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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.

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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.

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Thank you very much for the hint! – Italo Jan 9 '11 at 13:41

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