MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Hi, i'm reading the proof of the fact that $C^{\infty}(M,N)$ is dense in the sobolev space $W^{1,m}(M,N)$, where $M,N$ are compact riemannian manifolds of dimension respectively $m,n$. I recall quickly the definition of $W^{1,m}(M,N)$. Embedding $N$ isometrically in a $\mathbb{R}^J$ we define $$W^{1,m}(M,N)= \left\{ f \in W^{1,m}(M,\mathbb{R}^J)\quad \mid \quad f(x)\in N\quad \textrm{for a.e.}\quad x\in M \right\} $$ The proof goes as follows: embed $M$ isometrically in $\mathbb{R}^K$, consider a tubular neighborhood $T_M$ sufficiently small to have a smooth "nearest point" projection $\pi_M:T_M\to M$. Let $f\in W^{1,m}(M,N)$ consider $F=f\circ\pi_M$ and then its mollifications $F_{\epsilon} = F \ast \phi_{\epsilon}$ . We want that values of $ F_{\epsilon}$ stay in a tubular neighborhood of $N$ sufficiently small to have a smooth "nearest point" projection $\pi_N:T_N\to N$. To do this we want to estimate $dist(F_{\epsilon}(x),N)$ and show that it goes to zero as $\epsilon$ goes to zero.

Now comes the part that i don't understand, the author says that follows from poincare inequality that $$\frac{1}{\mu(B(x,\epsilon))}\int_{B(x,\epsilon)}|F(y)-F_{\epsilon}(x)|^m dy\leq \frac{C\epsilon^m}{\mu(B(x,\epsilon))}\int_{B(x,\epsilon)}|\nabla F(y)|^m dy$$ With C a positive constant. But to have an inequality like this don't i need that the mean of $F(y)-F_{\epsilon}(x)$ is zero on $B(x,\epsilon)$?

This is the crucial estimate because then it is enough to consider $\pi_N\circ F_{\epsilon}$ and i get a sequence of smooth maps converging to $f$ in the sobolev norm. Please can anyone help? Or anyone does know another proof or reference to a proof of the density of $C^{\infty}(M,N)$ in $W^{1,m}(M,N)$?

Thank you in advance.

share|cite|improve this question
I agree that the inequality does not look right. The two sides do not scale the same, do they? Are you supposed to have the $1/\mu(B(x,\epsilon))$ factor on both sides or just on the right? – Deane Yang Jan 7 '11 at 15:40
And have you tried to prove this inequality directly yourself using the fundamental theorem of calculus and the definition of $F_\epsilon$? What goes wrong? – Deane Yang Jan 7 '11 at 15:41
The inequality is false twice. First, it is not homogeneous in $\epsilon$. The fraction is the right-hand side should be deleted. Now the second integral must be taken on a larger ball $B(x;2\epsilon)$, assuming that the support of $\phi_\epsilon$ is $B(x;\epsilon)$. – Denis Serre Jan 7 '11 at 15:43
I'm really sorry i made a copy and paste error, now i fixed it – Italo Jan 7 '11 at 15:59
On the left, is the "$F_\epsilon(x)$" correct or should it be $F_\epsilon(y)$? – Deane Yang Jan 7 '11 at 16:05
up vote 1 down vote accepted

I don't have the time to work out a real answer to this question but estimates like this can usually be proved using the following:

$ F(y) - F_{\epsilon}(x) = \int \phi_\epsilon(x,z)(F(y) - F(z)) dz $

where $\phi_\epsilon$ is the mollifier kernel function and then substituting in

$ F(y) - F(z) = \int_0^1 \gamma'(t)\cdot\nabla F(\gamma(t)) dt $

where $\gamma$ is a constant speed geodesic joining $z$ to $y$.

You just need to estimate the $L_m$ norm of the right side of the first equation with the second equation substituted in. This should lead to something close to what you want.

share|cite|improve this answer
Thank you very much for the hint! – Italo Jan 9 '11 at 13:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.