on $\partial M$ whose lengths are no greater than $e_\epsilon$. The interior of the set $\operatorname{exp}(N)$ is foliated by geodesics. Edit- Note this is not necessarily true if the manifold has inward corners.

Given a smooth, continuous function $u$ on $M$, define $\bar{u}$ to be the restriction of $u$ in the complement of $\operatorname{exp}(N)$ and on each geodesic leaf of the interior of $\operatorname{exp}(N)$ define $\bar{u}$ to take the value that $u$ takes on the inward limit of that leaf.

Here's trouble. As $\bar{u}$ need not be smooth, I'd like to mollify it. Edit- I had a detail incorrect. A standard mollifier produces a function defined on compact subsets of the interior of $M$. So the mollifier has to be modified. One idea I'm following up on is varying the support of the mollifying function based on distance to $\partial M$. I'm skeptical, as varying the mollifying function will add another component to the gradient, but if it works I'll post as an answer.

$\endgroup$
  • 1
    $\begingroup$ I think the word "vanishing" has vanished in the first paragraph (currently: "with normal derivative"). Also "smooth, continuous" is redundant right? $\endgroup$ – Dap Oct 8 '17 at 23:58
  • $\begingroup$ @Dap Good catch, thanks. I mean to indicate those functions which are continuous on all of $M$ and smooth on the interior, as opposed to functions which are smooth on the interior but don't necessarily extend to continuous functions on $M$. $\endgroup$ – Neal Oct 9 '17 at 1:16
  • $\begingroup$ I think this would be appropriate for mathoverflow as well. $\endgroup$ – Dap Oct 10 '17 at 5:29
  • 2
    $\begingroup$ Why are you sure this result holds? Especially with inward corners $\endgroup$ – Bananach Oct 10 '17 at 18:13
  • 1
    $\begingroup$ @MoisheCohen No, it works on $[0,1]. $ Just approximate $f'$ in $L^2$ by a smooth $g$ with support in $(0,1).$ Then consider $G(x)=f(0)+ \int_0^x g.$ $\endgroup$ – zhw. Oct 14 '17 at 18:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.